R SVEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH STUDY MATERIAL TRANSMISSION LINES AND WAVEGUIDES DEPARTMENT OF ECE JUNE – 2010 Vel Tech Vel Tech Multi Tech Dr.Rangarajan Dr.Sakunthala Engineering College Vel Tech High Tech Dr. Rangarajan Dr.Sakunthala Engineering College VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH SEM - V INDEX UNITS PAGE NO. I. Filters 06 II. Transmission Line Parameters 51 III. The Line at Radio Frequency 95 IV. Guided Waves Between Parallel Planes 138 V. Waveguides 179 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH # 42 & 60, Avadi – Veltech Road, Avadi, Chennai – 62. Phone : 044 26840603 email :
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Companies Such as TCS, INFOSYS TECHNOLOGIES, IBM, WIPRO TECHNOLOGIES, KEANE SOFTWARE & T INFOTECH, ACCENTURE, HCL TECHNOLOGIES, TCE Consulting Engineers, SIEMENS, BIRLASOFT, MPHASIS(EDS), APOLLO HOSPITALS, CLAYTON, ASHOK LEYLAND, IDEA AE & E, SATYAM VENTURES, UNITED ENGINEERS, ETA-ASCON, CARBORANDUM UNIVERSAL, CIPLA, FUTURE GROUP, DELPHI-TVS DIESEL SYSTEMS, ICICI PRULIFE, ICICI LOMBARD, HWASHIN, HYUNDAI, TATA CHEMICAL LTD, RECKITT BENKIZER, MURUGAPPA GROUP, POLARIS, FOXCONN, LIONBRIDGE, USHA FIRE SAFETY, MALCO, YOUTELECOM, HONEYWELL, MANDOBRAKES, DEXTERITY, HEXAWARE, TEMENOS, RBS, NAVIA MARKETS, EUREKHA FORBES, RELIANCE INFOCOMM, NUMERIC POWER SYSTEMS, ORCHID CHEMICALS, JEEVAN DIESEL, AMALGAMATION CLUTCH VALEO, SAINT GOBAIN, SONA GROUP, NOKIA, NICHOLAS PHARIMAL, SKH METALS, ASIA MOTOR WORKS, PEROT, BRITANNIA, YOKAGAWA FED BY, JEEVAN DIESEL visit our campus annually to recruit our final year Engineering, Diploma, Medical and Management Students. Preface to the First Edition This edition is a sincere and co-ordinated effort which we hope has made a great difference in the quality of the material. “Giving the best to VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 3 R S VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH the students, making optimum use of available technical facilities & intellectual strength” has always been the motto of our institutions. In this edition the best staff across the group of colleges has been chosen to develop specific units. Hence the material, as a whole is the merge of the intellectual capacities of our faculties across the group of Institutions. 45 to 60, two mark questions and 15 to 20, sixteen mark questions for each unit are available in this material. Prepared By : Ms. S. Jalaja Asst. Professor. Mr. S. Jebasingh. Lecturer. EC2305 TRANSMISSION LINES AND WAVEGUIDES UNIT I FILTERS 9 The neper - the decibel - Characteristic impedance of Symmetrical Networks – Current and voltage ratios - Propogation constant, - Properties of Symmetrical Networks – Filter fundamentals – Pass and Stop bands. Behaviour of the Characteristic impedance. Constant K Filters - Low pass, High pass band, pass band elimination filters - m -derived sections – Filter circuit design – Filter performance – Crystal Filters. UNIT II TRANSMISSION LINE PARAMETERS 9 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 4 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH A line of cascaded T sections - Transmission lines - General Solution, Physical Significance of the equations, the infinite line, wavelength, velocity, propagation, Distortion line, the telephone cable, Reflection on a line not terminated in Zo, Reflection Coefficient, Open and short circuited lines, Insertion loss. UNIT III THE LINE AT RADIO FREQUENCY 9 Parameters of open wire line and Coaxial cable at RF – Line constants for dissipation - voltages and currents on the dissipation less line - standing waves – nodes – standing wave ratio - input impedance of open and short circuited lines - power and impedance measurement on lines – / 4 line, Impedance matching – single and double-stub matching circle diagram, smith chart and its applications – Problem solving using Smith chart. UNIT IV GUIDED WAVES BETWEEN PARALLEL PLANES 9 Application of the restrictions to Maxwell’s equations – transmission of TM waves between Parallel plans – Transmission of TE waves between Parallel planes. Transmission of TEM waves between Parallel planes – Manner of wave travel. Velocities of the waves – characteristic impedance - Attenuators UNIT V WAVEGUIDES 9 Application of Maxwell’s equations to the rectangular waveguide. TM waves in Rectangular guide. TE waves in Rectangular waveguide – Cylindrical waveguides. The TEM wave in coaxial lines. Excitation of wave guides. Guide termination and resonant cavities. TEXT BOOK: 1. John D.Ryder, "Networks, lines and fields", Prentice Hall of India, 2nd Edition, 2006. REFERENCES: 1. E.C.Jordan, K.G. Balmain: “E.M.Waves & Radiating Systems”, Pearson Education, 2006. 2. Joseph Edminister, Schaum’s Series, Electromegnetics, TMH, 2007. 3. G S N Raju, Electromagnetic Field Theory and Transmission Lines, Pearson Education, 2006. UNIT – I PART – A 1. Define Filter? A reactive network that will freely pass desired bands of frequencies while almost totally suppressing other band of frequencies are called as filters. 2. What do you mean by ideal filter? An ideal filter would pass all frequencies in a given without reduction in VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 5 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH magnitude and totally suppressing all other frequencies. 3. What is cutoff frequency? The frequency which separates a pass band and an attenuation band are called as cutoff frequency. 4. Define the unit of attenuation (or) Define Neper? Attenuation is expressed in decibels or nepers. Nepers is defined as the natural logarithm of the ratio of input voltage or current to the output voltage or current provided the network is properly terminated with Z0. 5. Define Decibel? Decibel is defined as the ten times common logarithms of the input power to the output power. 6. Give the relation between two units of attenuation? The relationship between two units of attenuation can 1 db = 0.115 neper. 7. Give the common types of filters? The four common types of filters are, High pass filter 1. Low pass filter 2. Band pas filter 3. Band stop filter 8. What are the characteristics of ideal filter? The characteristics of ideal filter are: 1. Transmit pass band frequencies without any attenuation. 2. Provide infinite attenuation 3. The transition region between the stop band and pass band would be very small. 4. Throughout the pass band characteristic impedance of the filter match circuit to which it is connected which prevents reflection loss. 9. When networks are said to be symmetric network? When two series arms of a T network are equal or when two shunt arms of a network are equal then the network is said to be symmetrical network. 10. Draw a symmetrical network in ‘T’ and Π sections. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 6 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 11. Give the characteristic impedance of low pass constant – K filter The characteristic impedance is given as ( ) 2 1 1 2 2 / 1 / 4 OT OT Z Z Z Z Z L C CL ω · + · − 12. Give the formula to calculate the cutoff frequency for low pass constant – k 1/ c f LC π · 13. What are the constant – k filters? A constant – k filters is a ‘T’ or ‘π ’ network in which the series and shunt impedance Z 1 and Z z are connected by the relation. Z 1 – Z 2 = R 2 K. Where ‘Rk’ is a real constant i.e., a resistant independent of frequency. 14. Why constant – k filters are also called as prototype filter? Constant – k filters are also called as prototype filter because other complex networks can be derived from it. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 7 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 15. For a low pass filter what is the condition for which the characteristic impedance Z0 is real? The characteristic impedance Z0 is real if (ω 2 LC/4)<1 16. Give the expression for the cutoff frequency of constant – k high pass filter. 1/ 4 c f LC π · 17. What are the phase shift and attenuation of constant –k low pass filter? The phase shift of constant – k low pass filer is given as, ( ) 1 2sin / C radians β ω ω − · The attenuation of constant low pass filter is given as ( ) 1 2cos / c nepers α ω ω − · 18. What are the phase shift and attenuation of constant –k high pass filter? The phase shift of constant – k high pass filter is given as β = 2 sin -1 ( ω c 2 /ω 2 ) radians. The attenuation of constant high pass filter is given as ( ) 2 1 2cos / c α ω ω − · nepers 19. For a high pass filter what is the conditions for which the characteristic impedance Z OT is real and imaginary. For a high pass filter, the characteristic impedance Z OT is real if (1/4ω 2 LC ) < 1 And Imaginary if (1/4ω 2 LC) >1 20. Give the characteristic impedance of high pass filter constant – k filter. The characteristic impedance of high pass filter constant- k filter is given as ( ) 2 / 1 1/ 4 OT Z L C LC ω · − 21. Draw the T- type and π type low pass constant – k filter. The prototype low pass filters are as shown below. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 8 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 22. Draw the T- type and Π type high pass constant –k filter 23. What are the advantages of m- derived filters? The advantages of m – derived filters are, 1. A sharper cutoff characteristic with steeper rise at f c , the slope of the rise being adjustable by fixing the distance between f c and f. 2. ‘Zo’ of the filter will be more uniform within the pass and when m – derived half section having m = 0.6 are connected at the ends. 3. m- derived filters makes it possible to construct composite filters to have any desired attenuation characteristics. 24. What is composite filter? A filter designed using one or more prototype constant – k filters and m – derived filters to have an attenuation between low pass and high filters is called as composite filters. 25. Mention the different sections of a composite filters?
[email protected] The different sections of a composite filters are 1. One or more prototype constant filters are 2. One or more m – derived sections 3. Two terminating m – derived half sections with m = 0.6 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 9 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 26. What are the drawbacks of constant – k filters (or) what are the disadvantages of constant – k filters? 1. The attenuation does not increase rapidly beyond cutoff frequencies. 2. Characteristic impedance varies widely in the transmission or pass band from the derived value. 27. Why m–derived half section is used as terminating section? Two m – derived half section with m = 0.6 is used as a terminating section to give constant input and output impedance. 28. Distinguish between low pass and high pass filter. S. No Low Pass Filter High Pass filter 1 This filter passes the frequencies without attenuation upto a cutoff frequency fc and attenuates all other frequencies greater than fc. It transmits frequencies above a designed cutoff frequency but attenuates frequencies below this 2 An ideal low pass filter is as shown below An ideal high pass filter is as shown below 29. Differentiate between band pass filter and band elimination filter S.No Bandpass filter Band elimination filter 1 This filter passes the frequencies between two designated cutoff frequencies and attenuates all other frequencies It transmits all frequencies while attenuates a band of frequency. 2. An ideal band pass filter is as shown below An ideal band elimination filter is as shown below. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 10 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 30. Draw a block diagram of a composite filter? 31. Draw the T-type and Π type low pass m – derived filter The m – derived low pass filters are as shown below. 32. Draw the type and Π type high pass m- derived filter The m- derived high pass filters are as shown below VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 11 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 33. Give the value of ‘m’ in m – derived low pass and high pass filters. The value of ‘m’ in the case of low pass filter is given as below m = ( ) 2 1 / fc f − ∞ The value of ‘m’ in the case of high pass filter is as below m = ( ) 2 1 / f fc − ∞ PART – B 1. Define the characteristic impedance of symmetrical networks. When Z 1 = Z 2 or the two series arms of a T network are equal, or Z a = Z c and the shunt arms of a π network are equal, the networks are said to be symmetrical. Filter networks are ordinarily set up as symmetrical sections, basically of the T or π type, such as shown at (b) and (d), Fig. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 12 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Figure: The T and π sections as derived from unsymmetrical L sections, showing notation used in symmetrical network analysis. Attention is called to the peculiarities of notation employed on the variouos arms. This peculiarity is largely dictated by custom, arising from the fact that both T and π networks can be considered as built of unsymmetrical L half sections, connected together in one fashion for the T network, and oppositely for the π network as at (a) and (c), Fig. A series connection of several T or π networks leads to so- called “ladder networks,” which are indistinguishable one from the other except for the end or terminating L half sections, as can be seen in fig. For a symmetrical network the image impedances Z 1i and Z 2i , are equal to each other, and the image impedance is then called the characteristic impedance or the iterative impedance, Z 0 . That is, if a symmetrical T network is terminated in Z 0 , its input impedance will also be Z 0 , or its impedance transformation ratio is unity. The term iterative impedance is apparent if the terminating impedance Z 0 is considered as the input impedance of a chain of similar networks, in which case Z 0 is iterated at the input to each network. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 13 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Figure: (a) Ladder network made from T section; (b) ladder network built from π sections. The parallel shunt arms will be combined. The value of Z 0 for a symmetrical network can be easily determined. For the T network of Fig. (a), terminated in an impedance Z 0 , the input impedance is ( ) 2 1 0 1 1in 1 2 0 Z Z / 2 Z Z Z 2 Z / 2 Z Z + · + + + It can be assumed that if Z 0 is properly chosen in terms of the network arms, it should be possible to make Z 1in equal to Z 0 . Requiring this equality gives 2 2 2 1 2 2 0 1 0 1 0 1 2 0 1 1 2 0 Z / 4 Z Z Z Z Z Z / 2 Z Z / 2 Z Z Z Z Z Z 4 + + + · + + · + For the symmetrical T section, then, 2 1 1 0T 1 2 1 2 2 Z Z Z Z Z Z Z 1 4 4Z ¸ _ · + · + ¸ , becomes the characteristic impedance. This result could also have been immediately obtained from eq. and for the image VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 14 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Figure: Determination of Z 0 : (a) for a T section; (b) for a π section impedance of a T section, by using the values of the arms of Fig. Similarly, for the π section of Fig. (b) the input impedance is 2 0 1 2 2 0 1in 2 0 1 2 2 0 2Z Z Z 2Z 2Z Z Z 2Z Z Z 2Z 2Z Z 1 ¸ _ + 1 + ¸ , ¸ ] · + + + Requiring that 1in 0 Z Z leads to · 1 2 0 1 2 Z Z Z 1 Z / 4Z π · + which is the characteristic impedance of the symmetrical π symmetrical π section. Certain information concerning networks was developed from measurements of Z and Z . ∞ ∞ If these measurements are made on the T section of (a), Fig. exclusive of the load Z 0 , then 2 2 1 1 2 1 1 2 1 1 2 1 1 2 0T Z Z Z Z 2 Z Z Z / 2 Z Z 2 Z / 2 Z Z Z Z Z Z Z 4 ∞ ∞ ∞ ∞ ∞ ∞ · · + · · + + · + · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 15 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Similar work for the π section leads to 2 2 1 2 0 1 2 4Z Z Z Z Z Z 4Z ∞ ∞ π · · + Therefore, for a symmetrical network, 0 Z Z Z ∞ ∞ · This result could have been directly obtained from the image impedance relations of section. It is a valuable relationship, since it supplies an easy experimental means of determining the Z 0 of any symmetrical network. 2. Explain the current and voltage ratios as exponentials; the propagation constant. Under the assumption of equal input and output impedances, which may now be interpreted as a Z 0 termination on the network, the absolute value of the ratio of input current to output current of a given symmetrical network was defined as an exponential function,* for the purpose of simplifying network calculations. obviously, the magnitude ratio does not express the * In the general case of unsymmetrical 4-terminal networks, terminated on an image basis, it is customary to define a transfer constant θ , by 2 1 1 2 2 input volt-amperes E I E I output volt-amperes θ ε · · Or 1 1 2 2 1 E I ln 2 E I θ · where θ is in general a complex number. For symmetrical networks Z 14 = Z 2 = Z 0 , and with a termination of Z 0 , the above discussion follows, with γ customarily replacing θ and implying symmetry and Z 0 termination. complete network performance, the phase angle between the currents being needed as well. the use of the exponential can be extended to include the phasor current ratio if it be defined that, under the condition of Z 0 termination, 1 2 I I γ · ε where γ is a complex number defined as j γ · α+ β VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 16 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Hence j 1 2 I I α+ β γ · ε · ε To illustrate further, if 1 2 I / I A ,then · β 1 2 j A I / I α β · · ε β · ε Since the input and output impedances are equal under the Z 0 termination, it is also true that 1 2 V V γ · ε The term γ has been given the name propagation constant. The exponent α is known as the attenuation constant, since it determines the magnitude ratio between input and output quantities, or the attenuation produced in passing through the network. The units of α are nepers. The exponent β is the phase constant as it determines the phase angle between input and output quantities, or the shift in phase introduced by the network. The units of β are radians. If a number of sections all having a common Z 0 value are cascaded, the ratio of currents is 3 1 2 1 2 3 4 n I I I I .. I I I I × × × · from which 1 2 3 n ... γ γ γ γ ε ×ε ×ε × · ε and taking the natural logarithm, 1 2 3 n ... γ + γ + γ + · γ Thus the over-all propagation constant is equal to the sum of the individual propagation constants. 3. Explain the properties of symmetrical networks. Use of the definition of γ , and the introduction of ε γ as the current ratio for a Z 0 terminated network, leads to further useful results. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 17 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Figure: Symmetrical network with generator and load. In Fig. the T network is considered equivalent to any connected symmetrical network, and is terminated in a load Z 0 . The mesh equation are 1 1 2 2 2 1 1 2 2 2 0 Z E I Z I Z 2 Z 0 I Z I Z Z 2 ¸ _ · + − ¸ , ¸ _ · − + + + ¸ , The current ratio for the two meshes, which is equal to ε γ by definition, can be obtained from the second equation as 1 2 0 1 2 2 Z / 2 Z Z I I Z γ + + · · ε After thus introducing ε γ , the above may be written ( ) 1 0 2 Z Z Z 1 2 γ · ε − − From Eq. for the characteristic impedance, 2 2 1 1 2 0 Z Z Z Z 4 · + If Z 0 is eliminated by use of Eq. in Eq. there results ( ) 2 2 1 2 1 2 1 2 1 2 Z 1 Z 0 Z 2 1 Z Z 1 2 2Z Z cosh 1 2Z γ γ ε γ γ γ γ −γ ε − − · ε − ε + · ε ε + ε · + γ · + Equation and its other derived forms will be of considerable value in the study of VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 18 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH filters. By use of the identify, Eq., that 2 2 cosh sinh 1 γ − γ · it is possible to write 0 2 Z sinh Z γ · Combining Eqs. and leads to 0 1 2 Z tanh Z / 2 Z γ · + By use of Eq. it is possible to write 1 2 1 2 1 Z sinh 1 1 2 2 2Z Z 4Z ¸ _ γ · + − ¸ , · an expression which will serve to predict filter performance. The propagation constant γ can be related to the network parameters by use of Eq. for Z 0T , in Eq. as 2 1 1 1 2 2 2 Z Z Z 1 2Z 2Z Z γ ¸ _ ε · + + + ¸ , Taking the natural logarithm 2 1 1 1 2 2 2 Z Z Z ln 1 2Z 2Z Z 1 ¸ _ 1 γ · + + + 1 ¸ , ¸ ] For a network of pure reactances this is not difficult to compute. For an impedance it may be noted that the logarithm of a complex quantity B lnB j . α · + α The input impedance of any T network, terminated in any impedance Z R , may also be written in terms of hyperbolic functions of γ . Writing 2 12 in 11 22 Z Z Z Z · − and substituting the required mesh relations from Fig. with Z 0 replaced by Z R , then VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 19 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) 2 2 1 2 in 2 1 2 R 1 2 1 2 R 1 1 2 R Z Z Z Z 2 Z / 2 Z Z Z / 4 Z Z Z / 2 Z Z Z / 2 Z Z · + · + + + + + · + + Use of Eqs. and leads to 2 R 0 0 in 0 R R 0 0 0 R Z Z Z / tanh Z Z / tanh Z Z cosh Z sinh Z Z cosh Z sinh + γ · γ + ¸ _ γ + γ · γ + γ ¸ , This is the input impedance of a symmetrical T network terminated in a load Z R , in terms of the propagation constant and Z 0 of the network. For a short-circuited network Z R = 0. The input impedance is then Z ∝ where, from the above equation, 0 Z Z tanh ∞ · γ For an open circuit Z R = ∝ in the limit, and Z ∝ is then Z 0 ZR Z lim tanh ∞ →∞ · γ From these two equations it can be seen that sc oc Z tanh Z γ · and 0 Z Z Z ∞ ∞ · which has already been proved from the properties of the characteristic impedance. In chapter 1, open-circuit and short-circuit measurements were used to describe the performance of a network. in this chapter, two new parameters, the characteristic impedance Z 0 , and the propagation constant γ , have been introduced, and the properties of the network have been developed in terms of these new parameters. The last few equations are relations between the two sets of parameters. 4. Explain the constant-k low-pass filter. If Z 1 and Z 2 of a reactance network are unlike reactance arms, then VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 20 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 1 2 Z Z k · where k is a constant independent of frequency. Networks or filter sections for which this relation holds are called constant-k filters. As a special case, let 1 2 Z j L and Z j/ C, · ω · − ω then the product 2 1 2 k L Z Z R C · · The term R k is used since k must be real if Z 1 and Z 2 are of opposite type. A T section so designed would appear as at (a), Fig. Figure: (a) Low-pass filter section; (b) reactance curves demonstrating that (a) is a low-pass section or has a pass band between Z 1 = 0 and Z 1 = -4Z 2 . The reactances of Z 1 and 4Z 2 will vary with frequency as sketched at (b), Fig. The curve representing -4Z 2 may be drawn and compared with the curve for Z 1 . It has been shown by Eq. that a pass band starts at the frequency at which Z 1 = 0 and runs to the frequency at which 1 2 Z 4Z . · Thus the reactance curves show that a pass band starts at f = 0 and continues to some higher frequency f c . All frequencies above f c lie in a stop, or attenuation, band. Thus the network is called a low-pass filter. The cutoff frequency f c may be readily determined, since at that point 1 2 c c c 4j Z 4Z , j L C 1 f LC · − ω · ω · This expression may be used to develop certain relations applicable to the low- pass network. then sinh / 2 γ may be evaluated as 2 1 2 j LC Z LC sinh 2 4Z 4 2 ω γ −ω · · · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 21 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH and in view of Eq. this is c f sinh j 2 f γ · Then if the frequency f is in the pass band or f/f c < 1, so that -1 < Z 1 /4Z 2 < 0, then -1 c c f f 1, =0, =2sin f f ¸ _ < α β ¸ , Figure: Variation of α and β with frequency for the low-pass section whereas if frequency f is in the attention band or f/f c > 1, so that Z 1 /4Z 2 < - 1, then -1 c f f 1, =2cosh , = f f ¸ _ > α β π ¸ , thereby allowing determination of α and β . The variation of α and β is plotted in Fig. as a function of f/f c . This method shows that the attenuation α is zero throughout the pass band but rises gradually from the cutoff frequency at f/f c = 1.0 to a value of ∝ at infinite frequency. The phase shift β is zero at zero frequency and increases gradually through the pass band, reaching π at f c and remaining at π for all higher frequencies. The characteristic impedance of a T section was obtained as 1 0T 1 2 2 Z Z Z Z 1 4Z ¸ _ · + ¸ , which becomes 2 0T L LC Z 1 C 4 ¸ _ ω · − ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 22 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH for the low-pass constant-k section under discussion. by use of Eq. the characteristic impedance of a low-pass filter may be stated as 2 0T c 2 k c L f Z 1 C f f R 1 f 1 ¸ _ · − 1 1 ¸ , ¸ ] ¸ _ · − ¸ , in accordance with the definition of R k in Eq. Values of Z 0T /R k are plotted against f/f c in Fig. It may be seen that cutoff, then becomes imaginary in the attenuation band, rising to infinite reactance at infinite frequency. A low-pass filter may be designed from a knowledge of the cutoff frequency desired and the load resistance to be supplied. It is desirable that the Z 0 in the pass band match the load; but because of the nature of the Z 0 curve in Fig., this result can occur at only one frequency. This match may be arranged to occur at any frequency which it is desired to favor by an impedance match. For reasons which will appear in section, the load is chosen as k L R R , C · · which will favor zero frequency for a low-pass filter. The design of a low-pass filter may be readily carried out. From Figure: Variation of 0T k Z / R with frequency for the low-pass section. the relation that at cutoff 1 2 Z 4Z · − VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 23 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH it is seen that e 0 4 L C ω · ω Using the cutoff frequency equation changes this to 2 2 c f LC 1 π · and use of the relation L R C · gives for the value of the shunt capacitance arm c 1 C f R · π By similar methods the inductance for Z 1 is obtained as c R L f · π Since the design is based on an impedance match at zero frequency only, power transfer only, power transfer to a matched load will drop at higher pass-band frequencies. This condition may be undesirable in certain applications, and a remedy will be discussed in section. A network such as is described here is called a prototype section. It may be employed when a sharp cutoff is not required, although cutoff may be sharpened by using a number of such networks in cascade. This is not usually an economic use of circuit elements, and introduces excessive losses over other available methods of raising the attenuation near the cutoff frequency. 5. Explain the constant-k high-pass filter. If the positions of inductance and capacitance are interchanged to make 1 Z j/ C · − ω and 2 Z j L, · ω then Z 1 Z 2 will still be given by 2 1 2 Z Z k · and the filter design obtained will be of the constant-k type. The T section will then appear as in (a), Fig. The reactances of Z 1 and Z 2 are sketched as functions of frequency in (b), and Z 1 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 24 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Figure: (a) High-pass filter section; (b) reactance curves demonstrating that (a) is a high-pass section or has a pass band between Z 1 = 0 and Z 1 = -4Z 2 . Is compared with -4Z 2 , showing a cutoff frequency at the point at which Z 1 equals -4Z 2 , with a pass band from that frequency to infinity where Z 1 = 0. The network is thus a high-pass filter. All frequencies below f c lie in an attenuation, or stop, band. The cutoff frequency is determined as the frequency at which 1 2 Z 4Z , or · − 2 0 c 0 c j j4 L, 4 LC 1 C 1 f 4 LC − · − ω ω · ω · π Using the above expression c 1 2 2 j f Z 1 sinh j 2 4Z 4 LC f 2 LC − γ · · − · · − ω ω The region in which c f / f 1 < is a pass band, so that the variation of γ inside and outside the pass band will be indentical with the values for the low-pass filter, and the curves of fig. will apply if the abscissa be considered as calibrated in terms of f c /f, except that the phase angle β will be negative, changing from 0 at infinite frequency or f c /f = 0, to -π at cutoff or f c /f = 1. The high-pass filter may be designed by again choosing a resistive load R equal to R k such that k L R R C · · From the relation that at cutoff 1 2 Z 4Z · − it was shown that 2 c 4 LC 1 ω · and again L/C = R 2 , so that the value of the capacitance for Z 1 , the series element, is VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 25 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH c 1 C 4 f R · π It should be noted that since Z 1 /2 is the value of each series arm, the capacity use din each series Z 1 /2 element should be 2C. By similar methods the value for the inductance for Z 2 , the shunt arm, is c R L 4 f · π The characteristic impedance for the high-pass filter may be transformed to 2 c 0T k f Z R 1 f ¸ _ · − ¸ , 6. Explain the m-derived T section filter. The constant-k prototype filter section, though simple, has two major disadvantages. The attenuation does not rise very rapidly at cutoff, so that frequencies just outside the pass band are not appreciably attenuated with respect to frequencies just inside the pass band. Also, the characteristic impedance varies widely over the pass band, so that a satisfactory impedance match is not possible. In cases where an impedance match is not important, the attenuation may be built up near cutoff by cascading or connecting a number of constant-k sections in series. It is more economical to attempt to raise the attenuation near cutoff by other means. Consider first the circuit of (a), Fig. The reactance curves sketched at (b) show that this circuit is alow pass filter. However, it can be seen that the shunt arm is a series circuit resonant at a frequency above f c . At this resonant frequency the shunt arm appears as a short circuit on the network, or the attenuation becomes infinite. This frequency of infinite or high attenuation is called f ∝ , will always be higher in value than f c . If, then, f ∝ can be chosen arbitrarily close to f c , the attenuation near cutoff may be made high. Figure: (a) Derivation of a low-pass section having a sharp cutoff action; (b) reactance curves for (a). VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 26 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The attenuation above f ∝ will fall to low values, so that if high attenuation is desired over the whole attenuation band, it is necessary to use a section such as in Fig. for high attenuation near cutoff, in series with a prototype section to provide high attenuation at frequencies well removed from cutoff. For satisfactory matching of several such types of filters in series, it is necessary that the Z 0 of all be identical at all points in the pass band. They will consequently also all have the same pass band. The network of fig. may be derived by assuming that ' 1 1 Z mZ · the primes indicating the derived section. It is then necessary to find the value for Z 2 ’ such that ' 0 0 Z Z . · Setting the characteristic impedances equal, ( ) 0 0 2 2 1 1 1 2 1 2 2 ' 2 2 1 Z ' Z mZ Z mZ Z ' Z Z 4 4 Z 1 m Z Z m 4m · + · + − · + It then appears that the shunt arm 2 Z ' consists of two impedances in series, as shown in Fig. As required, the characteristic impedance and f c remain equal to those of the T section prototype containing Z 1 and Z 2 values. Figure: The m-derived low-pass filter. Since m is arbitrary, it is possible to design an infinite variety of filter networks meeting the required conditions on Z 0 and f c . However, Z 2 will be opposite in sign to Z 1 , and it is desired that this relation continue in the two series impedances given by Eq. for the 2 Z ' arm. Equation then indicates that (1-m 2 )/4m must be positive, forcing the terms 1 – m 2 and m always to be positive. Thus m must always be chosen so that < 1 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 27 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Filter sections obtained in this manner are called m-derived sections. The shunt arm is to be chosen so that it is resonant at some frequency f ∝ above f c . This means that at the resonant frequency and for the low-pass filter ( ) 2 2 1 1 m 2 f L 2 f mC 4m 1 f 1 m LC ∞ ∞ ∞ − · π π · π − Since the cutoff frequency for the low-pass filter is c 1 f LC · π the frequency of infinite attenuation will be c 2 f f 1 m ∞ · − from which ( ) 2 c m 1 f f ∞ · − − This equation determines the m to be used for a particular f ∝ . Similar relations for the high-pass filter can be derived as 2 c f f 1 m ∞ · − and ( ) 2 c m 1 f / f ∞ · − The m-derived section is designed following the design of the prototype T section. The use of a prototype and one or more m-derived sections in series results in a composite filter. If a sharp cutoff is desired, an m-derived section may be used with f ∝ near f c , followed by as many m-derived sections as desired to place frequencies of high attenuation where needed to suppress various signal components or to produce a high attenuation over the entire attenuation band. The variation of attenuation over the attenuation band for a low-pass m-derived section in the stop band is dependent on the sign of the reactances or 1 -1 1 1 2 2 c Z Z 2cosh or =2sinh 4Z 4Z f f f f f − ∞ ∞ α · α < < < For 1 2 Z j L and Z j/ C for the prototype, then · ω · − ω VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 28 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) 1 2 2 Z m L 4Z 4 1/ m C L 1 m / 4m ω · 1 ω −ω − ¸ ] so that for c f f f ∞ < < 1 c 2 2 mf / f 2cosh 1 f / f − ∞ α · − Figure: Variation of attenuation for the prototype and m-derived sections, and the composite result of the two in series. and for f ∝ < f 1 c 2 2 mf / f 2sinh f / f 1 − ∞ α · − The value of α may be determined from the expression. Figure is a plot of a against f/f c for m=0.6, which gives a value of f ∝ equal to 1.25 times the cutoff frequency f c . The great increase in sharpness of cutoff for the m-derived section over the prototype is apparent. The higher attenuation over the whole attenuation band obtained by use of a prototype section and an m-derived section in series as a composite filter is also readily seen. Again following the procedure of section, the phase shift constant β may be determined, in the pass band, from ( ) ( ) 1 1 2 1 c 2 2 2 c Z 2sin 4Z mf / f 2sin 1 f / f 1 m − − β · · − − VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 29 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH In the attenuation band, up to f ∝ , β has the value π . Above f ∝ the value of β drop to zero, because the shunt arm becomes inductive above resonance. The phase shift of the m-derived section is plotted as a function of f/f c in fig. Figure: Variation of phase shift β , for the m-derived filter This material demonstrates the ability of the m-derived section to overcome the lack of a sharp cutoff in the simple prototype filter. Although it may be note that the sharpness of cutoff increases for small values of m, the attenuation beyond the point of peak attenuation becomes smaller for small m. This emphasizes the necessity of supplementing the m-derived section with a prototype section in series to raise the attenuation for frequencies well removed from cutoff. 7. Explain the derived π section filter. An m-derived π section may also be obtained. The characteristic impedance of the π section is ( ) 1 2 0 1 2 1 2 Z Z Z Z Z 1 Z / 4Z π · + The characteristic impedances of the prototype and m-derived sections are to be equal so that they may be joined without mismatch. By use of the transformation for the shunt arm, 2 2 Z Z ' m · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 30 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Figure: (a) Usual symmetrical π section; (b) the m-derived π filter. it is possible to equal the characteristic impedances as ( ) ( ) ( ) 1 2 1 2 1 2 1 2 1 2 1 2 Z 'Z / m Z Z Z Z / m 1 Z 'm/ 4Z Z Z 1 Z / 4Z · + + from which 1 1 2 2 1 Z ' 1 1 4m mZ Z 1 m · + − It is apparent that the series arm 1 Z ' is represented by two impedances in parallel, one being mZ 1 , the other being ( ) 2 2 4m/ 1 m Z − in value. Equations and thus give the values to be used in designing the m-derived π section. The circuit is drawn in fig. 8. Explain the Band-pass filters. Occasionally it is desirable to pass a band of frequencies and to attenuate frequencies on both sides of the pass band. The action might be thought of as that of low-pass and high-pass filters in series, in which the cutoff frequency of the low-pass filter is above the cutoff frequency of the high-pass filter, the overlap thus allowing only a band of frequencies to pass. Although such a design would function, it is more economical to combine the low-and high-pass functions into a single filter section. Figure: (a) Band-pass filter network; (b) reactance curves showing VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 31 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH possibility of two bands. Consider the circuit of (a), Fig. with a series-resonant series arm and an antiresonant shunt arm. In general, the reactance curves show that two pass bands might exist. If, however, the antiresonant frequency of the shunt arm is made to correspond to the resonant frequency of the series arm, the reactance curves become as shown in fig. and only one pass band appears. For this condition of equal resonant frequencies, Figure: Reactance curves for the band-pass network when resonant and antiresonant frequencies are properly adjusted. 2 2 0 1 1 0 2 2 1 1 2 2 L C 1 L C or L C L C ω · · ω · The impedances of the arms are ( ) ( ) ( ) 2 1 1 1 1 1 1 2 2 2 1 2 2 2 2 2 L C 1 1 Z j L j C C j L j/ C j L Z j L 1/ C 1 L C ω − ¸ _ · ω − · ω ω ¸ , ω − ω ω · · ω − ω −ω That a network such as (a), Fig. is still a constant-k filter is easily shown as ( ) ( ) 2 2 2 1 1 2 2 1 2 2 L L C 1 Z Z C 1 L C ω − · − −ω and if 1 1 2 2 L C L C , then · 2 2 1 1 2 k 1 2 L L Z Z R C C · · · Thus the previously developed theory still applies VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 32 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH At the cutoff frequencies, 1 2 Z 4Z · − Multiplying by Z 1 gives 2 2 1 1 2 k Z 4Z Z 4R · − · − from which the value of Z 1 at the cutoff frequencies is obtained as 1 k Z j2R · t so that 1 Z at lower cutoff f 1 = - Z 1 at upper cutoff f 2 The reactance of the series arm at the cutoff frequencies then can be written by use of the above as ( ) 1 1 2 1 1 1 2 1 2 2 1 1 1 1 2 1 1 2 1 1 L L C C 1 L C L C 1 −ω · ω − ω ω ω −ω · ω − ω Now from eq. 1 1 2 0 1 L C · ω so that eq. may be written as ( ) ( ) 2 2 1 1 2 2 2 0 2 0 2 0 1 2 1 2 1 2 1 2 f f f 1 1 f f f f f f f f f f f f f ¸ _ − · − ¸ , + · + · or the frequency of resonance of the individual arms should be the geometric mean of the two frequencies of cutoff. If the filter is terminated in a load R = R k , as is customary, then the values of the circuit components can be determined in terms of R and the cutoff frequencies f 1 and f 2 . At the lower cutoff frequency, VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 33 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 1 1 1 2 1 1 1 2 0 1 L 2R C f 1 4 Rf C f −ω · ω − · π In view of Eq., the expression for C 1 becomes 2 1 1 1 2 f f C 4 Rf f − · π It follows, then, from eqs. and that ( ) 1 2 1 R L f f · π − From equation it is possible to obtain the values for the shunt arm as ( ) ( ) 2 1 2 2 1 1 2 1 2 2 2 1 R f f L C R 4 f f L 1 C R R f f − · · π · · π − This completes the design of the prototype band-pass filter. Figure: m-derived band-pass section An m-derived band-pass section is also possible. Use of the transformation relation developed in section leads to a network of the form of fig. The shunt arm then consists of series-resonant and antiresonant circuits in series. Plotting reactance curves for these two circuits and adding to obtain the reactance variation of the shunt arm, Z 2 of the filter, gives the dashed curve of fig. The antiresonant frequency of the arm as a whole must, by previous reasoning, be f 0 of the filter. The reactance curve for Z 2 then shows that the shunt arm becomes resonant at a frequency below f 0 and VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 34 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH again at a frequency above f 0 . Figure: Reactance curves for the shunt arm of the m-derived band-pass section. At these frequencies the network is short-circuited, and thus they are frequencies of high attenuation, f ∝ . These frequencies of high attenuation are placed on each side of the pass band, and the m-derived section may be used to increase the attenuation near cutoff, as for the high-or low-pass cases. At one f ∝ , the reactances X r and X ar are equal and opposite, so that ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 2 2 2 1 2 2 1 2 1 1 2 2 2 j L / m j/ mC j 1 m j L 4m j 1 / mC L / m 4m/ 1 m C L C 1 m L C 1 4 L C 1 ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ω − ω ¸ _ − ω − · − ω −ω 1 ω − ¸ , ¸ ] ω − ω − · ω − In view of the fact that 1 1 2 2 2 0 1 L C L C · · ω L Equation becomes 2 2 2 2 2 2 1 2 0 1 m f 1 4 f L C 4 f ∞ ∞ ¸ _ − − · π ¸ , The term L 2 C 1 can be evaluated as a function of f 1 and f 2 from Equations and ( ) ( ) 2 2 1 2 1 2 1 2 1 2 4 1 2 1 2 0 f f R f f f f L C 4 Rf f 4 f f 16 f − − 1 − · · 1 π π π ¸ ] Equation then reduces to VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 35 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) ( ) ( ) 2 2 2 2 2 1 2 2 1 2 1 2 1 2 2 1 m f f f f f f f f f f f 0 1 m ∞ ∞ ∞ − − · − − − − · − Solving for the values of the frequencies of peak attenuation, ( ) ( ) 2 2 1 2 1 1 2 2 2 f f f f f f f 4 1 m 2 1 m ∞ − − · t + − − It is apparent that the radical is larger than ( ) 2 2 1 f f 2 1 m , − − and thus one root would appear as a negative frequency that has no physical significance here. Thus the expression for f ∝ should be reversed so that the two frequencies of peak attenuation are ( ) ( ) ( ) ( ) 2 2 1 2 1 1 1 2 2 2 2 2 1 2 1 2 1 2 2 2 f f f f f f f 4 1 m 2 1 m f f f f f f f 4 1 m 2 1 m ∞ ∞ − − · + − − − − − · + + − − Equation may be solved to determine the value of m, giving ( ) ( ) ( ) 2 2 1 2 1 2 2 2 2 2 1 2 2 1 2 f f f m 1 f f f 1 f f f f f f f ∞ ∞ ∞ ∞ ∞ − 1 · − 1 − ¸ ] − − − · − The value of m may be chosen to place either one of the two frequencies of peak attenuation at a desired point, the other frequency of peak attenuation then being definitely fixed. That this is true may be chosen by forming the product for 1 2 f f ∞ ∞ from Equations and: ( ) 2 2 2 2 2 1 1 1 2 2 1 2 1 1 2 1 2 2 1 2 0 f 2f f 4m f f f 2f f f f f f f 4 1 m f f f 2 ∞ ∞ ∞ ∞ + + − − + − · · − · Thus f 0 is the geometric mean of the frequencies of peak attenuation and, by Equation, of the cutoff frequencies as well. If m is selected to place f 01 at a desired point, then by Equation, 2 f ∞ and 2 f ∞ by the use of two m’s or an mm’-derived filter, as shown by Zobel (Reference 2). VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 36 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH An m-derived T section, rearranged as a π , may be split into two half sections and used as terminating half sections. If m is given the value 0.6, then satisfactory impedance matching conditions are maintained over the pass band. This usage follows the previously developed theory for low-or high-pass sections. 9. Explain the band-elimination filters. If the series and parallel-tuned arms of the band-pass filter are interchanged, the result is the band-elimination filter of (a), fig. That this circuit does eliminate or attenuate a given frequency band is shown by the reactance curves for Z 1 and -4Z 2 at (b). The action may be thought of as that of a low-pass filter in parallel with a high- pass section, in which the cut-off frequency of the low-pass filter is below that of the high-pass filter. Figure: (a) Band-elimination filter; (b) reactance showing action of band- elimination section. As for the band-pass filter, the series and shunt arms are made antiresonant and resonant at the same frequency f 0 . Again, it is possible to show that 2 1 2 k 2 1 L L R C C · · and that 0 1 2 f f f · At the cutoff frequencies, 2 2 1 2 1 2 2 k k 2 Z 4Z , Z Z 4Z R jR Z 2 · − · · · t If the filter is terminated in a load R = R k , then at the lower cutoff frequency, 2 1 2 1 2 jR 1 Z j L C 2 ¸ _ · −ω · ω ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 37 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Since 2 2 2 0 1 L C , · ω 2 1 1 2 2 0 f 1 f C R f − · π 2 1 2 1 2 1 f f C R f f ¸ _ − · π ¸ , In view of the fact that 0 1 2 2 2 1 f f f 2 L C · · π then ( ) 2 2 1 R L 4 f f · π − By use of Equation, the values for the series arm are obtained as ( ) ( ) 2 1 1 1 2 1 2 1 R f f L f f 1 C 4 R f f − · π · π − Section of the m-derived form may also be obtained. 10. Explain the crystal filters. The lattice structure can also be shown to have filter properties. Considering the network of fig. Figure: Lattice filter section VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 38 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 Z / 2 2Z Z 2 Z / 2 2Z Z Z 4 Z Z Z Z Z Z / 2 2Z Z / 2 2Z Z Z Z / 4 Z ∞ ∞ + · + · + · + + + · + The characteristic impedance of the fig. Lattice filter section. Lattice section then is 0L sc 1 2 Z Z Z Z Z ∞ · · Thus if the section elements are reactive, Z 0L is real, or a pass band exists for frequencies for which Z 1 and Z 2 are of opposite sign. Over ranges where Z 1 and Z 2 have the same sign, an attenuation band exists. Propagation can be investigated further by noting that sc 1 oc 2 1 2 Z Z 1 tanh Z Z 1 Z / 4Z 1 γ · · 1 + ¸ ] It may be noted that Z 0L depends on the product of Z 1 and Z 2 , whereas γ depends on the ratio of Z 1 to Z 2 . This feature permits somewhat greater versatility in design of the lattice section over the T or π section, especially for filters in which certain of the elements are constructed of piezoelectric crystals. These crystals have a resonant frequency of mechanical vibration dependent on certain of their dimensions; and because of the very high equivalent Q of the crystals, it is possible to make very narrow band filters and filters in which the attenuation rises very rapidly at cutoff. The equivalent electric circuit of a quartz mechanical-filter crystal is shown in fig. (a), which shows a possibility of both resonance and antiresonance occurring. The inductance L x is very large, being in henrys for crystals resonating near 500 kc, so that while Rx may approximate a few hundred or few thousand ohms, the effective Q may be in the range of 10,000 to 30,000. Considering the properties of resonant circuits, such as Q would provide a band width of 20 to 50 cycles at 500 kc. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 39 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Figure: (a) Equivalent electrical circuit for a piezoelectric crystal; (b) reactance curves for the circuit of (a). The resistance of the crystal is due largely to mechanical damping introduced by the electrodes and by the surrounding atmosphere. By placing a crystal in an evacuated container, the value of Q can be notably increased. The electrodes are normally electroplated onto the crystal faces and need not introduce much damping. Capacitance C s is the equivalent series capacitance of the crystal forming a resonant by the crystal electrodes. The values of C s , and C p are such that C p >> C s , so that resonant and anti-resonant frequencies of the circuit lie very close together, differing by a fraction of 1 per cent of the resonant frequency. The reactance curve sketch of fig. (b) shows the resonant frequency below the antiresonant one. By placing adjustable capacitors in parallel with the crystal, C p can be increased, resulting in the antiresonant frequency being moved closer to the resonant point. Since the crystal represents either a resonant or antiresonant circuit, it may be used to replace the normal elements of the band-pass or band-elimination filter. as previously shown for band-pass action, the resonant frequency of one arm must equal the antiresonant frequency of the other arm. Figure: (a) Circuit of a lattice crystal filter with series inductors and parallel capacitors; (b) the electrical equivalent of (a). The pass band with crystal elements will then be found to extend from the VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 40 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH lowest crystal resonant frequency to the highest crystal antiresonant frequency, or a width of pass band equal to twice the separation of the resonant and antiresonant frequencies of one crystal. This range will result in a pass band a fraction of 1 per cent wide. The band width can be reduced by putting adjustable capacitors in parallel with the crystal, furnishing a means of adjustment of the width of the pass band. By the addition of coils in series with the crystals the pass bands may be widened. Since the added coils have Q values very much below those of the crystals, there will be some loss in sharpness at cutoff. A circuit including series coils is shown in fig. (a), with its equivalent drawn at (b). The reactance curves for the A and B portions of this circuit are drawn in fig. (a), which shows how the resonances and antiresonances are arranged. The presence of the series coil adds an additional resonance, and the pass band exists from the lowest resonance of one crystal to the highest resonance of the other. If f 1 and f 2 are the frequencies of resonance of one of the circuits and f R is that of the antiresonance, then s 1,2 R p C f f 1 C · m The separation of f 1 and f 2 represents two-thirds of the pass band and is seen to depend on the s p C / C ratio. Since s p C / C may be of the order of 0.01, it can be seen that the separation of f 1 and f 2 be of the order of 0.10 f R , or 10 per cent of the resonant frequency. By placing coils in series with the crystals, it has been possible to widen the pass band considerably. By adjustment of C p it is then possible to narrow the band to any desired amount. Figure: (a) Reactacne curves for the circuit of fig. (a); (b) attenuation curves for that circuit. Thus the use of coils permits the bands to be widened to pass speech frequencies, and crystal filters are quite generally used to separate the various VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 41 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH channels in carrier telephone circuits, in the range above 50 kilocycles. 11. Design a composite low pass filter to meet the following specifications. The fitler is to be terminated in 500 ohms resistance and it is to have a cutoff frequency of 1000 Hz with very high attenuation at 1065 and 1250 Hz. Solution: Given R= 500 ; fc = 1000 Hz; f∞ = 1065 Hz. f∞2 = 1250 Hz. i) Design of low pass constant – kT section 500 0.159 1000 1 1 0.63 . 1000 500 R I H fc C f fc R µ · · · ∏ ∏× · · · ∏ ∏× × The assembly of this filter is a shown below with inductance L/2 in each series arm. ii. Design of m – derived low pass T- section (Ref fig (b) a) for 2 2 2 1 1 1065 1000 m 1 1 0.344 1065 t f fc m f ω ∞ · 1 1 · − · − · · 1 1 ¸ ] ¸ ] The components of m- derived lowpass T – section filter for m = 0.344 are VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 42 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) 2 2 0.344 1.59 0.0273 2 2 0.344 0.636 0.218 0.344 1 . 0.159 4 4 0.344 = 0.102 H mL H mc f m L m µ × · · · × · − − · × The assembly of this m – derived lowpass t – section filter is as shown below 2 2 2 2 1000 1 1250 fc m f ∞ 1 1 · − · − 1 1 ¸ ] ¸ ] The components of m – derived low pass T – section filter for m = 0.6 are ( ) ( ) 2 2 0.6 0.159 0.048 2 2 0.6 0.636 0.3816 1 0.6 1 0.159 4 4 0.6 = 0.0424 H mL H mc f m m µ × · · · × · − − ⊥ × The assembly of the m – derived low pass T – section filter for m = 0.6 as shown below. Assembling all the three sections we will get the desired composite filter. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 43 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH In the 3 rd section i.e., for m = 0.6, the filter is divided into sections so these value get changed. ( ) 2 1 2 . 0.0424 2 0.0848 4 0.218 0.109 2 2 m L m mc f and f µ µ − × · × · · · Combine the elements with ever possible. The series inductors may be added and the resulting final design I as shown below. 12. i. Describe a prototype T section band stop filter. ii. Determine the formulae required for designing band stop filter. iii. Explain the advantages of m – derived band stop fitler. i. Band stop filter: A band stop or bad elimination filter attenuates a certain range of frequencies and passes all other frequencies there fore a band stop action may be thought of as that of a low pass filter in parallel with a high pass filter in which the cut off frequency of low pass filter is below that of high pass filter. ii. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 44 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH T-type ( ) ( ) ( ) 1 1 1 1 1 1 2 2 1 1 2 1 1 1 1 1 2 2 2 2 2 2 2 2 1 1 1 1/ 1/ 1/ 1/ 1 LC LC LC L C Z j L j C j L C j LC C j L j C j L j L j C L C ω ω ω ω ω ω ω ω ω ω ω ω ω · · · · + · − · − · · + − If the filter is to be constant K type Z 1 Z 2 = RK 2 2 2 1 1 2 2 1 2 2 1 1 2 2 2 2 1 1 2 1 1 2 1 2 2 2 1 1 2 1 1 1; 4 4 4 LC j L j RK C L C LC L C L L RK C C Z Z Z L Z Z Z Z RK ω ω ω ω 1 − · · 1 − ¸ ] · · − · · − · − · − Z 1 at lower cut off fi = Z 1 at upper cut off f 1 2 1 2 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 j L j L j C j C L L C C ω ω ω ω ω ω ω ω 1 − + · − + 1 ¸ ] − − · + VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 45 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) ( ) 2 2 1 1 1 2 1 1 2 1 1 2 0 2 2 1 1 2 2 2 0 2 0 2 2 1 1 2 2 2 0 0 0 2 0 1 2 1 2 1 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 2 1 2 k k LC LC LC f f f f f f f f f f f f f fr f f Z JR j L jR j C ω ω ω ω ω ω ω ω ω ω ω ω ω − · − · 1 − · − 1 ¸ ] 1 − · − 1 ¸ ] + + + · · − + · − 1 1 1 1 2 1 1 1 1 1 2 1 2 k k j L jR C C L R C ω ω ω ω 1 − + · 1 ¸ ] − · 2 1 1 1 2 0 2 1 1 1 2 2 4 k k r R f C f f R f C f ω ω − · ∏ · ∏ 2 1 1 1 2 4 k f f C R f f − · ∏ ( ) ( ) ( ) 1 1 2 2 0 1 2 1 2 1 2 1 2 2 1 2 2 1 2 2 2 1 1 1 4 4 1 k k k k k LC f f R L f f R f f L C R f f L C R R f f ω π · · ∏ · ∏ − · · ∏ · · − iii) Advantage of m – derived filters VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 46 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1. The attenuation does not increase rapidly beyond the cut off frequencies. 2. Characteristic impedance varies widely in the transmission or pass band, from the desired value the design impedance R k . 13. For a given T – section low pass filter, determine the cut-ff frequency and normal characteristic impedance Re. Solution: Given : L= 80 mH ; c = 0.02 µf 1 c f LC · ∏ Cut off frequency ( ) ( ) 3 6 3 1 80 10 0.02 10 7.962 10 c c f Hz f Hz − · ∏ × × · × 3 0 6 3 0 80 10 0.02 10 2 10 L R C R ohms − − × · · × · × 14. Design a constant – k low pass T and II – sections fitlers having cut-off frequency = 3000 Hz and nominal characteristic impedance R0= 600 ohm. Solution: Given f c = 3000 Hz; Ro = 600 0 0 600 1 3000 5 63.68 1 1 600 3000 0.1753 c c R L f L mH C R f C f µ × · · ∏ ∏× × · · · ∏ ∏× × · Hence the require T and II- section lowpass filter are 63.68 mH VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 47 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 15. Design a constant – k low pass filter having Fc= 2000 Hz and nominal characteristic impedance R0=600Ω. Also find the frequency at which this filter offers attenuation of 19.1 dB. Solution: Given: f c = 2000 Hz: R0 = 600 ohm ; α = attenuation = 19.1 dB. 0 600 600 2000 95.54 c R L f L mH · · ∏ ×∏× · 0 1 1 600 2000 0.263 c C F R f C F µ · · ∏ ∏× × · Attenuation of 19.1 dB is expressed in nepers as 19.1 2.2 nepers 8.686 α · · ( ) -1 2 cosh / 2.2 cosh : 1.6685 2 1.6685 2000 3337 c f fc f f fc f F f Hz α · · · ∴ · × · 16. Design a T- section constant – K high pass filter having cut off frequency of 10KHz and nominal characteristic resistance of Ro=600 om. Find i) its characteristic impedance and phase constants at 25 KHz and ii) Attenuation 5KHz. Solution: Given Fc = 10 KHz ; R0 = 600 ohm 0 600 4 4 10 103 4.77 c R L f L mH · · ∏ ∏× × · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 48 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 3 0 1 1 4 . . 4 10 10 600 0.0132663 c C F R f C F µ · · ∏ ∏× × × · Each capacitor in the series arm T – section is 2C= 0.02652 µF and the inductor in shunt arm is L=4.777 mH. At f= 25 KHz ( ) ( ) 2 0 2 -1 1 10 600 1 25 545 2 sin / 2sin 10/ 25 47.2 =47.52 /180=0.0824 radians OT OT OT c fc Z R f Z Z f f or β β β − 1 · − 1 ¸ ] 1 · − 1 ¸ ] · Ω · · · ×∏ In the attenuation band, α is given as, ( ) ( ) -1 -1 2 cosh / 2 cosh 10/ 5 2.6 nepers c f f nepers nepers α α · · · 17. Determine a prototype band pass fitler section having cut off frequencies of 2000 Hz and 500 Hz and nominal characteristic impedance of 600 ohms. Solution: ( ) ( ) ( ) 0 1 2 1 1 1 2 1 1 1 2 600 5000 2000 63.68 31.84 2 5000 2000 4 , , 4 600 5000 2000 0.0381 0.0762 a R L f f L mH L mH f f C F F R f f C F ICI F µ µ · · ∏ − ∏ − · · − − · · ∏ ∏× × × · · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 49 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) ( ) ( ) 0 2 1 2 0 2 2 0 2 1 2 600 5000 2000 4 . . 4 5000 2000 14.33 1 1 600 5000 6000 0.1769 F c R f f L H R f L mH C R f f C µ − × − · · ∏ ∏× × · · · ∏ − ∏× × − · Hence the band pass filter is as shown below. UNIT – II PART – A 1. What is transmission line? Energy can be transmitted either by the radiation of free electromagnetic waves as in the radio or it can be constrained to move or carried in various conductor arrangement known as transmission line. It is a conductive method of guiding electrical energy from one place to another. 2. What are lumped parameters and distributed parameters? The parameters which are physically separable and can be shown to be at one place in the circuit in the lumped form are lumped parameters. The parameters which are not physical be separable and are distributed all over the length of the circuit like transmission line are called distributed parameters. 3. State the properties of infinite line. 1. No waves will ever reach receiving end hence there is no reflection. 2. The Zo of the sending end decides the current flowing when voltage is applied VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 50 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Z R has no effect on the sending current. 4. What is short line? The short line means a practical line with finite length and the word short does not reflect anything about the actual length of the line. 5. Sketch the group of Zo agains w. Zo = o 0 R+jwL G+jwC when w 0, R L Z and w Z G C → → →∞ → Practically R L is alwas higher then G C 6. What is called an infinite line? The analysis of the transmission of the electric waves along any uniform and symmetrical transmission line can be done in-terms of the result existing for an imaginary line of infinite length, having electrical constants per unit length identical to that of the line under consideration. 7. Discuss the importance of smooth line. A line terminated in its characteristic impedance Ro is called properly terminated line which acts as a smooth line. Because of proper termination, there is no mismatch of impedance. Hence no reflection takes place. Thus no standing waves are produced. Then the maximum power transfer from generator to load is possible. 8. Draw the equivalent circuit of a unit length of transmission line. 9. Define characteristic impedance. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 51 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH When a finite transmission line is terminated with ZO and the input impedance is also ZO, then ZO is known as characteristic impedance. i.e the input impedance of an infinite line is characteristic impedance of line, 10. Define return loss. Return loss is defined as the ratio of power at the receiving end due to incident wave to power due to reflected wave by the load. Returns loss = 20 log R o R o Z Z Z Z + − db 11. Define reflection factor? Reflection factor is defined as the ratio which indicates the change in current in the load due to reflection at the mismatched junction. K= 2 R o R o Z Z Z Z + 12. Express reflection factor in terms of impedance. Reflection factor = 1 2 1 2 2 Z Z Z Z + where Z1 and Z2 are the impedance seen looking both ways at any junction. 13. Find out the value of Reflection coefficient (K) for the following; i) Properly matched condition - |K| = 1 ii) Short circuited line - |K| = 0 iii) Open circuited line -|K| = 0 14. What is Campell’s equation? It makes possible the calculation of the effects of loading coils in reducing attenuation and distortion. 1 sinh 2 Zc CoshN N CoshN Zo λ λ λ · + when ZC → loading coil impedance Zo → Characteristic impedance N → Length of Transmission line section λ → Propagation constant 1 λ → Modified propagation constant. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 52 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 15. What are the basic application of transmission lines? i. It is used to transmit energy ii. It can be used a circuit elements like L>C. and resonant circuits. iii. It can be used as filters, transformers, measuring devices. 16. State the condition for minimum attenuation with L and C variable. With ‘L’ variable , L = CR G With ‘C; variable , C = LG R In general, RC = LG 17. Give the types of transmission line The commonly used transmission lines are: i) Open wire line. ii) Co-axial line iii) Strip line iv) Waveguides v) Optical fibers. 18. Write short notes on transmission line. The open wire transmission line consists of two conductors spaced at a certain distance apart. The spacing between the conductors is large in comparison to the diameter of the line conductors. As a result these lines can operate at higher frequencies. This type of line is used in transmission of electric power, telegraphy and telephony, The conductor of open wire line is as shown below. 19. Give the advantages of open wire transmission line. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 53 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The advantages of open wire line are: (i) Simple to construct and low cost. (ii) Insulation between the line and conductor is air so, the dielectric loss is extremely small. (iii) It is balanced with respect to ground 20. What are the disadvantages of open wire transmission line? (i) These lines are unsuitable for use at frequencies above 100MHz. Because energy loss takes place due to radiation. 21. Write short notes on co-axial line (or) Coaxial cable. To avoid radiation losses taking place in open wire lines at high frequencies a closed field configuration is used by surrounding the inner conductor with an outer cylindrical hollow conductor and the arrangement is termed as a coaxial cable. 22. Give the advantages and disadvantages of co-axial transmission line. Advantages: (i) The electric and magnetic fields are confined with in the outer conductor so the radiation losses are eliminated. (ii) It provides outer shielding from outer interfacing signals. (iii) The co-axial line can be used up to the frequency range of about 1 GHz for transmission of signals. Disadvantages: (i) They are costlier than open wire line. (ii) Beyond 1GHz these cables cannot be used because losses in the dielectric increases with frequency. 23. What do you mean by Waveguide? Wave guides are hollow conducting tubes of uniform cross section used for U. H. F. Transmission by continuous reflection from the inner walls of the guide. 24. Give the application of microwaves (i) Satellite communication (ii) Telemetry (iii) Transmission of video signals (iv) Microwave oven. 25. What are the advantages and disadvantages of Waveguide? The advantages of waveguides are: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 54 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH i) In wave guides, no power is lost through radiation because the electric and magnetic fields are confined to the space within the guides. ii) The dielectric loss is negligible. iii) Frequencies of the wave higher than 3GHz can be easily transmitted iv) Several modes of electromagnetic waves can be propagated with in a single Waveguide. The Disadvantages of wave guides are: (i) Cost of the wave guide is so high (ii) The wave guide walls should be specially plated to reduce resistance to avoid skin effect and power loss. 26. What is an optical fiber? An optical fiber is a dielectric wave guide that operates at optical frequencies. It confines electromagnetic energy in the form of light within its surfaces and guides the light in a direction parallel to its axis. 27. What are the advantages and disadvantages of optical fiber? Advantages: (i) Low transmission loss and very high band width (ii) Small size and weight (iii) No radio frequency and electromagnetic interference (iv) Ruggedness and flexibility Disadvantages: (i) It is difficult to run cables where the bending occurs. (ii) Different specialized techniques have to be followed to join ends of two cables. 28. What are the four important parameters of a transmission line? The main four parameters of transmission line are Resistance - R Inductance - L Capacitance - C Conductance - G 29. Define propagation constant of uniform line. The propagation constant per unit length of a uniform line is defined as the natural VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 55 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH logarithm of steady state vector ratio of current or voltage at any point, to that at a point unit distance further from source, when the line is infinitely long 30. When a transmission line is said to be uniform? A line is said to be uniform, when the primary constants R, L, C and G are uniformly distributed along the entire length of transmission. 31. Name the secondary constants of transmission line. Characteristic impedance Zo and Propagation constant P (or) γ are the Secondary constants of transmission line 32. What is the value of characteristic impedance of open-wire line? The characteristic impedance of open-wire line is Z o = 276 log 10 S r ohms Where ’S ‘ is the spacing between two wires-centre to centre ‘r’ is radius of either of the wire. 33. What is the value of characteristic impedance of coaxial cable? The characteristic impedance of co-axial cable Is 10 138log D Zo d · ohms Where ‘D’ is inner diameter of outer conductor ‘d’ is diameter of inner conductor 34. Define velocity of propagation. Velocity of propagation is defined as the velocity with which a signal of single frequency propagates along the line at a particular frequency ‘f’ . It is denoted as Vp and its unit is km/sec. 35. Define group velocity. Group velocity is defined as the velocity of envelope of a complex signal. (or) It is a velocity with which a signal produced by variation of a steady-state wave or by introduction of group frequencies. It is denoted as Vg. 36. What is loading of a transmission line? VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 56 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The process of achieving the condition RC = LG either by artificially increasing ‘L’ or decreasing ‘C’ is called loading of a line. 37. What are the advantages of lumped loading? The advantages of lumped loading are • There is no practical limit to the value by which the inductance can be increased. • Cost is small • Hysterisis and eddy current losses are small. 38. What are the disadvantages of reflection? 1. Reduction in efficiency, 2. If the attenuation is not large, then the reflected wave appears as echo at the sending end. 3. The part of received energy is rejected by the load, hence output reduces. These are the disadvantages of reflection. 39. What is frequency distortion? A complex applied voltage, such as a voice voltage containing many frequency will not have all frequencies transmitted with equal attenuation of the received wave form will not be identical with the input wave form at the sending end. This variation is called frequency distortion. 40. What is phase or Delay Distortion? All frequencies applied to a transmission line will not have the same time of transmission, some frequencies being delayed more than others. For an applied voice- voltage wave, the received wave will not be identical with the input wave form at the sending end, since some components will be delays more than the other. This phenomenon is called Delay distortion. 41. Define Reflection co-efficient. The ratio of amplitudes of the reflected of incident voltage waves at the receiving end of the line is called reflection co-efficient. Reflected voltage at load K Incident voltage at load · 42. What do you mean by Insertion loss? VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 57 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH It is defined as number of repers or decibels by which the current in the load is changed by the insertion. 43. What is return loss? Return loss is defined as the ratio of power at receiving end due to incident wave and power due to reflected wave in the load. R o R o 2 2 Reflection Loss = 2 2 + − It is the reciprocal of Reflection co-efficient 44. Write down the expression for transfer impedance ( ) l l s R o T R T R o E 2 2 Z e Ke I 2 Z z cos h l+z sin h l γ −γ + · · + · γ γ 45. What do you mean by Lumped circuits? The network where in the resistance, inductance and capacitance are individually concentrated or lumped at discrete points in the circuit is called Lumped circuit. 46. Write short notes on co-axial cable. One conductor is a hollow tube, the second conductor being located inside of co-axial with the tube. PART – B 1. Derive the General solution a transmission line. A transmission line is a circuit with distributed parameters hence the method of analyzing such circuit is different than the method of analysis of a circuit with lumped parameters. It is seen that the current and voltage varies from point to point along VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 58 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH the transmission line. The general solution of a transmission line includes the expressions for current and voltage at any point along a line of any length having uniformly distributed constants. The various notations used in this derivation are, R = Series resistance, ohms per unit length, including both the wires. L = Series inductance, henrys per unit length. C = Capacitance between the conductors, farads per unit length. G = Shunt leakage conductance between the conductors, mhos per unit length. ω L= Series reactance per unit length. ω C = Shunt susceptance in mhos per unit length Z = R + J ω L = Series impedance in ohms per unit length. Y = G + j ω C = shunt admittance in mhos per unit length. S = Distance out to point of consideration, measured from receiving end. I = Current in the line at any point. E= Voltage between the conductors at any point. l = Length of the line. The transmission line of length l can be considered to be made up of infinitesimal T section. One such section of length ds is shown in the Fig.4.17. It carries a current I. The point under consideration is at a distance a from the receiving end. The length of section is ds hence its series impedance is Zds and shunt admittance is Yds. The current is I and voltage is E at this section. The elemental voltage drop in the length ds is dE = I Zds dE ds ∴ · I Z ………(1) the leakage current flowing through shunt admittance from one conductor to other is given by dI = EY ds dI ds ∴ = EY …….(2) Differentiating equation (1) and (2) with respect to S we get VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 59 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 1 2 d E dI Z ds ds and d I dE Y ds ds · · This is because both E and I are function of S. 2 2 ..........(3) d E ZEY ds · IZ 2 2 .....(4) and d I YIZ ds · The equations (3) and (4) are the second order differential equations describing the transmission line having distributed constants, all along its length. It is necessary to solve these equations to obtain the expression of E and I. Replace the operation d/dS by m hence we get. (m 2 –ZY) E = 0 but E ≠ 0 m = ZY t …….. (5) Same result is true for the current equation. So, there exists two solutions for positive sign of m and negative sign of m. The general solution of the equations for E and I are, ..........(6) ( 1) ...........(7) E Ae ZYS Be ZYS I e ZYS e ZYS · + − · + − Where A, B, C and arbitrary constants of integration. It is now necessary to obtain the values of A, ,B, C and D. As distance is measured from the receiving end S = 0 indicates the receiving end E = IE R and I = I R at S = 0 Substituting in the solution, VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 60 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH E R = A + B ….. (8(a)) I R = C + D …..(9(b)) Same condition can be used in the equations obtained by differentiating the equations (6) and (7) with respect to S. ( ) ( ) dE A ZYe ZYs B ZY e ZYs ds and dI C ZYe ZYs D ZY e ZYs ds But dE dI IZand Ey ds ds · + − − · + − − · · ( ) ...(9) ( ) ..(10) ( ) IZ A ZYe Zys B Zy e Zys andEY C ZYe ZYs D ZY e ZYs A B I ZYe Zys ZY e ZYs Z Z ∴ · − − − · − − − · − − − . . ....(11) .....(12) Y Y i e I A e ZYS B e ZYS Z Z Z Z andE C e ZYS D ZYS Y Y · − − · − − SHORT SECTION PQ, distance x from the sending end of transmission line. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 61 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Now use S = 0 , E = E R and I - I R ............(13( ) .........(13( ) R R Y Y I A B a Z Z Y Y andE C D b Z Z ∴ · − · − The equation 8a, 13a, 13b are to be solved simultaneously to obtain the values of the constants A, B, C and D. Now while solving these equations use the results, R R o R E R j L Z Z andZ I G j C Y ω ω + · · · + Hence the various constants obtained, after solving the equations simultaneously are, 1 .......(14) 2 2 2 1 ........(15) 2 2 2 1 .........(16) 2 2 2 O R R R R O R R R R R R R R O Z E I E Z A Y Z Z E I E Z B Y Z I E I Z Y C Z Z ¸ _ · + · + ¸ , ¸ _ · − · − ¸ , ¸ _ · + · + ¸ , 1 ............(17) 2 2 2 R R R R O I E I Z Y D Z Z ¸ _ · − · − ¸ , Hence the general solution of the differential equation is, 1 1 ..(18) 2 1 1 ..(19) 2 O O R R R R R R O O Z Z E E e ZYS e ZYS Z Z I Z Z E e ZYS e ZYS Z Z 1 ¸ _ ¸ _ · + + − 1 ¸ , ¸ , ¸ ] 1 ¸ _ ¸ _ · + + − − 1 ¸ , ¸ , ¸ ] VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 62 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Taking LCM as Z R and taking R O R Z Z Z + out from equation (18) ( ) ( ) ..(20) 2 2 ( ) R R O R R O R R R O E Z Z E Z Z I e ZYs e ZYs Z Z Z Z 1 + − · − 1 + ¸ ] Taking LCM as Zo and taking R O R Z Z Z + out from equation (19) ( ) ( ) ..(21) 2 ( ) R R O R O O R O I Z Z Z Z I e ZYS e ZYS Z Z Z 1 + − · − 1 + ¸ ] The negative sign is used to convert Z o – Z R to Z R - Z o The equation (20) and (21) is the general solution of a transmission line. Another way of representing the equation is ( ) Re Re ( ) 2 2 R R O R O R R Oe R Oe E E Z Z e ZYs Z Z e ZYs Z E E Z ZYs Z ZYs Z ZYs Z Z ZYs 1 · + + − ¸ ] 1 · + + − − ¸ ] − But 2 R R R R R R R E E Z hence I I Z e ZYs e ZYs E E · · 1 + − ∴ · + 1 ¸ ] ......(22) 2 R O e ZYs e ZYs I Z 1 + − 1 ¸ ] VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 63 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH and 2 .............(23) 2 R R O e ZYs e ZYs I I E e ZYs e ZYs Z 1 + − · + 1 ¸ ] 1 + − 1 ¸ ] But ( ) ( ) 2 2 1 ....(24) R R O e ZYs e ZYs COSh ZYsand e ZYs e ZYs Sinh ZYs E E Cosh ZYs Z Sinh ZYs + − · + − · ∴ · + and I ( ) ( ) / sinh ..........(25) R R O I Cosh ZYs E Z ZYs The equation (24) and (25) give the values of E and I at any point along the length of the line. Important Note: The similar equations can be obtained in terms of sending and voltage Es and Is. If X is the distance measured down the line from the sending end then, X = 1 – s And the equation (24) and (25) get transferred in term Es and Is as ( ) ( ) ( ) ( ) / S S O S S O E E Cosh ZYx I Z Sinh ZYx I I Cosh ZY x E Z Sinh ZYx · + · + And ZY γ · as derived earlier and hence equation can be written in terms of propagation constant γ . Summarizing. If receiving end parameters are known and s is distance measure from the receiving end then, E= E R cos h ( γ s ) +I R Zo sin h ( γ s) VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 64 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH I = I R cosh ( γ s) +ER /Zo sin h ( γ s) And if sending end parameters are known and X is distance measure from the receiving end then, E = Es cosh ( γ x ) + Is Zo sinh ( γ x ) I = Is cosh ( γ x ) +Es/Zo sinh( γ x) Any set of equations can be used to solve the problems depending on the values given. 2. Explain the physical significance of General solution. From the qeuesed solutions, the sending end current can be obtained by substituting S = I measured from the receiving end. Es = E R cosh ( γ I ) + I R Zo sinh ( γ I ) ………(1) Is = I R [cos h ( γ I) + E R /Zo sin h ( γ I ) ……..(2) Now Z R = E R / I R ∴Is = [ I R cos h ( γ I) + Z R /Zo I R sin h ( γ I ) ] ∴ Is = [ I R cos h ( γ I )+ Z R / Zo sinh ( γ I ) ] ……(3) Now if the line is terminated in its characteristic impedance Zo then, Is = [ I R cosh ( γ I ) + sinh ( γ I ) ]…. As Z R = Zo Is / I R = [ cosh ( γ I ) + sinh ( γ I ) ]= e W …….(4) This is the equation which is already derived for the line terminated in Zo. Using E R = I R Z R in equation (1), Es = Z R I R cosh ( γ I ) + I R Zo sinh ( γ I ) Es = I R [ (Z R cos h ( γ I ) + Zo sinh ( γ I ) ]…..(5) Dividing (5) by (3), [ ] [ ] cosh( ) sinh( ) cosh( ) / sinh( ) R R O s s R R O I Z I Z I E I I I Z Z I γ γ γ γ + · + But Es / Is = Zs VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 65 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH [ ] [ ] cosh( ) sinh( ) cosh( ) sinh( ) O R O S O R Z Z I Z I Z Z I Z I γ γ γ γ + ∴ · + When the line is terminated in Zo then Z R = Zo SO substituting in equation (6) we get Zs = Zo This shows that for a line terminated in its characteristic impedance, it input impedance is also its characteristic impedance. Now consider an infinite line I → ∞ Using this in equation (6) we get, [ ] [ ] tanh( ) tanh( ) tanh( ) ........(8) O R O S O R S O Z Z Z I Z Z Z I and I IasI Z Z γ γ γ + · + → →∞ ∴ · This shows that finite line terminated in its characteristic impedance behaves as an infinite line, to the sending end generator. Thus the equations for Ex and Ix are applicable for the finite line terminated in Zo. The equations are reproduced here for the convenience of the reader. Ex = E s e –yx and I x = Is e –yx If in practice instruments are connected along the line then the instruments will show the magnitude Es e –yx and Is e –yx while the phase angles cannot be obtained. If the graph for Ex or Ix is plotted against x then it can be shown. This is the physical significance of the general solution of a transmission line . Its use will be more clear by studying the various cases of the line. 3. State and explain different types of distortions in line. When the received signal is not the exact replication of the transmitted signal then the signal is said to be distorted. There exists some kind of distortion in the signal. There are three types of distortions present in the transmitted wave along the transmission line. 1. Due to variation of characteristic impedance Zo with frequency. 2. Frequency distortion due to the variation of attenuation constant α with frequency. 3. Phase distortion due to the variation of phase constant with frequency. Distortion due to Zo varying with Frequency: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 66 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The characteristic impedance Zo of the line varies with the frequency while the line is terminated in an impedance with does not vary with frequency in similar fashion as that of Zo. This causes the distortion. The power is absorbed at certain frequencies while its gets reflected for certain frequencies. So there exists the selective power absorption, due to this type of distortion. It is known that, (1 / ) (1 / ) O R j L R j L R Z G j C G j C G ω ω ω ω + + + · · + + If for the line, the condition LG += CR is satisfied the L/R = C/G and hence (1 / ) (1 / ) / / O J L R J C G Z R G O L C O ω ω + · + ∴ · < · < Ω o o For such a line Zo does not vary with frequency ω and it is purely resistive in nature. Such a line can be easily and correctly terminated in an impedance which matches with Z o at the frequencies for such a line / / R Z R Gor L G · . This eliminates the distortion and hence selective power absorption. 4. Write brief notes on lumped loading. In this type of loading the inductors are introduced in lumps at the uniform distances in the line. Such inductors are called lumped inductors. The inductors are introduced in the limbs to keep the line as balanced circuit. The lumped inductors are in the form of coils called loading coils. The lumped loading is preferred for the open wire lines and cables for the transmission improvement. The loading coil design is very much important in this method. The core of the coil is usually Toro dial in shape and made if dimensions, very low eddy current losses and negligible external field which restricts the interference with neighboring circuits. The loading coil is wound of the largest gauge of wire consistent with small size. Each winding is divided into equal parts, so that exactly half the inductance can be VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 67 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH inserted in to each leg of the circuit. These are built into steel sizes to accommodate one or more coils. The pots protect the coils from external magnetic fields, weather and mechanical damage. The fig. shows the construction of loading coils. White installing the coils, the care must be taken so that the circuit balance is maintained. No winding is reversed. If winding is, it will neutralized the inductance of other winding reduced the overall inductance. In the case of lumped loading. The line behaves properly provided spacing is uniform and loading is balanced, up to a certain frequency called cutoff frequency of the line. Upto this frequency, the added inductance behaves as if it is distributed uniformly along the line. But above this cut-off frequency the attenuation constant increases rapidly. The line acts as low pass fitter. The graph of ∝ against the frequency called the attenuation frequency characteristics of the line show in the fig. It can be see that for continuous loading the attenuation is independent if frequency while for lumped loading it increases rapidly after the cut-off frequency. If the loading section distance is d than keeping inductance LS of the loading coil constant, cut off frequency is found to be proportional to the *. Hence to get the higher cut off frequency, small lumped inductance must be used at smaller distances. 5. Write short notes on different types of transmission lines. Transmission is the process it transmitting some signals from one place to another. Here it can be data as in the case of transmission of computer data along telephone lies or it can be audit/video signals from radio or television broad cost. Electrical energy can be transmitted from one point to another by one of the two methods namely 1. By radiation of electromagnetic waves through free space. 2. By use of electrical conductor arrangement known as transmission line. “Transmission line is a conductive method of guiding electrical energy from one place to another”. In communication these lines are used as link between transmitter and receiver. Transmission lines may be grouped as lumped lines or distributed lines while in distributed liens, the electrical parameters like inductance, capacitance, resistance and conductance are distributed uniformly across the entire line length white in lumped lines these parameters are jumped at intervals along the line. The commonly used transmission lines are 1. Open wire line. 2. Coaxial line. 3. Strip line. 4. Wave guides 5. Optical fibers. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 68 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1. Open wire line: These lines consist of two conductors spaced at a certain distance apart. This type of line is used in transmission of electric power, telegraphy and telephony signals. The construction of open wire line is as shown below. These lines have distributed sonic resistance and inductance, shunt capacitance and conductance. The electric and magnetic fields existing in the lines are shown below. The spacing between the conductors is large in comparison to the diameter of the line conductors. As a result these lines can operate at high voltage. However when operating at higher frequency the larger spacing proves to be a disadvantages because radiation of energy from the open wire line takes place. However the radiation loss can be minimized by reducing the spacing between wires. The primary constants of a open wire transmission line are, resistance, inductance, capacitance and conductance per unit length can be given as, R = 2p /r π 2 ohms. L = µ ο / π log e (S / r ) Henries. Where, P – Specific resistance r – radios S – Spacing between conductors. Advantages of open wire lines: 1. Simple to construct and low cost 2. Insulation between the line conductors is air. As a result the electric loss is extremely small 3. It is balanced with respect to ground. Disadvantages: 1. These lines are unsuitable for use at frequencies above 100 Mhz because energy loss takes place due to radiation. 2. Coaxial lines (or) Coaxial cables: The avoid radiation losses taking place in open wire lines at high frequencies, a closed field configuration is used by surrounding the inner conductor with an outer cylindrical hollow conductor and the arrangement is termed as a coaxial cable. The construction of coaxial cable, the electric magnetic field existing in a coaxial cable s as shown above and below. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 69 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH A typical air inner copper conductor held in position by insulating discs Flexible cables have polythene dielectric and outer conductor in the form of copper braided or flexibility. It ‘D’ and ‘d’ are diameters of the outer and inner conductor, and also µ and ε be the permeability and permittivity of the insulting medium, then the characteristics impedance P of the cable is given by 10 138 / log ( / ) o Z r r d d µ · ∈ the velocity of propagation 8 3*10 / V r r µ · ∈ Advantages: 1. The electric and magnetic fields are continued with in the conductor there by eliminating radiation losses. 2. It provides effective shielding from outer interfering signals. Disadvantages: 1. They are costlier then open wire lines. 2. Beyond 1Ghz these cables cannot be used because losses in the dielectric increases with frequency. STRIP LINES VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 70 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH A form of line using finite plates and an inverting dielectric medium is turned as strip line. Strip lines are two types: 1. Trip late 2. Micro strip. .Trip late Line: This line resembles as a co-axial line in which side conductors have been removed. The energy propagation in this line is in the form of TEM waves, provided. The distance b/w the centre and outer plate is smaller compared with the wave length of the signal Disadvantages: 1. It is costlier and it also requires a manufacturing skill. Micro strip line: The micro strip line has a narrow conductor supported by a dielectric. The bottom conducting plates serve as earth plate. The micro strip line is widely used in microwave integrated circuits these components find wide application in couplers circulators and receiver. Micro strip lines are fabricated on fiber glass or polystyrenes printed circuit boards as about 1.5mm thickness with copper strips. 3mm wide. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 71 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Wave guides: Wave guides are hollow conducting takes a uniform gross section used for U. H. F. transform by continuous reflection from the inner walls of the guide. Wave guide are used to minimize losses and for high power transmission at microwave freq. The shape of the wave guide may be rectangle or cylindrical thro which electromagnetic waves are propagated. The propagation takes place in open wire and co-axial lines propagation takes place in the form d transverse electric (TE) and transverse magnetic ™ waves. Rectangular waveguide Elliptical waveguide. Advantages: 1. In wave guide no power is cost throughout, because she elastic and magnetic fields are confined to the space with in the guides. 2. The dielectric loss is negligible. 3. Several modes of electro magnetic waves can be propagated with in a single wave guide. 4. Frequency of the wave higher than 39 hz can be easily transmitted. Disadvantages: 1. Cost of the wave guide is very high. 2. Wave guide walls should be specially plated to reduce resistance to avoid skin effect and power less. Optical fibers: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 72 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The entrance appearance of optical fibers are similar to C0-axial cables. Here the copper cores are replaced by highly put e – glass (or) silica which is used to carry modulated light energy similar to micro wave energy. An optical fiber is a dielectric wave guide that operated at optical frequency. It confines electromagnetic energy in the form a light to within its surfaces and guide the light in a direction 11 d to the axis. The structure of a time cable is show below. Advantages: 1.Low transmission loss and high band width. 2. Electrical isolation is there. 3. No radio frequency and electromagnetic item. 4. High degree of data securing is afforded. 5. Small sue and weight. Disadvantages: 1. The cost of fiber able is very high (Rs.7000/- per meter). 2. It is difficult to run the cables where the bending occurs. 3. Different specializes technique have to the followed to join ends of two cables. 6. Explain about transmission line parameter: Transmission line parameter: In a open wire line, when a current is passed through it, a magnetic fields are produced around the conductors and the voltage drop occurs along the line similarly when a voltage is passed through a open wire line, an electric field is produced b/w the two conductor. The magnetic field proportional to the usual indicate that the line L as series inductance ‘L’ and the voltage drop indicate the presented series resistance ‘R’ Similarly the electric field proportion to the voltage indicates that the line contains VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 73 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH shunt capacitance ‘C’ and this capacitance is new useless (or) perfect so, the line also contains conductance ‘G’ These four parameter R, L, C and G are distributed along the whole length of the line. The four line parameters, R, L, C. and G are termed as primary customs of a transmission line. They are defined Resistance ‘R’ Resistance ‘R’ is designed as loop resistance per unit length of line. They it is the sun of resistance of both the wire for unit line length. It is ohm/kn. Inductance ‘L’ Inductance t ‘L’ is defined as loop inductance per unit length of line. Thus it is the sum of inductance of both wire for unit length. Its unit is HCGrier /km. Conductance ‘G’ Conductance ‘G’ is define as shunt conductance b/w the two wires per unit length of line. Its unit is mhos / km. Capacitance: Capacitance is defined as shunt capacitance b/w the two wipes per unit length: It’s unit is farad / km Impedance ‘Z’ The series impedance of a transmission line per unit length is given as , Z = R + j ω L , Where, R – line resistance, and j ω L – Line reactance. Admittance: ‘Y’: The shunt admittance. ‘Y’ of a transmission line per unit length is given at y = G’ + j ω c. Where, G – line conductance and j ω C – line susceptance. Characteristic Impedance ‘Zo’: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 74 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Characteristic impedance is defines as the ratio of square root of impedance to admittances. / ( ) /( ) o o Z Z Y Z R j L G j C ω ω · · + + Propagation constant ’Y’: Propagation constant is defined at the product of square root of impedance and admittance. ( )( ) r ZY R j L G j L ω ω · · + + These characteristic impedance ‘Z o ’ and ‘r’ are called as secondary constants of transmission line. In addition to primary and secondary constants of a transmission line there are 3 more units of transmission line theory. Wave length: It is defined as the distance that the wave travels along the line in order that the total shift is 2 π radiance. It is denoted by ‘ λ ‘ and its unit is meter. λ = 2 π / β Group Velocity: The group velocity is defined as the velocity of the develop is a complete signal. It is denoted at √g. 2 1 2 1 ( ) /( ) g ω ω β β · − − Velocity of propagation: It is defined as the velocity with which a signal of single frequency propagation along the line at a particular frequency ‘f’. √g = λ f √g = 2 ∏ f/ β VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 75 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 7. Derive the expression for telephone cables: Telephone cables (or) telex lines are used as low frequencies transmission line. A telephone cable is formed by two wines insulated from each other by a layer oil impregnated paper and then twisted in pains. A large no of each pairs form an underground cable. Such transmission lines are called as telephone cables. At low frequencies, the series inductance reactance is quite negligible as negligible as compared to line resistance ‘R’. Similarly line conductance is also vary small as compared to susceptance substituting the condition in general equation for propagation constant (γ ) and the characteristic input (Zo) given. γ = ( )( )..........(1) 0; 0 XY R j L G j c subG L γ ω ω · + + · · γ = j RC ω γ = 45 ..........(2) cos 45 sin 45 Rc j RC j RC ω α β ω ω + · + o o o γ = / 2 / 2 / 2............(3) / 2............(4) RC j RC RC RC ω ω α ω β ω + · · Characteristic impedance ( )( ) o Z R j L G j L ω ω · + + 0 0 / / 45 ............(5) Z R j C Z R C and ω ω · · < o Velocity of propagation V p is given as γ p 4= ω / β VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 76 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH γ p = w/ / 2 2 / ...........(6) p RC V RC ω ω ⇒ · From the above equation its observed that α and √ are independent on frequency. This the higher frequency are attenuated move and travel faster than the lower frequency resulting in frequency and delay distortion. Hence large distortion occurs at higher frequencies in a telephone lines. 8. Explain insertion loss in detail and derive the expression for the same. Insertion loss occurs due to insertion of a network or a line in between source and load. If input impedance Z s is not equal to generator impedance Z g , then reflection loss occurs at terminals 1-1 1 . If Z R is not equal to Z o , then second reflection loss occurs at terminals 2-2 1 . The overall effect of insertion of a line is to change the current through the load and hence power delivered to load is less compared too power delivered to load when it was directly connected to generator. Thus insertion loss of a line or a network is defined as the number of ropers or dB by which the current in the load is changed by insertion of a line or a network between the load and the source. Consider the circuit in which generator of impedance Z g is connected to a load of impedance Z R . 1 (1) R g R E I Z Z · → + The line is inserted between load and generator. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 77 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Let Zs be input impedance of a line which different than Zg, (2) S g S E I Z Z · → + We know that input impedance of line is (3) (3) (2) ( ) ( ) l l S O l l S l l g O l l s l l l l g o e ke Z Z e ke sub in E I e ke Z Z e ke E I Z e ke Z e ke γ γ γ γ γ γ γ γ γ γ γ γ − − − − − − 1 + · → 1 − ¸ ] · 1 + + 1 − ¸ ] ∴ · − + + we know that ( ) ( ) (4) 2 l l R R o s o I Z Z I e ke Z γ γ − + · − → This equation is obtained from general solution of line by substituting s = l 2 (5) ( ) O S R l l R O Z I I Z Z e ke γ γ − · → 1 + − ¸ ] 0 2 ( ) (6) ( ) ( ) ( )( ) γ γ γ γ γ γ γ γ − − − − − · → − + + + − l l o l l l l g l l R o Z E e ke Z e ke Z e ke Z Z e ke we know that ref. 10eff. K= (7) R o R o Z Z Z Z − → + 0 0 2 ( ) R l l l l R R o R o o g R o R o Z E I Z Z Z Z Z Z Z e e Z e e Z Z Z Z γ γ γ γ − − ∴ · ¹ ¹ ¸ _ ¸ _ ¸ _ − − ¹ ¹ + + + − ' ) + + ¸ , ¸ , ¹ ¹ ¸ , ¹ ¹ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 78 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 ( ) o R l l l l R o R o R o o o g g R o R o Z E I Z Z Z Z Z Z Z e Z e Z e Z e Z Z Z Z γ γ γ γ · − − ¹ ¹ ¸ _ ¸ _ − − ¹ ¹ + + + − ' ) + + ¹ ¹ ¸ , ¸ , ¹ ¹ 0 2 (8) ( )( ) ( ) R l l R O o g R o Z E I Z Z Z Z e Z Z e γ γ − ∴ · → + + + − 1 1 0 . 2 ( )( ) ( _ )( ) R R g R R o R l l R o g R o o g I Insertionloss I E Z Z I I loss Z E I Z Z Z Z e Z Z Z Z e γ γ − · + ∴ · · + + + − 1 ( )( ) ( )( ) (9) 2 ( ) l l R o o g R o o g R R o g R Z Z Z Z e Z Z Z Z e I I Z Z z γ γ − + + + − − · → + The length of line is usually very large hence 0 l e γ − → ∴2 nd term in numerator can be neglected. 1 0 ( )( ) (10) 2 ( ) l R o g R R o g R Z Z Z Z e I I Z Z Z γ + + ∴ · → + 0 ( )( ) 2 ( ) l j l R o o g g R Z Z Z Z e e Z Z Z α β + + · + ( ) j γ α β · + Q But insertion loss has too be calculated as a function of ratio of current magnitudes and hence e j β l can be neglected. 2 (11) 2 | || | 2 , l R o o g R R o g R g R Z Z Z Z e I I Z Z Z Xand by Z Z α + + ∴ · → + ÷ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 79 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 1 2 | || | 4 | || | | || | 2 (12) 2 2 | | α α + + · + + + ∴ · → + L g R R O O R R g R O R g O R O R l R R g O R O R Z Z Z Z Z Zg e I I Z Z Z Zg Z Z Z Z Z ZgZ I e I Z Z Z Z Zg Z All the terms on RHS are reflection factors. Let Ks = 2 | | g o g o Z Z Z Z + = reflection factor at source side →(13) 2 | | R o R R o Z Z k Z Z · + = reflection factor at load side →(14) 2 | | g R SR g R Z Z k Z Z · + = refl factor for direct connection →(15) l e α indicates loss in the line. 1 l SR R R S R k I e I k k α ∴ · Insertion loss = 1 1 1 1 . R R S R SR I lu lu lu l Nepers I k k k α 1 · + − + 1 ¸ ] (or) Insertion loss = 20 1 1 1 log log log 0.4343 S R SR l dB k k k α 1 + − + 1 ¸ ] The term corresponding to k SR is negative. It is the loss if generator and load would have been directly connected. It is not related to insertion hence it is subtracted from overall loss. 9. Prove that 0 . ( ) ( ) tanh oc SC sc oc i Z Z Z Z ii Z γ · · (i)Consider a short line terminated I its characteristic impedance Z o . The short line is a symmetrical network and hence can be represented by equivalent T – section. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 80 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH We know that finite line terminated in Z o behaves as an infinite line, hence Z o must be Z o The i/p impedance Z in of equivalent T network is 1 1 2 1 2 0 1 1 2 0 || 2 2 2 2 2 in o in in o Z Z Z Z Z Z z Z Z Z Z Z Z ButZ Z ¹ ¹ ¸ _ · + + ' ) ¸ , ¹ ¹ ¸ _ + ¸ , · + + + · 1 2 1 0 1 2 2 2 2 o o Z Z Z Z Z Z Z Z ¸ _ + ¸ , · + + + 1 1 1 2 0 1 2 2 0 2 2 1 1 2 2 2 1 1 2 2 1 1 2 2 2 2 2 2 2 2 2 4 4 o o o o o Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z 1 ¸ _ ¸ _ + · + + + + 1 ¸ ] ¸ , ¸ , ∴ · + ⇒ · + ∴ · + to obtain Z 1 and Z 2 we open circuit and short circuit the network In open circuit the line is kept open and input impedance is measured. 1 2 2 oc Z Z Z · + In short circuit the second end of line is short ed and input impedance is measured. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 81 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 1 2 1 2 1 1 2 || 2 2 2 2 2 sc Z Z Z Z Z Z Z Z Z 1 · + 1 ¸ ] · + + 2 1 1 2 1 2 1 2 4 2 2 Z Z Z Z Z Z Z + + · + 2 2 . . o sc c o o oc sc o sc oc Z Z Z Z Z Z Z Z Z ∴ · · ⇒ · (iii) we know that 1 2 2 2 1 1 1 2 2 2 2 2 1 1 1 2 2 2 1 2 1 2 4 1 (1) 2 2 o o Z Z e Z Z subZ Z Z Z e Z Z Z Z Z Z e Z Z Z γ γ γ · + + · + + + ¸ _ ∴ · + + + → ¸ , mathematically 2 1 1 1 2 2 2 1 2 1 2 1 (2) 2 2 (1) (2) 2 1 2 2 Z Z Z e Z Z Z Z e e Z Z e e Z γ γ γ γ γ − − − ¸ _ + − + → ¸ , · + · + + ∴ · + ; VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 82 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 2 1 2. 1 1 2 2 2 cosh 1 2 cosh sinh (cosh sinh ) cosh 1 2 sinh 1 1 2 2 o Z Z Nowe Z e Z Z Z Z Z Z Z γ γ γ γ γ γ γ γ γ ∴ · + · + ¸ _ ∴ + − · − + ¸ , 1 ∴ · + + − + 1 ¸ ] Sinh 2 0 2 1 1 2 2 1 2 0 sinh tanh cosh 1 2 2 . 2 o o o sc oc c Z Z Z Z Z Z Z Z Z Z ButZ Z Z andZ Z γ γ γ γ · · · · + + · + · tanh tanh sc oc oc sc oc Z Z Z Z Z γ γ ∴ · ⇒ · 10. Write a note on Reflection factor and Reflection loss and derive the same. Reflection occurs due to improper termination at the receiving end. This concept can be extended to the function of any two impedances . Let a source of voltage Es and impedance 1 Z is connected to a load of impedance 2 Z . If 2 Z is not equal to 1 Z , reflection of energy takes place resulting in a change in the ratio of V to current and alteration in the distribution of energy between the Electric and magnetic field. The energy transferred to 2 Z is less than that with impedance matching. A reflection is said to have occurred. The magnitude of this loss can be computed by taking the ratio of current actually flowing into the load to the load to the current that would have flown if the impedance is were matched. The matching of impedance is called as image matching and can be obtained on a live by connecting a transformer. According to the transformer theory, VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 83 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 2 2 1 I z I z · → (1) For matching, the magnitude of 1 Z can be made equal to 2 Z by choosing proper transformer ratio. The current which flows through the generator is 1 1 1 (2) 2 I E I Z · → The current 1 2 I (Under image matching condition) which flows through the load 1 1 2 1 2 1 1 2 2 1 1 2 1 2 2 (3) Z I I Z Z E I Z Z E I Z Z · • · • · → Without image matching, the current flowing through the load is 2 2 1 2 . (4) I E I Z Z ∴ · → + The ratio of the current actually flowing into the load to that current flowing under image condition is 1 2 2 1 2 1 2 2 (5) Z Z I R I Z Z · · → This is called as Reflection factor (R) Reflection loss It is defined as the no.of repers or dB by which the current in the load under image matched condition would exceed the current actually flowing in the load. Then The reflection loss in repers is 1 2 1 2 1 2 1 2 2 20 log 2 loss loss Z Z R lu in lepers Z Z Z Z R dB Z Z 1 + · 1 + 1 ¸ ] 1 + · − 1 + 1 ¸ ] VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 84 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 11. Explain the Reflection of a line not terminated by Z0. Referring to Equation 0 ( ) ( ) 2 ( ) ( ) ( ) 2 ( ) zys R o R o R o zys o R R R o Z Z ER E Z Z e ZYS e ZR Z Z Z Z IR I Z Z e ZYS e ZO Z Z − − 1 − · + + 1 + ¸ ] 1 − · + + 1 + ¸ ] There are current and voltage relationships derived for the lines which are terminated in o Z But if a line is not terminated in o Z (or) it is joined to same impedance other than o Z then part of the wave is reflected back phenomenon exists for a line which is not terminated in o Z . Reflection is maximum when the line oCi, eZR= α Reflection is maximum S c i eZ R =O Reflection is zero when R o Z Z S · → distance measured from the receiving end and treated positive. When R o Z Z ≠ 1) One part varying exponentially with positive S 2) One part varying exponentially with negative S ( ) ( ) 85 85 0 ) 2 2 2 ( ) ( ) 2 2 2 ( ) ( ) 2 2 + ¸ _ − · + ¸ , + + ⇒ · + − − + − · − R R o R o R R R o R o R R o e o rly R o R R o o E Z Z Z Z E e zys e ys Z Z ER Z Z ZR ER Z Z E E e zys Z Z ys Z ZR IR Z Z I Z Z III I e e Z Z The first component of E or I which varies exponentially with TS is called incident wave which flows from the sending end to the receiving end. S=O at the receiving and maximum m(s=l) at the sending end. Thus as incident wave travels from the sending end to the receiving end, its amplification decreases. A wave which flows from sending end to the receiving end, with decreasing amplitude is the incident wave. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 85 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The second component of voltage or current which travels from the receiving end to the sending end which varies 85 e and its amplitude decreases as its progress towards the sending end. The total instantaneous voltage at any point on the line is the vector sum of incident and the reflected currents are in out of phase with each other. R o If z Z · it can be seen that the reflected wave is absent end there is not reflection. Such a line is uniform and there is no discontinuity existing to send the reflected wave back along the line. Similarly along the line and the energy is absorbed wave. Such a finite line terminated in o Z without having any reflection is called a smooth line. 12. Derive an expression for input impedance and transfer impedance and of transmission line terminated by an impedance. From the general solution ( ) ( ) (81) (81) (81) (81) o R o s o R Z Z wsh Z Sinh Zin Z Z Z wsh Z Sinh τ τ + + Dividing by ( cosh (81) ) . R Z both Nr and Dr We get Z 81 81 81 81 81 81 81 81 81 81 81 81 1 (81) (81) (81) ( ) (81) (3) (81) ( / ) ( / ) o R o o R R o s o R R o s o o R Z Jauh Z Zs Z Z Jauh Z Z Z Jauh or z Z Z Jauh e e or equ jauh e e Z Z e e e e Z Z Z Z e e e e τ τ τ − − − − − − + + 1 + 1 + ¸ ] − + 1 + − − ∴ 1 + + + ¸ ] Equ. (3) and (4) are the I/P Impedance of a live terminated by an impedance. Let s T R E Z I · = Transfer impedance of a live VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 86 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Now cosh(81) (81) cosh (81) sinh(81) cosh (81) sinh(81) cosh(81) sinh(81) cosh(81) sinh(81) τ τ τ τ − − − − − · s R o s T R o R s o s R s R s o E I Sinh Z xIR I Z I I Z while E E Z Es I Zo E E E I Z Sub in equ (5) 2 2 2 2 2 cosh(81) cosh(81) 1 sinh(81) sinh(81) cosh (81) cosh(81) sinh (81) 1 sinh(81) sinh(81) cosh(81) cosh(81) sinh (81) (cosh (81) (81)) cosh(81) (1) cosh(81) τ τ τ τ τ 1 − · − 1 ¸ ] − − − − − − − ∴ s R T R o o R T o o T R T T R T R E E Z I Z Z Z Z Z Z Z Z Z Z Sinh Z Z Z This is the reg. transfer impedence. 13. Derive an expression for the Input impedance of a lossless line. I/P impedance of a lossless live of any length or obtained from equ. tanh(81) ______(1) tanh(81) , α τ β + · · + · R o in o o o R Z Z Z Z Z Z Z for loss live o therefore P j will become jB only Hence, VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 87 81 81 81 81 cosh(81) sinh(81) 2 2 T R o T R o Z Z Z e e e e Z Z Z τ τ − − + 1 1 + + + 1 1 ¸ ] ¸ ] VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH tanh( ) _____(2) tanh( ) tanh tan R o in o o R Z Z j l Z Z Z Z j l But j l l β β β τ β + · · Therefore tan tan( ) 2 2 tan ______(3) 2 tan R o o o R R o o o R Z jZ l Zin Z Z Z j l l Since l Z jZ Zin Z l Z Z β β π β λ π λ π λ 1 + · 1 + ¸ ] · 1 + 1 · 1 1 + ¸ ] Again for a lossless live, the resistive component of the live i.e., R and G will be equal to zero Thus ( )( ) P j R jwl G jwc O j jwLxjwc j jw LC w LC α β β β β · + · + + + · · · ⇒ (or) If f is the frequency of operation and terminating impedance is a pure resistance R R Equ (3) will become Therefore, I/P Impedance of a lossy and lossless line. 14. Derive the equation for T and π section equivalent to lines. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 88 w LC β · tan 2 tan 2 R o in o R R jZ f LCl Z Z JR f LCl π π + · + VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 10c 1 3 2 3 15c 1 2 3 1 10c 3 2 20c 3 Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z · + · + + · · · − 3 2 3 1 Z Z Z Z · + 2 3 3 1 2 3 2 3 Z Z Z Z Z Z = Z Z ¸ _ + − + + ¸ , + ( ) ( ) 3 2 3 2 3 2 3 Z Z Z Z Z Z Z + − + ( ) ( ) 2 3 3 2 3 2 3 3 3 3 20c 10c 15c Z Z Z Z Z Z Z Z Z Z 2 Z ¸ _ ¸ , · + − · ∴ · − The input impedance of oc and SC lines are, l l R o R o o 10c o l l R o R o Z Z e e Z Z Z Z Z tan h l Z Z e e Z Z γ −γ γ −γ ¸ _ ¸ _ + + ¸ , · · γ ¸ _ − + ¸ , ¸ , for oc, Z R =∞ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 89 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH l l 10c o l l l l ly 13C l l e e 2 Z e e e e III Z 20tan h l=20 e e γ −γ γ −γ γ −γ γ −γ ¸ _ + ∴ · + ¸ , ¸ _ + · γ + ¸ , since the line is symmetrical Z 10c =Z 20c o o 3 o 2 o o o Z Z Z Z tan hrl tan h l tan h l Z Z Z tan h l = tan h l tan h l ¸ _ ∴ · − γ γ ¸ , 1 − γ 1 γ γ ¸ ] 2 o 2 o o o Z = 1 tan h l tan h l Z Z 1 = sec h l cosh l tan h l sin h l cos h h l Z = sin h l − γ γ γ · × γ × γ γ γ γ ( ) ( ) ( ) l l 1 2 10c 3 o 3 l l 2 l l l l l l o 2 3 o 2 l l e e Z Z Z Z Z Z e e Z e e e e e e Z Z e e γ −γ γ −γ γ −γ γ −γ γ −γ γ −γ ¸ _ + · · − · · + ¸ , ¸ _ + + + · − + ¸ , ( ) l l e e γ −γ + ( ) l e e γ + ( ) l l l e e −γ γ −γ + ( ) ( ) ( ) ( ) ( ) l l 2 3 o l l l l 2 l 2 l 2 o l l e e Z Z 1 e e e e 2 e e 2 Z e e 2 γ −γ γ −γ γ −γ γ − γ γ −γ ¸ _ + · − + ¸ , ¸ _ + + − − − · + ¸ , ( ) 2 o o 2 l l l l 4 2 z z e e e e γ −γ γ −γ ¸ _ ¸ _ · · − ¸ , − ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 90 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) l l 10c 2 o l l l l l l o l l e e 2 Z Z Z e e e e e e 2 Z e e γ −γ γ −γ γ −γ γ −γ γ −γ ¸ _ + − · − + + ¸ , 1 + − 1 · 1 + ¸ ] ( ) ( ) ( ) 2 l l / 2 o l / 2 l / 2 l / 2 l / 2 l / 2 l / 2 o l / 2 l / 2 1 2 o o o e e Z e e e e e e Z e e z z z tan h l/2 Z tan h l/2 Z tanh l/2 γ −γ γ −γ γ −γ γ −γ γ −γ 1 − 1 · 1 − + 1 ¸ ] ¸ _ − · − ¸ , · · γ ∴ γ γ π - Section equivalent :- 1 2 2 3 3 1 A B c A 10c 2 A B c 1 2 2 3 3 1 c A B B 20c 3 A B c 1 2 2 3 3 1 A B C 15c 1 A B Z Z Z Z Z Z Z (Z Z ) Z Z Z Z Z Z Z Z Z Z Z Z Z (Z Z ) Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z + + + · · + + + + + · · + + + + · · + Since the line is symmetrical VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 91 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Z A =Z c = ( ) 20C 15c 20c 20c 20c 10c 15c z z Z Z Z Z Z + − − − l l o l l e e Z e e γ −γ γ −γ + · + l l o l l l l o o l l l l e e Z e e 2Z e e Z e e e e γ −γ γ −γ γ −γ γ −γ γ −γ ¸ _ ¸ _ + × + ¸ , ¸ , ¸ _ + · − − − ¸ , 2 o l l o o l l l l l / 2 l / 2 o Z 2Z e e Z e e e e Z e e γ −γ γ −γ γ −γ γ −γ · · ¸ _ + − − − ¸ , − · ( ) ( ) ( ) l / 2 l / 2 2 l / 2 l / 2 e e e e γ −γ γ −γ − − ( ) o o A c 2oc 15c B 20c 10c 15c 2 2 o o o o Z Z Z Z tan h l tan h l Z .Z Z Z Z Z Z Z sin hrl = Z / sin h l Z · · γ γ · − · γ o Z sin h l · γ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 92 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH UNIT – III THE LINE AT RADIO FREQUENCIES PART – A 1. What is dissipation line? A line for which the effect of resistance ‘R’ is completely neglected is called dissipation less line. Dissipation less there is used for transmission of power at high frequency in which losses are neglected completely. 2. What is the nature and value of Z o for dissipation less line? For dissipation less line, Z o is purely resistive and is given by o o L Z R C · · 3. What is the value of ∝ and β for a dissipation less line? The value of attenuation constant ‘ ∝ ‘ for a dissipation less line is zero and the value of phase constant ‘ β ‘ for a dissipation less line is ω LC radian / m. 4. What are nodes and antinodes on a line? Nodes are points of zero voltage or current in a standing wave system and Antinodes are points of maximum voltage or current in a standing wave system. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 93 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 5. Draw the graph between standing wave ratio (s) and reflection co efficient (k) 6. Give the expression for skin depth. Skin depth or nominal depth of penetration is given by / S meters f m · ∏ l where ‘p’ is the resistivity of conductor in Ω /m ‘f’ is the frequency in HZ ‘m’ is absolute magnetic permeability of conductor in H/M 7. What are the advantages of dissipation less line? The advantages of dissipation less line are • The line acts as a smooth line • No reflection takes place at receiving end • Standing waves are not produces. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 94 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 8. Calculate the reflection co efficient if VSWR of the line is 1.5. VSWR = 1 | | 1 | | 1 | | 1.5 1 | | 1 | | 1.5 1.5| | 0.5 | | 1.5| | 0.5 | | 2.5 | | 0.2 K K K K K K K K K K + − + · − + · − · + · ∴ · 9. Define standing wave ratio. The ratio of maximum and minimum magnitude of current or voltage on a line having standing waves is called standing wave ratio. max max min | | | | . | | | | miin V I i eSWR V I · · 10. What are standing waves? When a transmission line is not terminated in its characteristic impedance, the traveling electromagnetic wave from generator at sending end is reflected completely or partially at the terminating end. The combination of incident and reflected wave gives rise to standing waves of current and voltage with definite maxima and minima along the line. 11. Give the relationship between VSWR and reflection coefficient for a transmission line The relationship between VSWR(S), and reflection coefficient ’K’ is 1 | | 1 | | K S K + · − 12. Define reflection coefficient. Reflection coefficient is defined as ratio of reflected voltage or current to the incident voltage or current. It is denoted as ‘K’. ( ) r r i i V I K or K V I · · − The current ratio is negative because the reflected current suffer a 180° phase shift at VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 95 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH the receiving end while the reflected voltage does not. 13. Give the velocity of propagation of open wire dissipation less line. The velocity of propagation for open wire dissipation less line is given by V = 3X10 8 m/sec 14. List out the features of power frequency line. • Power transmission lines are electrically short in length, with the length not exceeding 10 λ • The power efficiency of power transmission line is very high as compared to other energy sources. • Power transmission lines are operated at constant output voltage. 15. Write the expression for input impedance of RF line. The input impedance of dissipation less RF line is given by tan tan s R o in o o R s Z jR Z R R jZ β β 1 + · 1 + ¸ ] where R o is characteristic impedance Z R is terminating impedance β is phase constant ‘s’ is length of the line. 16. Give the expression for input impedance of short circuited line. The input impedance of short circuited line is 2 tan sc o s Z jR π λ ¸ _ · ¸ , where λ is wavelength s is length of the line R o is characteristic impedance 17. Give the expression for input impedance of open circuited line. The input impedance of open circuited line is 2 cot oc o S Z jR λ Π ¸ _ · − ¸ , where λ is wave length s is length of the line VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 96 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH R o is characteristic impedance 18. Define dissipation factor. Dissipation factor is defined as the ratio of energy dissipated to energy stored in dielectric per cycle. 19. Name the device used for measuring standing wave. The derive used for measuring standing wave is directional coupler. 20. What is the maximum resistive input impedance of a dissipation less line? The maximum resistive input impedance of a dissipation less line is R max = S R o Where S is standing wave ratio and Ro is characteristic impedance. 21. What is the minimum resistive input impedance of a dissipation less line? The minimum resistive input impedance of dissipation less line is min s R R S · where ‘S’ is standing wave ratio R o is characteristic impedance 22. A 50 Ω line is terminated in load Z R = (90 + j60) Ω . Determine VSWR due to this load. 1 | | 1 | | 90 60 50 90 60 50 1 1 R O R O K S K Z Z j whereK Z Z j s + · − − + − · · · + + + + ∴ · − 23. A lossless line of 300 Ω characteristic impedance is terminated in a pure resistance of 200 Ω Find the value of SWR? Z o = 300 Ω ; Z R =200 Ω VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 97 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 | | 1 | | 200 300 200 300 R O R O K S K Z Z whereK Z Z + · − − − · · + + 100 500 0.2 k · − · − 1 0.2 1.2 1 0.2 0.8 1.5 S S + ∴ · · − · 24. Sketch the standing waves on a dissipation less line terminated in a load not equal to Ro. 25. Sketch the standing waves on a line having open-or- short circuit termination. 26. What is directional coupler? Directional coupler is a device which is used to measure standing waves. It consists of Coaxial transmission line having two small holes in the outer sheath spaced by 1 4 wavelength clamped over these holes is a small section of line, terminated in its Ro value at both ends to prevent reflection. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 98 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 27. State the use of half wave line. The expression for the input impedance of the line is given as ZS = ZR. Thus the line represent repeats its terminating impedance. Hence it is operated as one to one transformer. The main application of a half wave line is to connect a load to a source where both of them cannot be made adjacent. 28. What are the uses of quarter wave line? The expression for input impedance of quarter wave line is given by 2 O R R Zs Z · This equation is similar to the equation for impedance matching using transformer. Hence the quarter wave line is considered as transformer to match impedance of Z R & Z S . It is used as an impedance matching section. It is used to couple a transmission line to a resistive load such as antenna. 29. What do you mean by reflection loss? When there is mismatch b/w the line and load, the reflection takes place. Because of this the energy delivered to the load by the line is less it composition with the power delivered to the load by a properly terminated line. This loss in power is called reflection loss. 30. Explain how smith chart can be used as an admittance chart. If the smith chart is to be used for admittance, the ‘ri’ axis becomes ‘gi’ axis, while’Xi’ axis becomes ‘bi’ axis. Then above real axis, the susceptance is inductive which is negative. The extreme left point on the real axis represents zero conductance while the extreme right point on the real axis represent infinite conductance. 31. What is the practical value of SWR we can achieve by double stub matching? VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 99 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH SWR = 1.2 32. What is smith chart? Smith chart is an impedance (or) admittance chart which is used to calculate all the parameters of transmission line. It consists of two sets of circles. 33. What is stub matching? A section of transmission line is used as a matching by inserting then b/w load and source. This is called sub matching. 34. What are the advantages of stub matching? 1. The length and characteristic impedance of the line remain unaltered. 2. From mechanical stand point, adjustable susceptance are added in shunt with the line. 35. Give type of stub matching. 1. Single stub matching 2. Double stub matching. 36. Name the impedance transformer that are used at higher frequencies? At higher frequency the impedance transformers consists as a section of transmission line in various arrangements as listed below. 1. Quarter wave transformer (impedance inverter) 2. Stub matching. 1. Single stub matching 2. Double stub matching. 37. What are the advantage and disadvantages as quarter transformer? Advantages: 1. It is very useful device because d its simplicity. 2. Its behaviour can be easily calculated. Disadvantages: 1. It is so sensitive to change in freq. 38. What do you mean by impedance circle diagram? VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 100 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH If impedance are plotted in the form of R – X diagram it turns that for a loss less line terminated in some fixed impedance ZF, the locus of the i/p impedance. Z in as electrical length pl is varied as a circle. This circle diagram is known as impedance circle diagram. 39. Draw the family of constant S – Circle diagram. 40. Explain the direction of movement towards generator or load in circle diagram. In circle diagram, the movement in the clockwise corresponds to transverse from the load towards the generator and the movement in the anti clockwise direction corresponds to transverse from the generator towards the load. 41. What are the advantages and disadvantages * circle diagram. Advantages: 1. It is very useful in calculating line impedance and admittances. Disadvantages: 1. ‘S’ and ‘pl’ circles are not concentric making interpolation difficult. 2. Only limited range impedance value can be contained in nc chart. 42. What are the differences between circle diagram and smith chart? The basic difference between circle diagram and smith charts are: 1. In circle diagram, the resistance component * an impedance represented in rectangular form . 2. In smith chart, the resistive component ‘p’ and resistive component ’x’ and impedance are refines in circular form. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 101 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 43. Draw the family of constant – R circles in smith chart. 44. Draw the family of constant x – circle in smith chart. 45. Give the properties or smith chart: The properties of smith chart are: 1. Normalizing impedance. 2.Plotting as an impedance 3.Determination of K in magnitude and direction 4.Determination of swr 5.Movement along the periphery of the chart. 46. Give some applications of smith chart: 1. Smith chart can be used as an admittance, diagram 2. Used for convey r + aj impedance into admittance. 3. Any value of input impedance can be easily determined. 4. Smith chart can also be used to determined load and impedance VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 102 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 5. The input impedance and admittance of shot circuited line and open circuited line can be easily calculated. 47. Draw a diagram showing how a quarter wave transformer can be used for matching two lines. 48. What are the disadvantages of single stub marks? 1. The single stub matching system is useful only for fixed frequencies. 2. Final adjustment of the sub has to the moved along the line shifts. 49. Why short circuit stub is used in single stub matching? The short circuit is invariably used because. 1. It radiates ups power and 2. It effective length may be varied by means of a shorting bar which normally takes the shapes of shorting plugs. 50. A loss less line has a characteristic impedance of 400 Ω . Determine the stands wave ratio with the following receiving and impedance Z1 = 70 + 0.0Ω Solution: O R R O O R R O Z Z Z Z K Z Z Z Z − − · + + Z R = 70 ; Zo = 400 70 400 33/ 47 70 400 k − · ⇒ − + VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 103 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH | | 33/ 47 1 | | 1 33/ 47 5.71 1 | | 1 33/ 47 k k s k ∴ · + + · · · − − 51. Give the formula to calculate the position and length of a short circuited stub, The position of the stub can be calculated using the formula. 1 1 1 tan 2 20 tan 2 2 O Z Z S R λ λ − − · Π · Π 52. What is dissipation less line? A line for which the effect of resistance R is completely neglected is called dissipation less line. 53. What is the nature of value of Z o for the dissipation less line? For the dissipation less line, Z o is purely resistive of given by o o L Z 1Z C · · 54. What is the range of values of standing wave ratio? The range of values of standing wave ratio is theoretically 1 to ∞. 55. Determine K of a line for with Z R =200u, Z o =692 o 12 − ( ) ( ) ( ) ( ) o R o o R o o o 200 692 12 200 676.8 j143.8 Z Z K Z Z 200 676 j143.8 200 692 12 489.4 162.91 K= 888.51 9.31 K=0.55 153.6 − − − − − · · · + + − + − − − − 56. What is the need for stub matching in transmission lines? When line at high frequency is terminated into its characteristic impedance R o , then the line operates as smooth line. Under this conditions, losses are absent, hence VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 104 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH maximum power is delivered with increased performance. But practically R o of the line termination are not matching. So to provide impedance matching between line of its termination, stub matching is used. 57. Why are short circuited stubs preferred over open circuited stub? A high frequencies, open circuited stubs radiated some energy which is not the case with short circuited stub. Hence over open circuited stubs, short circuited stubs are preferred. 58. What are the advantages of dissipation less line? i) The line acts as a smooth line ii) No reflection takes place at the receiving and iii) The standing waves are not produced. 59. If using of line is 1.5 then calculate its reflection co-efficient . 1 1K1 VSWR 1.5 1 1K1 + · · − 1+1K1=1.5-1.51K1 2.51K1=1.5-1=.5 1K1=0.2. 60. Give the expression for L & C for open-wire line at high frequency? 7 0 / ( ) 9.21 10 log / 2 12.07 / h d d L n henrys m or L henrys m a a C f m d n a µµ − · · × ∏ · l l 61. Give the expression for L & C for coaxial line at high frequency 7 2 10 henrys/m 2 C= / b n a b L n a farads m − · × ∏∈ l l PART – B 1. Write short notes on reflection losses on unmatched line. If a line is not matched to its load then the energy delivered by the line of the load VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 105 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH is less than the energy delivered by the matched line to load. Due to this unmatched system reflected waves and standing waves are produced. The voltage at a maximum voltage point is due to the in phase sum of the incident and reflected waves. In measurement of power and impedance on a Tx line we found that max (1| | .........(1) 2 R R O I Z R E k + · Now this equation can be writes as max | | | | | (1 | | ....(2) 2 R R O i r I Z R E E E k + · + · + The minimum voltage is due to the difference of the incident and reflected waves and it is given as , max | | | | (1 | | ....(3) 2 R R O i r I Z R E E E k + · − · − Hence the standing wave ratio is max min | | | | ..........(4) | | | | i r i r E E E S E E E + · · − The total power transmitted along the line and delivered to the load is given as ( ) max min 2 2 | | | | (| | | |) | | . | | | | | | ..........(5) i r i r o o i r o E E E E E E P R R E E R + − · · − · From the above expression we can recognize the transmitted power as the difference of two powers. One power Pi being transmitted in the incident wave and the other power Pr traveling back in the reflected wave. The ratio of the power ‘P’ delivered to the load to the power transmitted by the incident wave is 2 2 2 2 | | | | 1 | | i r i r r i i i i P P E E E P P P E E − − · · · − VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 106 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 1 | | K · − 2 2 1 1 1 4 ............(6) ( 1) S S S S − 1 · − 1 + ¸ ] · + Now the ratio power absorbed by the load to the power transmitted is plotted as a function of ‘S’ as shown in fig. 2. Explain Eighth wave line and half wave line. Eighth Wave Line: The input impedance of a line of length s = λ /8 is tan (1) tan tan( / 4) (2) tan( / 4 2 / β β β π λ 1 + · → 1 + ¸ ] 1 + Π · → 1 + Π ¸ ] · R O S O O R R O S O O R Z jR s Z R R jZ s Z jR Z R R jZ Where 2 . / 4 8 λ β λ Π · · Π (3) R O S O O R Z jR Z R R jZ 1 + · → 1 + ¸ ] If the line is terminated in a pure resistance R R , then (4) R O S O O R R jR Z R R jR 1 + · → 1 + ¸ ] In equation (4) the numerator and denominator have identical magnitudes so equ ($) becomes Zs = Ro ……(5) VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 107 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Thus an eighth – wave line can be used to transform any resistance to an impedance with a magnitude equal to Ro of the line. (or) to obtain a magnitude match between a resistance of any value and a source of R0 internal resistance. Half – Wave Line: When a length of line having s = λ / 2 is used, the input impedance is tan (1) tan 0 tan ..........(2) R O S O O R S R Z jR Z R Since R jZ Z Z 1 + Π · → Π · 1 + Π ¸ ] · Thus a half wave length of line may be considered as a one to one transformer. *It has greater utility in connecting a load to a source in cases where the load and source cannot be made adjacent. *A group of capacitors may be placed in parallel by connecting them with sections of line n half waves in length. As a result insulators on a high frequency line should not be spaced at half wave intervals, since their effect would then be cumulative, lowering the insulation resistance of the line. 3. Explain the principle and application of Quarter wave transformer for impedance matching (or) what are the features of a Quarter wave transformer? Quarter wave line Impedance Matching: The expression for the input impedance of dissipation less line is given as tan (1) tan SR O S O O R Z jR s Z R R jZ s β β 1 + · → 1 + ¸ ] Equ (1) is rearranged as tan (2) tan R O S O O R Z jR s Z R R jZ s β β 1 + 1 · → 1 1 + 1 ¸ ] For a Quarter wave line , s = λ / 4 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 108 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ∴equ (2) becomes tan / 2 tan / 2 R O S O O R Z jR Z R R jZ 1 + 1 Π · 1 1 + Π ¸ ] 2 0 .............(3) S R R Z Z · Since tan Π /2 = infinity i.e. the input impedance of the line is equal to the square of Ro of the line divided by the load impedance. A Quarter wave line acts as a transformer to match a load of Z R ohms. Such a match can be obtained if the characteristic impedance Ro of the matching Quarter wave section of line is chosen as, 0 .........(4) S R R Z Z · A Quarter wave line may be considered as an impedance inverter, here it transforms a low impedance into a high impedance and vice versa. An application of the Quarter wave matching section is to couple a transmission line to a resistive load such as a antenna. The Quarter wave matching section is designed to have a characteristic impedance Ro chosen that the antenna resistance R A is transformed to a value equal to the characteristic, impedance R o of the T x line The characteristic impedance Ro of the matching section should be ...........(5) o A O R R R · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 109 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The transformers also a single frequency or narrow band device. The bandwidth may be increased by using two or more Quarter wave sections in series each accomplishing part of the total transformation. A Quarter wave transformer may also be used if the load is not pure resistance. It should then be connected between points corresponding to I max or E min at which places the transmission line has resistive impedances given by Ro /s or s/Ro. For step down in impedances from the line value of Ro, the matching transformer characteristic impedance should be,] . / / O O O O R R R S R S · · Another application of the short circuited Quarter wave line is as an insulator to support an open wire line or the center conductor of a coaxial line. This application is illustrated is as shown below. These lines are sometimes referred as copper insulator. 4. A lossless line having Ro = 300 ohms is terminated by a load resistance of 78 ohm. The frequency of operations is 40MHz. What type of single stub will be required to provide impedance when placed nearest to the load? Calculate its length and find its location. Solution : The reflection coefficient is given by 78 300 0.587 78 300 R O R O R O R Z Z Z R K k Z Z Z R − − − · · · · − + + + | | 0.587 k ∴ · The first voltage minimum occurs at y2 = λ / 2, where 300 7.5 40 Meters λ · · The location of the stub nearest to the load is given by 1 1 cos | | cos 0.587 7.5 . . 4 4 K d λ − − · · Π Π d = 0.563 meters Hence the stub is located at a distance of Y 1 = 3.75 – 0.563 Y 1 = 3.187 meters The susceptance of the line at the location of the stub will be positive. Hence a VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 110 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH short circuited stub will be needed to provide the impedance match. The length of this short circuited stub is given by 2 1 1 1 1 | | tan 2 2 | | 1 (0.587) 7.5 7.5 tan tan (0.689) 2 2 0.587 2 0.72 λ − − − 1 − · 1 Π 1 ¸ ] 1 − · · 1 Π × Π 1 ¸ ] · K L K L L meter 5. A load of ZR = 140 ohms is to be connected to a line RO = 100 ohms by a quarter wave matching transformer. a. Find ZO of the matching transformer, b. What is the ‘S’ for the transformer? c. If the input voltage to the line is 100v. Find the load voltage. Solution min max ) . 14,100 ) / 140/100 118.322 . 1.1832 ) 100 ; 1.1832 100 118.32 · · · · · · · · × · O R O R O O load a Z Z R b S Z Z Z OHMS S c V v V V volts d) The location of V max on the N4 line is located at the load. 6. A 150 ohm transmission line is terminated by a load of 200 – j300. Determine the location and length of a short circuited matching stub. Solution: Given Z o = 150 ohm Z R = Z L = 200 – j 300 ohm The single stub is constructed as follows a) Calculate 200 300 150 1.33 2.00 R r O r Z Z Z Z j − · · · − VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 111 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH b) Mark this point as ‘A’ in the smith chart. c) Draw the SWR circle through ‘A’ the ‘S’ value is 4.9 d)Extend the radial line from point ‘A’ through the center to the other side of the SWR circle and mark the point as ‘B’ and to the outer perimeter as point ‘C’ at 0.0057 λ The point ‘B’ is the normalized load admittance (0.23 +j 0.36) equal to the reciprocal of the normalized load impedance at point A (1.33 – j 200) d) Find the intersection of SWR circle and G = 1 circle mark it as point ‘D’ e) Extend the radial line from the centre to ‘D’ continue the line to the perimeter ‘E’ at 0.183 λ The distance from load to the junction of the transmission line and the stub is I 1 = 0.183 λ - 0.057 λ ; I 1 = 0.126 λ f) Final step is the determination of the length of the stub. The susceptance at point ‘D’ is 0 = j 1.8. So the susceptance contributed by the stub must be j 1.8. Point ‘F’ corresponds to this for a short circuited line at 0.33 λ . The length of the matching stub is from 0.25 λ to 0.33 λ i.e., (0.33 – 0.25 ) λ = 0.08λ = 1 2 7. Explain in details the constant ‘R’ circles and constant ’X’ circles in a smith chart. A modified form of a circle diagram for the dissipation less line is the smith chart developed by P.H smith. This chart consists of two circles. 1. ‘R’ circles and 2. ‘X’ circles (For R circles & X circles diagram refer part A question) This chart is also know as circular chart. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 112 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH / 1 . . . / 1 1 / 1 R O R O R O R O r r R o r Z Z Z Z W K T K Z Z Z Z Z K WhereZ Z Z Z − − · · − + − · · + Z r – Normalized terminating impedance Hence Zr = 1 1 K K + − Since ‘Zr” and ‘K’ are complete quantities we have 1 1 ( ) r x r x K jK R jx K jK + + + · − + Rationalizing right hand side we get, 2 2 2 2 1 2 (1 ) r x r x K K jK R jx K K − − + + · − + Equating real and imaginary parts 2 2 2 2 2 2 1 ..............(1) (1 ) 2 ..............(2) (1 ) r x r x x r x K K R K K K X K K − − · − + · − + When equation (1) and (2) are solved, we get two sets of circles. Equation (1) will yield a family of circle called R – circles and equation (2) will yield a family of circles called x – circles. The constant ‘R’ circles: Consider equ (1) and cross multiply 2 2 2 2 1 (1 ) r x r x K K R K K − − · − + We get 2 2 2 2 2 2 2 (1 2 ) 1 2 1 0 r r x r x r r x r r R K K K K K R K R RK RK K K + − + · − − + − + − + + · 2 2 ( 1) ( 1) 2 1 r r r K R K R K R R + + + − · − VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 113 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 . ( 1) 1 ( 1) 2 . 1 1 1 r r x r r x K R R K K R R K R R K K R R 1 · + − · − 1 + ¸ ] − + − · + + Adding 2 2 (1 ) R R + on both sides to make it a perfect square we have . 2 2 2 2 2 2 2 2 2 2 2 . (1 ) 1 (1 ) (1 ) (1 ) 1 1 1 (1 ) r r x x r K R R R R K K R R R R R R R K K R R R − + − + · + + + + + − 1 + − + + 1 + + + ¸ ] 2 2 2 1 ............(3) 1 1 x r R K K R R 1 1 + − − 1 1 + + ¸ ] ¸ ] This equation represents the family of circles on the reflection coefficient plane, these circles are called constant –R circles having radius 1/1 + R and centre (R / 1 + R, O ). These circles have their centers on the positive Kr axis and are contained in the region ‘0’ to ‘1’ as shown in the figure below. R = o corresponds to a circle with center (0,0( on the plane K – plane. This circle forms the periphery of the smith chart. All constant ‘R’ circles touch the point ( 1, 0 ) Including that R = ∞ which is the same as the point itself. Constant X – circles: Consider the equation (2) and cross multiplying, we get 2 2 2 2 2 (1 ) 2 (1 ) 0 x r x r x K K K X K K K X − + · − + · · Adding (1/x 2 ) on both sides in order to make the ‘Kx” terms a perfect square we get, 2 2 ( 1) ( 1/ ) (1/ ....................(4) r x K K X X − + − · Equation (4) represents another family of circles called constant – X circles with centre, (1, 1/x ) and radius (1/x on the k – plane as shown in figure below. ‘X’ being the reactance can be positive or negative whenever ‘X’ is positive the circle lies above the horizontal line. On the other hand when ‘X’ is negative the circle lies below the real axis Kx = 0. When X=0 , the circle degenerates into a straight line because straight line is VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 114 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH circle whose radius is infinity and for X = 0, the radius 1/x will be infinity. All the circles touch the point (1, 0 ) 8. Derive an expression for the position of attachment and length of short circuited stub will remove the standing wave on a large potion of a transmission line. Consider a transmission line having a characteristic admittance y o terminated in a pure conductance y R as shown in the figure. Since we connect stub in parallel with the main line it is easier to deal with the admittance as they can be added up. We know that y R is different from y o standing waves are set up. When we along the line from the load towards the source (generator), the input admittance will be varying for a maximum conductance through a parallel combination of conductance and inductance a minimum conductance and so on this cycle repeats for every λ / 2. When the line is traversed from the point of maximum (or minimum) conductance to that of minimum (or maximum) conductance, there will be a point at which the real part of the admittance is equal to the characteristic admittance. If a suitable susceptance, obtained by using an appropriate length of a short circuited or open circuited line called stub is added in shunt at this point so as to obtain and 0 resonance with the susceptance already existing, then up to that point matching has been achieved. The input impedance of a transmission line at any point is given as [ ] tan tan ..........(1) tan R O R O in O O R Z jZ s Z jZ s Z Z Z jZ s β β β + + · + Convert impedance to admittance tan tan R O in O O R Y jY s Y Y Y jY s β β 1 + · 1 + ¸ ] tan ..........(2) 1 tan O r in O R Y Y j s Y Y jY s β β 1 + · 1 + ¸ ] Where R r o Y Y Y · (Normalized load admittance) VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 115 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 0 IN in Y Y Y · (Normalized input admittance ………..(3) Rationalizing equ (2) 2 2 2 2 tan (1 tan ) 1 tan (1 tan ) (1 tan ) (1 ) tan (1 tan ) β β β β β β β + − · + − + + − · + r r in r r r r r Y j s jY s Y jY s jY s Y s j Y s Y s For on reflection Y in = 1 Thus, the stub has too be located at a point where the real part is equal to unity. 2 2 (1 tan ) 1 1 tan r r Y s Y s β β + + · + 2 2 2 2 tan ( ) 1 tan 1 tan 1/ r r r r r s Y Y Y Y s s Y β β β − · − · · tan .................(4) o R Y s Y β · This equation gives the location of the stub ‘S’ and can further simplified as 1 tan / o R s Y Y β − · 1 1 2 . tan / . tan / .................(5) 2 R o R o s Z Z s Z Z λ λ − − Π · ∴ · Π VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 116 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 (1 ) tan ............(6) (1 ) tan s r o r b Y s Y Y s β β − · − Substitute equation (4) in (6) 2 2 2 2 2 2 (1 / ) / .................(7) 1 / . / (1 / ) / (1 / ) / 1 / − · + − · · − + r o o R s o r o o R r o o R R o o R R o Y Y Y Y b Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y ( ) o R o s o R Y Y Y b Y Y − · Advantages of stub matching: 1. It radiates less power. 2. Its effective length may be varied means of a shorting the bars. Disadvantages: 1. Single stub matching is used only for a fixed frequency because as the frequency changes, the location of the stub will too be changed. 2. For final adjustment the stub has to be moved along the line slightly. This is possibly only in open wire line and on co – axial single stub matching may become inaccurate in practice. 9. Determine the maximum value of conductance that can be matched by a double stub tuner with one stub at the load and the other stub at 3/8 back from the load. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 117 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH In order to overcome the two disadvantages of the single stub matching. Two short circuit stubs are connected. The lengths of these stubs are adjustable but the positions are fixed. Let the first stub whose length is It 1 be located at the point ‘A’ at the distance of 'S 1 ’ from load end then the normalized input admittance at that point will be 1 1 1 1 1 1 2 2 1 1 2 2 1 tan 1 tan tan 1 tan 1 tan 1 tan (1 tan ) (1 ) tan 1 tan β β β β β β β β β + · · + + − · × + − + + − · + · + A r A O r r r A r r r r A A A Y Y j S Y Y jY S Y j S jy S Y jyr S jy s S Y S j y S y S y s jb Where, 2 2 1 1 2 2 2 2 1 1 (1 ) tan ; 1 tan 1 tan r r A A r r Y Sec S Y S S b Y S Y S β β β β − · · + + When a stub having a susceptance value is altered b 1 is added at this point, the new admittance value will be, A A A y s jb · + Since only the susceptance value is altered by the addition of the stub the conductance part remains unchanged The stub length at ‘B’ is adjusted such that the zero value y A is equal to ‘I’ How the location of the stub can be encountered in practice. • The distance s 1 can never be more than or equal to λ / 2. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 118 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH • The distance chosen will be either λ / v or 3 λ / 8. • Since matching is obtained between the point ‘B’ and generator, we have reflection loss occurring to the right of ‘B’ due to mismatch. In order to avoid this loss, sometimes the first stub is located at the load itself. In common practice the distance s 1 is of the order of 0.1λ to 0.15λ 10. Give the method of constructing single stub matching using Smith Chart. A single Stub matching is constructing by following the procedure as below. Step 1 : Calculate the normalized impedance of admittance. Step 2 : Mark this point as ‘A’ in the smith chart, Step 3 : Draw the SWR circle through ‘A’ Step 4 : Extend the radial line from point ‘A’ through the centre to the other side of the SWR . Circle and mark this point as ‘B’ Step 5 : Extend the ‘AB’ radial line towards outer perimeter and mark it as ‘C’ Step 6 : Find the intersection of SWR circle and G = 1 circle mark it as ‘D’ Step 7 : Extend the radial line from the center to ‘D’ and continue the line to the perimeter and mark it as ‘E’ Step 8 : The difference between the points ‘C’ and ‘E’ gives the distance of the stub to load. Step 9 : Find the susceptance at the point ‘D’ to cancel this susceptance mark it in opposite direction and mark it as ‘F’. Step 10 : Find the difference between the values of 0.25 and the value at ‘F’ point. This gives the length of the stub. This is the procedure to construct a single stub matching using smith chart. 11. A 9/16λ long lossless line has Zs/Ro = 1.5 + j0.9. Where Zs is the impedance and Ro is the characteristic resistance. Find the load impedance normalized to R0 and also the standing wave ratio. Solution The given values are, 1.5 0.0 9/16 o o Z j R S λ · + · Load impedance may be calculated using the formula. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 119 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH tan tan 2 9 / tan . 16 2 9 1 tan . 16 β β π λ π λ + · + 1 + 1 ¸ ] · 1 + 1 ¸ ] O R O S O O R O S R O Z Z jR R R jZR s Z R j Z Z R j R 9 / tan 8 1.5 0.9 9 1 tan 8 R O R Z R j j Z j R π π ¸ _ + ¸ , + · ¸ _ + ¸ , 9 (1.5 0.9)1 . tan / tan(9 / 8) 8 (1.5 0.9)1 1 .(0.4142) / (0.4142) (0.5 0.9) / (0.37278 0.6213 1) (0.4142) 0.4142 (1.5 0.9) 0.6272 0.6213 R R O O R R O O R O R O jZ j Z R j R jZ j Z R j R j Z R j j Z j j R j π π + + · + ¸ _ + + + · + ¸ , + + − − · − + · − + 0.4142 (1.5 0.9) 1.5 0.4858 0.6272 0.6213 0.6272 0.6213 0.8198 1.5867 R O j j j j j Z j R − + − − · · − + − + · + Reflection coefficient / 1 / 1 0.8198 1.5867 1 0.8198 1.5857 1 0.180 1.5867 0.37569 0.544 1.8198 1.58 | | 0.3757 − − · · + + + − · · + + − + · + + · R O R O R O R O Z R Z R K Z R Z R j k j j j j k Standing wave ratio, S = 1 | | 1 0.3757 1 | | 1 0.3757 K K + + · − − VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 120 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH S = 2.20 12. If a lossless line has R o =200 ohms. What length of the line will be required to obtain at the input of an inductance of 8 micro hertz at frequency of 70 mHz with far end short circuited? Repeat the calculation of open circuited received end. Solution The input impedance of a short circuit 4d line is a pure reactance is given by, 0 tan(2 / ). ac Z jR S π λ · Substitute the values 6 6 1 2 2 70 10 8 10 200tan 2 15 70 tan 200 2 tan 17.6 s s J j s s π π λ π π λ π λ − − ¸ _ × × × × · ¸ , × ¸ _ · ¸ , · 86.75 180 radians π × · 0.241 S λ · The impedance of an open circuited line is given by 2 cot oc o s Z jR π λ ¸ _ · ¸ , Substitute the values, ( ) 6 6 1 2 2 70 10 80 10 200cot 2 2 560 cot 17.6 200 cot 2 / 17.6 2 / cos 17.6 s J j s s π π λ π π λ π π λ π π λ − ¸ _ × × × × · − ¸ , − × ¸ _ · · − ¸ , − · − · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 121 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 3.25 180 (1 2 / ) 3.25 /180 radians s π π λ π × · − · × 2 3.25 1 ; 0.981. / 2 180 s s λ λ − · − · 0.4905 s λ ∴ · 13) Explain the terms standing waves , Nodes, standing wave Ratio If the voltage magnitudes are measured along the length of a line terminated in a load other than o R the plotted values will appear as Figure: Resistive load of value not equal to RO In the case of either OC(or)SC lines current magnitudes will be same except there will be a / λ ∆ shift of maxima and minima. Nodes – Minimum (i.e) are points of zero voltage (or)zero currents. Antinodes – maxima (i.e) are points of maxima (voltage or current). For a open circuited like the voltage modes occur at a distance of / 4, 3 / 4, 5 / 4.... λ λ λ current mode occurs at a distance of o, / 2, / 2, ..... λ λ λ For a short circuited line, this modal pts get shifted by a distance of / 4 λ and voltage modes occur 0. / 2, ..... λ λ and current modes occurs at / 4, 3 / 4, 5 / 4...... λ λ λ The ratio of maximum to minimum magnitude of voltage or current in a line having standing waves is called Standing Wave Ratio. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 122 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 14) Derive the interrelation between reflection co-efficient and standing wave Ratio(SWR). From the SWR it is clear that the points of voltage maxima occur at points where the incident and reflected waves are in phase with each other. / max/ | | | | _______(1) E Ei Er · + The voltage minimum occurs at pts. Where with each other. /Emin/=/Ei/-/Er/________(2) From max | | | | | | min | | | | E Ei Er S E Ei Er + · · − i by E ÷ on both sides | | | | | | 1 | | | | | | | | 1 | | 1 | | ( | ) _____(4) 1 | | 1 | | ( ) 1 | | Ei Ei Er Ei k S Ei Ei Er Ei k k S k Er Ei k k s k + + · · − − + · ∴ · − − · + ⇒ 1k1= ma Im 1 1 min Im max Im 1 ( ) 1 min Im E ax E in E ax or E in · − − + + 15) Draw the phaser diagram for the i/pimpedance on the line and concenent on it . (or) Derive the I/Pimpedance of an dissipationless line. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 123 max Im min Im E ax Scors E in · · 1 1 1 1 S k S − · + max Im Im / / min Im Im E ax in K or E ax in − · + VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 cos sin 2 2 cos sin cos sin cos sin π π λ λ π π λ λ β β β β + · · + 1 + 1 ¸ ] · 1 + 1 ¸ ] R R o s s R s R o R o s o R s s E jI R E Z E s s I I j R IR E s j R s ER Z ER IR j s R I divide by divide by os 1 1 1 1 β β β β β 1 + 1 1 · 1 + 1 ¸ ] 1 + 1 1 · 1 + 1 ¸ ] R o R s R o o s R o c s I R j jan s E ER Z ER I j jan s R IR R j tan s ZR Z Z ZR j jan s R __________(2) Another form of I/P Impedance Equ. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 124 tan tan R o s o o Z jR s Z R R jzR s β β 1 + · 1 + ¸ ] VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 0 2 2 1 1 ( ) ( ) 2 ( ) 2 sin / / sin | | 1 sin tan tan tan cos β β β β β β β β β β β β β β τ β β τβ β φ − − + + · · · + ¸ _ − − ¸ , 1 + − · 1 − − ¸ ] · + · + · · s R R o s s s s s s E Z Z ej ke j s E zR z zR Zo I I ej ke j s Ro ej ke j s R ej ke j s ej s COS s j s ej s COS s s s s s angleof reflection coefficient 1 1 1 ________(3) 1 1 1 o s k s R s k s β φ β β φβ 1 + − 1 · − 1 ¸ ] 1 1 1 2 1 1 1 2 o k s R k s φ β φ β 1 + − 1 − − 1 ¸ ] I/P Impedance is maximum at a distance of 2 ________(1) . . 2 1 1 1 (max) 1 1 1 s o o s i e s k z Z sz k φ β φ β τ · · + 1 · 1 − ¸ ] If we travel a distance of / 4 λ from pt. where impedance is maximum, we get a point of minimum impedance. / i p ∴ impedance is min if, VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 125 (max) s z szo · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 / 4 ( ) 2 ) 2 2 S s s φ π λ λ β β φ π β β φ π β · + ∴ · · + + ∴ · 1 1 1 1 1 1 min 1 1 1 1 1 1 o o k k Z Z Z k k π π 1 + − − 1 ∴ · · 1 1 − − + ¸ ] ¸ ] ∴ 16) Derive the I/P Impedance expression for a lossless line terminated one a)short circuit b)open circuit a)Short circuit I/P impedance of a dissipation less live is tan tan R o s o o Z jR s Z R R jZR s β β 1 + · 1 + ¸ ] for a short circuit live ZR = 0, so that ZSC = Ro tan o o jR s R β 1 1 ¸ ] As Z is purely reactive or imaginary then Let 2 tan s Z j s s jxs jRo λ π λ · ¸ _ ∴ · ¸ , 2 tan s o x s R π λ · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 126 Z min = o z s tan o ZSC JR s β · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH In this s o X is R called as normalized impedance. The variation of sc o o Z x R R · with length of line S may be as for ; ; 4 ; 2 3 ; 4 ; s J o s J o s J o s J o s J o X S Z X S Z X S o Z X S Z X S Z λ λ λ λ · ∂ · ∞ · · ∞ · ∂ For an open-circuited live, R Z · ∞ 1 tan tan o R oc o R R j S Z Z R R j S Z β β ∂ 1 + 1 1 · 1 + 1 ¸ ] VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 127 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 tan 2 cot tan 2 cot oc o oc o o s o Z R j S j S Z R jR S X X S j R β π β π λ 1 · 1 ¸ ] 1 − · · − 1 ¸ ] · − Similarly for open circuit. S = 0, , , 3 , , , , 4 2 4 s o X o o Z λ λ λ · ∞ ∞ Sc and Oc impedance is purely reactance value. For the first quarter wavelength, shoot ckt, line acts as inductance and next wavelength it acts as capacitance. However, the curves are for the ideal dissipatedness live. In a practical live there will be a small resistance component of 1 P o impedance indicating source power loss; and zero or infinite impedance are never achieved the actual values treading to minima and maxima. 17. A load of admittance o 1/ R 1.25 j.25. G · + Find the length of location of single stub funec short circuited. The normalized load admittance is given by · + R o Y 1.25 j0.25 G VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 128 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) · + − − · · + − + R o o o R o R o R R o 1/ 2 1.25 j.25 1/ R R 1 R 2 R 2 R 2 2 R 1 2R 1- 1.25+j0.25 = 1 1.25+j.25 o o o o -.25-j.25 = -2.25+j.25 0.3535-135 = 2.26386.34 =0.1561-141.34 =0.1561-2.466 Calculating value of cos -1 (1K1) Cos -1 (1k1)=cos -1 (0.1561)=1.414 Calculation for length of location of stub:- Case(1) ( ) ( ) 1 1 1 cos 1k1 S 2 cos 1k1 2 2 -2.466+-1.414 = . I 0.0587 4 − − φ+ π− · β φ+ π− · π ¸ _ λ ¸ , π λ − λ π Length of the stub, VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 129 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) 2 1 2 1 1 1k1 L tan 2 21k1 1 .3535 = tan 2 2 0.3535 =0.1469 − − 1 λ − 1 · π 1 ¸ ] 1 − λ 1 1 π ¸ ] λ Case (2) ( ) 1 1 1 cos 1k1 S . 4 2.466 3.142 1.414 . 4 S 0.1662 − φ+ π− · λ π − + + · λ π · λ length of the stub, ( ) ( ) 2 1 2 1 1 1k1 L tan 2 21k1 1 0.3535 = tan 2 2 0.3535 − − 1 λ − 1 · π − 1 ¸ ] 1 − λ 1 1 π − ¸ ] ( ) ( ) 1 1 = tan -1.3231 2 = tan 1.013231 2 L 0.353 − − λ π λ 1 π− ¸ ] π · λ 18. A loss less RF line has Z o of 600 Ω of is connected to a resistive load of 75Ω . Find the position of length of short circulated stub of same VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 130 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH construction as line which would enable the main length of a line to be correctly terminated at 150MHz. Given , f=150MHz R o =600Ω Z R =75Ω Finding λ , f.λ =c 2 6 C 3 10 2m f 150 10 × λ · · · × The reflection co-efficient is given by, R o R o R o R o c o Z Z Z R K Z Z Z R 75 600 525 K 0.777 75 600 675 K 0.7777 0.7777180 − − · · + + − − · · · − + · π · Case(1) ( ) ( ) ( ) 1 1 1 1 -1 cos 1k1 S 2 2 cos 1k1 S . 4 +-cos 0.7777 = .2 4 =0.8918M − − φ+ π− · β π β · λ φ+ π− · λ π π π π The length of the stub is given by − 1 λ − 1 · π 1 ¸ ] 2 1 1 1k1 L tan 2 21k1 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 131 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) − 1 − 1 1 π ¸ ] 2 1 1 0.7777 2 L= tan 2 2 0.7777 =0.1222m Case (2) ( ) ( ) 1 1 -1 cos 1k1 S 2 + cos 0.7777 = . 4 =1.108m − φ+ π+ · β π π+ λ ×π The length of the stub is given by ( ) ( ) ( ) ( ) 2 2 1 1 1 1. 0.7777 1 1k1 L= tan = tan 2 21k1 2 2 0.7777 1 = tan 0.4041 = 0.384 2 0.8777m − − − 1 1 λ − λ 1 1 1 π − π − 1 ¸ ] ¸ ] λ − π− π π · Selecting a point located nearest to the load. Hence the stub location nearest to the load is calculated in case1. The stub must be located at a distance 0.8918m from the load of the length of the stub required is 0.1222m. 19. Design a quarter wave transformer to match a load of 200Ω to a source resistance of 500Ω . Operating frequency is 200 MHz. For a quarter wave transformer, the input impedance is given by. 2 o in R R Z 2s 2 · · The source impedance Z s = 500 Ω Load impedance =2Ω = 200Ω VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 132 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 o 2 o 2 o o R 500 200 R 500 200 R 100000 R 316.22 f 200MHz · · × · · Ω · wave length f.λ =c 8 6 C 3 10 1.5m f 200 10 × λ · · · × ∴ The length of quarter wave line is given by 1.5 S 0.375m 4 4 λ · · · C-s=λ /4=.345m. 20. A loss less transmission line with Z o = 75Ω of electrical length l=0.3λ is terminated with load impedance of 2Ω =(40+j20)Ω . Determine the reflection co-efficient at load, SWR of line, input impedance of the line. Solution: Given Z o =R o =75Ω Z r =(40+J20)Ω Reflection co-efficient is given by VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 133 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) R o R o o o o Z R K Z R 40+j20 75 = 40 j20 75 -35+j20 = 115 j20 40.31129.74 = 116.7219.86 =0.345319.88 − · + − + + + Standing wave ratio is 1 1k1 S 1 1k1 1+0.3453 = 1 0.3453 =2.0548 + · − − Input impedance of the line is given by Z in = R o ( ) R o o 2 s Z jR tan R j2Rtan 2 s/ π 1 ¸ _ + 1 λ ¸ , 1 + π λ 1 1 ¸ ] =75 ( ) ( ) 2 0.3 40 j20 j75tan 2 0.3 75 j 40 j20 tan π× λ 1 ¸ _ + + 1 λ ¸ , 1 π× λ ¸ _ 1 + + 1 λ ¸ , ¸ ] =75 ( ) ( ) ( ) 40 j20 j 230.82 75 j123.1 61.55 1 + + − 1 + − + 1 ¸ ] 1 − · 1 − ¸ ] 40 j210.82 75 136.55 j123.1 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 134 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) 1 − · 1 − ¸ ] · − · − Ω o o o 214.58 79.25 75 183.84 42.03 75 1.167 37.222 69.7 j52.95 21. A line with zero dissipation has R= 0.006Ω /m, L=2.5µ H/m d C=4.45PF/m. If the line is operated at 10 MHz, find i) R o ,ii)α iii) β iv) ν v) λ Given R=0.006Ω /m L=2.5 × 10 -6 H/m C= 4.45 PF/m F=10MHz At f=10MHz, WL=2π fL=2× π fL=2× π × 10× 10 -6 × 2.5 × 10 -6 =15.708Ω ∴ WL > > R at 10 MHz So according to standard assumption for the dissipation less line, we can neglect R. i) Characteristic impedance 6 o o 12 L 2.5 10 Z R 749.53 C 4.45 10 − − × · · · · Ω × ii) Propagation constant ( ) 6 6 12 j 0 jw LC j 0 j 2 10 10 2.5 10 4.45 10 j 0 j0.2095/ m =0 =0.2093 rad/m − − − ν · α+ β · + γ · α+ β · + π× × × × γ · α+ β · + ∴ α β iii) velocity of propagation 8 6 12 1 1 V 2.998 10 m/ sec LC 2.5 10 4.45 10 − − · · · × × × × iv) wave length is given by 2 2 29.13m 0.2095 π π γ · · · β UNIT – IV VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 135 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH PART – A 1. What are guided waves? Give examples. The electromagnetic waves that are guided along or over conducting or dielectric surface are called guided waves. Examples: Parallel wire and transmission lines. 2. What is cut-off frequency? The frequency (f c ) at which the wave motion ceases, is called the cut-off frequency of the wave guide. 3. Give the expression for guide wavelength when the wave transmitted in between two parallel plates. The guide wavelength 2 2 2 c m a π λ π ω µε · ¸ _ − ¸ , 4. What is TEM wave or principal wave? TEM wave is a special type of TM wave in which an electric field E along the direction of propagation is also zero. [OR] The transverse electromagnetic (TEM) waves are waves in which both electric and magnetic fields are transverse entirely but have no components of E z and H z . it is referred to as principal wave. 5. Mention the characteristic of TEM waves. It is a special type of TM wave. It does not have either E z or H z component. It velocity is independent of frequency. Its cut-off frequency is zero. 6. Define attenuation factor. Attenuation factor power lost /unit length 2 power transmitted α · × 7. Distinguish TE and TM waves. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 136 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH TE TM Electric field strength E is entire transverse. Magnetic field strength H is entirely transverse. It has z component of magnetic field H z . It has z component of electric field E z (direction of propagation) It has no z component of electric field E z (E z = 0) It has no component of magnetic field H z . (H z = 0) 8. Define wave impedance. Wave impedance is defined as the ratio of electric to magnetic field strength in the positive direction in the negative direction x xy y x xy y E Z H E Z H + − · · − 9. What are the characteristics of TEM waves? TEM wave is a special type of transverse magnetic wave in which the electric field E along the direction of propagation is also zero. 10. Give some examples of guided waves. 1. Electromagnetic waves along ordinary parallel wire 2. Wave in wave guides 3. Wave guided along the earth surface from a radio parameter to the receives. 11. What do you mean by cutoff – frequency? Cut – off frequency can be defined as the frequency at which the propagation constant changes form being real to imaginary. c m f 2a M · ∑ PART – B 1. Derive the field components of the wave propagating between parallel plates. Consider an electromagnetic wave propagating between a pair of parallel perfectly conducting planes of infinite in the y and z directions as shown in Fig 2.1. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 137 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Fig Parallel conducting guides Maxwell’s equations will be solved to determine the electromagnetic field configurations in the rectangular region. Maxwell’s equations for a non-conducting rectangular region and given as H j E E j H ωε ωµ ∇× · − ∇× · − x y z x y z a a a H x y z H H H ∂ ∂ ∂ ∇× · ∂ ∂ ∂ j a a a y x x z z z x y x x y z x y z H H H H H H a a a x z z x x y E E E ωε ∂ ¸ _ ∂ ∂ ∂ ∂ ∂ ¸ _ ¸ _ · − + − + − ∂ ∂ ∂ ∂ ∂ ∂ ¸ , ¸ , ¸ , 1 · + + ¸ ] Equating x ,y and z components on both sides, y z x x z y y x z H H j E y z H H j E z x H H j E z y ωε ωε ωε ∂ ¹ ∂ − · ¹ ∂ ∂ ¹ ¹ ∂ ∂ − · ) ∂ ∂ ¹ ∂ ¹ ∂ − · ¹ ∂ ∂ ¹ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 138 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH , x y z x y z a a a Similarly H x y z E E E ∂ ∂ ∂ ∇× · ∂ ∂ ∂ j a a a y x x z Y z x y x x y z x y z E E E e E E a a a y z z x x y H H H ωµ ∂ ¸ _ ¸ _ ∂ ∂ ∂ ∂ ∂ ¸ _ · − + − + − ∂ ∂ ∂ ∂ ∂ ∂ ¸ , ¸ , ¸ , 1 · + + ¸ ] Equating x, y and z components on both sides y z x x z y y x z E E j H y z E E j H z x E E j H X y ωµ ωµ ωε ∂ ¹ ∂ − · ¹ ∂ ∂ ¹ ¹ ∂ ∂ − · ) ∂ ∂ ¹ ∂ ¹ ∂ − · ¹ ∂ ∂ ¹ The wave equation is given by 2 2 2 2 2 =( +j ) (j ) E E H H Where γ γ γ σ ωε ωµ ∇ · ∇ · For a non-conducting medium, it becomes 2 2 2 2 E E H H ω µε ω µε ∇ · − ∇ · − 2 2 2 2 2 2 2 2 2 2 2 2 2 2 .....(3) E E E E x y z H H H H x y z ω µε ω µε ¹ ∂ ∂ ∂ + + · − ¹ ∂ ∂ ∂ ¹ ) ∂ ∂ ∂ ¹ + + · − ¹ ∂ ∂ ∂ ¹ It is assumed that the propagation is in the z direction an the variation of field components in this z direction may be expressed in the form e -yz , Where γ is propagation constant γ =α +jβ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 139 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH If α =0, wave propagates without attenuation. If α is real i.e. β =0, there is no wave motion but only an exponential decrease in amplitude. 0 y y 0 x 0 y y H H z H , z E E z yz y yz y y x yz y y Let H e H e H Similarly H Let E e E Similarly γ γ γ γ − − − · ∂ · − · − ∂ ∂ · − ∂ · ∂ · − ∂ x E z x E γ ∂ · − ∂ There is no variation in the direction i.e., derivative of y is zero substituting the values of z derivatives and y derivatives in the equation (1), (2) and (3). y y .....(4) H x .....(5) x γ ωε γ ωε ωε γ ωµ γ ωε ωε ¹ ¹ · − ¹ ∂ ¹ − − · ) ∂ ¹ ∂ ¹ · ¹ ∂ ¹ ¹ ¹ · − ¹ ∂ ¹ − − · − ) ∂ ¹ ∂ ¹ · − ¹ ∂ ¹ y x z x y z y x z x y z H j E H H j E x j E E j H E E j H x E j H 2 2 2 2 2 2 2 2 2 2 2 2 2 2 .....(6) Where and γ ω µε γ ω µε γ γ ¹ ∂ + · − ¹ ¹ ∂ ) ∂ ¹ + · − ¹ ∂ ¹ ∂ ∂ · · ∂ ∂ E E E x H H H x E H E H x x Solving the equation (4) and (5) , the fields H x , H y , E x and E y can be found out. To solve H x’ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 140 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH x y the above equation . H 1 E z x y y x y z x H H j E x E j H From E j H H j x γ ωε γ ωµ γ ωµ γ ωε ∂ − − · ∂ · − − · ∂ 1 · + 1 ∂ ¸ ] Substituting the value of E y in the above equation. 2 2 2 2 2 1 1 1 z x x z x x z x z x H H H j j j x H H H j x H H x H H x γ γ ωµ ωε ωε γ γ ωµ γ γ ω µε ω µε ω µε γ γ 1 ∂ − ¸ _ · − − + 1 ∂ ¸ , ¸ ] ∂ − 1 · − + 1 ∂ ¸ ] 1 ∂ − 1 · + · 1 1 ∂ ¸ ] ¸ ] ∂ 1 + · − ¸ ] ∂ 2 2 2 2 2 y' x y h solve h E H Z x z y x H H x where To E j E x j E γ ω µε γ γ ω µε γ ωε γ ωε ∂ − · + ∂ · + ∂ + · ∂ · y x the above equations, H 1 E x z y From j E E j H x ωε γ ωε γ · ∂ 1 · − 1 ∂ ¸ ] Substituting the value of Ex in the above equation. 1 . z y y E j H j H x ωε ωε γ γ ∂ 1 · − 1 ∂ ¸ ] VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 141 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 2 2 2 2 1 ( ) y y z y z y E j H H x E j H x E H y j x ω µε ωε γ γ ω µε ωε γ γ ω µε ωε ∂ − · − ∂ ¸ _ ∂ + · ∂ ¸ , ∂ + · ∂ y 2 2 2 2 2 H ( ) h x E j x ωε γ ω µε γ ω µε ∂ − · + ∂ · + To solve E x , z x y y z x , E E x H the value of H in the above equation, E E x x y x x x E j H j E Substituting j j E E γ γ ωε ωε γ ωε γ ωε γ ωµε γ ∂ · ∂ · 1 ∂ · 1 ∂ ¸ ] − · 2 2 2 2 2 2 [ ] z x x x x E E E x E E x E E x ω µε γ γ ω µε γ γ γ ω µε γ ∂ + · ∂ 1 ∂ + · − 1 ∂ ¸ ] ∂ + · − ∂ To solve E y , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 142 2 2 y E j H h x ωε ∂ − · ∂ 2 z x E E h x γ ∂ − · ∂ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 [ ] z x y z y z y H H j E x H E j j x H E j x γ ωε γ ωε ωµ γ ω µε ωµ ∂ · · − ∂ 1 ∂ − − · − 1 ∂ ¸ ] ∂ + · ∂ 2 2 2 h γ ω µε · + Where The components of electric and magnetic field strength (E x ,E y ,H x and H y ) are expressed in terms of Ez and Hz. It is observed that there must be a z components of either e or H; otherwise all the components would be zero. Although in general case Ez and Hz may be present at the same time, it is convenient to divide the solutions into two sets. In the first case, there is a component of E in the direction of propagation (Ez), but no component of H in this direction . Such waves are called E or Transverse magnetic ™ waves. In the second case, there is a component of H in the direction of propagation (Hz), but no component of E in this direction. Such waves are called H waves or transverse Electric (TE) waves. 2. Derive the electro Magnetic field for TE waves. TRANSVERSE ELECTRIC WAVES: Transverse electric (TE) waves are waves in which the electric field strength E is entirely transverse. It has a magnetic field strength Hz in the direction of propagation and no component of electric field E z in the same direction (E z =0) Substituting the value of Ez =0 in the following equations. 2 x y 2 2 x y E and H E 0 and H 0 E E j h x h x Then γ ωε ∂ ∂ − · · ∂ ∂ · · The wave equation for the components E y VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 143 2 z y H j E h x ωµ ∂ · ∂ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 2 2 2 2 2 2 2 2 =-( ) h Y Y y y y E E E x E E E x E But y γ ω µε ω µε γ ω µε γ ω µε ∂ + · − ∂ ∂ · − − ∂ + · + This is a differential equation of simple harmonic motion. The solution of his equation is given by E y =C 1 sin hx + C 2 cos hx Where C 1 and C 2 are arbitrary constants. If Ey is expressed in time and direction 0 ( ) yz y y E E e − · then the solution becomes E y =(C 1 sin hx + C 2 cos hx)e -yz The arbitrary constants C1 and C2 are determined from the boundary conditions. The tangential components of E is zero at the surface of conductors for all values of z. E y =0 at x=0 E y = 0 at x=a Applying the first boundary condition (x=0) 0=0+C 2 C 2 =0 Then E y =C 1 sin hx e -yz Applying the second boundary condition (x=a) VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 144 2 2 2 0 y y E h E x ∂ + · ∂ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 y 1 sinh 0 m h= a m=1,2,3........ Therefore, Ey=C E cos x yz yz a Where m Sin x e a m m C x e a a π π π π − − · ¸ _ ¸ , ∂ ¸ _ ∴ · ∂ ¸ , Equations (5) are y x y x 1 E x the first equation , H the value of E in the above equation: H sin y x z y x E j H j H E From j Substituting m C x e j a γ γ ωµ ωε γ ωµ γ π ωµ − · − ∂ · − ∂ · − − ¸ _ · ¸ , From the second equation , 1 y z E H j x ωε ∂ · − ∂ Substituting the value of E y , in the above equation. 1 z 1 y 1 x -m = j a -m H = j a field strengths for TE waves between parallel planes are E yz yz yz m C Cos x e a m C Cos x e a The m C Sin x e a H j π π ωµ π π ωµ π γ − − − ¸ _ ¸ , ¸ _ ¸ , ¸ _ · ¸ , − · 1 z 1 .......(7) yz yz m C Sin x e a m m H C Sin x e j a a π ωµ π π ωµ − − ¹ ¹ ¹ ¹ ¹ ¸ _ ) ¸ , ¹ ¹ − ¸ _ ¹ · ¹ ¸ , ¹ Each value of m specifies a particular field of configuration or mode and the wave associated with integer m is designated as TE m0 .wave or TE m0 mode. The second subscript refers to another integer which varies with y. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 145 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH If m =0 , then all the fields becomes zero, E y =0,H x =0, H z =0. Therefore, the lowest value of m =1. The lowest order mode is TE 10 . This is called dominant mode in TE waves. The propagation constant γ =α +jβ . If the wave propagates without attenuation, α =0, only phase shift exists. γ =jβ Then the fields strengths for TE waves. 1 1 1 j z y j z x j z z m E C Sin x e a m H C Sin x e j a jm m H C Cos x e a a β β β π β π ωµ π π ωµ − − − ¸ _ · ¸ , − ¸ _ · ¸ , ¸ _ · ¸ , The field distributions for TE 10 mode between parallel planes are shown in fig. Fig. Electric and magnetic fields between parallel planes for the TE 10 . 3. Derive the Electromagnetic fields expression for TM waves. TRANSVERSE MAGNETIC WAVES: Transverse magnetic (TM) waves are in which the magnetic field strength H is VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 146 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH entirely traverse. It has an electric field strength E z in the direction of propagation and no component of magnetic field Hz in the same direction (H z =0) Substituting the value of H z =0 in the following equations. y 2 2 x y z and E H 0 and E 0 [ H 0] z z x H H j H h x h x Then γ ωµ ∂ ∂ − · · ∂ ∂ · · · Q The wave equation for the component H y 2 2 2 2 2 2 2 2 2 2 2 ( ) h y y y y y H H H x H H x But γ ω µε ω µε γ γ ω µε ∂ + · ∂ ∂ · − + ∂ · + This is also a differential equation of simple harmonic motion. The solution of this equation is 3 4 sinh +C cosh y H C x x · Where C3 and C4 are arbitrary constants. If H y is expressed in time and direction , then the solution becomes. 3 4 ( sinh +C cosh ) yz y H C x x e − · The boundary conditions cannot be applied directly to H y , to determine the arbitrary constants C3 and C4 because the tangential component of H I not zero at the surface of a conductor . However E z can be obtained in terms of H z . VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 147 2 2 2 0 y y H h H x ∂ + · ∂ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH z 3 4 z 3 -yz 4 [eqn.(4)] 1 E h = [ cos h sinh ] j the first boundary condition (E =0 at x=0) C 0 sin hx e Y z y yz z H j E x H j x C x C x e Applying h Then E C j ωε ωε ωε ωε − ∂ · ∂ ∂ · ∂ − · − · Applying the second boundary condition (E z =0 at x=a) m h= a m is a mode m=1,2,3......... where π 4 4 y 4 , sin sin H cos yz z yz yz m m Therefore E C x e j a a jm m C x e a a m C x e a π π ωε π π ωε π − − − ¸ _ · − ¸ , ¸ _ · − ¸ , ¸ _ · ¸ , y z 4 , H E = cos j x y yz But j E H j m C x e a γ ωε γ ωε γ π ωε − · · ¸ _ ¸ , The field strengths for TM waves between parallel planes are 4 4 4 cos sin yz y yz x yz z m H C x e a m E C x e j a jm m E C x e a a π γ π ωε π π ωε − − − ¸ _ · ¸ , ¸ _ · ¸ , ¸ _ · ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 148 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The transverse magnetic wave associated with the integer m is designated as TM m0 wave or TM m0 mode. If m=0 all the fields will not be equal to zero i.e., E x and H y exist and only E z =0. In the case of TM wave there is a possibility of m=0 If the wave propagates without attenuation (α =0), the propagation constant become γ =jβ . The field strengths for TM waves between parallel conducting planes are: 4 cos j z y m H C x e a β π − ¸ _ · ¸ , 4 4 cos cos j z x j z x m E C x e a jm m E C x e a β β β π ωε π π ωε − − ¸ _ · ¸ , ¸ _ · ¸ , The field distributions for TM 10 wave between parallel planes shown in fig 2.3 Fig The TM 10 wave between parallel planes. 4. Describe the field expression for TEM wave guided by parallel conducting plane. TRANSVERSE ELECTROMAGNETIC WAVES: It is a special type of transverse magnetic wave in which electric field E along the direction of propagation is also zero. The transverse electromagnetic (TEM) waves in which both electric and magnetic fields are transverse entirely VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 149 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH but has not component of E z and H z . It is referred to as principal waves. The field strength for TM waves are: 4 4 cos cos j z y j z x m H C x e a m E C x e a β β π β π ωε − − ¸ _ · ¸ , ¸ _ · ¸ , 4 4 4 sin TEM waves E=0 and the minimum value of m=0 0 j z z j z y j z x z jm m E C x e A a for H C e E C e E β β β π π ωε β ωε − − − ¸ _ · ¸ , · · · The fields are not only entirely transverse , but they are constant in amplitude between the parallel planes. Characteristics: For, lowest value m=0 and dielectric is air. Propagation constant 2 0 0 0 0 0 0 0 0 0 =j 1 v= 2 = Velocity c c Wavelength f γ ω µ ε ω µ ε β ω µ ε ω β µ ε π λ β · − · · · · Unlike TE and TM waves, the velocity of TEM wave independent of frequency and has the value c=3 x 10 8 m/sec. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 150 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Fig. The TEM wave between parallel planes The cut-off frequency for TEM wave is zero 0 (m=0) 2 c m f a µε · · This means that for TEM wave, all frequencies down to zero can propagate along the guide, the ratio of E to H between the parallel planes for a traveling wave is 0 0 E H µ ε · The fields distribution are shown. 5. Describe the different of various of propagation between 2 plates and prove that Vg Vp=C2. VELOCITY OF PROPAGATION: The velocity with which the energy propagates along a guide is called group velocity g d v d ω β · If the frequency spread of the group is small enough d d ω β VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 151 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH may be considered to be a constant through the group. It is always less then the free space velocity c. Phase velocity is defined as the velocity of propagation of equiphase surfaces along a guide. It is denoted by p d v d ω β · It is always greater then the free space velocity c. The phase-shift is given by 2 2 2 2 2 on both sides 2 2 0 m a m Squaring a Differentiate d d π β ω µε π β ω µε β β ω ωµε ¸ _ · − ¸ , ¸ _ · − ¸ , · − 1 d d ω β β ωµε ω µε β · − ¸ _ ¸ , Where group velocity phase velocity 1 free space velocity g p d v d v v ω β ω β µε · · · The product of group velocity and phase velocity is the square of free space velocity 2 2 g p v v v c · · 6. Obtain the attenuation factor in parallel plane. The field strengths between parallel conducting planes for TE, TM and TEM waves have been obtained without any loss. In actual wave guides, there are some losses. These losses will modify the field strength by the introduction of multiplying factor e - α z . The attenuation factor α that is caused by losses in the walls of the VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 152 2 g p v v v · 2 g p v v c · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH guide is determined as follows. The voltage and current phasors in waveguide are 0 0 V = V I = I z j z z j z e e e e α β α β − − − − The average power transmitted is [ ] av 0 0 * 2 0 0 1 W Real part of VI* 2 1 = Real part of V 2 1 = Real part of V 2 z j z z j z z e e I e e I e α β α β α − − − − · 1 ¸ ] 1 ¸ ] Where I* is the complex conjugate of I * 0 * z j z I I e e α β − · The rate of decrease of transmitted power * 2 0 0 av 1 2 Re V I 2 = 2 W z av W al e z α α α − −∂ 1 1 · ¸ ] 1 ∂ ¸ ] This is the power lost per unit length or power dissipated per unit length. av av 2 W Power lost/unit length 2 power transmitted W The attenuation factor is α α α ∴ · · This is the attenuation factor for more general case of guided wave transmission. The α can be determined for TE, TM and TEM waves. 7. Derive an expression for attenuation Factor for TE Waves. The electric and magnetic field strengths between perfectly conduction parallel planes for TE waves are VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 153 Power lost/unit length 2 power transmitted α · × VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 1 1 sin sin cos β β β π α β π ωµ α π π ωµα α − − − ¸ _ · ¸ , − ¸ _ · ¸ , ¸ _ · ¸ , j z y j z x j z x m E C x e m H C x e jm m H C x e The amplitude of linear current density in the conduction planes will be equal to the tangential component of H at x=0 and x = a, 1 zy z J H m C π ωµα · · The power loss in each conduction plane is 2 zy 1 J 2 s R Where R s is the surface resistance. 2 s R ωµ α · The power loss in two conducting (upper and lower) planes 2 2 2 2 1 zy 2 2 2 / 2 1 2 J 2 s m C R π ωµ α ω µ α × · The power transmitted in the z direction is Power transmitted/unit area = * 2 2 1 1 Re 2 1 2 sin 2 y x al E H E H c m x a β π ωµ 1 · × ¸ ] 1 · − ¸ ] ¸ _ · ¸ , Power transmitted in z direction for a guide 1 metre wide with a spacing between conductors of metres is VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 154 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 1 1 0 2 2 1 0 sin sin 2 2 sin = 2 2 4 x a x a c c m m x dx x dx a a m x c x a m a β β π π ωµ ωµ π β π ωµ · · ¸ _ ¸ _ · ¸ , ¸ , 1 ¸ _ 1 ¸ , 1 − 1 1 ¸ ] ∫ 2 1 4 c a β ωµ · Power transmitted/unit length = 2 1 4 C a β ωµ · The attenuation factor 2 2 2 2 2 2 2 1 power lost/unit length 2 power transmitted m / 2 = 2 4 α π ωµ σ ω µ β ωµ · × × C a C a Substitute the value of 2 2 m a π β ω µε ¸ _ · − ¸ , The attenuation factor α decreases from infinity at cut-off frequency to a low value at higher frequency. 2 2 2 m 2 2 a m a a π ωµ σ α π ωµ ω µε ¸ _ ¸ , · ¸ _ − ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 155 2 2 3 2m / 2 a π ωµ σ α β ωµ · 2 2 2 3 2 2m / 2 m a a π ωµ σ α π ωµ ω µε · ¸ _ − ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 m 2 2 s R a where Rs m a a π ωµ σ π ωµ ω µε ¸ _ ¸ , · · ¸ _ − ¸ , At cut- off frequency, 2 2 m a π ω µε ¸ _ − ¸ , 2 2 2 2 c m a π ω µε ω µε ω µε ¸ _ − · − ¸ , 2 2 1 1 c c f f ω ω µε ω ω µε ¸ _ · − ¸ , ¸ _ · − ¸ , Substituting this value, 2 2 2 2 2 2 1 2 = 1 c s c c s c R f f R f a f ω µε α ω µα µε ω ω µ ε · ¸ _ − ¸ , ¸ _ ¸ , ¸ _ − ¸ , 2 2 2 = = 1 c s c f R f f a f µ α η ε η ¸ _ 1 ¸ , 1 ¸ ] ¸ _ − ¸ , Q 8. Derive on Expression for Attenuation Factor for TM Waves. The Expressions for E and H for the Transverse magnetic waves between perfectly conducting parallel planes are VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 156 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH y 4 y 4 z 4 H =C cos = C cos =- C sin j z j z j z m x e a m E x e a m m E x e j a a β β β π β π ωε π π ωε − − − ¸ _ ¸ , ¸ _ ¸ , ¸ _ ¸ , The amplitude of the current density in each plane is J = C 4 The power loss per unit length in each conducting plane is 2 2 4 s 1 1 j 2 2 R 2 s s m m R C R ω µ σ · · The power transmitted down the guide per unit are x y 4 4 2 2 4 1 Re (E H*) 2 1 = (E H ) 2 1 = cos cos 2 1 = cos 2 al C m m x C x a a C m x a β π π ωε β π ωε · × 1 ¸ _ ¸ _ 1 ¸ , ¸ , ¸ ] ¸ _ ¸ , Power transmitted in the z direction for a guide 1 metre wide with a spacing between conductors of ‘s’ metre is 2 2 2 4 4 0 1 1 cos 2 4 Power lost/ unit lengh Attenuation factor ( ) = 2 power transmitted a x C C m x dx a a β β π ωε ωε α · ¸ _ · ¸ , × ∫ 2 4 2 4 C 2 2 4 m m C a ω µ σ α β ωε · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 157 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 m m a m But a ω µ ωε σ β π β ω µε · ¸ _ · − ¸ , The attenuation factor for TM waves is 2 2 2 / 2 m m TM m a a ω ωµ σ α π ω µε · ¸ _ − ¸ , The attenuation factor for TE waves is 2 2 2 3 2 2 2 m m TE m m a a ωµ π σ α π ωµ ω µε · ¸ _ − ¸ , Dividing α TE by α TM ( ) 2 2 2 3 2 2 2 = 2 TE TM m m a a a m a f π π α ωµ ωε α ω µε π π µε ¸ _ ¸ , · · ¸ _ ¸ , 2 2 1 2 1 . 2 c m a f m But f a µε µε ¸ _ ¸ , · · 2 2 c TE TM f f α α · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 158 2 c TE TM f f α α ¸ _ · ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The attenuation factor for TM waves is s 2 2 2 2 2 R substituting the value of TM c m a a m a ωε α π ω µε π ω µε · ¸ _ − ¸ , ¸ _ · ¸ , s 2 2 s 2 2 R 2 R = 1 TM c c a a ωε α ω µε ω µε ωε ω ω µε ω · − ¸ _ − ¸ , s 2 s 2 2R = 1 2R = 1 c c f a f f a f µ ε µ η ε η ¸ _ − ¸ , 1 · ∴ 1 ] ¸ ¸ _ − ¸ , Where c f = f m s m c m m f R f π µ σ π µ σ · 2 2 1 m m c TM c f f f a f f π µ σ α η ∴ · ¸ _ − ¸ , 2 2 1 c m m TM c c f or f f a f f π µ σ α η · 1 ¸ _ ¸ _ − 1 ¸ , ¸ , 1 ¸ ] VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 159 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The minimum value of α TM is determined by equating derivative of α with respect to , zero. c f to f ¸ _ ¸ , Let K = 2 c m m c f a f x f π µ σ η · 2 3 2 = K 1 1 = 1 x x x K x α ¸ _ − ¸ , − 1 3 2 2 = 1 x K x ¸ _ − ¸ , Differentiating with respect to ‘x’, ( ) 2 2 3 1 2 2 3 2 2 d 3 ( 1) (2 ) = dx 1 2 1 K x x x x x x x α 1 − − 1 1 − ¸ _ ¸ ] − ¸ , = ( ) 2 4 2 4 2 3 2 1 3 3 2 2 1 K x x x x x x 1 − − − 1 1 − ¸ ] Equating 0 d dx α · ( ) ( ) 2 4 2 2 3 2 4 2 2 2 1 3 0 2 1 3 0 1 K x x x x x x x x 1 − − 1 · 1 − ¸ ] 1 − 1 · 1 − ¸ ] x 4 = 3 x 2 Dividing by x 2 , x 2 = 3 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 160 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH c c x = 3 f x = f f 3 f But ∴ t · t Take only positive value of frequency The attenuation α TM reaches a minimum value at a frequency of 3 times the cut-off frequency and then increases with frequency. After substituting this value, s s min 2 2 R 2.5R = = a 1 a 1- 3 α η η ¸ _ ¸ , 9. Define Wave Impedance. Obtain the wave impedance expressions for TE, TEM and TM waves. WAVE IMPEDANCES In transmission-line theory power is propagated along one axis only, and only one impedance constant is involved. However, in the three dimensional wave propagation power may be transmitted along three axes of the coordinate system and consequently three impedance constants must be defined. Wave impedances are defined by the following ratios of electric to magnetic field strengths for the positive directions of the coordinates. ; ; ; ; y x z xy yz zx y z x y x z yx zy xz x y z E E E Z Z Z H H H E E E Z Z Z H H H + + + + + + · · · · − · − · − The wave impedances for the negative directions of the coordinates are VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 161 3 c f f · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ; ; ; ; 1 y x z xy yz zx y z x y x z yx zy xz y z E E E Z Z Z H H H E E E Z Z Z x H H − − − − − − · − · − · − · · · + For TE waves, the wave impedance is given by = y yx x E Z H j j ωµ β ωµ β + · − · 2 2 yx Z m a ωµ π ω µε + · ¸ _ − ¸ , 2 2 1 / 1 1 c c m a m a f f ωµ π ω µε ϖ µε µ π ω ε µε · ¸ _ − ¸ , 1 · · 1 1 ¸ ] ¸ _ − ¸ , Q 2 1 yx c Z f f η + · ¸ _ − ¸ , At cut-off frequency 2 2 m a π ω µε ¸ _ · ¸ , , wave impedance Z + yx becomes infinity. At very high frequency (greater than cut-off frequency) wave impedance becomes, 2 2 2 yx m Z a ωµ π ω µε ω µε µ µ η ε µε + 1 ¸ _ · >> 1 ¸ , 1 ¸ ] · · · Q VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 162 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH At ω >> ω c , the wave impedance is equal to the intrinsic impedance. Z + yx = η For TM waves, the wave impedance is given by 2 2 2 2 1 / = 1 x xy y c m a E Z H m a π ω µε β ωε ωε π ω µε ω µε ωε ω µ ε ω + ¸ _ − ¸ , · · · ¸ _ − ¸ , ¸ _ · − ¸ , 2 1 c xy f Z f η + ¸ _ · − ¸ , At cut-off frequency 2 2 m a π ω µε 1 ¸ _ · 1 ¸ , 1 ¸ ] , the wave impedance becomes zero. At very high frequency (greater than cut-off frequency), the wave impedance becomes 2 xy xy Z Z ω µε ωε µ η ε η + + · · · · For TEM wave, the wave impedance is x xy y o o E Z H β ωε µ ε + · · · = η o intrinsic impedance VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 163 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH xy o Z η + · The wave impedances for TE, TM and TEM waves between parallel planes are shown as functions of frequency in figure. Wave impedance versus frequency characteristics of waves between parallel conducting plane 10. A parallel plane wave guide consists of two sheets of good conductor separated by 10 cm. Find the propagation constant at frequencies of 100 MHz and 10 GHz, when the wave guide is operated in TE 10 mode. Does the propagation take place in each case. Given: TE 10 mode : m = 1, n = 0 a = 10 cm = 0.1 m f = 100 MHz, 10 GHz For free space µ = µ o and ε = ε o Free space velocity c = 8 1 3 10 / sec o o m µ ε · × Propagation constant is given by 2 2 2 2 2 2 2 [ 2 ] 1 2 m a f f a c f a c π γ ω µε π π ω π π ¸ _ · − ¸ , ¸ _ ¸ _ · − · ¸ , ¸ , ¸ _ ¸ _ · − ¸ , ¸ , Q VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 164 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH For f = 100 MHz 2 6 2 8 2 1 2 100 10 (0.1) 3 10 2 100 3 31.346 / Nepers metre π π ¸ _ × × · − × ¸ , ¸ _ · − ¸ , · Here propagation constant γ has real value. i.e., γ = α Hence no propagation takes place at 100 MHz. For f = 10 GHz 2 9 2 8 2 1 2 10 10 (0.1) 3 10 200 100 3 207.07 j γ π π ¸ _ × × · − × ¸ , ¸ _ · − ¸ , · Here γ has imaginary value i.e., γ = jβ . Hence propagation takes place at 10 GHz. 11. A pair of perfectly conducting planes are separated 8 cm in air. For frequency of 5000 MHz with the TM 1 mode excited find the following. (i) cut-off frequency (ii) characteristic impedance (iii) attenuation constant for f = 0.95 f c (iv) phase shift (v) phase velocity and group velocity (vi) wavelength measured along the guiding walls Given : TM 1 mode : m = 1 a = 8 cm = 0.08 m 9 1 / 36 10 o F m ε π · × f = 5000 MHz 7 4 10 / o H m µ π − · × (i) Cut-off frequency: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 165 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 1 . 1 [ , 2 ] 2 c c c m a m a m a m f v v f a π γ ω µε π ω µε π ω µε ω π µε ¸ _ · − ¸ , ¸ _ · ¸ , · · · · Q But v = c = 3 × 10 8 m/sec f c = 8 1 3 10 2 0.08 × × × 8 18.75 10 1.875 Hz GHz · × · (ii) Characteristic impedance: z = 120 o o µ π ε · 120 η π · ohms or 377 ohms (iii) Propagation constant becomes attenuation constant (i.e., real value) if the operating frequency is less than the cut-off frequency. f = 0.95 f c 2 2 2 2 2 0.95 c m a m f a v f f π γ α ω µε π π ¸ _ · · − ¸ , ¸ _ ¸ _ · − ¸ , ¸ , · Cut-off wave length 2 c c v a f m λ · · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 166 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 2 0.95 2 0.95 1 (0.95) 39.27 0.0975 12.26 / π α π π π π ¸ _ ¸ _ · − ¸ , ¸ , ¸ _ ¸ _ · − ¸ , ¸ , · − · · m m a a m m a a m a Nepers m (iv) Phase shift: 2 2 2 2 5000 2 π β ω µε π π β ¸ _ · − ¸ , · ¸ _ ¸ _ · − ¸ , ¸ , m a f MHz f m v a 2 6 8 2 2 5000 10 1 3 10 0.08 100 156.25 3 97.08 π π ¸ _ × × ¸ _ · − × ¸ , ¸ , ¸ _ · − ¸ , · (v) Phase velocity: 6 8 2 2 5000 10 97.08 3.236 10 / sec p v f m ω β π β π · · × × · · × or VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 167 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 8 2 9 6 8 1 3 10 1.875 10 1 5000 10 3.236 10 / sec p c c v f f m · ¸ _ − ¸ , × · ¸ _ × − × ¸ , · × Group velocity 2 g p c v v · 8 2 8 (3 10 ) 3.236 10 2.78 / sec. m × · × · (vi) Wave guide wavelength: 2 1 g c λ λ λ λ · ¸ _ − ¸ , But 8 8 3 10 0.06 50 10 m λ × · · × 12. For a frequency of 6000 MHz and plane separation = 7 cm, find the following for the TE 1 mode. (i) cut-off frequency (ii) angle of incidence of the planes (iii) phase velocity and group velocity is it possible to propagate TE 3 mode? Given : TE 1 mode : m = 1, a = 0.07 m f = 6000 MHz (i) Cut-off: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 168 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 8 8 1 . 2 2 1 3 10 2 0.07 21.4286 10 c c c m a m f a m v a f Hz π ω µε µε ¸ _ · ¸ , · · · × × × · × (ii) Phase velocity 2 8 8 6 8 1 3 10 21.4286 10 1 6000 10 3.2118 10 / sec p c c v f f m · ¸ _ − ¸ , × · ¸ _ × − × ¸ , · × Group velocity 2 g p c v v · 8 2 8 (3 10 ) 3.2118 10 2.802 / sec. m × · × · (iii) Angle of incidence 8 8 1 cos 3 10 cos 3.2118 10 cos (0.934) 20.92 p p c v c v θ θ θ − · × · · × · · ° VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 169 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 2 2 2 6 8 2 2 3 2 6000 10 0.07 3 10 π γ ω µε π π π π ¸ _ · − ¸ , ¸ _ ¸ _ · − ¸ , ¸ , ¸ _ ¸ _ · − ¸ , ¸ , ¸ _ × × ¸ _ · − × ¸ , ¸ , m a m f a v m f a v 1836.735 16000 236.735 48.337 / Nepers m π π · − · · Propagation constant has real value is v = α . Propagation is not possible for TE 3 mode. 13. Consider a parallel plate wave guide with plate separation 20 cm with the TE 10 mode excited at 1 GHz. Find the propagation constant, the cut-off frequency and guide wavelength assuming ε r = 4 for medium of propagation in the guide. Given: TE 10 mode : m = 1, n = 0, a = 0.2 m f = 1 GHz Propagation constant 2 2 2 2 9 8 1 2 10 2 0.2 3 10 1 1 39.3 / o o r m a v j radians m π γ ω µε π µε µ ε ε ¸ _ · − ¸ , ¸ _ × × ¸ _ · − × ¸ , ¸ , 1 · · 1 1 ¸ ] · Q Cut-off Frequency VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 170 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 8 1 2 1 2 4 3 10 2 0.2 2 375 c r m f a m c c a MHz µε µε ε · · · × · × × · Guide wavelength 2 1 g c λ λ λ λ · ¸ _ − ¸ , 8 9 2 2 2 0.2 0.4 3 10 1 10 0.3 0.3 0.3 1 0.4 0.16 c g a m m m m λ λ λ · · × · × · × · · ¸ _ − ¸ , · 14. A 4 GHz signal is propagated in a rectangular wave guide with internal dimensions of 2.5 × 5 cm. Assuming the dominant mode, calculate: (i) cut-off wavelength (ii) guide wavelength (iii) group velocity (iv) phase velocity and (v) wave impedance Given : TE 10 mode : m = 1, n = 0 f = 1 GHz a = 0.05 m b = 0.025 m (i) Cut-off wave length VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 171 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 0.05 0.10 c a m m λ · · × · (ii) Guide wave length : 2 1 g c c f λ λ λ λ λ · ¸ _ − ¸ , · 8 9 2 3 10 0.075 4 10 0.075 0.1134 0.075 1 0.1 g m m λ × · · × · · ¸ _ − ¸ , (iii) Group velocity 8 8 0.075 3 10 0.1134 1.984 10 / sec. g g v c m λ λ ¸ _ · ¸ , ¸ _ · × ¸ , · × (iv) Phase velocity VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 172 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 8 8 8 2 8 9 8 1 . 2 2 1 3 10 0.1 30 10 3 10 30 10 1 4 10 4.5356 10 / sec. · ¸ _ − ¸ , · · · × × · × × · ¸ _ × − × ¸ , · × p c c c p c v f f mv m f c a a f Hz v m 8 8 or 0.1134 3 10 0.075 4.5356 10 / sec. g p v c m λ λ ¸ _ · ¸ , ¸ _ · × × ¸ , · × (iv) Wave impedance 0 2 1 TE c z z λ λ · ¸ _ − ¸ , Intrinsic impedance 0 2 120 ohms 120 0.075 1 0.1 = 570 ohms. TE z z π π · · ¸ _ − ¸ , 15. An uniform plane wave at 2.45 GHz is transmitted through a medium having σ =2.17 s/m ∑=47∑ o µ =µ o . Find the complex propagation constant, phase velocity and the wave impedance of the medium. If the electric field mag is 10V/m, find the time average power flow per unit area. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 173 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Given : F=2.45GHz=2.45× 10 9 Hz W=2π f=2π × 2.45× 10 9 =15.394× 10 9 σ =2.17s/m ∑=47∑ o =47× 8.854× 10 -12 =416.14× 10 -12 F/m µ =µ o =4π × 10 -7 H/m ( ) 2 o j j = j - = -123922.51+j41977.94 =361.580.63 γ · ωµ σ+ ω∑ ωµσ ω µ∑ γ =58.85+j356.65 α =58.85N/m β =356.65 rad/m V p =ω /β =4316× 10 8 m/s wave impedance Z =54.24 ωµ · β Ω Intrinsic impedance o j j ωµ η · σ+ ω∑ Time average power flow / unit area P av = 1 2 EH = 1 2 × 10× 192=96w/m 2 16. What are the characteristics of TE and TM waves? (1) Propagation constant VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 174 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 2 2 2 2 2 2 2 2 h h r h m sub h= a m r a j · γ + ω µ ∑ γ · −ω µ ∑ · −ω µ ∑ π π ¸ _ · −ω µ∑ ¸ , γ · α+ β at lower freq. ω 2 µ ∑ is < 2 m , a π ¸ _ ∴γ ¸ , becomes real with value of α , and β =0. ∴ there is only attenuation, without any propagation. At higher frequencies value of ω 2 µ ∑ becomes greater than 2 m , a π ¸ _ ¸ , making γ .imaginary. For f<f c 2 m , a π ¸ _ ¸ , ω 2 µ ∑, =0 γ · α β For f<f c 2 m , a π ¸ _ ¸ , < ω 2 µ ∑, =0 γ · β α For f<f c 2 m , a π ¸ _ ¸ , ω 2 µ ∑, (ii) Cut – off frequency:- at f= f c 2 2 2 2 c c c m o a m a m a m 2 f a m f 2a π ¸ _ · −ω µ∑ ¸ , π ¸ _ ω µ∑ · ¸ , π ω µ∑ · π π µ∑ · · µ∑ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 175 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH (iii) Phase – constant :- 2 2 2 2 m m sub c in eqn of = a a π π ¸ _ ¸ _ ω µ∑ · γ −ω µ∑ ¸ , ¸ , ( ) 2 2 2 2 2 2 2 2 2 2 c c c j c j f fc j 2 f fc γ · ω µ∑−ω µ∑ γ · µ∑− ω −ω γ · µ∑ ω −ω γ · µ∑ − · β β · π µ∑ − (iv) cut-off wave length λ · V c fc ¸ _ · · µ∑ µ∑ ¸ , m 1 mv fc but v= 2a 2a c c 2a j1 m 2a m µ∑ λ · × µ∑ λ · (v) Group wavelength:- VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 176 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH g 2 2 2 2 g 2 2 2 2 v f 2 f f f 2 m a m sub c a 2 c 1 2 = c 2 f 1 ω · · λ β / ω π λ · · / β β π λ · π ¸ _ ω µ∑− ¸ , π ¸ _ · ω µ∑ ¸ , π λ · ω ω µ∑ − − ω π ω π − ω g 2 2 V f c 1 λ · ω − ω vi) Velocity of propagation:- ω ω · · β π ¸ _ ω µ∑− ¸ , 2 2 V m a π ¸ _ · ω µ∑ ¸ , ω · −ω ω µ∑ − ω · ¸ _ − ¸ , 2 2 2 2 o 2 c m sub c a V c 1 V V f 1 f UNIT – V VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 177 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH PART – A 1. Write down the Maxwell’s equation for the loss-free region within the guide? ∂H z + γ H v = jω ε E x , z V z E E jw H y γ µ ∂ + · − ∂ z x V H H jw E x γ ∂ + · − ∈ ∂ , z x v E E jw H y γ µ ∂ + · ∂ , y x z H H jw E x y ∂ −∂ · ∈ ∂ ∂ y x z E E jw H x y µ ∂ −∂ · − ∂ ∂ 2. What is the wave equation for E 2 and H 2 ? 2 2 2 2 2 2 z z z z E E E E x y γ ω µε ∂ ∂ + + · − ∂ ∂ 2 2 2 2 2 2 z z z z H H H H x y γ ω µε ∂ ∂ + + · − ∂ ∂ 3. In waves between parallel plates what are the classification by field configurations? 1. Transverse Magnetic Waves (TM) 2. Transverse Electric waves (TE) 4. For Rectangular guide shown in figure what is the boundary condition? E x = E z = 0 At y = 0, and y = b E v = E z = 0 at x = 0 and x = a. 5. Write down the propagation constant for a rectangular guide for TM waves. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 178 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 2 2 2 = A m = a G B n b ν ω µε ω µε π π ω µε · − + − ¸ _ ¸ _ + − ¸ , ¸ , 6. What is the propagation constant for an ordinary transmission line? j γ α β · + α → Attenuation constant β → Phase constant. 7. What is the expression of attenuation constant, propagation constant for a perfectly conducting wall? α = 0 such that ω > ω c 2 2 2 m = a n b π π β ω µε ¸ _ ¸ _ − − ¸ , ¸ , The value of ω c = 2 2 1 m = a n b π π µε ¸ _ ¸ _ + ¸ , ¸ , 8. The Cut-off frequency that is the frequency below which wave propagation will not occur, is 2 2 1 m = a 2 c n f b π π π µε ¸ _ ¸ _ · + ¸ , ¸ , 9. The cur-off wavelength, that is the cutoff wavelength which wave propagation will not occur is 0 2 2 2 = m a n b λ ¸ _ ¸ _ + ¸ , ¸ , 10. What is dominant wave? The wave which has the lowest cut-off frequency is called the dominant wave. 11. For TE 10 wave, what is cut-off frequency? Cut-off frequency is that frequency for which the corresponding half wavelength is equal to the width of the guide, the cut-off frequency is independent of the VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 179 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH dimension b. 12. What is a waveguide? A hallow conducting metallic tube of uniform section (rectangular or circular) is used for propagating electromagnetic waves. Waves that are guided along the surface (walls) of the tube is called waveguide. 13. What do you mean by propagation of waveguide? Propagation of waveguide can be considered as a phenomenon in which waves are reflected from wall to wall and hence pass down the waveguide in a zig-zag fashion. 14. Write down the maxwell’s equation for non-conducting medium. [ ] = j E =0 x y z xH a a a xH x y z Hx Hy Hz ωξ σ ∇ ∴ ∂ ∂ ∂ ∇ · ∂ ∂ ∂ = jω ε ( a xEx+ a yEy+ a z E z ] 15. Write down the wave equation for rectangular waveguide. 2 2 2 2 2 2 z z z z E E E E x y γ ω µε ∂ ∂ + + · − ∂ ∂ 16. What is Dominant mode? The lowest mode for TE wave is TE 10 (m = 1, n = 0) whereas the lowest mode for TM wave is TM 11 (m = 1, n=1). This wave has the lowest cut-off frequency. Hence the TE 10 mode is the dominant mode of a rectangular waveguide. 17. What is wave impedance? The wave impedances defined as the ratio of electric field intensity to magnetic field intensity are ; ; y x z xy yx zx y x x E E E Z Z Z H H H + + + · · · 18. What are the wave impedance for different modes? For TM, 2 1 fc z f η ¸ _ · − ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 180 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH For TE, 2 1 z fc f η · ¸ _ − ¸ , For TE M , z µ η ε · · 19. What is characteristic impedance? For transmission lines the integrated characteristic impedance Z o can be defined as in terms of the voltage current or interms of the power transmitted for a given voltage or given current. i.e., Z o (V, I) = V I 20. What are the sources of attenuation in wave guide? Attenuation for propagating modes results when there are losses in the dielectric and in the imperfectly conducting guide walls. 21. What is attenuation constant in propagation? It is given by Power lost per unit length 2 x power transmitted. α · 22. How will you calculate power? The power transmitted is obtained by integrating the axial component of the pointing vector through the cross-section of the guide the pointing vector P z is given by P t = ½ |E trans ||H trans| 23. What is attenuation constant in terms of power? The attenuation constant is lost 2 x power transmitted. Power α · 24. What is attenuation constant for TM 11 mode? VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 181 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 2 1 1 1 , b a Rs a b fc ab f a b α η ¸ _ + ¸ , · ¸ _ ¸ _ − ¸ , ¸ , 25. Draw the variation in attenuation with frequency due to wall losses in a rectangular waveguide? 26. Write the field expression for rectangular TM waves. E x 0 2 j c h β − · B cos Bx sin Ay E y 0 2 j c h β − · A sin Bx cos Ay H x 0 2 j c h ωε · A sin Bx cos Ay H y 0 2 j c h ωε − · B cos Bx sin Ay 27. Write the field expression of TE wave guide. H x 0 2 j h β · C B sin Bx cos Ay H y 0 2 j h β · C A cos Bx sin Ay H z 0 = C cos Ay cos Bx E x 0 2 j h ωµ · C A cos Bx sin Ay E y 0 2 j h ωµ − · C B sin Bx cos Ay 28. What are the types of waveguides? 1. Rectangular waveguide 2. Circular waveguide VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 182 F TM 11 TE 10 α i n ( d B | m ) VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 3. Elliptical waveguide 29. Mention the application of waveguides. The waveguides are employed for transmission of energy at every high frequencies where the attenuation caused by waveguide is smaller. Waveguide are used in microwave transmission circular waveguides used as attenuation and phase shifters. 30. Why are rectangular waveguides preferred over circular waveguides? Rectangular waveguides are preferred over circular waveguides because of the following reasons. 1. Rectangular waveguide is smaller in size than a circular waveguide of the same operating frequency. 2. It does not maintain its polarization through the circular waveguide. 31. Why is rectangular or circular form used as waveguides? Waveguides usually take the form of rectangular or circular cylinders because of its simpler form in use and less expensive to manufacture. 32. For an air filled copper x-band waveguide with dimension a = 2.286 cms and b = 1.016 cms determine the cut off frequencies for TE 11 and TM 11 modes? a = 2.28 6 cm = 2.286 x 10 -2 m b = 1.016 cm = 1.016 x 10 -2 m For TE 11 mode, m = 1, n = 1 Cut-off frequency, 2 2 1 m a 2 c n f b π π π µε ¸ _ ¸ _ · + ¸ , ¸ , 2 2 c m = 2 a n b ¸ _ ¸ _ + ¸ , ¸ , 2 2 8 2 2 3 10 10 10 2 2.286 1.016 ¸ _ ¸ _ × · + ¸ , ¸ , F C = 16.156 GH z . The cut of frequency for TE 11 mode is same as that of TM 11 . 33. What is an evanescent mode? When the operating frequency is lower than the cut-off frequency the propagation constant becomes real i.e., Y = α . The wave cannot be propagated. This non-propagating mode is known as evanescent mode. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 183 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 34. Which are the non-zero field components for the TE 10 mode in a rectangular waveguide? H x , H z and E y 35. Which are the non-zero field components for the TM 11 mode in a rectangular waveguide? E x , H y E y and E z 36. Draw a neat sketch showing the variation of wave impedance with frequency for TE and TM waves in a waveguide. 37. What is the cut-off wavelength and cut-off frequency of the TE 10 mode in a rectangular waveguide? Cut-off wavelength, λ c = 22 Cut-off frequency, f c = 22 C 38. What is the cut-off wavelength and cut-off frequency of the TM 11 mode in a rectangular waveguide? Cut-off wavelength, λ c = 2 2 2 1 1 a b ¸ _ ¸ _ + ¸ , ¸ , Cut-off frequency, f c = 2 2 1 1 1 2 a b µε ¸ _ ¸ _ + ¸ , ¸ , = 2 2 1 1 2 a b ν ¸ _ ¸ _ + ¸ , ¸ , 39. What is the wave impedance of TEM waves in waveguide? VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 184 f c Region of number Propagation TM TE W a v e i m p e d a n c e , η VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Wave impedance of TEM becomes the intrinsic impedance of the medium Z µ η ε · · 40. Write down the expression for wave impedance of TE mode? 2 1 c Z f f η · ¸ _ − ¸ , 41. Write down the expression for wave impedance of TM mode? 2 1 c f Z f η ¸ _ · − ¸ , 42. Write down the expression for phase velocity in a waveguide? 2 2 2 = m a n b ω ν π π ω µε ¸ _ ¸ _ − − ¸ , ¸ , 43. Define character is the impedance in a waveguide. For transmission lines the integrated characteristic impedance can be defined as in terms of the voltage current ratio or in terms power transmitted for a given voltage or a given current. i.e., Z o (V, I) = V I ( ) 2 , * o w Z w I II · ( ) * , 2 o VV Z w I W · Where V and I are peak phasors. W is the power transmitted. * indicates complex conjugate. 44. What are the types of power loss in waveguides? There are two types of power loss occurs in waveguides 1. Loss due to attenuation of signals below cut-off frequency. 2. Loss due to dissipation with in the waveguide walls and the dielectric with in the waveguide. 45. The larger dimension of the cross section of a rectangular waveguide is VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 185 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2cm. Find the cut off frequency and wavelength for the domain and TE mode. The given data are a = 2 cm. We know the dominant TE mode is TE 10 mode, thus m = 1 and n = 0. The cut-off frequency of rectangular waveguides is 2 2 2 2 c v m n v m f a a a ¸ _ ¸ _ · + · ¸ , ¸ , Since n = 0 8 8 9 3 10 1 1.5 10 7.5 10 7.5 2 0.02 0.02 GHz × × ¸ _ · · · × · ¸ , Cutoff wavelength λ c = 2 2 2 2 m a a m n b π π π · ¸ _ ¸ _ − ¸ , ¸ , since n = 0 ( ) 2 0.02 0.04 1 c mts λ · · 46. Explain why TEM waves are not possible in rectangular waveguide. (Apr. 2004.) Since TEM wave does not have actual component of a E or H. it cannot be propagated with in a single conductor wave guide. There fore for a displacement current the guide requires an axial component of E, which is not present in TEM waves. 47. A rectangular wave guide has the following dimensions l=2.54cm b=1.27cm, wave guide thickness = 0.127 cm. Calculate the cut off frequency for TE 11 mode. 2 2 C m n fc 2 a b π π π ¸ _ ¸ _ · + ¸ , ¸ , 48. Explain why TM 01 and TM 10 modes in a rectangular wave guide do not exist. (May 2006) For the modes TM 01 and TM 10 , Fe > f Where f → frequency of the wave to be propagated ∴ The dominant mode in TM mode VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 186 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH is TM 11 49. Define the difference between the wave impedance and the intrinsic impedance. The intrinsic impedance η is given by the ratio between the permeability and the permittivity. For free space, M 377 η ε · Ω ; for cu. The wave impedance is the radio between the electrostatic energy and the magnetic field energy. y x x y E E Z H H ∴ · · 50. Define surface impedance. The surface impedance is defined by the conductivity. s w R 2 µ σ · 51. Give the attenuation factor for TM waves Power lost / length 2 Power transmitted/length α · × where the power loss in the 4 walls of the guide is the sum of losses in x=0 and y=0. 52. Which mode is the dominant mode in a rectangular wave? (Nov 2005) The dominant modes are, TE 10 and TM 11 But the lowest mode is TE 10 53. Show the excitation method of TE11 and TM11 modes is a rectangular wave guide. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 187 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Fig. (a) TE11 (b) TM11 54. Give the attenuation factor for TM wave is a rectangular wave guide. (Apr. 2005) 2 2 s TM 2 2 2 s b a 2R a b ab 1 1 fc 1 a b c w where R 2 α η µ σ ¸ _ + ¸ , · ¸ _ ¸ _ + − ¸ , ¸ , ¸ · 55. What is a guided wavelength? g 2 c 1 λ λ λ λ · ¸ _ − ¸ , where c c and f λ λ · = cut off wave length. 56. Mention the applications of circular waveguides. Circular waveguides are used as attenuators and phase shifters. 57. Which mode in a circular waveguides has alternation effect decreasing with increasing in frequency? TE 01 58. Mention the dominant modes in rectangular and circular waveguides. For a rectangular waveguide, the dominant mode TE 10 , For a circular waveguide VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 188 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH the dominant mode TE 11 . 59. Write down the expression for cut-off frequency in a circular waveguide. 2 nm c h f π µε · where h nm = ( ) nm ha a 60. Calculate the cut off frequency of copper tube with 3 cm diameter inside with air filled, in TE 11 mode. 2 nm c h f m π µ · ; 2 3 10 2 a m − · × ( ) 11 11 a h h a · 2 3.85 2 3 10 − × · × [ (ha) 11 = 3.85] = 2.566 x 10 2 2 8 2.566 10 3 10 2 c f π × · × × 1 c µε 1 · 1 ¸ ] Q = 12.25 GH z 61. Determine the cut-off frequency of a circular waveguide with a diameter of 2.36 cms operating in the dominant mode. a = 2 2.36 10 . 2 m − × Dominant mode is TE 11 , h nm = ( ) nm ha a ( ) 11 a h a · 2 3.85 2 10 2.36 × × · = 3.263 x 10 2 2 nm c h f π µε · 2 8 3.263 10 3 10 2π × · × × 8 1 3 10 µε 1 · × 1 ¸ ] Q VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 189 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH f c = 15.58 GH z . 62. Why is TM 01 mode prepared to the TE 01 mode in a circular waveguide? TM 01 mode is preferred to the TE 01 mode, since it requires a smaller diameter for the same cut-off wavelength. 63. Define Q of a waveguide. Quality factor Q is given by Q = stored/unit length lost / unit length / second. energy energy ω · 64. Give the relation b/w quality factor and attenuation factor of a waveguide? 2 2 Q Vg ω · 65. Which of the following wave guide is easier to manufacture? ⇒ Circular ⇒ Rectangular ⇒ Elliptical ⇒ none Ans (a) circular 66. Which of the following waveguide / transmission line would offer the maximum attenuation? a) Coaxial b) Rectangular waveguide c) Circular waveguide. Ans: (a) coaxial. 67. A circular waveguide will behave like a) Low pass filter b) Band pass filter c) High pass filter d) Non of the above Ans: (c) High pass filter. 68. What are the uses of circular wave guides? Circular waveguides are used as attenuators and phase shifters. 69. Draw the variation of attenuation as a function of frequency for different VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 190 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH modes. 70. What is the wave impedance of a circular waveguide? The wave impedance of a circular waveguide is the ratio of the resultant transverse electric fields to the transverse magnetic field. For TM waves 2 TM = E H 71. What is Bessel function? In solving for the electromagnetic fields within guides of circular cross section, a differential equation known as Bessel’s Equation is encountered. P = C 1 ( ) ( ) 2 2 0 1 2 1 ! r r r r ρ ∞ · ¸ _ ¸ , − ∑ This series is convergent for all values of p 1 either real complex. It is called Bessel function of first kind of order zero d is denoted by J 0 (P). 72. What the cut-off freq or critical frequency below which the transmission of a wave will not occur? f c = 2 nm h π µε where h nm = ( ) nm ha a 73. What is the expression for TM waves in circular guides? H 2 0 = C n J n (hp) cos n φ VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 191 TE 11 TM 01 A t t e n u a t o r TE 11 Frequency VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH H p 0 = n j C h β − J n (hp) cos n φ H φ 0 = 2 n jn C h p β J n (hp) sin n φ 0 0 p p F H ωµ β · 0 0 p E H φ ωµ β − · 74. What are the boundry conditions for TM waves in circular guides? The boundry conditions to be met for TM waves are that E φ = q at P = a. 75. What is the wave impedances at a point? z y z x zy x y z y x z xy y zx H H H y z x yx xz H H H E E E Z Z Z E E E Z Z Z + + + + + + · · · − − · · − · 76. What is the work impedence for waves guided by transmission lines / wave guides? Z 2 = 2 2 2 2 x y trans trans x y E E E H H H + · + Z z = ZPφ = -Zφ p = β ωε ( ) 2 2 2 c 2 1 = 1- c z w Z TM w µ ε ω η ω ¸ _ · − ¸ , ¸ _ ¸ , DE β · 77. Draw the characteristic between wave impedance and frequency. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 192 Wave impedance TE TM fc f VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 78. Define wave impedance. The wave impedances are defined by the following rations of electric to magnetic field strengths for the positive direction as well as negative direction of the co-ordinates. + + + · · · · · · - xy yx - yz zy - zx xz Ex Ey Z Z Hy Hx Ey Ez Z Z Hx Hy Ez Ex Z Z Hx Hz 79. Write condition for minimum attenuation for TM waves. ¸ _ ¸ , TM dα =0 f d fc 2wμ wε α = . σ βa ¸ _ ¸ , 2 2wμ 2πfε = . σ mπ ωμε- .a a simplifying we get, f= 3fc 80. Write the wave impedances for TE, TM Q TEM at cut off frequency. * For TE waves at fc 1 ZTE =α For TM waves at fc 1 ZTM = 0 For TEM waves at fc 1 ZTEM = η 81. Relationship between α TM and α TE: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 193 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 TE 2 TM α fc = α f ¸ _ ¸ , The ratio between the attenuation factors of TM and TE waves is given by the ratio between the cutoff frequency and the wave frequency. 82. Give the equation for the power loss at the magnetic field. Power loss = I 2 R = (JYZ) 2 . Rs = (Hy) 2 . Rs (Hy) = C4 2 4 ωμ \ Power loss=C 2σ PART – B 1. Write the expression for transverse magnetic waves in rectangular wave guides. The wave equations are partial differential equations that can be solved by the usual technique of assuming a product, solution. This procedure leads to two ordinary differential the solutions of which are known. Nothing that E z (x, y, z) = E z 0 (x, y) e -y2 E z 0 = xy Where X is a function of x alone, and y is a function of y alone. 2 2 2 2 2 2 d x d y y x y xy xy dx dy ω µε + + · − Putting h 2 = 2 y + ω 2 µ ε as before VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 194 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 0 d x dy y x h xy dx dy + + · Divide by XY 2 2 2 2 2 1 1 d x d y h X dx y dy + · − The above equation equates a function of x alone to a function of y alone. The only way in which such a relation can hold for all values of x and y is due to have each of these function equal to some constant. 2 2 2 2 1 d x h A X dx + · 2 2 2 1 d y A y dy · − The solution of equation is X = c 1 cos B x + c 2 sin B x B 2 = h 2 – A 2 The above solution is y = c 3 cos Ay + c 4 sin Ay This gives E 2 0 = xy = c 1 c 3 cos Bx cos Ay+c 1 c 4 cos Bx sin Ay + c 2 c 3 sin Bx cos Ay + c 2 c 4 sin Bx sin Ay. The constants c 1 , c 2 , c 3 , c 4 , A and B must now be selected to fit the boundary condition. E 2 0 = 0 when x = 0, x = a, y = 0 y = b. If X = 0 the general expression. E 2 0 = c 1 c 3 cos A y + c 1 c 4 sin Ay E 2 0 = c 2 c 3 sin Bx cos A y + c 2 c 4 sin Bx sin Ay When y = 0 reduces to E 2 0 = c 2 c 3 sin Bx For this to be zero for all values of x if it possible to have either c 2 (or) c 3 equal VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 195 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH to zero (assuming B ≠ 0) putting c 2 =0 would make E 2 0 identically zero. So instead c 3 will be put equal to zero. E 2 0 = c 2 c 3 sin Bx sin Ay In addition to the amplitude constant c = c 2 c 4 If x = a E 2 0 = c sin B a sin A y . In order for this to vanish for all values of y (and assuming A ≠ 0) (Because A = 0 would make E 2 0 identically zero) The constant B must have m B a π · when m = 1, 2, 3 Again if y = b; E 2 0 = c sin m a π x sin Ab A = n b π where n = 1, 2, 3. Therefore the final expression for E 2 0 is = c sin m a π x sin n y b π y j β · E x 0 = 2 j c h β − B cos Bx sin Ay E y 0 = 2 j c h β − A sin Bx cos Ay 0 2 sin cos z j c H A Bx Ay h ωε · 0 2 cos sin y j c H B Bx Ay h ωε − · n and A = b m B a π π · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 196 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH These expressions show how each of the components of electric and magnetic field strengths varies with x and y. The variation with time and along the axis of the guide, that is M the z direction, is shown by putting back intro each of these expressions the factor e jwt- xz and then taking the real port. In the division of the fields it was found necessary to restrict the constants A and B to the values given by expressions. A 2 + B 2 = h 2 h 2 = y 2 + ωµ ε 2 2 2 2 2 2 2 2 = m = a y h A B n b ω µε ω µε π π ω µε · − + − ¸ _ ¸ _ + − ¸ , ¸ , The above equation defines the propagation constant for a rectangular guide for TM waves. For low frequencies where ω 2 µ ε is small y will be a real number. This propagation constant met with in ordinary transmission-line theory is a complex number. That is y a j β · + where a is the mathematics constant and β is the phase shift constant. If y is real, β must be zero and there can be no phase shift along the tube 2 2 2 m = a n b π π β ω µε ¸ _ ¸ _ − − ¸ , ¸ , This means there can be no wave motion along the tube for low frequencies y = jβ . The attenuation constant a is zero for all frequencies such that w > w c . 2 2 c 1 m = a n b π π ω µε ¸ _ ¸ _ + ¸ , ¸ , 2 2 c 1 m f = a 2 n b π π π µε ¸ _ ¸ _ + ¸ , ¸ , c 2 2 2 = m a n b λ ¸ _ ¸ _ + ¸ , ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 197 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH f c λ c = v 0 2 2 2 v= = m a n b ω ω β π π ω µε ¸ _ ¸ _ − − ¸ , ¸ , This last expressions indicates than the velocity of propagation of the wave in the guide is greater than the phase velocity in free space. Since the wave length in the guide is given is given by v L f · it will be longer than the corresponding free space wave length. 2 2 2 2 L = m a n b π π π ω µε ¸ _ ¸ _ − − ¸ , ¸ , In the above expressions the only restriction on m and n is that they be integers. However from the above equations it is seen that is either m or n is zero fields will all be identically zero. Therefore the lowest possible values for the either m or n is unity. The lowest cut off of frequency will occur for m = n = 1 This frequency TM waves which can be propagated through the guide. This particular wave is called the TM 11 wave for obvious reasons. High order waves require higher frequencies in order to be propagated along a guide of given dimensions. 2. Derive the Expression for Transverse Electric waves in Rectangular waveguides. The wave equation, in a rectangular waveguide is given by 2 2 2 2 2 2 z z z z H H H H x y γ ω µε ∂ ∂ + + · − ∂ ∂ The solution of the equation is, H z (x, y) = H z 0 (x, y) e -vz H z 0 (x, y) = xy Let Where x → function of x only. Y → function of y only. Substituting the value of H 2 in wave equation, VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 198 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 2 2 2 2 2 2 0 d x d y y x xy xy dx dy d x d y y x h xy dx dy γ ω µε + + · − + + · where G 2 = γ + ω 2 µ ε Dividing by xy 2 2 2 2 2 2 2 2 2 2 1 1 0 1 1 d x d y h X dx y dy d x d y h X dx y dy + + · + · − The equation relates a function of x alone to a function of y alone and this can be equated to a constant. 2 2 2 2 2 2 2 2 1 1 0 d x h A X dx d x h A X dx + · + − · let 2 2 2 1 d x B o X dx + · The solution of this equation is, X = C 1 cos Bx + C 2 sin Bx Similarly 2 2 2 2 2 2 1 1 0 d y A y dy d y A y dy − · + · The solution of this equation is y = C 3 cos Ay + C 4 sin Ay But H 2 0 = xy = (c 1 cos Bx + c 2 sin Bx) (c 3 cos Ay + c 4 sin Ay) VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 199 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH = c 1 c 3 cos Ay cos Bx + c 2 c 3 cos Ay sin Bx + c 1 c 4 cos Bx sin Ay + c 2 c 4 s in Ay sin Bx It is known that, 2 2 2 2 z x E H j E h x h y γ ωµ ∂ ∂ · − − ∂ ∂ For TE waves E z = 0 Ex = 2 z H j h y ωµ ∂ − ∂ \ = 2 j h ωµ − [ -c 1 c 3 A sin Ay cos Bx – c 2 c 3 A sin Ay sin Bx + c 1 c 4 A cos Bx cos Ay + c 2 c 4 A cos Ay sin Bx] Applying Boundary conditions. E 2 = 0, when y = 0, y = b If y = 0, the general solution is Ex = 2 j h ωµ − (c 1 c 4 A cos Bx + c 2 c 4 A sin Bx ] = 0 For E x = 0, C 4 = 0 ( c 4 is common) Then the general solution is Ex = 2 j h ωµ − [ - c 1 c 3 A sin Ay sin Bx – c 2 c 3 A sin Ay Sin Bx] If y = b, Ex = 0 For E x = 0, it is possible either B = 0 or A = n b π , It B = 0, the above solution is identically new, so it is bitter to select A = n b π . The general solution is E x 0 = 2 j h ωµ [ c 1 c 3 Ay cos Bx + c 2 c 3 A sin Ay + Sin Bx] VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 200 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Similarly for E y 2 2 2 2 z y E H j E h y h x γ ωµ ∂ ∂ · − + ∂ ∂ = 2 z H j h x ωµ ∂ ∂ = 2 j h ωµ [ -c 1 c 3 B cos Ay Sin Bx + c 2 c 3 cos Ay cos Bx -c 1 c 4 B sin Bx sin Ay + c 2 c 4 B sin B sin Ay cos Bx] Applying boundry conditions, E y = 0; x = 0 and x = a E y 0 = 2 j h ωµ [c 2 c 3 B cos Ay+c 2 c 3 B cos Ay + c 2 c 4 B Sin Ay] For E y 0 =0, c 2 = 0 Then the general solution is E y 0 = 2 j h ωµ [- c 1 c 3 B cos Ay sin Bx – c 1 c 4 B sin Bx Sin Ay] If x = a, then E y 0 = 0 E y 0 = 2 j h ωµ Bc 1 sin Bx [c 3 cos Ay + c 4 Sin Ay] For E y 0 = 0, either A = 0 or B = m a π Since A = 0, will make Ey identically zero, it is better to take B = m a π E y 0 = 2 j h ωµ − [ c 1 c 3 B sin Bx cos Ay + c 1 c 4 B sin Bx Sin Ay] E x 0 = 2 j h ωµ [c 1 c 3 A sin Ay cos Bx + c 2 c 3 A sin Ay Sin Bx] Substituting the value of c 2 = c 4 = 0 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 201 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH E x 0 = 2 j h ωµ c 1 c 3 A sin A cos Bx E y 0 = 2 j h ωµ − c 1 c 3 B sin Bx cos Ay Let C = c 1 c 2 Where A = n b π , B = m a π Similarly for H x 0 2 2 2 o z z x z H E j H h x h y H h x γ ωε γ ∂ ∂ − · + ∂ ∂ ∂ − · ∂ for propagation γ = jβ , 2 o z x H j H h x β ∂ − · ∂ but Ey = 2 z H j h x ωµ ∂ ∂ 2 z y H h E x j ωµ ∂ · ⋅ ∂ ∂ Substituting the value of z H x ∂ ∂ in the above H x 0 equation. 2 2 0 y o o x y h j H E h j E β ωµ β ωµ − · ⋅ − · Substituting the value of E y 0 in the above H x 0 equation. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 202 E x 0 = 2 j h ωµ CA sin A cos Bx E y 0 = 2 j h ωµ − CB sin Bx cos Ay VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH β ωµ − · 2 j h ωµ − ¸ CB sin Bx cos Ay] Similarly for H y o 2 2 2 o z z y z H E j H h y h x H h y γ ωε γ ∂ ∂ · − − ∂ ∂ ∂ − · ∂ [∴ E 2 = 0] For propagation γ = jβ 0 2 z y H j H h y β ∂ − · ∂ But Ex = 2 z H j h x ωµ ∂ − ∂ 2 z x H h E y jωµ ∂ − · ⋅ ∂ Substituting this value of z H y ∂ ∂ in the above H y 0 equation. ( ) 2 2 0 x o o y x h j H E h j E β ωµ β ωµ − − · ⋅ · Substituting the value of E x in the above equation H y 0 . H y 0 β ωµ · 2 j h ωµ ¸ CA sin Ay cos Bx] H y 0 = 2 j h β CA sin Ay cos Bx H 2 0 = xy = c 1 c 3 B cos Ay cos Bx + c 2 c 3 cos Ay sin Bx +c 1 c 4 cos Bx sin Ay + c 2 c 4 sin Ay VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 203 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH sin Bx. But c 2 = c 4 = 0 H z 0 = c 1 c 3 cos Ay cos Bx C = c 1 c 3 H z 0 = C cos Ay cos Bx The field equation of TE waves are as follows, H x 0 = 2 j h β CB sin Bx cos Ay H y 0 = 2 j h β CA cos Bx sin Ay H z 0 = C cos Ay cos Bx E x 0 = 2 j h ωµ CA cos Bx sin Ay E y 0 = 2 j h ωµ − CB sin Bx cos Ay Where A = n b π , B = m a π For TE waves the equation for β , f c , λ c iv and λ are found to be identical to those of TM waves. 2 2 2 m = a n b π π β ω µε ¸ _ ¸ _ − − ¸ , ¸ , 2 2 1 m a c n b π π ω µε ¸ _ ¸ _ · + ¸ , ¸ , 2 2 1 m a 2 c n f b π π π µε ¸ _ ¸ _ · + ¸ , ¸ , The corresponding cut off wavelength is 2 2 2 m a c n b λ · ¸ _ ¸ _ + ¸ , ¸ , The velocity of propagation. v ω β · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 204 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 m a n b ω π π ω µε · ¸ _ ¸ _ − − ¸ , ¸ , 2 2 2 2 m a n b π λ π π ω µε · ¸ _ ¸ _ − − ¸ , ¸ , 3. Explain about dominant mode in rectangular waveguide. The lowest mode for TE wave is TE 10 (m = 1, n =0) whereas the lowest mode for TM waves is TM 11 (m = 1, n = 1). This wave has the lowest cut-off frequency. Hence the TE 10 mode is the dominant mode of a rectangular wave guide. Because the TE 10 mode has the lowest attenuation of all modes in a rectangular wave guide and its electric field is definitely polarized in one direction every where. For TE 10 mode m = 1, n = 0 2 2 2 m a 0 n h b a a π π π π ¸ _ ¸ _ · − ¸ , ¸ , ¸ _ · + ¸ , · The field expressions are H x 0 = 2 j h β CB sin Bx cos Ay H x 0 = ( ) 2 j a β π C ( ) a π sin ( ) x a π j ac β π · sin ( ) x a π H y 0 = 0 H z 0 = C cos ( ) x a π E x 0 = 0 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 205 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH E y 0 = ( ) 2 j a ωµ π − C ( ) a π sin ( ) x a π = j wac µ π − sin ( ) x a π The instantaneous field expressions for the dominant TE 10 mode are obtained by the phasor expressions in above equations with e j(wt - β z) and the taking the real part of the product. H x 0 (x, y, z, t) = ac β π − sin ( ) x a π sin (wt - β z) H z 0 (x, y, z, t) = C cos ( ) x a π cos (wt - β z) E y 0 (x, y, z, t) = w a µ π C sin ( ) x a π sin (wt - β z) E x 0 = H y 0 = 0 For TE 10 mode m = 1, n = 0 2 1 a . c C a π ω µε π ¸ _ · ¸ , · 1 a 2 2 c f c a π π µε ¸ _ · ¸ , · where 0 0 1 C µ ε · - 3 x 10 8 m/sec 2 2 1 2 . c c a a λ λ · ¸ _ ¸ , · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 206 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH For TE 10 mode the cutoff wave length is equal to twice the width of the guide. Its cut off frequency is independent of the dimension ‘b’ the field configurations for the lower order TE waves in rectangular wave guider. TE 10 Wave Electric and magnetic field configuration for the lower order mode in a rectangular wave guide. The surface current density on surface waveguide walls is given by J s = n a x H At t = 0 When x = 0 J s = y a − H z VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 207 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH J s (x = 0) = y a − C cos (0 - β z ) When x = a J s = y a H z J s (x = a) = y a C cos β z When y = 0 J s = x a H z y a − H x J s (y = 0) = ax C cos ( ) x a π cos β z - z a ac β π sin ( ) a π x sin β z When y = b J s (y = 0) = J 3 (y = b) The surface currents on side walls at x = 0 and at y = b are selected the below figure. Surface currents on wave guide walls for TE 10 mode in rectangular wave guide. 4. Explain Wave Impedance. In a Cartesian coordinate s/m three wave impedances (impedance constants) must be defined. The wave impedances defines as the ratio of electric field intensity to magnetic field intensity are ; ; y x z xy yx zx y x x E E E Z Z Z H H H + + + · · · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 208 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The wave impedance in the opposite directions are the negative of those given above. (i.e.) ; ; y x z xy yx zx y x x E E E Z Z Z H H H − − − · − · − · − For waves guided by transmission liens or wave guides, the wave impedance which is seen in the direction of propagation z is given by Z z = z xy = Z yz 2 2 2 2 x y z x y E E Z H H + · + For TM waves in a rectangular waveguide. y x TM y x E E Z H H β ωε − · · · It is known that, the propagation constant is 2 2 y h ω µε · − At cut-off frequency f c , λ = 0 2 2 0 c h ω µε − · h 2 – w c 2 µ ε = 0, w c 2 µ ε = h 2 ( ) 2 2 2 2 c 2 f 2 c h h w µε π µε · · The cutoff frequency is 2 c h f π µε · For propagation λ must be imaginary λ β λ µε β µε · − − 2 2 2 2 = h = w j w h VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 209 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH µε µε µε µε − ¸ _ − ¸ , ¸ _ − ¸ , 2 2 2 2 = w =w 1 = w 1 c c w wc w f f The wave impedance of TM waves 2 2 2 1 1 1 M c c c TM p Z f w f f f f Z f ωε µε ωε µ ε η + · ¸ _ − ¸ , · ¸ _ · − ¸ , ¸ _ · − ¸ , Where η is a characteristic impedance µ η ε · The wave impedance of propagating TM modes in a waveguide with a loss less dielectric is purely resistive and is always less than the intrinsic impedance of the dielectric medium. When the operating frequency is lower than the cut-off frequency. The propagation constant is real. W < w c 2 2 2 2 c 2 c 2 = h = w =w 1 (or) a =h 1 c c a w w w wc f f λ µε µε µε µε · − − ¸ _ − ¸ , ¸ _ − ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 210 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH For a given mode the waves with frequencies lower than the cutoff frequency cannot be propagated i.e. waves are attenuated as e -y2 = e -az with z. It propagates if the operating frequency is greater than the cutoff frequency. Therefore a waveguide exhibits the property of a high pass filter. The wave impedance in a non propagation mode is 2 2 c 1 1 f < f TM c c Z j f h f j jh f f for λ ωε ωε ωε · ¸ _ − ¸ , · ¸ _ − · − ¸ , wave impedance is purely reactive, indicating that there is no power flow for f < fc For TE waves in a rectangular wave guide y x TE y x E E Z H H ωµ β − · · · For propagation Y must be imaginary 2 2 2 2 2 2 2 2 = h = w = w =w 1 = w 1 c c y j y j w h w wc w f f β β µε β µε µε µε µε µε · · − − − ¸ _ − ¸ , ¸ _ − ¸ , The wave impedance of TE waves VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 211 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Z TE = µ ω β 2 2 2 1 1 1 c c TE c f w f f f Z f f ωµ µε µ ε η · ¸ _ − ¸ , · ¸ _ − ¸ , · ¸ _ − ¸ , Where η is the characteristic impedance µ η ε · The wave impedance of propagating TM modes in a waveguide with a loss less dielectric is purely resistive and is always larger than the intrinsic impedance of the dielectric medium. When the operating frequency is lower than the cut-off frequency. The propagation codes does not take place. i.e, propagation constant becomes real. 2 1 c f y h f ¸ _ · − ¸ , The wave impedance in a non-propagating mode c 2 f<f 1 TE c j Z j f h f ωµ λ ωµ · · ¸ _ − ¸ , Wave impedance is purely reactive, indicating that there is no power flow for f < f c . For TEM waves in a rectangular wave guide. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 212 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Wave impedance ( ) x TEM y TEM E j Z H or Z j ωµ λ λ ωε − · · · The propagation constant y = a + jβ = jβ jω µε · Substituting in wave impedance equation ( ) TEM TEM j Z j or j Z j ω µε µ ωε ε ωµ µ ε ω µε · · · · Wave impedances of TEM waves is the characteristic impedance of any medium Z TEM = η The variation of wave impedance with frequency is below. Wave impedance versus frequency characteristics of waves between parallel conducting plane. The wave impedance for different are given below. Mod e Wave Impedance TM Z = 2 1 c f f η ¸ _ · − ¸ , TE Z = 2 1 c f f η ¸ _ − ¸ , TEM z µ η ε · · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 213 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH The phase velocity in a waveguide is given by 2 2 2 2 1 =w 1 1 1 1 c c c o c f w f f where f r f f V V f f ω ω γ β µε β µε µε · · ¸ _ − ¸ , ¸ _ − ¸ , · ¸ _ − ¸ , · ¸ _ − ¸ , Where V o = 1 µε The wave length in the waveguide is 2 1 o c V V f f f f λ · · ¸ _ − ¸ , Since 2 2 c o c f f λ λ ¸ _ · ¸ , 2 1 o o c h h λ λ · ¸ _ − ¸ , 2 2 0 c o c λ λ λ λ λ · − or 2 2 2 2 2 0 o c c λ λ λ λ λ · − VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 214 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) 2 2 2 2 2 c o o c λ λ λ λ λ − · ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 2 2 0 c o c o c c o o c o c c c c c c λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λλ λ λ λ − · · + + · · + · + 5. Explain the Excitation methods for various modes? In order to launch a particular mode, a type of probe is chosen which will produce lines of E and H that are roughly parallel to the lines of E and H for that mode possible methods for feeding rectangular waveguides are shown. In figure the probe is parallel to that y – axis and so produces lines of E in the y direction and lines of H which the in the xz plane. This is the perfect field configuration for the TE 10 mode. In Figure the parallel produces fed with opposite phase tend to set up the TE 20 mode in figure the probes which are parallel to the z-axis produces electric field liens in the xy plane for TE 11 mode. In figure the probe parallel to the z-axis produce magnetic field lines in the xy plane. This is the perfect field configuration for the TM 11 mode. It is possible for several modes to exist simultaneously in waveguides, if the VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 215 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH frequency is above cut-off for these particular modes. However the waveguide dimensions are often chosen so that only the dominant mode can exist. 6. Derive the equation for Alternation in waveguides. For the practical consideration we assumed that the waveguides has infinite conductivity that the waveguides has infinite conductivity thus there is the no loss, but impractical the conductivity is not infinite, but some high value, due to this the is some less occurs in the waveguide, which is known as “waveguide attenuation”. Attenuation due to waveguide walls can be defines as 1 Power lost in guide walls 2 transmitted Power α 1 · 1 ¸ ] TM Waves The current induced in the waveguide alls depend on the magnitude of the H x and Hy at the surfaces these the power lost and be written as 2 2 2 0 0 1 2 a b s s x y P J Rs R H dx H dy 1 · · + 1 ¸ ] ∫ ∫ Where J s 2 = H, $ 2 2 x y n H i H j · + = linear current density per meter length per conducting wall. R s = Surface impedance = 2 m m ωµ σ 2 2 0 0 2 a b m x y m P H dx H dy ωµ σ 1 ∴ · + 1 ¸ ] ∫ ∫ Substitute the value of H x and H y integrate and simplify we get, 2 2 2 2 2 2 4 2 8 m m c v f c m a m b f b a ωµ σ η 1 ¸ _ · + 1 ¸ , ¸ ] Power transmitted, VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 216 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1 2 x y y x P E H E H ρ 1 · + ¸ ] 2 2 0 0 1 cos sin 2 a b c m c m x n y dxdy a a a π π π ωεβ ω µε 1 ¸ _ ¸ _ · 1 ¸ , ¸ , ¸ ] ∫ ∫ 2 2 2 0 0 1 sin cos 2 a b c n c m x n y dxdy b a a π π π ωεβ ω µε ¸ _ ¸ _ + ¸ , ¸ , ∫ ∫ 2 2 2 1 8 c T c f abc f P f f η ¸ _ ¸ _ · − ¸ , ¸ , watt 2 2 2 2 2 4 2 2 2 8 1 2 1 8 m c c c m v f c n a m b f b a f abc f f f ωµ σ η α η 1 ¸ _¸ _ 1 + 1 ¸ , ¸ , · 1 ¸ _ ¸ _ 1 − 1 ¸ , ¸ , ¸ ] 2 2 2 2 2 2 / 2 2 1 m c c n m m v f b a N m f f f ωµ α σ η ¸ _ ¸ _ + ¸ , ¸ , · ¸ _ − ¸ , for TE waves The power lost 2 2 1 1 2 2 sx sz J Rs dx J Rs dz · · ∫ ∫ 2 2 2 2 2 2 4 0 0 cos sin 2 a b s R m x m m x A dx A dx a h a a π β π π 1 ¸ _ ¸ _ ¸ _ · + 1 ¸ , ¸ , ¸ , 1 ¸ ] ∫ ∫ 2 2 2 4 2 2 2 2 4 . 2 plane. 2 2 2 . 2 plane 2 2 2 s s R A m a A for y h a R aA m a A for y h a π β π β π 1 ¸ _ · + 1 ¸ , ¸ ] 1 ¸ _ · + 1 ¸ , 1 ¸ ] Total power lost = ( ) 2 2 2 2 4 2 s R A m n a b h a a β π π 1 ¸ _ ¸ _ · + + + 1 ¸ , ¸ , 1 ¸ ] Power transmitted P T = ½ [E x H z + E y H x ] VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 217 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 4 0 0 1 cos sin 2 a b J A n m x n P y dxdy h b a a ωµβ π π π 1 ¸ _ ¸ _ ¸ _ · 1 ¸ , ¸ , ¸ , 1 ¸ ] ∫ ∫ 2 2 2 0 0 sin cos a b m m x n y dxdy a a a π π π 1 ¸ _ ¸ _ ¸ _ + 1 ¸ , ¸ , ¸ , 1 ¸ ] ∫ ∫ 2 2 2 2 1 8 8 c c f ab abA A f ωµβ ωµω µε ω µε ¸ _ · · + ¸ , 2 2 2 1 8 c c f f ab A f f µ ε ¸ _ ¸ _ · + ¸ , ¸ , 1 Power lost 2 Power transmitted α 1 − 1 ¸ ] ( ) 2 2 2 2 4 2 s R A m n a b h a a β π π 1 1 1 ¸ _ ¸ _ 1 · + + + 1 1 1 ¸ , ¸ , 1 1 ¸ ] ¸ ] ¸ ] After simplifications, 2 2 2 1 1 c c f m f a m b f f a c πµ α δ η 1 ¸ _ · + 1 ¸ , 1 ¸ ] ¸ _ − ¸ , nepers/m 7. Explain the characteristics of TE & TM waves. The propagation characteristics of TE & TM waves are obtained as follows, From the above analysis we got 2 2 2 2 2 2 2 2 2 2 2 2 2 P P = m n h p A B a b m n or a b m n or a b π π ω µω π π ω µω π π ωµ ¸ _ ¸ _ · + · + · ¸ , ¸ , ¸ _ ¸ _ · − ¸ , ¸ , ¸ _ ¸ _ − Σ ¸ , ¸ , We know ‘P’ is Q complex number I e VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 218 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 P j m n P j a b α β π π α β ω µ · + ¸ _ ¸ _ ∴ · + − Σ ¸ , ¸ , At low frequencies 2 2 2 m n a b π π ω µ ¸ _ ¸ _ Σ << + ¸ , ¸ , Thus ‘P’ becomes real i.e. p = α , the wave is hence the wave cannot propagate. Hence α = 2 2 2 m n a b π π ω ¸ _ ¸ _ + − Σ ¸ , ¸ , At high frequencies 2 2 2 m n a b π π ω µ ¸ _ ¸ _ Σ >> ¸ , ¸ , Thus ‘p’ becomes purely imaginary [i.e. α = 0] hence the wave propagates 2 2 2 m n j j a b π π β ω µ 1 ¸ _ ¸ _ · Σ − + 1 ¸ , ¸ , 1 ¸ ] At the transition ‘p’ becomes zero, the frequency at which ‘p’ becomes just zero, is defined as cut-off frequency’ At F = f c : P = 0 Hence the equation (1) becomes 2 2 2 2 2 2 2 2 2 c 2 2 c 2 2 0 1 1 f 2 v f= 2 c m n a b m n or a b m n or a b m n or a b m n or a b π π ω µ π π ω µ π π ω µ π π π µ ¸ _ ¸ _ · − Σ ¸ , ¸ , ¸ _ ¸ _ Σ · ¸ , ¸ , 1 ¸ _ ¸ _ · 1 Σ ¸ , ¸ , 1 ¸ ] ¸ _ ¸ _ · Σ ¸ , ¸ , 1 ¸ _ ¸ _ 1 + 1 ¸ , ¸ , ¸ ] Where 8 7 12 1 1 3 10 / sec 4 10 8.854 10 v m µ π − − · · · × Σ × × × The cutoff wavelength is VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 219 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) 2 2 2 2 2 2 c c v v ab f mb na v m n a b λ · · · 1 + ¸ _ ¸ _ 1 ¸ , ¸ , 1 ¸ ] Guide wavelength: It can be defined as the distance traveled by the wave in order to undergo a phase shift of 2π radians. It is denoted by, 2 2 2 2 2 ~ g m a b π π λ β π π ω µ · · 1 ¸ _ ¸ _ Σ − + 1 ¸ , ¸ , 1 ¸ ] Practically the guide wavelength is different from free space wavelength. W.K.T. 2 2 2 c m n a b π π ω µ ¸ _ ¸ _ Σ · + ¸ , ¸ , Thus 2 2 2 2 2 c 2 2 1 2 v = f 1- 1 1 f g c c c c v f f f π π λ ω ω µ ω µ ω µ ω π λ λ ω λ · · Σ − Σ ¸ _ Σ − ¸ , · · ¸ _ ¸ _ ¸ _ + − ¸ , ¸ , ¸ , W.K.T. 1 1 1 , & c c v f f λ α λα µ · Σ 2 1 c c λ λ λ λ ∴ · ¸ _ + ¸ , squaring on both sides we get, 2 2 2 2 2 1 1 1 1 g c c g λ λ λ λ λ λ λ ¸ _ ¸ _ · − ¸ , ¸ , · + VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 220 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Phase Velocity:- W.K.T. The wave propagates in the waveguide Wavelength g λ is greater than the free space wavelength λ , thus the velocity of propagation is defined as the rate at which the wavelength changes its phase in terms of g λ 2 2 2 2 2 . 2 2 (2 / ) = g p g g c f f v f m n a b π λ π ω λ π π λ β ω ω ω µ ω µ π π ω µ · · · · · Σ − Σ 1 ¸ _ ¸ _ Σ − 1 ¸ , ¸ , 1 ¸ ] Since at cut-off frequency 2 2 2 c m n a b π π ω µ ¸ _ ¸ _ Σ · + ¸ , ¸ , Then 2 2 1 1 1 p p v v c fc f v v velocity of light fc c ω ω ω µ ω · · 1 1 ¸ _ ¸ _ Σ − − 1 1 ¸ , ¸ ] ¸ , 1 ¸ ] · · ¸ _ − ¸ , The phase velocity is the velocity of TE & TM waves. Group velocity It can be defined as v g d dP ω · W.K.T. 2 2 2 2 2 2 2 ( ) c m n a b π π β ω µ ω µ ω µ µ ω ω ¸ _ ¸ _ ¸ _ · Σ − + ¸ , ¸ , ¸ , · Σ − Σ · Σ − Differentiate the above equation w.r.t. ‘w’ we get, VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 221 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 2 g 2 2 ( ) 1 1 d 1 v 1 c y d d c fc f fc or v d c d or v d µ β ωµ ω ω ω µ ω ω ω β µ ω λ β λ Σ Σ · · − Σ ¸ _ − ¸ , ¸ _ − 1 ¸ _ ¸ , 1 · · − 1 Σ ¸ , ¸ ] ¸ _ · · − ¸ , 8. For an a/r filled copper x –brand wave side with dimensions a=2.286cm, b=1.016cm determine the cut-off frequency of the first four propagating modes. What is the alteration for metre length of the guide when operating at the frequency of 10GHZ? Given a =2.286cm; b=1.016cm F = 10GHZ length = 1m 10 01 11 2 2 2 2 2 2 2 2 TE ,TE ,TE ,TE m n c a b 1 m n c a b M c m n fc 2 a b π π ¸ _ ¸ _ ω µ∈· + ¸ , ¸ , π π ¸ _ ¸ _ ω · + ∈ ¸ , ¸ , ¸ _ ¸ _ · + ¸ , ¸ , for TE 10 mode: m=1, n=0 8 2 c fc 2a 3 10 2 2.286 10 6.56 GHZ − · × · × × · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 222 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 01 8 2 11 2 2 2 2 8 2 2 for TE mode : m 0, n 1 c fc 2b 3 10 14.76 GHZ 2 1.016 10 for TE mode m=1, n=1 c 1 1 fc= 2 a b 3 10 10 10 2 2.286 1.016 − · · · × · · × × ¸ _ ¸ _ + ¸ , ¸ , ¸ _ ¸ _ × · + ¸ , ¸ , = 16.156 GHZ for TE 02 mode: c 8 2 2 2 2 2 2 2 2 m 0, n = 2 c 2 f 2 b c 3 10 = 29.53 GH b 1.016 10 Pr opagation constant, m n = M a b m n 2f a b c − · · × · · × π π ¸ _ ¸ _ γ + − ω ∈ ¸ , ¸ , ¸ _ ¸ _ ¸ _ · + ¸ , ¸ , ¸ , If the operating frequency is less than the cut-off frequency alternation takes place i.e., propagation does not take place. For TE 01 , TE 11, TE 02 modes propagation will not take place i.e., propagation constant γ · α . For TE 11 mode, m =1, n = 1 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 223 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 9 8 10 10 2 10 10 . 2.286 1.016 3 10 265.77 Nepers in Alternation = 337.4 1 = 337.4 Nepers ¸ _ ¸ _ ¸ _ × × ν · π + × ¸ , ¸ , ¸ , ν · α · · α × l for TE 01 mode; m=0, n=1 2 2 2 9 8 2 2 2 9 8 10 2 10 10 1.016 3 10 227.5 Nepers/ m Alternation 227.5 1 =227.5Nepers 2 10 2 10 10 = 1.016 3 10 581.88 Nepers/ m Alternation 581.88 1 58 ¸ _ ¸ _ × × γ · π − × ¸ , ¸ , α · · α · × ¸ _ ¸ _ × × × γ π − × ¸ , ¸ , α · · α · × · l l 1.88 Nepers 9. A rectangular waveguide has cross-section dimensions a = 7cm & b = 4cm. Determine all the modes which will propagate through the waveguide at frequency of 6 GHZ Given a = 7cm = 7 x 10 -2 m b = 4cm = 4 x 10 -2 m f = 6GHZ λ · ¸ _ ¸ _ + ¸ , ¸ , c 2 2 The cut-off wavelength 2 m n a b o 8 9 I method, c f 3 10 = 5cm 6 10 λ · × · × VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 224 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH · · λ · × 10 c for TE mode : m 1,n 0 2a = 2 7=14cm If λ c >λ o then the propagation takes place the cut off wavelength .λ c , should be greater then the minimum wavelength (λ o ) for propagation, since λ c ,λ o propagation is possible for TE 10 mode, 01 c c o 01 11 2 2 2 2 11 zo for TE mode : m 0;n 1 2b = 2 4=8cm ,propagation is possible for TE mode, for TE mode; m=1, n=1 10 10 = 1600 7 4 87.228 rad/ m propagation is possible for TE mode for TE mode · · λ · × λ > λ ¸ _ ¸ _ γ π + − ¸ , ¸ , · m = 2, n = 0 2 2 10 1600 7 j87.95rad/ m ¸ _ × γ · π − ¸ , · Propagation is possible for TE 20 mode For TE 02 mode: 2 2 10 1600 94.24 4 ¸ _ × γ · π − · ¸ , γ Becomes real value i.e., γ = α (∴β =0) Propagation will not take place in this made ∴ TE 10 , TE 01 , TE 11 , TE 20 modes will propagate VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 225 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 10. A rectangular waveguide measures 3 x 5 cm in equally of ahs a 10GHz signal propagated in it. Calculate the cut off wavelength the guide wavelength of the characteristic wave impedance for the TE 10 mode. Given a = 5 cm = 5 x 10 -2 cm b = 3 cm = 3 x 10 -2 cm f = 10 GHz, TE 10 , m = 1, n = 0 Cut – off wavelength C 2 -2 2a m =2 5 10 = 10 10 m − λ · × × × Cut – off frequency c C 8 2 c f 3 10 10 10 3GHz − · λ × · × · Guide wavelength 0 g e 0 8 9 -2 f 1 f c But f 3 10 10 10 3cm or 3 10 m 2 λ λ · ¸ _ · ¸ , λ · × · × · × g 2 2 3 3 1 10 3.145cm/ 2.145 10 m − λ · ¸ _ − ¸ , · × Characteristic wave impedance VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 226 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) 2 c 2 2z =120 f 1 f 120 3 1 10 395.19 η · η πΩ ¸ _ − ¸ , π · ¸ _ − ¸ , · Ω Q 11. Design a rectangular waveguide with the following specifications a) At a 7.5 GHz the guide wave length for TE 10 mode is 90% of the cut off wave length. B) TE 30 & TE 12 wave the same cut off frequency. Given g e 0 8 9 f 7.5GHz 0.9 c f 3 10 4cm 7.5 10 · λ · λ λ · × · · × 0 g 2 0 C 2 g 2 c 0 c 2 g s 0 c 1 1 1 cross multiplying 1 λ λ · ¸ _ λ + λ ¸ , λ ¸ _ · λ ¸ _ ¸ , λ − λ ¸ , λ ¸ _ ¸ _ λ − · λ λ ¸ , ¸ , 2 2 0 2 g 0 1 = 1+(0.9) 1.81 1.81 1.81 4 = 7.24 cm g g c λ λ λ λ λ λ ¸ _ ¸ _ · + ¸ , ¸ , · · · × VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 227 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH g c c 0.9 0.9 7.29 = 8.04 0.9 2 c g But cm a m λ λ λ λ λ · · · · 10 c c : λ =2a λ a = 2 8.04 a = 2 a = 4.02 cm. TE 30 c c 2 for TE : a = 2 a = 12.067cm a m m λ λ · 12 c 2 2 for TE : m = 1, n = 2 2 m n a b λ · ¸ _ ¸ _ + ¸ , ¸ , ( ) ( ) 2 2 2 2 2 2 4 4 1 2/b (8.04) 12.069 c m n a b λ ¸ _ ¸ _ + · ¸ , ¸ , · − 2 0.0619 0.00687 0.055 b 0.055 b = 8.5cm a = 12.067 cm A · − · · 2 2 2 g 1 = 1+(0.9) 1.81 1.81 0 1081 4 =7.24cm =7.24cm s o g o λ λ λ λ λ ¸ _ ¸ _ · + ¸ , ¸ , · · > · × VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 228 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH But 9 c 0.9 7.24 = 8.04 0.9 2 c cm a m λ λ λ · · · For TE 10 : 2 / 2 8.04 2 4.02 c c a a a a cm λ λ · · · · For TE 30: 2 2 12.067 . c a m xm a a cm λ · > · · For TE 12 : M=1, n=2. 2 2 2 2 2 2 2 2 2 4/ 2 4 1 (8.0 ~) (12.067) = 0.0619-0.00687 = 0.055 4 b 0.055 b = λ λ · ¸ _ ¸ _ ¸ , ¸ , ¸ _ ¸ _ + · ¸ , ¸ , ¸ _ · − ¸ , · c m n a b m n a b b 8.5 cm a =12.067 cm. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 229 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 12. An air filled hollow rectangular conducting waveguide has cross section dimensions of 8x10 cm. How many TE modes will this waveguide transmit at frequencies below 4 GHz? How these mode are designated & What are their cut-off frequencies? Given a = 0.1m, b = 0.08m, f = 4 GHz 8 9 / 3 10 = 0.075 . 4 10 c f m λ · × · × The cut-off wavelength 2 2 2 c m n a b λ · ¸ _ ¸ _ ¸ , ¸ , 2 2 2 0.1 0.08 m n · ¸ _ ¸ _ ¸ , ¸ , Propagation constant, 2 2 2 2 2 2 2 2 2 9 8 2 2 2 2 = 2 = 4 10 40 3 10 3 2 40 0.1 0.08 3 π π ν ω µ π π π π γ π ¸ _ ¸ _ · + − Σ ¸ , ¸ , ¸ _ ¸ _ ¸ _ + − ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ + − ¸ , ¸ , ¸ , × · · × × ¸ _ ¸ _ ¸ _ · + + ¸ , ¸ , ¸ , m n a b m n f a b c m n f a b c f c m n Let m = 1, n = 0 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 230 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 2 1 80 0 0.1 3 = j77.66rad/metre 1 80 m=0, n=1 = 0+ 0.08 3 = j74 rad/m m=1, n=1, 1 80 = 0 0.08 3 = j74rad/m m=1, n γ π γ π γ π ¸ _ ¸ _ · + − ¸ , ¸ , 1 ¸ _ − 1 ¸ ] ¸ , ¸ _ ¸ _ + + ¸ , ¸ , 2 2 2 =1, 1 1 80 = 0.1 0.08 3 =j64 rad/m. γ π ¸ _ ¸ _ ¸ _ + − ¸ , ¸ , ¸ , m = 2, n = 0, 2 2 2 2 2 80 0 0.1 3 = j 55.4 rad/m m=0, n=2 2 80 = 0+ 0.08 3 γ π γ π ¸ _ ¸ _ · + − ¸ , ¸ , ¸ _ ¸ _ + ¸ , ¸ , 2 2 2 =j39.1 rad/m m=1, n=2 1 2 80 = 1 0.8 3 γ π ¸ _ ¸ _ ¸ _ + − ¸ , ¸ , ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 231 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 = 11.708 nepers/m m=2, n=2 2 2 80 = 1 0.08 3 γ π ¸ _ ¸ _ ¸ _ + − − ¸ , ¸ , ¸ , It propagation constant γ is imaginary, propagation will take place. For TE 10, TE 01, TE 11, TE 20, TE 02, TE 21 modes, γ is imaginary. These are the nodes will be nodes will be propagated. The corresponding wavelength for each mode is given by. 2 2 10 c 2 8 c : f 2 3 10 1 f 0 2 1 = 1.5 GHz. ¸ _ ¸ _ · + ¸ , ¸ , × ¸ _ · + ¸ , c m n TE a b TE 01: 2 8 3 10 1 0 2 0.08 = 1.875 GHz c f × ¸ _ · + ¸ , TE 11: 2 2 8 3 10 1 1 = 2 0.1 0.08 = 2.5 GHz × ¸ _ ¸ _ ¸ , ¸ , TE 20: 8 3 10 2 0 2 0.1 = 3 GHz c f × ¸ _ · + ¸ , TE 02 : 2 8 3 10 2 0 2 0.08 c f × ¸ _ · + ¸ , TE 21 : 2 2 8 3 10 2 1 2 0.1 0.08 = 3.457 GHz c f × ¸ _ ¸ _ · ¸ , ¸ , 13. A rectangular air filled copper waveguide with dimension 2cm x 1cm VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 232 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH cross-section & 300m length is operation at a cuttz with a dominant mode find cut-off frequency, guide wavelength, phase velocity, characteristic impedance and atomization. Assume 7 5.8 10 σ · × for copper . Given 2 2 9 2 8 2 9 2 2 10 1 1 10 9 10 30 30 10 3 10 = 3.33 10 9 10 a cm m b cm m f Hz l cm m c f m λ − − − − · · × · · × · × · · × · × · × × The dominant mode is TE 10 , Cut-off frequency 8 2 2 3 10 = 2 2 10 7.5 c c c c f a f GHz λ − · · × × × · Guide wavelength 2 2 2 2 1 3.33 10 = 7.5 1 6.02 10 λ λ λ − − · ¸ _ − ¸ , × ¸ _ − ¸ , · × g fc f a m Phase velocity 2 1 c VP fc f · ¸ _ − ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 233 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 8 8 3 10 = 7.5 1 5.43 10 / sec a VP m × ¸ _ − ¸ , · × Characteristic impedance 2 2 7 1 120 = 7.5 1 9 682 . TE TE Z fc f Z ohms π · ¸ _ − ¸ , ¸ _ − ¸ , · Surface resistance 9 7 7 2 9 10 4 10 = 5: 8 10 2.475 10 . π µ σ π π − − − · × × × × × · × s s f R R ohms Alteration constant 2 2 2 -2 2 2 2 1 1 1 7.5 2.47 10 1 2 2 9 = 7.5 1 10 120 1 9 = 0.02 Nepeirs/m s o o b fc R a c fc b f α µ π − ¸ _ ¸ _ + ¸ , ¸ , · ¸ _ − Σ ¸ , 1 ¸ _ × × × × × 1 ¸ , 1 ¸ ] ¸ _ × × − ¸ , Total Alteration 0.02 0.30 0.006 . l nepers α · × · 14. What are the reasons for impossibilities of TEM mode in a rectangular VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 234 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH waveguide? TEM mode cannot propagate in a rectangular waveguide can be proved by the following argument: 1) By definition, TEM wave means that there is no variation of electric & magnetic fields in the transverse plane. If this is to be satisfied than the transferees field components Ex, Ey, Hxy & Hy have to be constant w.r.t x & y. if this so, then this will violate the boundary conditions E tan = 0< H normal =o. if the boundary conditions are to be satisfied for a rectangular waveguide, then this wave cannot be a TEM wave. 2) In the expression for the field components, if we put E 2 =H 2 =0 then all other field components Ex, Ey, Hx & Hy will also be zero as could be seen from equation. 11 2 11 2 X kx X y ky y · · ∴ There is no wave propagation if both E 2 & H 2 =0 since & x y ∂ ∂ ∂ ∂ are not zero for a rectangular waveguide. Hence no TEM wave propagation can take for a rectangular waveguide. 3) Let us assume that a TEM wave exist inside a rectangular waveguide which is a single conductor system. Existence of TEM wave means the magnetic field must the entirely in the transverse plane. For a magnetic field, div 0 H · that is the magnetic field lines must form closed loops in the x-y transverse plane inside the waveguide. If use apply amperes circuit law tot his magnetic field, the lines of this magnetic field aloud these closed path us must be equal to the current enclosed in the axial direction current or a displacement orient in the 2-direction existence of such displacement current will require an axial component of electric field E 2 B~A if E 2 is present then this wave cannot a TEM wave. Further, if instead of displacement current conduction current exist, then there should be a centre conductor to provide return path, which is not the case in a rectangular waveguide. This argument holds good for any ingle conductor waveguide. There no TEM wave can exist inside a single conductor waveguide. 15. A x-band waveguide which is over filled has inner dimensions of a =2.286 cm and b=1.016 cm. Calculate the cut-off frequencies of the following modes. TE 10 , TE 20, Tm 11 , TM 21 and TM 12. Also find our which of the modes will propagate along the waveguide and which of them will evince when the signal frequency is 10 GHz? The cut-off frequency for a rectangular waveguide equation. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 235 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1/ 2 2 2 , 1 2 m n m n fc a b π π π µε 1 ¸ _ ¸ _ · 1 ¸ , ¸ , 1 ¸ ] For the TE 10 mode put m=1 and n=0 in the above equation 2 10 2 3 10 / 6.56 2 9 2 2 2.286 o c c fc TE GHz a π π · × · · × · V For the TE 01 mode put m= 0 and n = 1. 1/ 2 2 10 01 2 1 1 3 10 / 2 1.016 2 2 c f TE b m b π π ε µε 1 × · · · 1 × ¸ ] For the TE 20 mode 1/ 2 20 10 4 2 / 2 2 3 10 = 13.12 2 2.286 c z c c c f TE a a a GHz π π π π 1 · · · 1 ¸ ] × · × For the TE 11 mode. 2 1/ 2 2 11 2 2 / 2 c c f TE a b π π π 1 · + 1 1 ¸ ] 1/ 2 1/ 2 2 2 10 2 2 2 2 2 2 1/ 2 10 3 10 (2.286) (1.016) 2 2 (2.286) (1.016) 3 10 2.5 16.16 2 2.32 1 1 + × + · · 1 1 ¸ ] ¸ ] × 1 · · 1 ¸ ] c a b a b GHz For the TM 11 mode 11 / c f TM will be same as above F c /TM 11 = 16.16 GHz For the TM 21 mode 1/ 2 1/ 2 10 10 21 2 2 2 2 10 1/ 2 3 10 4 1 4 1 / 1.5 10 2 (0.286) (1.06) =1.5 10 [0.765 0.887] 19.28 c f TM a b GHz 1 × 1 · + · × + 1 1 ¸ ] ¸ ] × + · For the TM 12 mode VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 236 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 1/ 2 10 12 2 2 1/ 2 10 2 2 10 1/ 2 3 10 1 4 / 2 1 4 = 1.5 10 (2.286) (1.016) = 1.5 10 (0.191 3.875) = 30.24 GHz. c f TM a b × 1 · + 1 ¸ ] 1 × + 1 ¸ ] × + Since the signal frequency is 10 GHz only the modes with cutoff frequencies less than 10 GHz will propagate and the others will evanesce. The waves that will propagate is only TE 10 mode. The modes that will evanesce are: TE 01, TE 20, TE 11 , TM 21 and TM 12 . 16. A 10GHZ signal is propagated is a dominant mode in a rectangular wave guide if vg is to be 90% of the free space velocity of the light, then what be the breath of the waveguide. Find the characteristic impedance also. Given: g TE V 0.9 C f 10GHZ a ? Z ? · · · · ( ) ( ) 2 g 2 g 2 2 fc i V C 1 c V sub. fc= 2a V fc 1 C c fc 0.9 1 c fc 4.358GHZ for f= 10 GHZ ¸ _ · − ¸ , ¸ _ · − ¸ , ¸ _ · − ¸ , ∴ · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 237 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) TE 2 TE V ii fc= 4.358GHZ 2a a 0.0344m and Z fc 1 b Z 418.868 η · ∴ · · ¸ _ − ¸ , · Ω 17. Design a rectangular wave guide with following specifications (a) at .5 GHZ the guide λ g for TE 10 is 90% of λ c (b) TE 30 & TE 12 have same fc. Given: g c g 2 f 7.5GHZ 0.9 1 c λ λ λ λ λ λ · · · ¸ _ − ¸ , 2 g 2 2 2 2 g g 1 1 c 1 c λ λ λ λ λ λ λ λ λ λ ¸ _ · ¸ , ¸ _ − ¸ , ¸ _ ¸ _ ¸ _ − · ¸ , ¸ , ¸ , ( ) 2 2 g 2 2 g 1 0.9 1.8 λ λ λ λ ¸ _ · + ¸ , · g g c 1.8 f 0.053m λ λ ¸ _ · ¸ , · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 238 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH g c c 0.058 0.9 WKT 2a 0.058=2a a = 0.029 m λ λ λ · · · ∴ ( ) 30 3V b TE fc 2a 3V fc= 2a 2a c 3 λ · · ∴ · ( ) 10 30 TE & TE have sane 0.024 fc & c c 30 0.059 c a 3 2 λ λ λ · ∴ · 9 a 0.08m 3V fc 5.08x10 Hz 2a · · · ( ) ( ) 2 2 1 m n fc a b 2 fc 30 fc 12 π π π µε ¸ _ ¸ _ · + ¸ , ¸ , · ( ) 12 12 2 2 2 at TE c m n a b sub a=0.08m, we get, b=2.063m π λ π π ⇒ · ¸ _ ¸ _ + ¸ , ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 239 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 18. Derive the Equation for Beset function: Beset Function : In solving for the electromagnetic fields with in guides or circular cross section a differential equation known as Bessel’s equation is encountered. The solution of the equation dents to Bessel functions. These functions will be considered briefly in this section in propagation for the following section on circular wave-guides. These same functions can be expelled to appear in any two dimensional problem in which these is circular symmetry. Examples of such problems are the vibrations of a circular membrane. The propagation of waves within a circular cylinder , and the electromagnetic field distribution about an in finitely long wire. The differential equation involved in there problems these the form. 2 2 2 2 d p 1 dp n 1 p 0 dp e de p ¸ _ + + − · ¸ , Where n is any integer. One solution to this equation can be obtained by assuming a power series solution. 2 0 1 2 p a a e a p ............. · + + + Substitution of this assumed solution back into and equating the coefficients of like power leads to a series solution for the direrential eqauation, for example in me special case when n=0 2 2 d p 1 dp p 0 dp p de + + · When the power series is interasted in and the sums of the coefficients of each power of p are equated to zero. The folowing series is obtained. ( ) ( ) 4 6 2 1 1 2 2 2 4 2 1 2 2 2 2 2 2 1 1 e e e 2 2 p p C 1 ....... 2 2! 3! P P P C 1 .... 2 2 4 2 4 6 1 ¸ _ ¸ _ 1 ¸ _ ¸ , ¸ , 1 · · − + − + 1 ¸ , 1 ¸ ] 1 − · − + + 1 ⋅ ⋅ ⋅ ¸ ] ( ) 2r r 1 1 T 0 (1/ 2e) C 1 (r ) α · · − ∑ this series is convergent for all values of p, either real or complex. iT is called bissels funtion of the first kind of order zero and is denoted by the symbol. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 240 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH J 0 (p) The zero order refers to the fact that if is the solution case of n=0. The coresponded solution for n=1,2,3…etc are designated. J 0 (p), J 2 (p), J 3 (p) where the subscript n donotes the order of the bessel function since second order differential equation, thermust be two linearly independent solutions for each value of n . The second solution may be obtained ina manner somewhat similar to thaeir used for the first but starting with a slightly different senes that is suitably manipulated to yield a solution. The second soulution is known as bessel’s function of the seond kind, or neuman’s function and is designated by the symbol N n (p) Where again n indicates the order of the funtion. In the zero order of this solution of the secend kind the following series is obtained ( ) ( ) ( ) 0 2r r 2 r 1 0 0 2 P N (p) ln r Jo(p) R 1/ 2p 2 1 1 1 1 1 .... 2 3 r r! p AJ (p) Bw (p) α · ¹ ¹ ¸ _ · + ' ) π ¸ , ¹ ¹ ¸ _ · − − + + + + π ¸ , · + ∑ A plot of J o (p) and N o (e) is shown is shown in figure. Because all the neumarn function become infinte at p=0; these second solutions cannot be used for any physical problem in which the origin is included us for example the hellow wave- guide problem VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 241 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH In is apperent the (except near the origin for No (p) ) these curves beat a marked similary to 2 ( ) cos( / 4) 2 ( ) sin( / 4) E J e P e No P P e π π π π → − → − 19. Derive the solution of the field equations of cylindrical co-ordinates. The method of solution of the electromagnetic equations for guided of-circular cross section is similar to that followed for rectangular guides however, in order to simplify the applications of the boundary conditions, it is expedient to express the field equations & the wave equations in the cylindrical co-ordinate system. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 242 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH In cylindrical co-ordinates in a non-conducting region mass ell’s equations are. 2 2 2 2 2 1 ( ) 1 ( ) γ ϕ ω ρ ρ ϕ γ φ ω φ ρ φ γ ρ ωµ φ ρ γ ρ µ φ ρ ρ φ ρ ω ρ ρ φ ρ φ ρ ωµ ρ ρ φ ∂Η + Η · Σ ∂ ∂ + · − Σ ∂ ∂ − − · ∂ ∂ − · ∂ ¸ _ ∂ ∂ − · Σ ∂ ∂ ¸ , ¸ _ ∂ ∂ − · − ∂ ∂ ¸ , j E E E j E E H j E E E jw H H H j E E E j Hz These equations can be combined to give 2 2 2 2 2 2 2 2 2 E h H j E h H j E H j h E ω ρ γ ρ φ ρ γ φ ω ρ ρ φ ωµ ρ υ ρ ρ φ ∂ ∂Η Σ · − ∂ ∂ ∂ ∂Η · − Σ − − ∂ ∂ ∂ ∂ · − − ∂ ∂ H 2 E ϕ 2 2 H j γ ωµ φ ρ ∂Σ ∂ · + ∂ ∂ l 2 2 2 h γ ω µ · + Σ The wave equation in cylindrical co-ordinates for E 2 is 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 f E E E ω µ ρ ρ φ ρ ρ ∂ ∂ ∂ ∂ + + + · ΣΕ ∂ ∂ ∂ ∂ proceeding in a manner similar to that followed in the rectangular case, let E 2 = P(P) e (φ ) 2 2 z c e E e γ γ · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 243 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Where P(p) is a function of P alone & Q(φ ) is a function of φ alone. Sub the expression for E 2 in the wave equation gives. 2 2 2 2 2 2 0 d P QdP d Q Q P PQ PQ dP Pdl dQ γ ω µ + + + + Σ · by PQ, 2 2 2 2 2 2 1 1 1 0 d P d Q h P dP P QP dQ ρ + + + + · As before equation above can be broken up in to two ordinary different equations 2 2 2 2 2 2 2 1 0 2 d Q n Q dQ d P dP n h P dP P dP P · − ¸ _ − + + · ¸ , Where n is a constant. The structure of above equation is, ( cos sin ) Q An b n φ ϕ · + ∩ Through by h 2 , equation (32) is founded in to 2 2 2 2 1 1 0 ( ) ( ) ( 2) d P dp n p d ph Ph d ph p 1 + + + · 1 ¸ ] This is a standard form of Bessel’s equation is term of (eh) using only the solution that is finite at (eh)=0, gives P(eh)=Jn(eh) Where Jn (ph) is Bessel’s function of the first kind of order n sub the solution of (3) & (5) in (2 E 2 = S -7 ( 2 ( )( cos n )e V h An ρ ϕ The solution of Hz will have exactly the save form as fel E 2 & can ∴ be written 2 2 ( )( cos sin ) V n n H Jn Ph c n D n e ϕ ϕ · + for Tm waves the remaining beld components can be obtained by inserting (b) into equation (b) for 7E waves (7) must be inserted into the set corresponding to(b). 20. Derive the equation for Tm & TE waves in circular waveguides. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 244 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH As in the case of rectangular guides, it is convenient to divide the possible solution for circular guides into transverse magnetic & transverse electric waves for the TM waves H 2 is identically zero and the wave equation for E 2 is used. The Bonn dry conditions require the E 2 must vanish at the surface of the guide from ∴ (6) Jn(ha) = 0 (8) Where a is the radius of the guide. There is an infinite number of possible TM waves corresponding to the infinitive number of roots of (8) As before h 2 = 2 2 γ ω µ + Σ & in the case of rectangular guides h 2 must be less than 2 2 ω µ for transmission to occur. There extract high frequencies will be required. This in turn means that only the first few roots of (8) will be of practical interest the first few roots are. (ha) 01 = 2.405 (ha) 11 = 3.85 (ha) 02 = 5.25 – (ha) 12 = 7.02 The first subscript refers to the value of n & the second refers to the roots n their order of magnitude. The various Tm waves will be referred to as Tm 0, & Tm 12 etc. Since 2 2 h γ ω µ · − Σ this gives for β β mn = 2 2 h nm ω µΣ− the cut-off or critical frequency below which transmission of a wave will not occur is 2 (ha)nm hnm= a c hnm f m Where π · Σ The phase velocity is 2 2 V h nm ω ω µ · Σ− from equation (b) the various components of TM waves can be computed of Tm waves can be computed in terms of E 2 . The expression for Tm waves in circular guides are E 2 0 =An Jn (h ρ )cos n ϕ H ρ ° = 1 2 (( ) sin ) jAn n Jn h n h ω ϕ ρ − Σ · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 245 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) cos jAn H jn h n h El H E H ω ϕ ρ ϕ β ϕ ω β ϕ ρ ω − Σ ° · ° · ° Σ ° · ° Σ The variation of each of there field components with time & in the 2 direction are shown by multiplying each of the expressions of (3) by the factor ( ) i t z e ω β − & taking the real part. In the original expression (6) for E 2 1 the arbitrary constant b n has been part equal to zero. The relative amplitudes of An & B n determine the orientation of the field in the guide, & for a circular guide & any particular value of n 1 the ϕ =0, can always i.e. oriented to make either An or B n equal to new. For transverse electric waves E 2 is identically zero & H 2 is given by equation (6), by substituting (6) into (5), the remaining for Tm waves in circular guides are. 2 2 ( ) cos '( ) cos ( ) sin H CnJn hl n j cn Hl Jn h n h jo cn H Jn h n h E H ϕ β ρ ϕ β ϕ ρ ϕ ρ ωµ ρ ρ β ° · − ° · ° · − ° · ° φ The boundary conditions to be met for Tm waves are that Eφ at 1 a ρ · from (b) Eφ is proportional to / Hz ρ ∂ ∂ & ∴ to Jn’ (h ρ )’ (h ρ ) where the prime denote the derivative write (h ρ ) ∴ for Tm waves the boundary conditions require that 1 ( ) Jn ha o − · & its the roots of (47) which must be determined. The first few of these roots are (ha) 1 01 = 3.83 (ha) 1 11 = 1.84 (l a) 1 02 = 7.02 (ha) 1 12 = 3.33 The corresponding TE waves are referred to as TE 01, TE 11 & so on The equations for f c, , , & β λ γ are identical to those for the Tm waves. It is understood, of course that the roots of equation (7) are to be used in connection with TE waves only. The equations shows that the wave having the lowest cut-off frequency is the VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 246 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH TE 11 wave. The wave having the next lowest cut-off frequency is the Tm 01 . 21. Derive the expression for wave impedance & characteristic impedance. The wave impedances at a point have been defined by equation already. For waves guided by transmission lines 1 wave guider, interest centers on the wave impedance which is seen when looking in the direction of propagation that is along the =axis. 2 2 2 2 y y x E Ex Ey H H Hx Hy β ω − − + − + · · · Σ + ∴ 2xy = 2yx = 2 z β ω · Σ The wave impedances looking in the 2-direction are equal & may be put equal to 2 z , where 2 z = 2 2 2 2 trans trans E x y H Hx Hy Σ +Σ · + is the ratio of the total transverse electric field transverse electric field straits to the total transverse magnetic field strength. A similar inspection of egn (b) for in waves in circular grids shows that for them also 2 z =2pφ = -2φ ι = β ωΣ If is seen that for in waves in rectangular a circle girder a indeed in cylindrical grids of any cress-section the wave impedance in the direction of propagation U Constant over the cross section of the guide, & is the same for girder of different shapes reaching that. 2 2 h β ω µ · − & that the cut-off angular frequency ω c has been defined is that frequency that wakes. 2 2 c h ω ΜΣ · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 247 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH it follows that 2 2 1 ( / ) M c β ω ω ω · Σ − Then from (3) a (+) the wave impedance is the 2-direction for Tm waves is 2 2 2 2 2 ( ) 1 ( / ) = 7 1-( c / ) z m Tm c ω ω ω ω · − Σ Thus for any cylindrical guide the wave impedance for Tm waves is dependent only on the intrinsic impedance of the dielectric & the ration of the frequency to the cut-off frequency For TE waves the same conclusion can be recharged. However for TE waves it is found that. 2 2 7 ( ) 1 ( / ) z c Z TE ωµ β ω ω · − for TEM waves between parallel planes or on ordinary parallel wire or co-trig transmission lines the cut-off frequency is zero, d the wave impedance reducer to, Z z (TEM) =7 The dependence of β on the ratic of frequency to cut-off frequency as shown by (3) effects the phase velocity & the wavelength in a corresponding manner thus the phase or wave velocity in a cylindrical guide of any cross section is given by. 2 2 2 2 1 1 1 ( / ) 1 ( / ) o u c c ω γ β µ ω ω ω ω · · · · Σ − − Where V o = 1 µΣ , d & µ Σ are the constants of the dielectric. The wavelength in the guide measured in the direction of propagation, is 2 2 2 2 V 2 1 1 ( / ) >o = 1-( c / ) f c π λ ρ β µ ω ω ω ω · · Σ − Where o λ is the wavelength of a TEM wave of frequency f in a dielectric having the constants & µ Σ . Since 2 2 / ω ω = 2 2 / o c λ λ it follows that VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 248 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 o c c o o c λ λ λ λ λ λ λλ · − · A quantity of great usefulness in correction with ordinary two-conductor transmission liner is the characteristic impeding, 20 of the line for such lines , 20 can be defined in terms of the voltage-current ratio or in terms of the power transmitted for a given voltage or a given current. That is for or infinitely long line. o 2 ; z ; 20= 2 o v v Z I ω ν π ω ∗ ∗ · · Where V & I are peak phasors. For ordinary transmission fives these definitions are equivalent but for wave guides they lead to three values that depend upon the guide dimensions in the same way but which differ by a constant. For example consider the three definitions given by (6) for the case of the TE 10 mode in a rectangular grids. The voltage will be taken as the maximum voltage from the lower face of the grids to the upper face this warms at x=9/2 & has a value, U n =-H x = sin j ac x a β π π − The total longitudinal current in the lower face is I = 2 2 2 2 a b j a c J dx β π − · ∫ Then the integrated characteristic impedance by the first definition 2 2 2 7 20( . ) 2 2 1 ( / ) 2 bGwm b b V I a a fc f a π π π β · · · − Terms of the second definition, the characteristic impendence for the TE 10 wave in a rectangular guide is found to be. Z o (w. I) = 2 ( , ) 8 z o b Z Z u I a x π π · Terms of the third definition the integrated characteristic impedance. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 249 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Z o (W, V) = 2 4 ( , ) z o b Z Z u I a π Explain the excitations of modes in circular waveguides:- TE modes here no 2 component of an Electric field , & TM modes have no 2 – component of magnetic field. If a device is used in a circular waveguide in such a wary that it excites only a 2 – component of electric field, the wave propagative through the guide will be in the TM mode, or the other hand if a device is placed in a circular waveguide in such a way that it exists only the 2-component of magnetic field, the traveling wave will be in the TE mode. The methods o excitation for various modes in circular waveguides are shown. A common method of excitation of TM modes in a circular waveguide by co-axial line is shown. At the end of the co-axial line a large magnetic field exists in the direction of propagation the magnetic field from the co-axial line will excide the TM modes in the guide however, when the guide is connected to the source by a co-axial a discontinuity problem. At the function will increase the eventually decrease the power transmission, it is after necessary to place a turning device around the function in order suppress the reflection. 22. Given a circular waveguide used for a signal at a frequency of 11GHz propagated in the TE 11 mode & the internal diameter is 4-5 cm, Calculate. 1) Cut-off wavelength 2) Group velocity 3) Grid wavelength 4) Phase velocity VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 250 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 5) Characteristic impedance Given f = 11 GHz d = 4.5 cm, a = 2.25 for TE 01, (ha) 11 = 1.84 λ = c/f = 8 9 3 10 11 10 × × = 0.02727 m. i) cut – off wavelength: 11 2 ( ) 2 2.25 = 1.84 π λ π · × c a ha = 7.68 cm / 0.0768 m. ii) Guide wavelength: 2 2 1 0.02727 = 0.02727 1- 0.0768 = 0.029 m. λ λ λ · ¸ _ − ¸ , ¸ _ ¸ , g A c iii) Phase velocity : VP 8 8 0.029 3 10 0.02727 0.029 10 / sec. λ λ 1 · 1 ¸ ] ¸ _ · × ¸ , · × g c m iv) Group velocity: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 251 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH V g 8 8 0.02727 3 10 0.029 2.80 10 / sec λ λ ¸ _ · ¸ , ¸ _ · × ¸ , · × g u c u m v) Characteristic impedance: Z 2 2 20 1 120 1108 ~ . 0.2727 1 0.029 c λ λ π · ¸ _ − ¸ , · · ¸ _ − ¸ , 23. Calculate the cut-off wavelength the guide wavelength & the characteristic wave impedance diameter is ucm for a g GHZ signal propagated in it is the TE 11 mode. Given: F = 9 GHZ D = 4cm; a=A/2 =2cm For TE 11 : (ha) 11 = 1.8~ cut-off wavelength X c ( ) 11 2 9 2 = 2 a = 6.8 / 0.068.3 1.8~ / 3 10 3.33 / 0.0333 , 9 10 a ha cm m x c f cm m π π β · · × · · × Guide wavelength, VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 252 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 9 2 2 1 ( / ) 3.33 = .33 1- 6.83 c λ λ λ λ · − Σ ¸ _ ¸ , = 3.81cm/ 0.0381 m. Diacharacteristic wave: m Polence Z z 2 2 = 1 ( / ) 120 = 3.35 1- 6.83 =435.7 ohms. c η π − > > ¸ _ ¸ , 24. Determine the cut-of frequencies of the first two propagating modes of circular waveguide with a=0.5 cm & 1 2.25 Σ · if the guide is 50cm in length operating at f=13 GHz determine the attenuation Given : A=0.5cm = 0.5x10 -2 m 2.25, 0 50 , 0.5 13 r l cm m f Hz µ µ Σ · · · · · ∠ Cut-off frequency: F c 11 ( ) 2 c ha a π · F x TE 01 mode: (ha) 0.1 = 3.832 for = 8 2 3.832 3 10 2 0.5 10 π × × × × =36.6 GHz for TE 11 mode: (ha) 11 =1.8~1 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 253 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH F c = 8 2 1.8 ~1 3 10 2 0.5 10 π − × × × × Propagator constant 2 2 2 ( ) 1 = nm nm h ha c a γ ω µ ω µ µ γ · − Σ 1 ¸ _ + Σ ∴ · 1 Σ Σ ¸ , 1 ¸ ] (or) TE 01 2 2 01 2 2 9 2 8 2 ( ) 3.832 2 13 10 2.25 = 0.5 10 3 10 648.5 / . π γ γ π α ¸ _ Σ ¸ _ · − ¸ , ¸ , ¸ _ × × × ¸ _ − × × ¸ , ¸ , · f ha a c Nepers m Propagation constant γ becomes real value i.e. v= α If the length of the waveguide is 0.5~ than the Alternation l α = 648.5 x 0.5 = 324.26 Nepers/m 25. A TE 11 mode is propagating through a circular waveguide. The radius of the guide is 1cm and the guide content an air dielectric a) Determine in the out-off frequency b) Determine the wavelength in the guide for an operating frequency of 3GHz. c) Determine the wave impedance in the guide. For TE 11 mode (ha) 11 = 1.8~1 A = 5 x 10 -2 m (a) Cut –off frequency: 11 3 2 ( ) 2 1.8 ~1 3 10 2 5 10 ha c a π π − · × × · × × b) The phase constant in the guide is . VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 254 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 9 7 12 11 3 10 4 10 8.854 10 ( ) h f ha h a β ω µ µ π − − · − · × · × Σ · × · 9 2 7 12 12 1.8 ~1 (2 3 10 ) 4 10 8.854 5 10 β π π − − − ¸ _ · × × × × − × ¸ , the wavelength in the guide is 2 2 50.9 12.3 . cm π λ β π λ λ · · · c) the wave impedance is 9 7 2 2 3 10 4 10 50.9 2 465 . TE TE M ohms ω β π π − · × × × × · · 26. An air filled circular loan guide having an inner radius of icon is excited in dominant mode at 10 GHz. Find the a) cut off frequency of dominant mode at 10GHz . Find the cut-off frequency: guide wavelength and wave impedance. Find the bandwidth. For operation in dominant mode only. The dominant mode is TE, For TE,, mode (ha) 11 = 1.84) F = 10x10 -9 H2 A = 1x10 -2 m a) unit off frequency: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 255 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 8 2 ( ), 2 1.841 3 10 2 1 10 c ha c f a π − · × × · × × b) Guide wavelength: 2 2 2 9 1 1 / 3 10 10 10 g fc c f λ λ λ − · ¸ _ − ¸ , · × · × 2 2 2 2 3 10 3 10 8.795 1 10 6.3 10 m g m λ − − − · × × · ¸ _ − ¸ , · × c) wave impedance: 2 2 1 120 8.795 1 10 792 TE z fc f η π · ¸ _ − ¸ , · ¸ _ − ¸ , · Bandwidth = cut-off frequency of TM01-cut-off frequency of TE,, Band width = 11.49 - 8.795 = 2.695 GHz RECTANGULAR WAVEGUIDES VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 256 ( ) 01 01 01 8 01 3 fc of TM 2 ( ) 2.405 1.405 3 10 fc of TM 2 1 10 11.49 π π − · · × × · × × × · ha c a ha GHz VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 27. Derive the field configuration, cut-off frequency and velocity of propagation for TM waves in rectangular wave guides. Consider the shape of the rectangular waveguides above with dimensions a and b (assume a > b) and the parameters e and m. For TM waves, H z =0 and E, should be solved from equation for TM mode; ° 2 0 2 0 0 xy z z N E h E + · Since E z (x,y,z) = ( ) 0 , gz z E x y e − , we get the following equation, ( ) 2 2 2 0 2 2 , 0 x h E x y x y ¸ _ ∂ ∂ + + · ∂ ∂ ¸ , If we use the method of separation of variables, that is ( ) ( ) ( ) 0 , , z E x y X x Y y · we get ( ) ( ) ( ) ( ) 2 2 2 2 2 1 1 d X x d Y y h X x dx Y y dy − · + Since the right side contains x terms only and the left side contains y terms only, they are both equal to a constant. Calling that constant as k x 2 , we get, ( ) ( ) ( ) ( ) 2 2 2 2 2 2 0 0 x y d X x k X x dx d Y y k Y y dy + · + · Where 2 2 2 y x k h k · + VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 257 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Now, we should solve for X and Y from the preceding equations. Also we have the boundary conditions of; ( ) ( ) ( ) ( ) 0 0 0 0 0, 0 , 0 , 0 0 , 0 z z z z E y E a y E x E x b · · · · From all these, we conclude that X(x) is in the form of sin k, x where k x = mp/a, m= 1,2,3… Y (y) is in the form of sin k y y, where k y = np/b, n=1,2,3…. So the solution of ( ) 0 , z E x y is ( ) 0 0 / , sin sin z V m m n E x y E x y b π π α ¸ _ ¸ _ · ¸ , ¸ , From 2 2 2 , have y z k h k we · − 2 2 m n h a b π π ¸ _ ¸ _ · + ¸ , ¸ , For TM waves, we have 0 0 2 0 0 2 z x z y E jW H h y E jW H h x ε ε ∂ · ∂ ∂ · − ∂ 0 0 2 0 0 2 z x z y E Y E h x E E h y γ ∂ · − ∂ ∂ · ∂ From these equations, we get VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 258 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) ( ) ( ) 0 0 2 0 2 0 2 0 2 , cos sin , sin cos , sin cos , cos sin x y o x o y o m m n E x y E x y h a a b n m n E x y E x y h b a b jw n m n H x y E x y h b a b jw m m n H x y E x y h b a b γ π π π γ π π π ε π π π ε π π π ¸ _ ¸ _ ¸ _ · − ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ · − ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ · − ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ · − ¸ , ¸ , ¸ , Where 2 2 2 m n j j w a b π π γ β µε ¸ _ ¸ _ · · − − ¸ , ¸ , Here, m and n represent possible modes and it is designated as the TM mn mode. M denotes the number of half cycle variations of the fields in the x-direction and n denotes the number of half cycle variations of the fields in the y-direction. When we observe the above equations we see that for TM modes in rectangular waveguides neither m nor n can be zero. This is because of the fact that the field expressions are identically zero if either m or n is zero. Therefore, the lowest mode for rectangular waveguide TM mode is TM 11 Here, the cut-off wave number is 2 2 c m n k a b π π ¸ _ ¸ _ · + ¸ , ¸ , and therefore, 2 2 c k k β · − The cut-off frequency is at the point where g vanishes. Therefore, ( ) 2 2 1 2 z m n f H a b εµ ¸ _ ¸ _ · + ¸ , ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 259 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH Since I= u/f, we have the cut-off wavelength ( ) 2 2 2 c m m n a b λ · ¸ _ ¸ _ + ¸ , ¸ , At a given operating frequency f, only those frequencies , which have f c < f will propagate. The modes with f < f c will lead to an imaginary b which means that the field components will decay exponentially and will not propagate. Such modes are called cut-ff or evanescent modes. The mode with the lowest cut-off frequency is called the dominant mode. Since TM modes for rectangular waveguides start from TM 11 mode, the dominant frequency is ( ) ( ) 2 2 11 1 1 1 2 c f Hz a b εµ ¸ _ ¸ _ · + ¸ , ¸ , The wave impedance is defined as the ratio of the transverse electric and magnetic fields, Therefore, we get from the expressions for E x and H y (see the equations above). x TM TM y E Y j j Z Z H jW jw jw k β β βη ε ε ε · · · · ⇒ · The guide wavelength is defined as the distance between two equal phase planes along the waveguide and it is equal to 2 2 k π π λ λ β · > · Which is thus greater than 1, the wavelength of a plane wave in the filling medium. The phase velocity is 1 p w w u k β µε · > · Which is greater than the speed of light (plane wave) in the filling material Attenuation for propagating modes results when there are losses in the dielectric and in the imperfectly conducting guide walls. The attenuation constant due to the losses in the dielectric can be found as follows: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 260 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2 2 2 2 2 1 1 1 c c c c f f f j j k k jk jw jw f f jw f σ γ β µ ε ¸ _ ¸ _ ¸ _ · · − · − · − · + − ¸ , ¸ , ¸ , 28. Derive the field configuration cut-off frequency and velocity of propagation for TE waves in rectangular wave guide. TE Modes Consider again the rectangular waveguide below the dimensions a and b (assume a>b) and the parameters e and m. For TE waves E z =0 and H z should be solved from equation of TE mode; Since H z (x,y,z) = ( ) 0 , , gz z H x y e − we get the following equation, ( ) 2 2 2 0 2 2 , 0 z h H x y x y ¸ _ ∂ ∂ + + · ∂ ∂ ¸ , If we use the method of separation of variables, that is H z 0 (x,y) =X(x), Y(y) we get, ( ) ( ) ( ) 2 2 2 2 2 1 ( ) 1 d Y y d X x h X x dx Y y dy − · + Since the right side contains x terms only and the left side contains y terms only, they are both equal to a constant. Calling that constant as k x 2 , we get; ( ) 2 2 2 ( ) 0 z d X x k X x dx + · VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 261 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) 2 2 2 0 y d Y y k Y y dy + · Where 2 2 2 y x k h k · − Here, we must solve for X and Y from the preceding equations. Aslo we have the following boundary conditions: ( ) ( ) ( ) ( ) 0 x=0 0 x=a 0 y=0 0 y = b 0 0 0 0 0 0 0 0 z y at z y at z x at z x at H E dx H E dx H E dy H E dx ∂ · · ∂ · · ∂ · · ∂ · · From al these, we get ( ) ( ) 0 0 / , cos cos z A m m n H x y H x y a b π π ¸ _ ¸ _ · ¸ , ¸ , From 2 2 2 , have; y z k h k we · − 2 2 2 m n h a b π π ¸ _ ¸ _ · + ¸ , ¸ , For TE waves, we have 0 0 2 0 0 2 z x z y H H h x H H h y γ γ ∂ · − ∂ ∂ · − ∂ 0 0 2 z x H jw E h y µ ∂ · ∂ 0 0 2 z y H jW E h x µ ∂ · ∂ From these equations, we obtain VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 262 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) ( ) ( ) ( ) 0 0 2 0 0 2 0 0 2 0 0 2 , cos sin , sin cos , sin cos , cos sin x y x y jw n m n E x y H x y h b a b jw m m n E x y H x y h b a b m m n H x y H x y h a a b n m n H x y H x y h b a b µ π π π µ π π π γ π π π γ π π π ¸ _ ¸ _ ¸ _ · ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ · ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ · ¸ , ¸ , ¸ , ¸ _ ¸ _ ¸ _ · ¸ , ¸ , ¸ , Where 2 2 2 m n j j w s a b π π γ β µ ¸ _ ¸ _ · · − − ¸ , ¸ , As explained before, m and n represent possible modes and it is shown as the TE mn mode. ,m denotes the number of half cycle variations of the fields in the x-direction and n denotes the number of half cycle variations of the fields in the y-direction. Here, the cut-off wave number is 2 2 c m n k a b π π ¸ _ ¸ _ · + ¸ , ¸ , And therefore, 2 2 c k k β · − The cut-off frequency is at the point where g vanishes, Therefore, ( ) 2 2 1 2 c z m n f H a b εµ ¸ _ ¸ _ · + ¸ , ¸ , Since I = u/f, we have the cut-off wavelength VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 263 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH ( ) 2 2 2 c m m n a b λ · ¸ _ ¸ _ + ¸ , ¸ , At a given operating frequency f, only those frequencies, which have f>f c will propagate. The modes with f<f c will not propagate The mode with the lowest cut-off frequency is called the dominant mode. Since TE 10 mode is the minimum possible mode that gives nonzero field expressions for rectangular waveguides, it is the dominant mode of a rectangular waveguide with a>b and so the dominant frequency is ( ) ( ) 10 1 2 c f Hz a µε · The wave impedance is defined as the ratio of the tranverse electric and magnetic fields. Therefore, we get from the expressions for E x and H y (see the equations above); x TE TE y E jw jw k Z Z H j µ µ η γ β β · · · ⇒ · The guide wavelength is defined as the distance between two equal phase planes along the waveguide and it is equal to 2 2 g k π π λ λ β · > · Which is thus greater than 1, the wavelength of a plane wave in the filling medium. The phase velocity is 1 p w w u k β µε · > · Which is greater than the speed of the plane wave in the filling material The attenuation constant due to the losses in the dielectric is obtained as follows: 2 2 2 2 2 1 1 1 c c c c f f f j j k k jk jw jw f f jw f σ γ β µ ε ¸ _ ¸ _ ¸ _ · · − · − · − · + − ¸ , ¸ , ¸ , VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 264 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH After some manipulation, we get 2 2 tan 2 2 1 d c cn k f f δ α β · · ¸ _ − ¸ , Example: Consider a length of air-filled copper x-band waveguide, with dimensions a=2.286 cm, b= 1.016 cm. find the cut-off frequencies of the first four propagating modes. Solution: From the formula for the cut-off frequency ( ) 2 2 2 2 1 2 2 air filled c m n c m n f Hz a b a b εµ − ¸ _ ¸ _ ¸ _ ¸ _ · + ÷÷÷÷→ + ¸ , ¸ , ¸ , ¸ , B.E./B.TECH. DEGREE EXAMINATION, NOVEMBER/DECEMBER 2008 FIFTH SEMESTER ELECTRONICS AND COMMUNICATION ENGINEERING EC 1305 – TRANSMISSION LINES AND WAVEGUIDES (COMMON TO B.E. (PART – TIME) FOURTH SEMESTER REGULATION 2005) PART - A 1. Briefly discuss the difference between wavelength and period of a sine wave. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 265 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 2. Find the attenuation and phase shift constant of a wave propagating along the line whose propagation constant is 1.048× 10 -4 ∠ 88.8 o . 3. Give the minimum and maximum value of SWR and reflection coefficient. 4. Why is the quarter wave line called as copper insulator? 5. Enumerate the properties of TEM waves between parallel planes of perfect conductors. 6. Plot the frequency – versus – attenuation characteristic curve of TM and TE waves guided between parallel conducting plates. 7. How is the TE 10 mode launched or initiated in rectangular wave guide using an open ended coaxial cable? 8. Calculate the cut-off frequency of a rectangular wave guide whose dimensions are ‘a’=2.5cm and ‘b’=1.5cm operating at TE 10 mode. 9. Why is the Bessel’s function of the second kind (neumann’s function applicable for the field analysis inside the circular wave guide? 10. Distinguish between wave guides and cavity resonator. PART – B 11. (a) Derive the general transmission line equations for voltage and current any point on a line. Or (b)(i) Write a brief notes on frequency and phase distortions. (ii) The characteristic impedance of a 805m-long transmission line 94∠ - 23.2 o Ω , the attenuation constant is 74.5× 10 -6 Np/m and the phase shift constant is 174× 10 -6 rad / m at 5KHz. Calculate the line parameters R, L, G and C per meter and the phase velocity on the line. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 266 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 12. (a) (i) A 75Ω loss less transmission line is to be matched to a resistive load impedance of Z L =100Ω via a quarter-wave section. Find the characteristic impedance of the quarter wave transformer. (ii) A 50Ω loss less transmission line is terminated in a load impedance of Z L =(25+j50)Ω . Use the SMITH chart to find. (1)Voltage reflection coefficient, (2)VSWR, (3) Input impedance of the line, given that the line is 3.3λ long and (4)Input admittance of the line. Or (b) A 50Ω loss less feeder line is to be matched to an antenna with Z L (75-j20)Ω at 100MHz using SINGLE shorted stub. Calculate the stub length and distance between the antenna and stub using smith chart. 13. (a)(i) Derive the components of Electric and Magnetic field strength between a pair of parallel perfectly conducting planes of infinite extent in the ‘Y’ and ‘Z’ directions. The planes are separated in X direction by “a” meter. (ii) A parallel perfectly conducting plates are separated by 5cmin air and carries a signal with frequency of 10 GHz in TM 11 mode. Find the cut-off frequency and Cut-off wave length. Or (b) (i) Discuss on the characteristics of TE, TM and TEM waves between parallel conducting planes. And also derive the expressions for the cut off frequency and phase velocity from the propagation constant. (ii) Describe the Velocity of propagation of wave between a pair of perfectly conducting plates. 14. (a) Derive the field configuration, cut of frequency and velocity of propagation for TE waves in rectangular wave guide. Or (b) A TE 10 wave at 10 GHz propagates in a X-band copper rectangular wave VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 267 VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH guide whose inner dimensions are ‘a’=2.3cm and ‘b’ =1cm, which is filled with Teflon ε r =2.1, µ r =1. Calculate the cut-off frequency, velocity of propagation, Phase velocity, Phase constant, Guide wave length and Wave impedance. 15. (a)(i) Derive the expression for TM wave components in circular wave guides using Bessel function. (ii) Write a brief note on excitation of modes in circular wave guides. Or (b) Derive the equation for Q – factor of a rectangular cavity resonator for TE 101 mode. *************** VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 268 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL SEM - V INDEX UNITS PAGE NO. I. Filters 06 II. Transmission Line Parameters 51 III. The Line at Radio Frequency 95 IV. Guided Waves Between Parallel Planes 138 V. Waveguides 179 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 2 VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL # 42 & 60, Avadi – Veltech Road, Avadi, Chennai – 62. Phone : 044 26840603 26841601 26840766 email :
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Companies Such as TCS, INFOSYS TECHNOLOGIES, IBM, WIPRO TECHNOLOGIES, KEANE SOFTWARE & T INFOTECH, ACCENTURE, HCL TECHNOLOGIES, TCE Consulting Engineers, SIEMENS, BIRLASOFT, MPHASIS(EDS), APOLLO HOSPITALS, CLAYTON, ASHOK LEYLAND, IDEA AE & E, SATYAM VENTURES, UNITED ENGINEERS, ETA-ASCON, CARBORANDUM UNIVERSAL, CIPLA, FUTURE GROUP, DELPHI-TVS DIESEL SYSTEMS, ICICI PRULIFE, ICICI LOMBARD, HWASHIN, HYUNDAI, TATA CHEMICAL LTD, RECKITT BENKIZER, MURUGAPPA GROUP, POLARIS, FOXCONN, LIONBRIDGE, USHA FIRE SAFETY, MALCO, YOUTELECOM, HONEYWELL, MANDOBRAKES, DEXTERITY, HEXAWARE, TEMENOS, RBS, NAVIA MARKETS, EUREKHA FORBES, RELIANCE INFOCOMM, NUMERIC POWER SYSTEMS, ORCHID CHEMICALS, JEEVAN DIESEL, AMALGAMATION CLUTCH VALEO, SAINT GOBAIN, SONA GROUP, NOKIA, NICHOLAS PHARIMAL, SKH METALS, ASIA MOTOR WORKS, PEROT, BRITANNIA, YOKAGAWA FED BY, JEEVAN DIESEL visit our campus annually to recruit our final year Engineering, Diploma, Medical and Management Students. Preface to the First Edition This edition is a sincere and co-ordinated effort which we hope has made a great difference in the quality of the material. “Giving the best to VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 3 VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL the students, making optimum use of available technical facilities & intellectual strength” has always been the motto of our institutions. In this edition the best staff across the group of colleges has been chosen to develop specific units. Hence the material, as a whole is the merge of the intellectual capacities of our faculties unit are available in this material. Prepared By : Ms. S. Jalaja Asst. Professor. Mr. S. Jebasingh. Lecturer. across the group of Institutions. 45 to 60, two mark questions and 15 to 20, sixteen mark questions for each EC2305 UNIT I TRANSMISSION LINES AND WAVEGUIDES FILTERS 9 The neper - the decibel - Characteristic impedance of Symmetrical Networks – Current and voltage ratios - Propogation constant, - Properties of Symmetrical Networks – Filter fundamentals – Pass and Stop bands. Behaviour of the Characteristic impedance. Constant K Filters - Low pass, High pass band, pass band elimination filters - m -derived sections – Filter circuit design – Filter performance – Crystal Filters. UNIT II 9 TRANSMISSION LINE PARAMETERS VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 4 VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL A line of cascaded T sections - Transmission lines - General Solution, Physical Significance of the equations, the infinite line, wavelength, velocity, propagation, Distortion line, the telephone cable, Reflection on a line not terminated in Zo, Reflection Coefficient, Open and short circuited lines, Insertion loss. UNIT III THE LINE AT RADIO FREQUENCY 9 Parameters of open wire line and Coaxial cable at RF – Line constants for dissipation voltages and currents on the dissipation less line - standing waves – nodes – standing wave ratio - input impedance of open and short circuited lines - power and impedance measurement on lines – / 4 line, Impedance matching – single and double-stub matching circle diagram, smith chart and its applications – Problem solving using Smith chart. UNIT IV GUIDED WAVES BETWEEN PARALLEL PLANES 9 Application of the restrictions to Maxwell’s equations – transmission of TM waves between Parallel plans – Transmission of TE waves between Parallel planes. Transmission of TEM waves between Parallel planes – Manner of wave travel. Velocities of the waves – characteristic impedance - Attenuators UNIT V WAVEGUIDES 9 Application of Maxwell’s equations to the rectangular waveguide. TM waves in Rectangular guide. TE waves in Rectangular waveguide – Cylindrical waveguides. The TEM wave in coaxial lines. Excitation of wave guides. Guide termination and resonant cavities. TEXT BOOK: 1. John D.Ryder, "Networks, lines and fields", Prentice Hall of India, 2nd Edition, 2006. REFERENCES: 1. E.C.Jordan, K.G. Balmain: “E.M.Waves & Radiating Systems”, Pearson Education, 2006. 2. Joseph Edminister, Schaum’s Series, Electromegnetics, TMH, 2007. 3. G S N Raju, Electromagnetic Field Theory and Transmission Lines, Pearson Education, 2006. UNIT – I PART – A 1. Define Filter? A reactive network that will freely pass desired bands of frequencies while almost totally suppressing other band of frequencies are called as filters. 2. What do you mean by ideal filter? An ideal filter would pass all frequencies in a given without reduction in VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 5 VEL Give the relation between two units of attenuation? The relationship between two units of attenuation can 1 db = 0. 10. 6.115 neper. Define the unit of attenuation (or) Define Neper? Attenuation is expressed in decibels or nepers. 9. Provide infinite attenuation 3. Band pas filter 3. Draw a symmetrical network in ‘T’ and Π sections. 3. 5. What are the characteristics of ideal filter? The characteristics of ideal filter are: 1. Nepers is defined as the natural logarithm of the ratio of input voltage or current to the output voltage or current provided the network is properly terminated with Z0. 7. When networks are said to be symmetric network? When two series arms of a T network are equal or when two shunt arms of a network are equal then the network is said to be symmetrical network. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 6 VEL . Give the common types of filters? The four common types of filters are. Band stop filter 8. The transition region between the stop band and pass band would be very small. Transmit pass band frequencies without any attenuation. 2. 4. Low pass filter 2. Define Decibel? Decibel is defined as the ten times common logarithms of the input power to the output power. High pass filter 1. 4. What is cutoff frequency? The frequency which separates a pass band and an attenuation band are called as cutoff frequency. Throughout the pass band characteristic impedance of the filter match circuit to which it is connected which prevents reflection loss.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL magnitude and totally suppressing all other frequencies. . Give the characteristic impedance of low pass constant – K filter The characteristic impedance is given as Z OT = (Z 2 1 + Z1 Z 2 ) Z OT = L / C 1 − ω 2CL / 4 12. 14. Give the formula to calculate the cutoff frequency for low pass constant –k f c = 1/ π LC 13. Why constant – k filters are also called as prototype filter? Constant – k filters are also called as prototype filter because other complex networks can be derived from it. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 7 VEL .e. Where ‘Rk’ is a real constant i. Z1 – Z2 = R2 K. What are the constant – k filters? A constant – k filters is a ‘T’ or ‘π ’ network in which the series and shunt impedance Z1 and Zz are connected by the relation. a resistant independent of frequency.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 11. For a high pass filter what is the conditions for which the characteristic impedance ZOT is real and imaginary.k filter is given as 2 LC ) < 1 21. Draw the T. f c = 1/ 4π LC 17. Z OT = L / C 1 − ( 1/ 4ω 2 LC ) The characteristic impedance of high pass filter constant. The attenuation of constant high pass filter is given as α = 2 cos −1 ( ω c / ω ) nepers 2 19. For a low pass filter what is the condition for which the characteristic impedance Z0 is real? The characteristic impedance Z0 is real if (ω 2 LC/4)<1 16. What are the phase shift and attenuation of constant –k high pass filter? The phase shift of constant – k high pass filter is given as β = 2 sin-1 ( ω c2/ω 2) radians. For a high pass filter. The prototype low pass filters are as shown below. What are the phase shift and attenuation of constant –k low pass filter? −1 The phase shift of constant – k low pass filer is given as.type and π type low pass constant – k filter. Give the expression for the cutoff frequency of constant – k high pass filter.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 15. Give the characteristic impedance of high pass filter constant – k filter. the characteristic impedance ZOT is real if (1/4ω And Imaginary if (1/4ω 2 LC) >1 20. β = 2sin ω / ωC radians −1 The attenuation of constant low pass filter is given as α = 2 cos ( ω / ω c ) nepers ( ) 18. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 8 VEL . VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 22.derived filters makes it possible to construct composite filters to have any desired attenuation characteristics. 24.type and Π type high pass constant –k filter 23.com The different sections of a composite filters are 1. One or more m – derived sections 3. Mention the different sections of a composite filters?
[email protected] are connected at the ends. What is composite filter? A filter designed using one or more prototype constant – k filters and m – derived filters to have an attenuation between low pass and high filters is called as composite filters. What are the advantages of m. ‘Zo’ of the filter will be more uniform within the pass and when m – derived half section having m = 0. m. 2. One or more prototype constant filters are 2. Two terminating m – derived half sections with m = 0. 25. A sharper cutoff characteristic with steeper rise at fc. Draw the T.6 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 9 VEL . 3. the slope of the rise being adjustable by fixing the distance between fc and f. 1.derived filters? The advantages of m – derived filters are. The attenuation does not increase rapidly beyond cutoff frequencies. Characteristic impedance varies widely in the transmission or pass band from the derived value. No 1 Low Pass Filter This filter passes the frequencies without attenuation upto a cutoff frequency fc and attenuates all other frequencies greater than fc. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 10 VEL .No 1 Bandpass filter This filter passes the frequencies between two designated cutoff frequencies and attenuates all other frequencies An ideal band pass filter is as shown below Band elimination filter It transmits all frequencies while attenuates a band of frequency. 2. 2. 27. Distinguish between low pass and high pass filter. S. Why m–derived half section is used as terminating section? Two m – derived half section with m = 0. An ideal low pass filter is as shown below High Pass filter It transmits frequencies above a designed cutoff frequency but attenuates frequencies below this An ideal high pass filter is as shown below 2 29. 28.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 26. Differentiate between band pass filter and band elimination filter S. An ideal band elimination filter is as shown below. What are the drawbacks of constant – k filters (or) what are the disadvantages of constant – k filters? 1.6 is used as a terminating section to give constant input and output impedance. Draw a block diagram of a composite filter? 31.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 30. Draw the T-type and Π type low pass m – derived filter The m – derived low pass filters are as shown below. 32.derived high pass filters are as shown below VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 11 VEL . Draw the type and Π type high pass m.derived filter The m. When Z1 = Z2 or the two series arms of a T network are equal. Filter networks are ordinarily set up as symmetrical sections.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 33. such as shown at (b) and (d). or Za = Zc and the shunt arms of a π network are equal. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 12 VEL . Give the value of ‘m’ in m – derived low pass and high pass filters. the networks are said to be symmetrical. Fig. basically of the T or π type. Define the characteristic impedance of symmetrical networks. The value of ‘m’ in the case of low pass filter is given as below m = The value of ‘m’ in the case of high pass filter is as below m = 1 − ( fc / f ∞ ) 2 2 1 − ( f ∞ / fc ) PART – B 1. Fig. showing notation used in symmetrical network analysis.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Figure: The T and π sections as derived from unsymmetrical L sections. arising from the fact that both T and π networks can be considered as built of unsymmetrical L half sections. Z0.” which are indistinguishable one from the other except for the end or terminating L half sections. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 13 VEL . are equal to each other. and the image impedance is then called the characteristic impedance or the iterative impedance. This peculiarity is largely dictated by custom. A series connection of several T or π networks leads to socalled “ladder networks. and oppositely for the π network as at (a) and (c). That is. The term iterative impedance is apparent if the terminating impedance Z0 is considered as the input impedance of a chain of similar networks. For a symmetrical network the image impedances Z1i and Z2i. connected together in one fashion for the T network. Attention is called to the peculiarities of notation employed on the variouos arms. in which case Z0 is iterated at the input to each network. or its impedance transformation ratio is unity. as can be seen in fig. if a symmetrical T network is terminated in Z0. its input impedance will also be Z0. the input impedance is Z1in = Z1 Z 2 ( Z1 / 2 + Z0 ) + 2 Z1 / 2 + Z2 + Z0 It can be assumed that if Z0 is properly chosen in terms of the network arms. it should be possible to make Z1in equal to Z0. For the T network of Fig. The value of Z0 for a symmetrical network can be easily determined. This result could also have been immediately obtained from eq. Requiring this equality gives Z0 = Z02 = Z12 / 4 + Z1Z2 + Z2Z0 + Z1Z0 / 2 Z1 / 2 + Z2 + Z0 Z12 4 + Z1Z2 For the symmetrical T section. The parallel shunt arms will be combined. terminated in an impedance Z0. then. (a). and for the image VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 14 VEL . (b) ladder network built from π sections.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Figure: (a) Ladder network made from T section. Z0T = Z12 Z + Z1Z2 = Z1Z2 1+ 1 4 4Z2 becomes the characteristic impedance. for the π section of Fig.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Figure: Determination of Z0: (a) for a T section. exclusive of the load Z0. by using the values of the arms of Fig. Fig. (b) for a π section impedance of a T section. then Z1 + Z2 2 Z Z Z /2 Z1∞ = Z∞ = 1 + 1 2 2 Z1 / 2 + Z2 Z2 Z∞ Z∞ = 1 + Z1Z2 = Z0T 2 4 Z1∞ = Z∞ = VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 15 VEL . Certain information concerning networks was developed from measurements of Z∞ and Z∞ . If these measurements are made on the T section of (a). Similarly. (b) the input impedance is 2Z2Z0 Z1 + 2Z2 2Z 2 + Z0 = 2Z2Z0 Z1 + + 2Z2 2Z 2 + Z0 Z1in Requiring that Z1in = Z0 leads to Z0π = Z1Z2 1+ Z1 / 4Z2 which is the characteristic impedance of the symmetrical π symmetrical π section. by input volt-amperes E1I 1 = E2I 2 output volt-amperes 1 EI θ = ln 1 1 2 E2I 2 ε 2θ = Or where θ is in general a complex number. Z0 = Z ∞ Z∞ This result could have been directly obtained from the image impedance relations of section. obviously. under the condition of Z0 termination. For symmetrical networks Z 14 = Z2 = Z0. for a symmetrical network.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Similar work for the π section leads to Z∞ Z∞ = 4Z22 Z1 Z1 + 4Z2 = Z0π2 Therefore. It is a valuable relationship. the use of the exponential can be extended to include the phasor current ratio if it be defined that. Explain the current and voltage ratios as exponentials. the absolute value of the ratio of input current to output current of a given symmetrical network was defined as an exponential function. with γ customarily replacing θ and implying symmetry and Z0 termination. which may now be interpreted as a Z0 termination on the network. I1 = εγ I2 where γ is a complex number defined as γ = α + jβ VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 16 VEL . it is customary to define a transfer constant θ . complete network performance. terminated on an image basis. and with a termination of Z0. the phase angle between the currents being needed as well. since it supplies an easy experimental means of determining the Z0 of any symmetrical network.* for the purpose of simplifying network calculations. the magnitude ratio does not express the * In the general case of unsymmetrical 4-terminal networks. the above discussion follows. 2. Under the assumption of equal input and output impedances. the propagation constant. The units of β are radians. = 1 I 2 I 3 I4 In from which ε γ 1 × ε γ 2 × εγ 3 × . or the attenuation produced in passing through the network. If a number of sections all having a common Z 0 value are cascaded. The exponent α is known as the attenuation constant. and the introduction of ε Z0 terminated network. if I 1 / I 2 = A β . Use of the definition of γ . leads to further useful results.. = γ n Thus the over-all propagation constant is equal to the sum of the individual propagation constants. = εγn and taking the natural logarithm. γ as the current ratio for a VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 17 VEL .. The exponent β is the phase constant as it determines the phase angle between input and output quantities. the ratio of currents is I1 I2 I 3 I × × × . since it determines the magnitude ratio between input and output quantities.then A = I 1 / I 2 = εα β = ε jβ Since the input and output impedances are equal under the Z 0 termination. γ 1 + γ 2 + γ 3 + . or the shift in phase introduced by the network..VEL TECH I1 = ε γ = εα+ jβ I2 VEL TECH MULTI TECH TECH HIGHTECH VEL Hence To illustrate further.. The units of α are nepers. Explain the properties of symmetrical networks. it is also true that V1 = εγ V2 The term γ has been given the name propagation constant. 3.. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Figure: Symmetrical network with generator and load. in Eq. and is terminated in a load Z0. for the characteristic impedance. Z02 = Z12 4 + Z1Z2 If Z0 is eliminated by use of Eq. In Fig. which is equal to ε obtained from the second equation as I 1 Z1 / 2 + Z2 + Z0 = = εγ I2 Z2 After thus introducing ε γ γ by definition. the above may be written Z1 2 Z0 = Z2 ( ε γ − 1) − From Eq. there results Z2 ( ε γ − 1) − Z1εγ = 0 2 ε 2γ − 2ε γ + 1= Z1 γ ε Z2 ε γ + ε− γ Z = 1+ 1 2 2Z2 Z cosh γ = 1+ 1 2Z2 Equation and its other derived forms will be of considerable value in the study of VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 18 VEL . The mesh equation are Z E = I 1 1 + Z2 − I 2Z2 2 Z 0 = −I 1Z2 + I 2 1 + Z2 + Z0 2 The current ratio for the two meshes. can be . the T network is considered equivalent to any connected symmetrical network. with Z0 replaced by ZR. for Z0T. VEL TECH MULTI TECH TECH HIGHTECH VEL By use of the identify. The propagation constant γ can be related to the network parameters by use of Eq. that cosh2 γ − sinh2 γ = 1 it is possible to write Z0 Z2 Combining Eqs.. it is possible to write tanh γ = γ 1 Z1 sinh = − 1 1+ 2 2 2Z2 = Z1 4Z2 an expression which will serve to predict filter performance. then VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 19 VEL . For an impedance it may be noted that the logarithm of a complex quantity B α = ln B + jα. Eq. may also be written in terms of hyperbolic functions of γ . and leads to sinh γ = Z0 Z1 / 2 + Z2 By use of Eq.VEL TECH filters. The input impedance of any T network. terminated in any impedance ZR. as Z Z Z ε = 1+ 1 + 1 + 1 2Z2 2Z2 Z2 Taking the natural logarithm γ 2 1+ Z1 + Z1 + Z1 γ = ln 2Z2 2Z2 Z2 2 For a network of pure reactances this is not difficult to compute. Writing Zin = Z11 − Z122 Z22 and substituting the required mesh relations from Fig. in Eq. and leads to Zin = Z02 + ZR Z0 / tanh γ Z0 / tanh γ + ZR Z cosh γ + Z0 sinh γ = Z0 R Z0 cosh γ + ZR sinh γ This is the input impedance of a symmetrical T network terminated in a load Z R. and the propagation constant γ . in this chapter. If Z1 and Z2 of a reactance network are unlike reactance arms. For a short-circuited network ZR = 0. The input impedance is then Z∝ where. from the above equation. the characteristic impedance Z0. In chapter 1. The last few equations are relations between the two sets of parameters. and the properties of the network have been developed in terms of these new parameters. then VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 20 VEL . in terms of the propagation constant and Z0 of the network. have been introduced. Explain the constant-k low-pass filter. and Z∝ is then limZ∞→∞ = ZR Z0 tanh γ From these two equations it can be seen that tanh γ = and Zsc Zoc Z0 = Z ∞ Z ∞ which has already been proved from the properties of the characteristic impedance. open-circuit and short-circuit measurements were used to describe the performance of a network. Z∞ = Z0 tanh γ For an open circuit ZR = ∝ in the limit. two new parameters.VEL TECH Zin = = VEL TECH MULTI TECH TECH HIGHTECH Z 22 Z1 + Z2 = 2 Z1 / 2 + Z2 + ZR Z1 / 2 + Z2 + ZR VEL Z12 / 4 + Z1Z2 + ( Z1 / 2 + Z2 ) ZR Use of Eqs. 4. A T section so designed would appear as at (a). As a special case. Fig. The curve representing -4Z2 may be drawn and compared with the curve for Z1. Networks or filter sections for which this relation holds are called constant-k filters. then the product Z1Z2 = L = R k2 C The term Rk is used since k must be real if Z1 and Z2 are of opposite type. Figure: (a) Low-pass filter section. The cutoff frequency fc may be readily determined. Thus the network is called a low-pass filter. Fig. since at that point Z1 = −4Z2 . then sinh γ / 2 may be evaluated as γ Z1 −ω2LC jω LC sinh = = = 2 4Z2 4 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 21 VEL . (b) reactance curves demonstrating that (a) is a low-pass section or has a pass band between Z1 = 0 and Z1 = -4Z2. The reactances of Z1 and 4Z2 will vary with frequency as sketched at (b). or attenuation. Thus the reactance curves show that a pass band starts at f = 0 and continues to some higher frequency fc. All frequencies above fc lie in a stop. jωcL = fc = 4j ωcC 1 LC This expression may be used to develop certain relations applicable to the lowpass network. let Z1 = jωL and Z2 = − j/ ωC. band. It has been shown by Eq. that a pass band starts at the frequency at which Z1 = 0 and runs to the frequency at which Z1 = 4Z2.VEL TECH Z1Z2 = k2 VEL TECH MULTI TECH TECH HIGHTECH VEL where k is a constant independent of frequency. reaching π at fc and remaining at π for all higher frequencies. this is γ f sinh = j 2 fc VEL TECH MULTI TECH TECH HIGHTECH VEL Then if the frequency f is in the pass band or f/fc < 1. The phase shift β is zero at zero frequency and increases gradually through the pass band. so that Z1/4Z2 < . α = 2cosh-1 . This method shows that the attenuation α is zero throughout the pass band but rises gradually from the cutoff frequency at f/f c = 1. α = 0. The variation of α and β is plotted in Fig. f fc β= π thereby allowing determination of α and β . The characteristic impedance of a T section was obtained as Z Z0T = Z1Z2 1+ 1 4Z 2 which becomes VEL TECH Z0T = L ω2LC 1− C 4 VEL TECH MULTI TECH TECH HIGHTECH 22 VEL . then f f > 1.1.VEL TECH and in view of Eq. as a function of f/fc. fc f β= 2sin-1 fc Figure: Variation of α and β with frequency for the low-pass section whereas if frequency f is in the attention band or f/fc > 1. then f < 1.0 to a value of ∝ at infinite frequency. so that -1 < Z1/4Z2 < 0. the characteristic impedance of a low-pass filter may be stated as L f = 1− C fc = Rk f 1− fc 2 2 Z0T in accordance with the definition of Rk in Eq. A low-pass filter may be designed from a knowledge of the cutoff frequency desired and the load resistance to be supplied. It is desirable that the Z0 in the pass band match the load. The design of a low-pass filter may be readily carried out. but because of the nature of the Z 0 curve in Fig. From Figure: Variation of Z0T / R k with frequency for the low-pass section. this result can occur at only one frequency.. Values of Z0T/Rk are plotted against f/fc in Fig. This match may be arranged to occur at any frequency which it is desired to favor by an impedance match. which will favor zero frequency for a low-pass filter. by use of Eq. then becomes imaginary in the attenuation band. For reasons which will appear in section. the load is chosen as R = R k = L C . rising to infinite reactance at infinite frequency. the relation that at cutoff Z1 = −4Z2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 23 VEL . It may be seen that cutoff.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL for the low-pass constant-k section under discussion. although cutoff may be sharpened by using a number of such networks in cascade.VEL TECH it is seen that ωeL = VEL TECH MULTI TECH TECH HIGHTECH 4 ω0C VEL Using the cutoff frequency equation changes this to π2fc2LC = 1 and use of the relation R = L C gives for the value of the shunt capacitance arm C= 1 πfcR By similar methods the inductance for Z1 is obtained as L= R πfc Since the design is based on an impedance match at zero frequency only. This is not usually an economic use of circuit elements. 5. power transfer only. and a remedy will be discussed in section. Explain the constant-k high-pass filter. and introduces excessive losses over other available methods of raising the attenuation near the cutoff frequency. power transfer to a matched load will drop at higher pass-band frequencies. then Z1Z2 will still be given by Z1Z2 = k2 and the filter design obtained will be of the constant-k type. A network such as is described here is called a prototype section. The reactances of Z1 and Z2 are sketched as functions of frequency in (b). and Z1 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 24 VEL . The T section will then appear as in (a). If the positions of inductance and capacitance are interchanged to make Z1 = − j/ ωC and Z2 = jωL . Fig. This condition may be undesirable in certain applications. It may be employed when a sharp cutoff is not required. and the curves of fig. is VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 25 VEL . Is compared with -4Z2. except that the phase angle β will be negative. will apply if the abscissa be considered as calibrated in terms of f c/f. so that the variation of γ inside and outside the pass band will be indentical with the values for the low-pass filter. The cutoff frequency is determined as the frequency at which Z1 = −4Z2 . All frequencies below fc lie in an attenuation. changing from 0 at infinite frequency or fc/f = 0. with a pass band from that frequency to infinity where Z 1 = 0. so that the value of the capacitance for Z1.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Figure: (a) High-pass filter section. The high-pass filter may be designed by again choosing a resistive load R equal to Rk such that L R = Rk = C From the relation that at cutoff Z1 = −4Z2 it was shown that 2 4ωcLC = 1 and again L/C = R2. (b) reactance curves demonstrating that (a) is a high-pass section or has a pass band between Z1 = 0 and Z1 = -4Z2. 4ωc LC = 1 ω0C 1 fc = 4π LC Using the above expression −j γ Z1 1 f sinh = = − 2 = = −j c 2 4Z2 4ω LC 2ω LC f The region in which fc / f < 1 is a pass band. showing a cutoff frequency at the point at which Z1 equals -4Z2. The network is thus a high-pass filter. or stop. the series element. or −j 2 = − j4ω0L. to -π at cutoff or fc/f = 1. band. However. so that frequencies just outside the pass band are not appreciably attenuated with respect to frequencies just inside the pass band. or the attenuation becomes infinite. the attenuation near cutoff may be made high. so that a satisfactory impedance match is not possible. the shunt arm. Consider first the circuit of (a). This frequency of infinite or high attenuation is called f∝. it can be seen that the shunt arm is a series circuit resonant at a frequency above fc. It is more economical to attempt to raise the attenuation near cutoff by other means. is R L= 4πfc The characteristic impedance for the high-pass filter may be transformed to Z0T = R k f 1− c f 2 6. though simple. The attenuation does not rise very rapidly at cutoff. Figure: (a) Derivation of a low-pass section having a sharp cutoff action. Also. By similar methods the value for the inductance for Z2.VEL TECH C= 1 4πfcR VEL TECH MULTI TECH TECH HIGHTECH VEL It should be noted that since Z1/2 is the value of each series arm. the capacity use din each series Z1/2 element should be 2C. Explain the m-derived T section filter. then. the characteristic impedance varies widely over the pass band. The constant-k prototype filter section. will always be higher in value than fc. the attenuation may be built up near cutoff by cascading or connecting a number of constant-k sections in series. At this resonant frequency the shunt arm appears as a short circuit on the network. The reactance curves sketched at (b) show that this circuit is alow pass filter. (b) reactance curves for (a). If. has two major disadvantages. f∝ can be chosen arbitrarily close to fc. Fig. In cases where an impedance match is not important. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 26 VEL . the characteristic impedance and fc remain equal to those of the T section prototype containing Z1 and Z2 values. Thus m must always be chosen so that <1 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 27 VEL . Z 0 ' = Z0 ( mZ1 ) 4 Z'2 = 2 + mZ1Z2 ' = 2 Z1 + Z1Z2 4 Z 2 1− m2 + Z1 m 4m It then appears that the shunt arm Z2 ' consists of two impedances in series. it is necessary to use a section such as in Fig. Figure: The m-derived low-pass filter. it is necessary that the Z0 of all be identical at all points in the pass band. and it is desired that this relation continue in the two series impedances given by Eq. As required. it is possible to design an infinite variety of filter networks meeting the required conditions on Z0 and fc. for high attenuation near cutoff. For satisfactory matching of several such types of filters in series. They will consequently also all have the same pass band. The network of fig. as shown in Fig. in series with a prototype section to provide high attenuation at frequencies well removed from cutoff. may be derived by assuming that Z'1 = mZ1 the primes indicating the derived section. Equation then indicates that (1-m2)/4m must be positive. It is then necessary to find the value for Z 2’ ' such that Z0 = Z0. Since m is arbitrary. forcing the terms 1 – m2 and m always to be positive. Z2 will be opposite in sign to Z1. Setting the characteristic impedances equal.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The attenuation above f∝ will fall to low values. However. for the Z2 ' arm. so that if high attenuation is desired over the whole attenuation band. Similar relations for the high-pass filter can be derived as f∞ = fc 1− m2 and m = 1− ( f∞ / fc ) 2 The m-derived section is designed following the design of the prototype T section. an m-derived section may be used with f∝ near fc.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Filter sections obtained in this manner are called m-derived sections. followed by as many m-derived sections as desired to place frequencies of high attenuation where needed to suppress various signal components or to produce a high attenuation over the entire attenuation band. This means that at the resonant frequency and for the low-pass filter 1 1− m2 = 2πf∞ L 2πf∞ mC 4m 1 f∞ = π ( 1− m2 ) LC Since the cutoff frequency for the low-pass filter is 1 fc = π LC the frequency of infinite attenuation will be f∞ = from which fc 1− m2 2 m = 1− ( fc − f∞ ) This equation determines the m to be used for a particular f ∝. The shunt arm is to be chosen so that it is resonant at some frequency f ∝ above fc. If a sharp cutoff is desired. The use of a prototype and one or more m-derived sections in series results in a composite filter. then VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 28 VEL . The variation of attenuation over the attenuation band for a low-pass m-derived section in the stop band is dependent on the sign of the reactances or α = 2cosh−1 fc < f < f∞ Z1 Z1 or α = 2sinh-1 4Z2 4Z2 f∞ < f For Z1 = jωL and Z2 = − j/ ωC for the prototype. and the composite result of the two in series. and for f∝ < f mf / f α = 2sinh−1 2 2c f / f∞ − 1 The value of α may be determined from the expression.6.25 times the cutoff frequency fc. from β = 2sin−1 = 2sin−1 Z1 4Z2 1− ( f 2 / fc2 ) ( 1− m2 ) VEL TECH MULTI TECH TECH HIGHTECH 29 VEL mf / fc VEL TECH . in the pass band. The higher attenuation over the whole attenuation band obtained by use of a prototype section and an m-derived section in series as a composite filter is also readily seen.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH Z1 mωL = 4Z2 41/ mωC − ωL ( 1− m2 ) / 4m VEL so that for fc < f < f∞ α = 2cosh−1 mf / fc 1− f 2 / f∞ 2 Figure: Variation of attenuation for the prototype and m-derived sections. Figure is a plot of a against f/fc for m=0. the phase shift constant β may be determined. which gives a value of f ∝ equal to 1. The great increase in sharpness of cutoff for the m-derived section over the prototype is apparent. Again following the procedure of section. By use of the transformation for the shunt arm. The characteristic impedance of the π section is Z1Z2 Z0π = Z1Z2 ( 1+ Z1 / 4Z2 ) The characteristic impedances of the prototype and m-derived sections are to be equal so that they may be joined without mismatch. Although it may be note that the sharpness of cutoff increases for small values of m. the attenuation beyond the point of peak attenuation becomes smaller for small m. The phase shift of the m-derived section is plotted as a function of f/fc in fig. because the shunt arm becomes inductive above resonance. This emphasizes the necessity of supplementing the m-derived section with a prototype section in series to raise the attenuation for frequencies well removed from cutoff. up to f∝. Figure: Variation of phase shift β . β has the value π . An m-derived π section may also be obtained. 7. for the m-derived filter This material demonstrates the ability of the m-derived section to overcome the lack of a sharp cutoff in the simple prototype filter. Z2 ' = Z2 m VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 30 VEL . Explain the derived π section filter. Above f∝ the value of β drop to zero.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL In the attenuation band. 2 one being mZ1. it is more economical to combine the low-and high-pass functions into a single filter section. (b) the m-derived π filter. 8. Explain the Band-pass filters.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Figure: (a) Usual symmetrical π section. the other being 4m/ ( 1− m ) Z2 in value. The circuit is drawn in fig. Although such a design would function. the overlap thus allowing only a band of frequencies to pass. in which the cutoff frequency of the low-pass filter is above the cutoff frequency of the high-pass filter. The action might be thought of as that of low-pass and high-pass filters in series. it is possible to equal the characteristic impedances as Z1 'Z2 / m Z1Z2 = Z1 ( Z2 / m) ( 1+ Z1 'm/ 4Z2 ) Z1Z2 ( 1+ Z1 / 4Z2 ) from which Z1 ' = 1 1 1 + 4m mZ1 Z2 1− m2 It is apparent that the series arm Z1 ' is represented by two impedances in parallel. (b) reactance curves showing VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 31 VEL . Occasionally it is desirable to pass a band of frequencies and to attenuate frequencies on both sides of the pass band. Equations and thus give the values to be used in designing the m-derived π section. Figure: (a) Band-pass filter network. If. then Z1Z2 = L 2 L1 = = R k2 C1 C 2 Thus the previously developed theory still applies VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 32 VEL . and only one pass band appears. the antiresonant frequency of the shunt arm is made to correspond to the resonant frequency of the series arm. Fig. VEL Consider the circuit of (a). the reactance curves become as shown in fig. ω02L 1C1 = 1 = ω02L 2C2 L 1C1 = L 2C2 or The impedances of the arms are 2 1 ( ω L 1C1 − 1) Z1 = j ωL 1 − =j ωC1 ωC1 jωL 2 ( − j/ ωC2 ) jωL 2 Z1 = = j( ωL 2 − 1/ ωC2 ) 1− ω2L 2C2 That a network such as (a). Fig. For this condition of equal resonant frequencies. with a series-resonant series arm and an antiresonant shunt arm. the reactance curves show that two pass bands might exist. however. is still a constant-k filter is easily shown as Z1Z2 = − C1 ( 1− ω2L 2C2 ) L 2 ( ω2L 2C1 − 1) and if L 1C1 = L 2C2 . Figure: Reactance curves for the band-pass network when resonant and antiresonant frequencies are properly adjusted. In general.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH possibility of two bands. as is customary. At the lower cutoff frequency. may be written as f12 f1 f22 1− 2 = 2 − 1 f0 f2 f0 2 f0 ( f1 + f2 ) = f1f2 ( f1 + f2 ) f = f1f2 or the frequency of resonance of the individual arms should be the geometric mean of the two frequencies of cutoff.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL At the cutoff frequencies. then the values of the circuit components can be determined in terms of R and the cutoff frequencies f1 and f2. Z1 = −4Z2 Multiplying by Z1 gives Z12 = −4Z1Z2 = −4Rk 2 from which the value of Z1 at the cutoff frequencies is obtained as Z1 = ± j2R k so that Z1 at lower cutoff f1 = . VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 33 VEL .Z1 at upper cutoff f2 The reactance of the series arm at the cutoff frequencies then can be written by use of the above as 1 1 − ω1L 1 = ω2L 1 − ω1C1 ω2C1 ω 1− ω12L 1C1 = 1 ( ω22L1C1 − 1) ω2 Now from eq. If the filter is terminated in a load R = Rk. 1 L 1C1 = 2 ω0 so that eq. The antiresonant frequency of the arm as a whole must. Plotting reactance curves for these two circuits and adding to obtain the reactance variation of the shunt arm. then. the expression for C1 becomes C1 = f2 − f1 4πRf1f2 It follows. gives the dashed curve of fig. Figure: m-derived band-pass section An m-derived band-pass section is also possible. The shunt arm then consists of series-resonant and antiresonant circuits in series. Z2 of the filter.. be f0 of the filter. The reactance curve for Z2 then shows that the shunt arm becomes resonant at a frequency below f 0 and VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 34 VEL .VEL TECH 1 − ω1L 1 = 2R ω1C1 1− f12 = 4πRf1C1 f02 VEL TECH MULTI TECH TECH HIGHTECH VEL In view of Eq. from eqs. by previous reasoning. and that L1 = R π ( f2 − f1 ) From equation it is possible to obtain the values for the shunt arm as R ( f2 − f1 ) 4πf1f2 L 1 C2 = 1 = 2 R πR ( f2 − f1 ) L 2 = C1R 2 = This completes the design of the prototype band-pass filter. Use of the transformation relation developed in section leads to a network of the form of fig. At one f∝. f∝. as for the high-or low-pass cases. VEL TECH MULTI TECH TECH HIGHTECH VEL Figure: Reactance curves for the shunt arm of the m-derived band-pass section. and the m-derived section may be used to increase the attenuation near cutoff.VEL TECH again at a frequency above f0. the reactances Xr and Xar are equal and opposite. At these frequencies the network is short-circuited. These frequencies of high attenuation are placed on each side of the pass band. so that j ( jω∞ L 2 / m) ( − j/ ω∞ mC2 ) 1− m2 jω∞ = L1 − 2 ω∞ 4m/ ( 1− m ) C1 j( 1− / ω∞ mC 2 − ω∞ L 2 / m) 4m ω L 2C1 1− m2 ( ω∞ 2L 1C1 − 1) = ω 2∞2 C − 1 4 ∞ L2 2 In view of the fact that L 1C1 = L 2C2 = Equation becomes 1− m2 f∞ 2 2 2 2 − 1 = 4π f∞ L 2C1 4 f0 The term L2C1 can be evaluated as a function of f1 and f2 from Equations and f − f ( f − f ) R ( f − f ) L 2C1 = 2 1 2 1 = 2 21 4 4πRf1f2 4πf1f2 16π f0 Equation then reduces to VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 35 VEL 2 2 1 L ω02 . and thus they are frequencies of high attenuation. f∞ = f2 − f1 2 1− m2 ± 4( 1− m2 ) ( f2 − f1 ) 2 + f1f2 It is apparent that the radical is larger than ( f2 − f1 ) 2 1− m2 . the other frequency of peak attenuation then being definitely fixed. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 36 VEL . Thus the expression for f∝ should be reversed so that the two frequencies of peak attenuation are f∞1 = f∞ 2 = 4( 1− m ( f2 − f1 ) 2 2 ) ) + f1f2 − + f1f2 + f2 − f1 2 1− m2 f2 − f1 2 1− m2 Equation may be solved to determine the value of m.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL ( 1− m ) ( f 2 ∞ 2 f∞ 2 − ( f2 − f1 ) − f1f2 ) = f∞ 2 ( f2 − f1 ) 2 2 1− m2 − f1f2 = 0 Solving for the values of the frequencies of peak attenuation. by Equation. giving f ( f − f ) m = 1− ∞ 2 2 1 f∞ − f1f2 = f∞ 2 − f1f2 2 4( 1− m ( f2 − f1 ) 2 2 1− ( f∞ 2 − f12 ) ( f∞ 2 − f22 ) The value of m may be chosen to place either one of the two frequencies of peak attenuation at a desired point. then by Equation. as shown by Zobel (Reference 2). and thus one root would appear as a negative frequency that has no physical significance here. That this is true may be chosen by forming the product for f∞1f∞ 2 from Equations and: f∞1f∞ 2 = f22 + 2f1 + f12 − 4m2f1f2 − f22 + 2f1f2 − f12 = f1 f2 4( 1− m2 ) f∞1f∞ 2 = f0 Thus f0 is the geometric mean of the frequencies of peak attenuation and. f∞ 2 and f∞ 2 by the use of two m’s or an mm’-derived filter. If m is selected to place f 01 at a desired point. of the cutoff frequencies as well. As for the band-pass filter. This usage follows the previously developed theory for low-or high-pass sections. That this circuit does eliminate or attenuate a given frequency band is shown by the reactance curves for Z1 and -4Z2 at (b). in which the cut-off frequency of the low-pass filter is below that of the high-pass filter. may be split into two half sections and used as terminating half sections. fig. Figure: (a) Band-elimination filter.6. If m is given the value 0. 1 jR Z2 = j − ω1L 2 = ω1C 2 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 37 VEL . then at the lower cutoff frequency. If the series and parallel-tuned arms of the band-pass filter are interchanged. the series and shunt arms are made antiresonant and resonant at the same frequency f0. it is possible to show that R k2 = L1 L2 = C 2 C1 f0 = f1f2 and that At the cutoff frequencies. Explain the band-elimination filters. Z1 = −4Z2 . then satisfactory impedance matching conditions are maintained over the pass band. Z1Z2 = 4Z2 2 = Rk 2 jR Z2 = ± k 2 If the filter is terminated in a load R = Rk. rearranged as a π . Again. The action may be thought of as that of a low-pass filter in parallel with a highpass section. (b) reactance showing action of bandelimination section.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL An m-derived T section. the result is the band-elimination filter of (a). 9. Explain the crystal filters. Figure: Lattice filter section VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 38 VEL . the values for the series arm are obtained as R ( f2 − f1 ) πf1f2 1 C1 = 4πR ( f2 − f1 ) L1 = Section of the m-derived form may also be obtained.VEL TECH 1 Since L 2C 2 = ω 2 . 0 1− f 12 f02 = πf1C 2R VEL TECH MULTI TECH TECH HIGHTECH VEL C2 = 1 f2 − f1 πR f1f2 In view of the fact that f0 = f1f2 = 1 2π L 2C 2 then L2 = R 4π ( f2 − f1 ) By use of Equation. 10. The lattice structure can also be shown to have filter properties. Considering the network of fig. being in henrys for crystals resonating near 500 kc. These crystals have a resonant frequency of mechanical vibration dependent on certain of their dimensions. so that while Rx may approximate a few hundred or few thousand ohms. Z0L is real.000.000 to 30. it is possible to make very narrow band filters and filters in which the attenuation rises very rapidly at cutoff. and because of the very high equivalent Q of the crystals. an attenuation band exists. Considering the properties of resonant circuits.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 2 VEL ( Z / 2 + 2Z2 ) Z∞ = 1 2( Z1 / 2 + 2Z2 ) Z1 + Z2 4 Z1Z 2 Z1Z2 Z∞ = + Z1 / 2 + 2Z2 Z1 / 2 + 2Z2 Z1Z2 = Z1 / 4 + Z2 = The characteristic impedance of the fig. especially for filters in which certain of the elements are constructed of piezoelectric crystals. whereas γ depends on the ratio of Z1 to Z2. or a pass band exists for frequencies for which Z1 and Z2 are of opposite sign. the effective Q may be in the range of 10. This feature permits somewhat greater versatility in design of the lattice section over the T or π section. such as Q would provide a band width of 20 to 50 cycles at 500 kc. (a). VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 39 VEL . Lattice filter section. Propagation can be investigated further by noting that tanh γ = Zsc Z1 1 = Zoc Z2 1+ Z1 / 4Z2 It may be noted that Z0L depends on the product of Z1 and Z2. The inductance Lx is very large. Lattice section then is Z0L = Z∞ Zsc = Z1Z2 Thus if the section elements are reactive. Over ranges where Z1 and Z2 have the same sign. which shows a possibility of both resonance and antiresonance occurring. The equivalent electric circuit of a quartz mechanical-filter crystal is shown in fig. By placing a crystal in an evacuated container. The values of Cs. Cp can be increased. Figure: (a) Circuit of a lattice crystal filter with series inductors and parallel capacitors. The resistance of the crystal is due largely to mechanical damping introduced by the electrodes and by the surrounding atmosphere. (b) the electrical equivalent of (a). differing by a fraction of 1 per cent of the resonant frequency. (b) shows the resonant frequency below the antiresonant one. By placing adjustable capacitors in parallel with the crystal. The reactance curve sketch of fig. The pass band with crystal elements will then be found to extend from the VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 40 VEL . the resonant frequency of one arm must equal the antiresonant frequency of the other arm. it may be used to replace the normal elements of the band-pass or band-elimination filter.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Figure: (a) Equivalent electrical circuit for a piezoelectric crystal. (b) reactance curves for the circuit of (a). Capacitance Cs is the equivalent series capacitance of the crystal forming a resonant by the crystal electrodes. as previously shown for band-pass action. Since the crystal represents either a resonant or antiresonant circuit. the value of Q can be notably increased. resulting in the antiresonant frequency being moved closer to the resonant point. The electrodes are normally electroplated onto the crystal faces and need not introduce much damping. so that resonant and anti-resonant frequencies of the circuit lie very close together. and Cp are such that Cp >> Cs. Since C s / C p may be of the order of 0. with its equivalent drawn at (b). which shows how the resonances and antiresonances are arranged. (b) attenuation curves for that circuit. Thus the use of coils permits the bands to be widened to pass speech frequencies.2 = fR 1m s Cp The separation of f1 and f2 represents two-thirds of the pass band and is seen to depend on the C s / C p ratio.10 fR. it can be seen that the separation of f1 and f2 be of the order of 0. (a). The presence of the series coil adds an additional resonance. By placing coils in series with the crystals. By the addition of coils in series with the crystals the pass bands may be widened. The band width can be reduced by putting adjustable capacitors in parallel with the crystal. By adjustment of Cp it is then possible to narrow the band to any desired amount. or a width of pass band equal to twice the separation of the resonant and antiresonant frequencies of one crystal. there will be some loss in sharpness at cutoff. Since the added coils have Q values very much below those of the crystals. then C f1. (a). furnishing a means of adjustment of the width of the pass band. A circuit including series coils is shown in fig. The reactance curves for the A and B portions of this circuit are drawn in fig. and the pass band exists from the lowest resonance of one crystal to the highest resonance of the other. If f1 and f2 are the frequencies of resonance of one of the circuits and fR is that of the antiresonance. and crystal filters are quite generally used to separate the various VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 41 . (a). This range will result in a pass band a fraction of 1 per cent wide.01. Figure: (a) Reactacne curves for the circuit of fig.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL lowest crystal resonant frequency to the highest crystal antiresonant frequency. or 10 per cent of the resonant frequency. it has been possible to widen the pass band considerably. f∞ = 1065 Hz. fc = 1000 Hz. 11. The fitler is to be terminated in 500 ohms resistance and it is to have a cutoff frequency of 1000 Hz with very high attenuation at 1065 and 1250 Hz. in the range above 50 kilocycles. Solution: Given R= 500 .63µ f ∏ fc.R ∏×1000 × 500 I= The assembly of this filter is a shown below with inductance L/2 in each series arm.344 are VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 42 VEL .section (Ref fig (b) a) for fω 2 = 1065 fc 1000 m1 = 1 − = 1 − = m1 = 0. Design a composite low pass filter to meet the following specifications.derived lowpass T – section filter for m = 0. Design of m – derived low pass T.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL channels in carrier telephone circuits.344 1065 f ∞t 2 2 The components of m. i) Design of low pass constant – kT section R 500 = = 0. ii. f∞2 = 1250 Hz.159 H ∏ fc ∏×1000 1 1 C= = = 0. 344 ) 2 VEL TECH MULTI TECH TECH HIGHTECH VEL = 0.3816 µ f 1 − m 2 1 − ( 0.0424 H The assembly of the m – derived low pass T – section filter for m = 0.159 4m 4 ( 0.59 = = 0.159 = = 0.344 × 1. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 43 VEL .048 H 2 2 mc = 0.6 as shown below.6 × 0.159 4m 4 ( 0.636 = 0.VEL TECH mL 0.636 = 0.6 ) 2 2 2 = 0.102 H The assembly of this m – derived lowpass t – section filter is as shown below fc 1000 m2 = 1 − = − 1250 f∞ 2 The components of m – derived low pass T – section filter for m = 0.6 × 0.0273H 2 2 mc = 0.6 are mL 0.344 ) 1 − m2 .218µ f − ( 0.6 ) ⊥ × 0.344 × 0.L = × 0. Assembling all the three sections we will get the desired composite filter. L = ( 0.e. for m = 0. the filter is divided into sections so these value get changed. 1 − m2 2× . Describe a prototype T section band stop filter. The series inductors may be added and the resulting final design I as shown below. i.6.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL In the 3rd section i. iii. Band stop filter: A band stop or bad elimination filter attenuates a certain range of frequencies and passes all other frequencies there fore a band stop action may be thought of as that of a low pass filter in parallel with a high pass filter in which the cut off frequency of low pass filter is below that of high pass filter. ii. ii. Explain the advantages of m – derived band stop fitler.109µ f 2 2 Combine the elements with ever possible.218µ f and = = 0. 12.. i.0848 4m mc 0.0424 ) × 2 = 0. Determine the formulae required for designing band stop filter. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 44 VEL . Z1 = −4 Z 2 L1Z 2 Z12 = −4 Z1 Z2 = −4 RK 2 Z1 at lower cut off fi = Z1 at upper cut off f −1 1 + jω L1 = −1 2 + jω2 L1 jω1C1 jω C1 1 −1 − ω1 L1 = + ω2 L1 ω1C1 ω2 C1 = j ( ω L1 − 1/ ωC1 ) = j ( ω 2 L1C1 − 1/ ωC1 ) VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 45 VEL .VEL TECH T-type 1 ω= = L1C1 L1C1 = L2 C2 Z1 = jω L1 + 1 jωC VEL TECH MULTI TECH TECH HIGHTECH VEL 1 L1C1 jω L2 ( 1/ jωC2 ) jω L2 = jω L2 + 1/ jωC2 1 − ω 2 L2 C2 If the filter is to be constant K type Z1Z2 = RK2 = ω 2 L1C1 − 1 jω L2 j= = RK 2 2 ωC1 1 − ω L2 C2 L1C1 = L2 C2 L2 L1 = − RK 2 C1 C2 Z1 = 1. VEL TECH 1 − ω 2 L1C1 = L1C1 = 1 ω02 VEL TECH MULTI TECH TECH HIGHTECH VEL ω1 2 ( ω2 L1C1 − 1) ω2 2 ω12 ω1 ω2 1− 2 = 2 − 1 ω0 ω2 ω0 f2 f f2 1 − 12 = 1 22 − 1 f0 f 0 f0 f 02 ( f1 + f2 ) + f1 f2 ( f1 + f2 ) fr = f2 f2 Z1 = −2 JRk 1 + jω1 L1 = −2 jRk jω1C1 −1 j + ω1 L1 = 2 jRk ω1C1 2 1 − ω1 C1 L1 = 2 Rk ωC1 1− ω12 = 2 Rk 2 ∏ f1C1 ω02 f 2 f1 = 4 ∏ Rk f1C fr f 2 − f1 4 ∏ Rk f1 f 2 1 1 = 2 2 ω0 4 ∏ f1 f2 C1 = L1C1 = L1 = Rk ∏ ( f 2 f1 ) Rk ( f 2 − f1 ) 4 ∏ f2 f2 L2 = C1 Rk2 = C2 = L1 1 = 2 Rk π Rk ( f 2 − f1 ) iii) Advantage of m – derived filters VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 46 VEL . Solution: Given fc = 3000 Hz. Solution: Given : L= 80 mH . determine the cut-ff frequency and normal characteristic impedance Re.section lowpass filter are 63. For a given T – section low pass filter.02 µf 1 ∏ LC Cut off frequency fc = fc = 1 ∏ ( 80 ×10−3 ) ( 0.02 ×106 ) Hz f c = 7. c = 0. from the desired value the design impedance Rk. Characteristic impedance varies widely in the transmission or pass band. Ro = 600 L= R0 600 × 1 = ∏ f c ∏×3000 × 5 L = 63.68 mH VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 47 VEL .1753µ f Hence the require T and II. 13.02 × 10−6 R0 = 2 ×103 ohms 14.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 1.68mH 1 1 C= = ∏ R0 f c ∏×600 × 3000 C = 0. Design a constant – k low pass T and II – sections fitlers having cut-off frequency = 3000 Hz and nominal characteristic impedance R0= 600 ohm. The attenuation does not increase rapidly beyond the cut off frequencies. 2.962 × 103 Hz L 80 ×10−3 R0 = = C 0. 2 nepers 8.77 mH VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 48 VEL .2 f = cosh : = 1. Solution: Given: fc = 2000 Hz: R0 = 600 ohm .1 dB. R 600 L= 0 = ∏ f c 600 × ∏×2000 L = 95. Find i) its characteristic impedance and phase constants at 25 KHz and ii) Attenuation 5KHz.686 α = 2 cosh -1 ( f / fc ) f 2. α = attenuation = 19.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 15.1 dB.1 α= = 2. Design a T.263µ F Attenuation of 19.6685 × 2000 f = 3337 Hz 16.54mH 1 1 C= = F ∏ R0 f c ∏×600 × 2000 C = 0.1 dB is expressed in nepers as 19. Design a constant – k low pass filter having Fc= 2000 Hz and nominal characteristic impedance R0=600Ω. R0 = 600 ohm R0 600 L= = 4 ∏ f c 4 ∏×10 ×103 L = 4.section constant – K high pass filter having cut off frequency of 10KHz and nominal characteristic resistance of Ro=600 om. Also find the frequency at which this filter offers attenuation of 19. Solution: Given Fc = 10 KHz .6685 fc 2 fc ∴ F = 1. α = 2 cosh -1 ( f c / f ) nepers = 2 cosh-1 ( 10 / 5 ) nepers α = 2.0381µ F ICI = 0.R0 .2 or β =47.0824 radians In the attenuation band. At f= 25 KHz Z OT Z OT fc = R0 1 − f 2 10 = 600 1 − 25 2 Z OT = 545Ω β = 2 sin -1 ( f c / f ) = 2sin − ( 10 / 25 ) β = 47.0762 µ F VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 49 VEL .VEL TECH C= 1 1 = F 4 ∏ . Determine a prototype band pass fitler section having cut off frequencies of 2000 Hz and 500 Hz and nominal characteristic impedance of 600 ohms. α is given as.68mH L1 = 31.84mH 2 ( 5000 − 2000 ) f 2 − f1 C1 = F= F 4 ∏ Ra .6 nepers 17. f c 4 ∏×10 ×103 × 600 VEL TECH MULTI TECH TECH HIGHTECH VEL C = 0. f 2 4 ∏×600 × 5000 × 2000 C = 0.52 × ∏ /180=0.02652 µF and the inductor in shunt arm is L=4.0132663µ F Each capacitor in the series arm T – section is 2C= 0. Solution: R0 600 L1 = = ∏ ( f 2 − f1 ) ∏ ( 5000 − 2000 ) L1 = 63. f1 .777 mH. What is transmission line? Energy can be transmitted either by the radiation of free electromagnetic waves as in the radio or it can be constrained to move or carried in various conductor arrangement known as transmission line.VEL TECH R0 ( f 2 − f1 ) 600 × ( 5000 − 2000 ) = H 4 ∏ . It is a conductive method of guiding electrical energy from one place to another. No waves will ever reach receiving end hence there is no reflection. The parameters which are not physical be separable and are distributed all over the length of the circuit like transmission line are called distributed parameters.1769 µ F Hence the band pass filter is as shown below. 3. f c 4 ∏×5000 × 2000 VEL TECH MULTI TECH TECH HIGHTECH VEL L2 = L2 = 14. What are lumped parameters and distributed parameters? The parameters which are physically separable and can be shown to be at one place in the circuit in the lumped form are lumped parameters. State the properties of infinite line. The Zo of the sending end decides the current flowing when voltage is applied VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 50 VEL . 2. 2.R0 . UNIT – II PART – A 1.33mH C2 = 1 1 = ∏ R0 ( f 2 − f1 ) ∏×600 × ( 5000 − 6000 ) C2 = 0. 1. Define characteristic impedance. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 51 VEL . Discuss the importance of smooth line.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL ZR has no effect on the sending current. Hence no reflection takes place. having electrical constants per unit length identical to that of the line under consideration. Then the maximum power transfer from generator to load is possible. 7. 8. 9. 4. there is no mismatch of impedance. R L Zo → and w → ∞Z0 → G C R Practically is alwas higher then G when Zo = L C 6. A line terminated in its characteristic impedance Ro is called properly terminated line which acts as a smooth line. Draw the equivalent circuit of a unit length of transmission line. Sketch the group of Zo agains w. R+jwL G+jwC w → 0. What is called an infinite line? The analysis of the transmission of the electric waves along any uniform and symmetrical transmission line can be done in-terms of the result existing for an imaginary line of infinite length. 5. Because of proper termination. What is short line? The short line means a practical line with finite length and the word short does not reflect anything about the actual length of the line. Thus no standing waves are produced. Returns loss = 20 log Z R + Zo Z R − Zo db 11. What is Campell’s equation? It makes possible the calculation of the effects of loading coils in reducing attenuation and distortion. 13. Define reflection factor? Reflection factor is defined as the ratio which indicates the change in current in the load due to reflection at the mismatched junction. 2 Z R Zo K= Z R + Zo 12. Reflection factor = 2 Z1Z 2 Z1 + Z 2 where Z1 and Z2 are the impedance seen looking both ways at any junction. Express reflection factor in terms of impedance. 10. Define return loss.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL When a finite transmission line is terminated with ZO and the input impedance is also ZO.|K| = 1 Short circuited line . Zc CoshN λ 1 = sinh N λ + CoshN λ 2 Zo when ZC → loading coil impedance Zo → Characteristic impedance N → Length of Transmission line section λ→ Propagation constant 1 λ → Modified propagation constant. Find out the value of Reflection coefficient (K) for the following. i.|K| = 0 Open circuited line -|K| = 0 14. Return loss is defined as the ratio of power at the receiving end due to incident wave to power due to reflected wave by the load. i) ii) iii) Properly matched condition .e the input impedance of an infinite line is characteristic impedance of line. then ZO is known as characteristic impedance. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 52 VEL . RC = LG 17. L = The commonly used transmission lines are: i) Open wire line. 18. iii.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 15. State the condition for minimum attenuation with L and C variable. As a result these lines can operate at higher frequencies. telegraphy and telephony. 16. The conductor of open wire line is as shown below. Give the types of transmission line With ‘L’ variable . What are the basic application of transmission lines? i. Write short notes on transmission line. variable . It is used to transmit energy It can be used a circuit elements like L>C. 19. This type of line is used in transmission of electric power. Give the advantages of open wire transmission line. ii) Co-axial line iii) Strip line iv) Waveguides v) Optical fibers. C = R In general. measuring devices. CR G LG With ‘C. The spacing between the conductors is large in comparison to the diameter of the line conductors. It can be used as filters. and resonant circuits. The open wire transmission line consists of two conductors spaced at a certain distance apart. ii. transformers. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 53 VEL . 25. 24. Give the application of microwaves (i) Satellite communication (ii) Telemetry (iii) Transmission of video signals (iv) Microwave oven. Transmission by continuous reflection from the inner walls of the guide. H. (ii) It provides outer shielding from outer interfacing signals. Because energy loss takes place due to radiation.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The advantages of open wire line are: (i) (ii) (iii) Simple to construct and low cost. 21. F. To avoid radiation losses taking place in open wire lines at high frequencies a closed field configuration is used by surrounding the inner conductor with an outer cylindrical hollow conductor and the arrangement is termed as a coaxial cable. Give the advantages and disadvantages of co-axial transmission line. Advantages: (i) The electric and magnetic fields are confined with in the outer conductor so the radiation losses are eliminated. Beyond 1GHz these cables cannot be used because losses in the dielectric increases with frequency. the dielectric loss is extremely small. Disadvantages: (i) (ii) They are costlier than open wire line. What are the advantages and disadvantages of Waveguide? The advantages of waveguides are: VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 54 VEL . It is balanced with respect to ground 20. What do you mean by Waveguide? Wave guides are hollow conducting tubes of uniform cross section used for U. (iii) The co-axial line can be used up to the frequency range of about 1 GHz for transmission of signals. Insulation between the line and conductor is air so. What are the disadvantages of open wire transmission line? These lines are unsuitable for use at frequencies above 100MHz. (i) 23. 22. Write short notes on co-axial line (or) Coaxial cable. no power is lost through radiation because the electric and magnetic fields are confined to the space within the guides. The propagation constant per unit length of a uniform line is defined as the natural VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 55 VEL . Define propagation constant of uniform line. 28.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL i) In wave guides. It confines electromagnetic energy in the form of light within its surfaces and guides the light in a direction parallel to its axis. 26. ii) The dielectric loss is negligible. What are the advantages and disadvantages of optical fiber? Advantages: (i) (ii) (iii) (iv) Low transmission loss and very high band width Small size and weight No radio frequency and electromagnetic interference Ruggedness and flexibility Disadvantages: (i) (ii) It is difficult to run cables where the bending occurs. The Disadvantages of wave guides are: (i) (ii) Cost of the wave guide is so high The wave guide walls should be specially plated to reduce resistance to avoid skin effect and power loss. What is an optical fiber? An optical fiber is a dielectric wave guide that operates at optical frequencies. Different specialized techniques have to be followed to join ends of two cables. 27. iii) Frequencies of the wave higher than 3GHz can be easily transmitted iv) Several modes of electromagnetic waves can be propagated with in a single Waveguide. What are the four important parameters of a transmission line? The main four parameters of transmission line are Resistance Inductance Capacitance Conductance R L C G 29. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL logarithm of steady state vector ratio of current or voltage at any point. when the primary constants R. 35. Name the secondary constants of transmission line. L. 31. when the line is infinitely long 30. D ohms d 36. Characteristic impedance Zo and Propagation constant P (or) Secondary constants of transmission line γ are the 32. What is the value of characteristic impedance of coaxial cable? The characteristic impedance of co-axial cable Is Zo = 138log10 Where ‘D’ is inner diameter of outer conductor ‘d’ is diameter of inner conductor 34. C and G are uniformly distributed along the entire length of transmission. (or) It is a velocity with which a signal produced by variation of a steady-state wave or by introduction of group frequencies. Define velocity of propagation. What is loading of a transmission line? VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 56 VEL . It is denoted as Vg. 33. When a transmission line is said to be uniform? A line is said to be uniform. It is denoted as Vp and its unit is km/sec. to that at a point unit distance further from source. Define group velocity. Velocity of propagation is defined as the velocity with which a signal of single frequency propagates along the line at a particular frequency ‘f’ . Group velocity is defined as the velocity of envelope of a complex signal. What is the value of characteristic impedance of open-wire line? The characteristic impedance of open-wire line is S Zo = 276 log 10 ohms r Where ’S ‘ is the spacing between two wires-centre to centre ‘r’ is radius of either of the wire. the received wave will not be identical with the input wave form at the sending end. some frequencies being delayed more than others. This variation is called frequency distortion. since some components will be delays more than the other. What are the disadvantages of reflection? 1. Reduction in efficiency. 41. hence output reduces. 3. If the attenuation is not large. 40. Define Reflection co-efficient. 37. then the reflected wave appears as echo at the sending end. 2. What is phase or Delay Distortion? All frequencies applied to a transmission line will not have the same time of transmission. These are the disadvantages of reflection. What is frequency distortion? A complex applied voltage. such as a voice voltage containing many frequency will not have all frequencies transmitted with equal attenuation of the received wave form will not be identical with the input wave form at the sending end. The ratio of amplitudes of the reflected of incident voltage waves at the receiving end of the line is called reflection co-efficient. What are the advantages of lumped loading? The advantages of lumped loading are • There is no practical limit to the value by which the inductance can be increased. This phenomenon is called Delay distortion. • Cost is small • Hysterisis and eddy current losses are small. For an applied voicevoltage wave. The part of received energy is rejected by the load. 39.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The process of achieving the condition RC = LG either by artificially increasing ‘L’ or decreasing ‘C’ is called loading of a line. 38. K= Reflected voltage at load Incident voltage at load 42. What do you mean by Insertion loss? VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 57 VEL . VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL It is defined as number of repers or decibels by which the current in the load is changed by the insertion. It is seen that the current and voltage varies from point to point along VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 58 VEL . Write short notes on co-axial cable. One conductor is a hollow tube. A transmission line is a circuit with distributed parameters hence the method of analyzing such circuit is different than the method of analysis of a circuit with lumped parameters. 2 + 2o Reflection Loss = R 2R − 2o It is the reciprocal of Reflection co-efficient 44. 46. What is return loss? Return loss is defined as the ratio of power at receiving end due to incident wave and power due to reflected wave in the load. the second conductor being located inside of co-axial with the tube. What do you mean by Lumped circuits? The network where in the resistance. Write down the expression for transfer impedance ZT = Es 2R + 2o γl = e + Ke−γl IR 2 ( ) ZT = zR cos h γl+ o sin h γl z 45. 43. inductance and capacitance are individually concentrated or lumped at discrete points in the circuit is called Lumped circuit. PART – B 1. Derive the General solution a transmission line. l = Length of the line. Y = G + j ω C = shunt admittance in mhos per unit length. The elemental voltage drop in the length ds is dE = I Zds dE ∴ = I Z ………(1) ds the leakage current flowing through shunt admittance from one conductor to other is given by dI = EY ds ∴ dI = EY ds ……. ω C = Shunt susceptance in mhos per unit length Z = R + J ω L = Series impedance in ohms per unit length. The current is I and voltage is E at this section. The length of section is ds hence its series impedance is Zds and shunt admittance is Yds. The various notations used in this derivation are.(2) Differentiating equation (1) and (2) with respect to S we get VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 59 VEL . farads per unit length. ohms per unit length. ω L= Series reactance per unit length. L = Series inductance. C = Capacitance between the conductors. S = Distance out to point of consideration. including both the wires. henrys per unit length.4. The point under consideration is at a distance a from the receiving end. R = Series resistance. measured from receiving end. One such section of length ds is shown in the Fig. I = Current in the line at any point. E= Voltage between the conductors at any point. The transmission line of length l can be considered to be made up of infinitesimal T section.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL the transmission line. The general solution of a transmission line includes the expressions for current and voltage at any point along a line of any length having uniformly distributed constants.17. G = Shunt leakage conductance between the conductors. It carries a current I. mhos per unit length. . there exists two solutions for positive sign of m and negative sign of m. B.. C and D.. The general solution of the equations for E and I are...VEL TECH d2 E dI =Z 2 ds ds and d1 I dE =Y 2 ds ds VEL TECH MULTI TECH TECH HIGHTECH VEL This is because both E and I are function of S. Replace the operation d/dS by m hence we get... It is now necessary to obtain the values of A...... . So.(4) ds 2 The equations (3) and (4) are the second order differential equations describing the transmission line having distributed constants... VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 60 VEL .(3) ds 2 IZ and d 2I = YIZ .. d 2E = ZEY ....(7) Where A... (m2 –ZY) E = 0 but E ≠ 0 m = ± ZY ……... E = Ae ZYS + Be − ZYS ..(6) I = (e ZYS + 1)e − ZYS . C and arbitrary constants of integration. all along its length. It is necessary to solve these equations to obtain the expression of E and I.. (5) Same result is true for the current equation.....B.. As distance is measured from the receiving end S = 0 indicates the receiving end E = IER and I = IR at S = 0 Substituting in the solution..... e...I = A andE = C SHORT SECTION PQ. distance x from the sending end of transmission line.(9) andEY = C ZY e ZYs − D( − ZY )e − ZYs .... VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 61 VEL ... (8(a)) IR = C + D ….(10) I= A B ZY e Zys − ( − ZY )e − ZYs Z Z Y Y e ZYS − B e − ZYS .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL ER = A + B …. dE = A ZY e ZY s + B( − ZY )e − ZY s ds and dI = C ZY e ZYs + D(− ZY )e − ZYs ds But dE dI = IZand = Ey ds ds ∴ IZ = A ZY e Zys − B (− Zy )e − Zys .(9(b)) Same condition can be used in the equations obtained by differentiating the equations (6) and (7) with respect to S...(11) Z Z Z Z e ZYS − D − ZYS ..(12) Y Y i... VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Now use S = 0 , E = ER and I - IR ∴ IR = A Y Y −B ............(13( a) Z Z Y Y −D .........(13(b) Z Z andER = C The equation 8a, 13a, 13b are to be solved simultaneously to obtain the values of the constants A, B, C and D. Now while solving these equations use the results, ZR = ER R + jω L Z andZo = = IR G + jω C Y Hence the various constants obtained, after solving the equations simultaneously are, A= B= C= ER I R + 2 2 ER I R − 2 2 I R ER + 2 2 I R ER − 2 2 Z ER Z O = 1 + .......(14) Y 2 ZR Z ER Z O = 1 − ........(15) Y 2 ZR Y IR = Z 2 Y IR = Z 2 ZR 1 + ZO .........(16) D= ZR 1 − ............(17) ZO Hence the general solution of the differential equation is, E= ER 2 Z O ZO 1 + e ZYS + 1 − ZR Z R e ZYS ..(18) e − ZYS ..(19) E= ZR I R Z R 1 + e ZYS + 1 − 2 Z O ZO VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 62 VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH Z R + ZO out from equation (18) ZR VEL Taking LCM as ZR and taking E ( Z + ZO ) E ( Z − ZO ) I = R R e ZYs R R e − ZYs ..(20) 2Z R 2 Z R ( ZR + ZO ) Taking LCM as Zo and taking Z R + ZO out from equation (19) ZR I ( Z + ZO ) ( Z − ZO ) I = R R e ZYS R e − ZY S ..(21) 2ZO ( Z R + ZO ) The negative sign is used to convert Zo – ZR to ZR - Zo The equation (20) and (21) is the general solution of a transmission line. Another way of representing the equation is E= ER ( Z R + ZO ) e ZYs + ( ZR − ZO )e ZY s 2Z R ER Z Re ZYs + ZOe ZYs + ZRe − ZYs − 2Z R E= Z Oe − ZYs ER E = Z R hence R = I R IR ZR But e ZYs + e − ZYs ∴ E = ER + 2 e ZYs + e − ZYs I R ZO ......(22) 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 63 VEL VEL TECH e ZYs + e − ZYs I = IR + 2 ER ZO VEL TECH MULTI TECH TECH HIGHTECH VEL and e ZYs + e − ZYs .............(23) 2 e ZYs + e − ZYs = COSh ZYsand 2 But e ZYs + e − ZYs = Sinh ZYs 2 ∴ E = ERCosh and I I R Cosh ( ZYs + 1R ZO Sinh ) ( ZYs ....(24) ) ( ZYs ER / ZO sinh ) ( ZYs ..........(25) ) The equation (24) and (25) give the values of E and I at any point along the length of the line. Important Note: The similar equations can be obtained in terms of sending and voltage Es and Is. If X is the distance measured down the line from the sending end then, X=1–s And the equation (24) and (25) get transferred in term Es and Is as E = ES Cosh I = I S Cosh ( ZYx + I S ZO Sinh ) ( ( ZYx ) ) ( ZY x + ES / ZO Sinh ) ZYx And ZY = γ as derived earlier and hence equation can be written in terms of propagation constant γ . Summarizing. If receiving end parameters are known and s is distance measure from the receiving end then, E= ER cos h ( γ s ) +IR Zo sin h ( γ s) VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 64 VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL I = IR cosh ( γ s) +ER /Zo sin h ( γ s) And if sending end parameters are known and X is distance measure from the receiving end then, E = Es cosh ( γ x ) + Is Zo sinh ( γ x ) I = Is cosh ( γ x ) +Es/Zo sinh( γ x) Any set of equations can be used to solve the problems depending on the values given. 2. Explain the physical significance of General solution. From the qeuesed solutions, the sending end current can be obtained by substituting S = I measured from the receiving end. Es = ER cosh ( γ Is = IR [cos h ( γ Now ZR = ER / IR ∴ Is = [ IR cos h ( γ I) + ZR /Zo IR sin h ( γ I ) ] ∴ Is = [ IR cos h ( γ I )+ ZR / Zo sinh ( γ I ) ] ……(3) Now if the line is terminated in its characteristic impedance Zo then, Is = [ IR cosh ( γ I ) + sinh ( γ I ) ]…. As ZR = Zo Is / IR = [ cosh ( γ I ) + sinh ( γ I ) ]= eW …….(4) This is the equation which is already derived for the line terminated in Zo. Using ER = IR ZR in equation (1), Es = ZR IR cosh ( γ I ) + IR Zo sinh ( γ I ) Es = IR [ (ZR cos h ( γ I ) + Zo sinh ( γ I ) ]…..(5) Dividing (5) by (3), I [ Z cosh(γ I ) + ZO sinh(γ I ) ] Es = R R I s I R [ cosh(γ I ) + ZR / ZO sinh(γ I )] But Es / Is = Zs I ) + IR Zo sinh ( γ I ) ………(1) I) + ER /Zo sin h ( γ I ) ……..(2) VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 65 VEL VEL TECH Z O [ Z R cosh(γ I ) + ZO sinh(γ I ) ] VEL TECH MULTI TECH TECH HIGHTECH VEL ∴ ZS = [ ZO cosh(γ I ) + Z R sinh(γ I )] When the line is terminated in Zo then ZR = Zo SO substituting in equation (6) we get Zs = Zo This shows that for a line terminated in its impedance is also its characteristic impedance. characteristic impedance, it input Now consider an infinite line I → ∞ Using this in equation (6) we get, ZS = Z O [ Z R + ZO tanh(γ I ) ] [ ZO + Z R tanh(γ I )] and tanh(γ I ) → IasI → ∞ ∴ Z S = ZO ........(8) This shows that finite line terminated in its characteristic impedance behaves as an infinite line, to the sending end generator. Thus the equations for Ex and Ix are applicable for the finite line terminated in Zo. The equations are reproduced here for the convenience of the reader. Ex = Es e –yx and Ix = Is e –yx If in practice instruments are connected along the line then the instruments will show the magnitude Es e –yx and Is e –yx while the phase angles cannot be obtained. If the graph for Ex or Ix is plotted against x then it can be shown. This is the physical significance of the general solution of a transmission line . Its use will be more clear by studying the various cases of the line. 3. State and explain different types of distortions in line. When the received signal is not the exact replication of the transmitted signal then the signal is said to be distorted. There exists some kind of distortion in the signal. There are three types of distortions present in the transmitted wave along the transmission line. 1. Due to variation of characteristic impedance Zo with frequency. 2. Frequency distortion due to the variation of attenuation constant α with frequency. 3. Phase distortion due to the variation of phase constant with frequency. Distortion due to Zo varying with Frequency: VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 66 VEL The power is absorbed at certain frequencies while its gets reflected for certain frequencies. The inductors are introduced in the limbs to keep the line as balanced circuit.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The characteristic impedance Zo of the line varies with the frequency while the line is terminated in an impedance with does not vary with frequency in similar fashion as that of Zo. Each winding is divided into equal parts. Such a line can be easily and correctly terminated in an impedance which matches with Zo at the frequencies for such a line Z R = R / Gor L / G . This causes the distortion. so that exactly half the inductance can be VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 67 VEL . This eliminates the distortion and hence selective power absorption. 4. The core of the coil is usually Toro dial in shape and made if dimensions. The loading coil is wound of the largest gauge of wire consistent with small size. very low eddy current losses and negligible external field which restricts the interference with neighboring circuits. It is known that. the condition LG += CR is satisfied the L/R = C/G and hence (1 + J ω L / R) = (1 + J ω C / G ) ∴ Z O = R / G < Oo = L / C < OoΩ For such a line Zo does not vary with frequency ω and it is purely resistive in nature. So there exists the selective power absorption. The loading coil design is very much important in this method. Write brief notes on lumped loading. Such inductors are called lumped inductors. ZO = R + jω L = G + jω C R + (1 + jω L / R ) G (1 + jω C / G ) If for the line. The lumped loading is preferred for the open wire lines and cables for the transmission improvement. In this type of loading the inductors are introduced in lumps at the uniform distances in the line. The lumped inductors are in the form of coils called loading coils. due to this type of distortion. resistance and conductance are distributed uniformly across the entire line length white in lumped lines these parameters are jumped at intervals along the line. By use of electrical conductor arrangement known as transmission line. 3. the care must be taken so that the circuit balance is maintained. If the loading section distance is d than keeping inductance LS of the loading coil constant. Wave guides 5. The pots protect the coils from external magnetic fields. Write short notes on different types of transmission lines. Strip line. But above this cut-off frequency the attenuation constant increases rapidly. up to a certain frequency called cutoff frequency of the line. weather and mechanical damage. In communication these lines are used as link between transmitter and receiver. Transmission is the process it transmitting some signals from one place to another. the electrical parameters like inductance. White installing the coils. Open wire line. Hence to get the higher cut off frequency. The graph of ∝ against the frequency called the attenuation frequency characteristics of the line show in the fig. The line acts as low pass fitter. Upto this frequency. Coaxial line. By radiation of electromagnetic waves through free space. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 68 VEL . 2. In the case of lumped loading. The fig.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL inserted in to each leg of the circuit. It can be see that for continuous loading the attenuation is independent if frequency while for lumped loading it increases rapidly after the cut-off frequency. These are built into steel sizes to accommodate one or more coils. No winding is reversed. If winding is. “Transmission line is a conductive method of guiding electrical energy from one place to another”. cut off frequency is found to be proportional to the *. Transmission lines may be grouped as lumped lines or distributed lines while in distributed liens. capacitance. 5. the added inductance behaves as if it is distributed uniformly along the line. The line behaves properly provided spacing is uniform and loading is balanced. it will neutralized the inductance of other winding reduced the overall inductance. shows the construction of loading coils. 2. 4. small lumped inductance must be used at smaller distances. Here it can be data as in the case of transmission of computer data along telephone lies or it can be audit/video signals from radio or television broad cost. The commonly used transmission lines are 1. Electrical energy can be transmitted from one point to another by one of the two methods namely 1. Optical fibers. The construction of open wire line is as shown below. Open wire line: These lines consist of two conductors spaced at a certain distance apart. The construction of coaxial cable. a closed field configuration is used by surrounding the inner conductor with an outer cylindrical hollow conductor and the arrangement is termed as a coaxial cable. Coaxial lines (or) Coaxial cables: The avoid radiation losses taking place in open wire lines at high frequencies. These lines are unsuitable for use at frequencies above 100 Mhz because energy loss takes place due to radiation. Simple to construct and low cost 2. However the radiation loss can be minimized by reducing the spacing between wires. The spacing between the conductors is large in comparison to the diameter of the line conductors. As a result these lines can operate at high voltage. resistance. However when operating at higher frequency the larger spacing proves to be a disadvantages because radiation of energy from the open wire line takes place. 2.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 1. It is balanced with respect to ground. Insulation between the line conductors is air. capacitance and conductance per unit length can be given as. Where. L = µ ο / π log e (S / r ) Henries. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 69 As a result the electric loss is . R = 2p /r π 2 ohms. The electric and magnetic fields existing in the lines are shown below. These lines have distributed sonic resistance and inductance. shunt capacitance and conductance. inductance. Disadvantages: 1. telegraphy and telephony signals. the electric magnetic field existing in a coaxial cable s as shown above and below. P – Specific resistance r – radios S – Spacing between conductors. The primary constants of a open wire transmission line are. This type of line is used in transmission of electric power. extremely small 3. Advantages of open wire lines: 1. Beyond 1Ghz these cables cannot be used because losses in the dielectric increases with frequency. Disadvantages: 1. STRIP LINES VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 70 VEL . It provides effective shielding from outer interfering signals.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL A typical air inner copper conductor held in position by insulating discs Flexible cables have polythene dielectric and outer conductor in the form of copper braided or flexibility. The electric and magnetic fields are continued with in the conductor there by eliminating radiation losses. 2. It ‘D’ and ‘d’ are diameters of the outer and inner conductor. 2. then the characteristics impedance P of the cable is given by Z o = 138 µ r / ∈ r log10 (d / d ) the velocity of propagation V = 3*108 / µ r ∈ r Advantages: 1. and also µ and ε be the permeability and permittivity of the insulting medium. They are costlier then open wire lines. Strip lines are two types: 1. The distance b/w the centre and outer plate is smaller compared with the wave length of the signal Disadvantages: 1. The energy propagation in this line is in the form of TEM waves. The bottom conducting plates serve as earth plate. Micro strip lines are fabricated on fiber glass or polystyrenes printed circuit boards as about 1.5mm thickness with copper strips. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 71 VEL . The micro strip line is widely used in microwave integrated circuits these components find wide application in couplers circulators and receiver.Trip late Line: This line resembles as a co-axial line in which side conductors have been removed. . Trip late 2. Micro strip line: The micro strip line has a narrow conductor supported by a dielectric.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL A form of line using finite plates and an inverting dielectric medium is turned as strip line. provided. Micro strip. 3mm wide. It is costlier and it also requires a manufacturing skill. The dielectric loss is negligible. The propagation takes place in open wire and co-axial lines propagation takes place in the form d transverse electric (TE) and transverse magnetic ™ waves. Disadvantages: 1. because she elastic and magnetic fields are confined to the space with in the guides. Optical fibers: VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 72 VEL . Wave guide walls should be specially plated to reduce resistance to avoid skin effect and power less. 1. Wave guide are used to minimize losses and for high power transmission at microwave freq. transform by continuous reflection from the inner walls of the guide. 4. Cost of the wave guide is very high. F. 3. In wave guide no power is cost throughout. Frequency of the wave higher than 39 hz can be easily transmitted. Several modes of electro magnetic waves can be propagated with in a single wave guide. The shape of the wave guide may be rectangle or cylindrical thro which electromagnetic waves are propagated. 2.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Wave guides: Wave guides are hollow conducting takes a uniform gross section used for U. Rectangular waveguide Advantages: Elliptical waveguide. 2. H. an electric field is produced b/w the two conductor. 2. The structure of a time cable is show below.7000/.Low transmission loss and high band width. The magnetic field proportional to the usual indicate that the line L as series inductance ‘L’ and the voltage drop indicate the presented series resistance ‘R’ Similarly the electric field proportion to the voltage indicates that the line contains VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 73 . when a current is passed through it. Electrical isolation is there. Small sue and weight. 5. Disadvantages: 1. It confines electromagnetic energy in the form a light to within its surfaces and guide the light in a direction 11 d to the axis. Here the copper cores are replaced by highly put e – glass (or) silica which is used to carry modulated light energy similar to micro wave energy.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The entrance appearance of optical fibers are similar to C0-axial cables.per meter). 3. 4. 2. Different specializes technique have to the followed to join ends of two cables. No radio frequency and electromagnetic item. An optical fiber is a dielectric wave guide that operated at optical frequency. Advantages: 1. It is difficult to run the cables where the bending occurs. 6. The cost of fiber able is very high (Rs. Explain about transmission line parameter: Transmission line parameter: In a open wire line. 3. High degree of data securing is afforded. a magnetic fields are produced around the conductors and the voltage drop occurs along the line similarly when a voltage is passed through a open wire line. Z=R+jωL. Thus it is the sum of inductance of both wire for unit length. They it is the sun of resistance of both the wire for unit line length. R. C. The four line parameters. Characteristic Impedance ‘Zo’: VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 74 VEL . and G are termed as primary customs of a transmission line. Its unit is mhos / km. Inductance ‘L’ Inductance t ‘L’ is defined as loop inductance per unit length of line. C and G are distributed along the whole length of the line. Conductance ‘G’ Conductance ‘G’ is define as shunt conductance b/w the two wires per unit length of line. Admittance: ‘Y’: The shunt admittance. R – line resistance.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL shunt capacitance ‘C’ and this capacitance is new useless (or) perfect so. Capacitance: Capacitance is defined as shunt capacitance b/w the two wipes per unit length: It’s unit is farad / km Impedance ‘Z’ The series impedance of a transmission line per unit length is given as . ‘Y’ of a transmission line per unit length is given at y = G’ + j ω c. L. Where. G – line conductance and j ω C – line susceptance. Its unit is HCGrier /km. the line also contains conductance ‘G’ These four parameter R. They are defined Resistance ‘R’ Resistance ‘R’ is designed as loop resistance per unit length of line. It is ohm/kn. L. and j ω L – Line reactance. Where. r = ZY = ( R + jω L)(G + jω L) These characteristic impedance ‘Zo’ and ‘r’ are called as secondary constants of transmission line. Zo = Z / Y Z o = ( R + jω L) /(G + jω C ) Propagation constant ’Y’: Propagation constant is defined at the product of square root of impedance and admittance.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Characteristic impedance is defines as the ratio of square root of impedance to admittances. √g = λ f √g = 2 ∏ f/ β VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 75 VEL . In addition to primary and secondary constants of a transmission line there are 3 more units of transmission line theory. It is denoted by ‘ λ ‘ and its unit is meter. It is denoted at √g. g = (ω 2 − ω1 ) /( β 2 − β1 ) Velocity of propagation: It is defined as the velocity with which a signal of single frequency propagation along the line at a particular frequency ‘f’. λ =2π / β Group Velocity: The group velocity is defined as the velocity of the develop is a complete signal. Wave length: It is defined as the distance that the wave travels along the line in order that the total shift is 2 π radiance. ..... γ = XY γ = ( R + jω L)(G + jω c)........VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 7... Similarly line conductance is also vary small as compared to susceptance substituting the condition in general equation for propagation constant (γ ) and the characteristic input (Zo) given. L = 0 γ = γ = jω RC ω Rc 45o.....(4) Characteristic impedance Z o = ( R + jω L)(G + jω L) Z 0 = R / jω C Z 0 = R / ω C < 45oand . the series inductance reactance is quite negligible as negligible as compared to line resistance ‘R’......... At low frequencies.... A large no of each pairs form an underground cable. Such transmission lines are called as telephone cables..(3) β = ω RC / 2.....(1) subG = 0.. Derive the expression for telephone cables: Telephone cables (or) telex lines are used as low frequencies transmission line.........(2) α + j β = ω RC cos 45o + j ω RC sin 45o ω RC / 2 + j ω RC / 2 γ = α = ω RC / 2......(5) Velocity of propagation Vp is given as γ p 4= ω / β VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 76 VEL . A telephone cable is formed by two wines insulated from each other by a layer oil impregnated paper and then twisted in pains.... The overall effect of insertion of a line is to change the current through the load and hence power delivered to load is less compared too power delivered to load when it was directly connected to generator. If ZR is not equal to Zo.. I R1 = E → (1) Zg + ZR The line is inserted between load and generator. Hence large distortion occurs at higher frequencies in a telephone lines. Insertion loss occurs due to insertion of a network or a line in between source and load. Consider the circuit in which generator of impedance Zg is connected to a load of impedance ZR. 8.. Thus insertion loss of a line or a network is defined as the number of ropers or dB by which the current in the load is changed by insertion of a line or a network between the load and the source. This the higher frequency are attenuated move and travel faster than the lower frequency resulting in frequency and delay distortion. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 77 VEL .(6) From the above equation its observed that α and √ are independent on frequency...... then second reflection loss occurs at terminals 2-21... then reflection loss occurs at terminals 1-11. Explain insertion loss in detail and derive the expression for the same.VEL TECH γ p = w/ VEL TECH MULTI TECH TECH HIGHTECH VEL ω RC / 2 ⇒ V p = 2ω / RC . If input impedance Zs is not equal to generator impedance Zg.. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Let Zs be input impedance of a line which different than Zg. IS = E → (2) Z g + ZS We know that input impedance of line is eγ l + ke−γ l Z S = ZO γ l → (3) −γ l e − ke sub(3)in(2) IS = E eγ l + ke−γ l Z g + ZO γ l −γ l e − ke Z g (e − ke γl −γ l ∴ Is = E ) + Zo (eγ l + ke−γ l ) we know that Is = I R ( Z R + Zo ) γ l (e − ke −γ l ) → (4) 2Z o This equation is obtained from general solution of line by substituting s = l IR = 2ZO I S → (5) ( Z R + ZO ) eγ l − ke −γ l 2 Z o E (eγ l − ke −γ l ) = → (6) Z g (eγ l − ke −γ l ) + Z 0 (eγ l + ke−γ l ) ( Z R + Z o )(eγ l − ke −γ l ) we know that ref. 10eff. K= Z R − Zo → (7) Z R + Zo ∴ IR = 2Z 0 E Z R − Z0 + Z R + Zo − ( Z R + Z o ) Z o γel lγ e Z l + g e γ Z R − Z R −o Z +o Z e− l γ VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 78 VEL . Z R + Z o Zo + Z g eα l I R2 ∴ = → (11) IR 2 | Zo || Z g + ZR | Xand ÷ by 2 Z g Z R . ∴ γl I R1 ( Z R + Z o )( Z 0 + Z g )e = → (10) IR 2Zo ( Z g + Z R ) = ( Z R + Z o )( Zo + Z g )eα l e j β l 2Z 0 ( Z g + Z R ) (Q γ = α + j β ) But insertion loss has too be calculated as a function of ratio of current magnitudes β and hence e j l can be neglected.loss = = 2Z o E IR γl ( Z R + Z o )( Z 0 + Z g )e + ( ZR _ Zo )( Zo − Zg )e −γ l 1 R γl −γ l I R1 ( Z R + Z o )( Zo + Z g )e + ( ZR − Zo )( Zo − Zg )e = → (9) IR 2 Zo ( Zg + zR ) The length of line is usually very large hence e −γ l → 0 ∴2nd term in numerator can be neglected. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 79 VEL .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL I R= 2Z o E γ ( Z R + Z o ) Z o e l Z − Zo − + R Zo e Z R + Zo lγ + g el γ Z Z −o Z −R ZR + o Z −l Z ge γ ∴ IR = 2Z 0 E → (8) ( Z R + ZO )( Zo + Z g )eγ l + ( ZR − Zo )e−γ l I R1 IR Insertionloss = E Z g + ZR I ∴ I . VEL TECH I R1 2 Z g Z R | Z R + ZO || ZO + Zg | e = IR 4 Z g Z R | ZO || Zg + Z R | ∴ αL VEL TECH MULTI TECH TECH HIGHTECH VEL I R1 | Z g + ZO || Z R + ZO | 2 Zg ZR α l = e → (12) IR 2 Z g ZO 2 Z R ZO | Zg + ZR | All the terms on RHS are reflection factors. 2 Z g Zo Let Ks = = reflection factor at source side →(13) | Z g + Zo | kR = 2 Z R Zo = reflection factor at load side →(14) | Z R + Zo | 2 Zg ZR | Zg + ZR | = refl factor for direct connection →(15) k SR = eα l indicates loss in the line. k I R1 ∴ = SR eα l IR kS k R Insertion loss = (or) I R1 1 1 1 = lu + lu − lu + α l Nepers. I R kS kR kSR 1 1 1 + 0.4343α l dB Insertion loss = 20 log + log − log kS kR kSR The term corresponding to kSR is negative. It is the loss if generator and load would have been directly connected. It is not related to insertion hence it is subtracted from overall loss. 9. Prove that (i ) Z 0 = Z oc. Z SC (ii ) tanh γ = Z sc Z oc (i)Consider a short line terminated I its characteristic impedance Z o. The short line is a symmetrical network and hence can be represented by equivalent T – section. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 80 VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL We know that finite line terminated in Zo behaves as an infinite line, hence Zo must be Zo The i/p impedance Z in of equivalent T network is Z Z Z in = 1 + Z 2 || 1 + Zo 2 2 Z z2 1 + Z 0 Z 2 Z in = 1 + 2 Z + Z1 + Z 2 0 2 ButZ in = Z o Z Z2 1 + Zo Z 2 Z0 = 1 + 2 Z + Z1 + Z 2 o 2 Z Z Z 2 Z o Z 2 + 1 Z0 = Z1 Z2 + 1 + Zo + 2 Z2 1 + Z0 2 2 2 2 Z ∴ 2 Z o 2 = 2 Z1 Z 2 + 1 2 2 Z ⇒ Z o 2 = 1 + Z1 Z 2 4 ∴ Zo = Z12 + Z1Z 2 4 to obtain Z1 and Z2 we open circuit and short circuit the network In open circuit the line is kept open and input impedance is measured. Z Z oc = 1 + Z 2 2 In short circuit the second end of line is short ed and input impedance is measured. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 81 VEL VEL TECH Z1 Z + Z 2 || 1 2 2 Z1 Z2 Z = 1+ 2 2 Z1 + Z 2 2 Z sc = Z12 Z1 Z 2 + + Z1Z 2 2 = 4 Z1 + Z2 2 Z2 ∴ Z sc = o c Zo Z o 2 = Z oc. Z sc ⇒ Zo = Zsc .Zoc (iii) we know that Z1 Z o + 2Z 2 Z 2 VEL TECH MULTI TECH TECH HIGHTECH VEL eγ = 1 + subZ o Z1 Z12 Z e = 1+ + + 1 2 2Z 2 4Z2 Z2 γ Z Z Z ∴ e = 1 + 1 + 1 + 1 → (1) 2Z 2 2Z 2 Z2 γ 2 mathematically Z Z Z e ; 1 + 1 − 1 + 1 → (2) 2Z 2 2Z 2 Z2 (1) = (2) Z eγ + e −γ = 2 + 1 Z2 −γ 2 ∴ eγ + e −γ Z = 1+ 1 2 2Z 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 82 VEL VEL TECH ∴ cosh γ = 1 + Z1 2Z 2 VEL TECH MULTI TECH TECH HIGHTECH VEL Noweγ = coshγ + sinhγ Z ∴ (cosh γ + sinh γ ) − cosh γ = eγ − 1 + 1 2 Z 2. Z Z Z ∴ sinh γ = 1 + 1 + o − 1 + 1 2 Z 2 Z 2 2 Z2 γ = Zo Z2 Zo Z0 sinh γ Z2 Sinh tanh γ = = = Z cosh γ 1 + Z1 Z2 + 1 2Z 2 2 ButZ o = Z sc .Zoc andZ 2 + Z sc Z oc Z oc Z sc Z oc Reflection loss and derive the Z1 = Z 0c 2 ∴ tanh γ = ⇒ tanh γ = 10. Write a note on Reflection factor and same. Reflection occurs due to improper termination at the receiving end. This concept can be extended to the function of any two impedances . Let a source of voltage Es and impedance Z1 is connected to a load of impedance Z 2 . If Z 2 is not equal to Z1 , reflection of energy takes place resulting in a change in the ratio of V to current and alteration in the distribution of energy between the Electric and magnetic field. The energy transferred to Z 2 is less than that with impedance matching. A reflection is said to have occurred. The magnitude of this loss can be computed by taking the ratio of current actually flowing into the load to the load to the current that would have flown if the impedance is were matched. The matching of impedance is called as image matching and can be obtained on a live by connecting a transformer. According to the transformer theory, VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 83 VEL VEL TECH I1 = I2 z2 z1 VEL TECH MULTI TECH TECH HIGHTECH (1) VEL → For matching, the magnitude of Z1 can be made equal to Z 2 by choosing proper transformer ratio. The current which flows through the generator is I1 I1 = E 2 Z1 → (2) 1 The current I 2 (Under image matching condition) which flows through the load 1 I2 = Z1 Z2 Z1 Z2 E Z1Z 2 • I1 • E 2 Z1 → (3) I2. 1 I2 = 1 I2 = E → (4) Z1 + Z 2 The ratio of the current actually flowing into the load to that current flowing under image condition Without image matching, the current flowing through the load is ∴ I2 = 2 Z1Z 2 I2 = = R → (5) 1 I2 Z1 Z2 is This is called as Reflection factor (R) Reflection loss It is defined as the no.of repers or dB by which the current in the load under image matched condition would exceed the current actually flowing in the load. Then The reflection loss in repers is Z + Z2 Rloss = lu 1 in lepers 2 Z1 + Z 2 Z + Z2 Rloss = 20 log 1 − dB 2 Z1 + Z 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 84 VEL Thus as incident wave travels from the sending end to the receiving end. with decreasing amplitude is the incident wave. S=O at the receiving and maximum m(s=l) at the sending end. eZR= α Reflection is maximum S c i eZ R =O Reflection is zero when Z R = Zo S → distance measured from the receiving end and treated positive. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 85 VEL .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 11. A wave which flows from sending end to the receiving end. When Z R ≠ Zo 1) One part varying exponentially with positive S 2) One part varying exponentially with negative S E= ER ( Z R + Z o ) Z − Zo ) e zys + R e 2 ys 2Z R ZR ER ( Z R + Z o ) 2 ZR ⇒E= ER ( Z R + Z o ) E e zys + R ( Z R − Z o )e − 2 ys 2Z o 2 ZR IR ( Z R + Z o ) 85 I R ( ZR − Z 0 ) 85 e − e 2Z o 2 Zo III rly I = The first component of E or I which varies exponentially with TS is called incident wave which flows from the sending end to the receiving end. Reflection is maximum when the line oCi. Explain the Reflection of a line not terminated by Z0. its amplification decreases. Referring to Equation ( Z − Z R ) − zys IR I= ( Z 0 + Z R ) e ZYS + o e 2ZO (Z R + Zo ) E= zys (Z − Zo ) − ER ( Z R + Z o ) e ZYS + R e 2 ZR (Z R + Zo ) There are current and voltage relationships derived for the lines which are terminated in Z o But if a line is not terminated in Z o (or) it is joined to same impedance other than Z o then part of the wave is reflected back phenomenon exists for a line which is not terminated in Z o . The total instantaneous voltage at any point on the line is the vector sum of incident and the reflected currents are in out of phase with each other. 12. We get 1+ Zsτ Z o Zo Jauh(81) ZR Zo + Jauh(81) ZR Z + Z o Jauh(81) (or ) zsτ R Z Z o + Z R Jauh(81) or equ (3) jauh (81) e81 − e −81 e81 + e −81 Z + Z o (e81 − e −81 / e81 − e−81 ) ∴ Z sτ Z o R 81 −81 81 −81 Z o + Z R (e + e / e + e ) Equ. Similarly along the line and the energy is absorbed wave. Such a line is uniform and there is no discontinuity existing to send the reflected wave back along the line. (3) and (4) are the I/P Impedance of a live terminated by an impedance. Derive an expression for input impedance and transfer impedance and of transmission line terminated by an impedance. From the general solution Zinτ Z s Z τ Z o ( Z R wsh(81) + Zo Sinh(81) ) ( Z o wsh(81) + ZR Sinh(81) ) Dividing by ( Z R cosh (81) )both Nr and Dr. Let ZT = Es = Transfer impedance of a live IR VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 86 VEL . If z R = Z o it can be seen that the reflected wave is absent end there is not reflection.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The second component of voltage or current which travels from the receiving end to the sending end which varies e85 and its amplitude decreases as its progress towards the sending end. Such a finite line terminated in Z o without having any reflection is called a smooth line. Derive an expression for the Input impedance of a lossless line. therefore Pτ j β will become jB only Hence. transfer impedence. Z in = Z o = Zo Z R + Z o tanh(81) Z o + Z R tanh(81) ______(1) for loss live α = o. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 87 VEL . I/P impedance of a lossless live of any length or obtained from equ. 13.VEL TECH I Rτ cosh(81) − Iτ Now Es Sinh(81) Z o xIR VEL TECH MULTI TECH TECH HIGHTECH VEL Is Z cosh (81) − T sinh(81) IR Zo while ERτ Es cosh (81) − Zo sinh(81) τ Es cosh(81) I s Zo sinh(81) − ER Is = Es cosh(81) − ER Z o sinh(81) Sub in equ (5) 1τ = 1τ cosh(81) IR Es cosh(81) − ER ZT sinh(81) − Zo sinh(81) Zo cosh 2 (81) − Z R cosh(81) − ZT sinh2 (81) Z o sinh(81) Z o sinh(81)τ ZT cosh(81) − ZR cosh(81) − ZT sinh 2 (81) τ ZT (cosh 2 (81) − Sinh2 (81)) − ZR cosh(81) τ ZT (1) − Z R cosh(81) ZTτ Z R cosh(81) + Zo sinh(81) e81 + e −81 e81 + e−81 ZT τ Z R + Zo 2 2 ∴ This is the reg. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 88 VEL .e.VEL TECH Z in = Zo = VEL TECH MULTI TECH TECH HIGHTECH VEL Z R + Z o tanh( j β l ) _____(2) Z o Z R tanh( j β l ) But tanh j β lτ tan β l Therefore Z + jZ o tan β l Zin = Z o R Z o + Z R tan( j β l ) 2π l Since β = λ 2π l Z R + jZ o tan λ Zin = Z o ______(3) 2π l Z o + Z R tan λ Again for a lossless live. R and G will be equal to zero P = α + j β = ( R + jwl )(G + jwc) Thus O + jβ = jwLxjwc j β = jw LC β = w LC ⇒ (or) β = w LC If f is the frequency of operation and terminating impedance is a pure resistance RR Equ (3) will become RR + jZ o tan 2π f LCl Z o + JRR tan 2π f LCl Z in = Therefore.. the resistive component of the live i. Derive the equation for T and π section equivalent to lines. I/P Impedance of a lossy and lossless line. 14. ZR=∞ VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 89 VEL . γl ZR Zo −γl e + e ZR + Z o Zo Z10c = = Zo γl Z Z −γl tan h γl e − R o e ZR + Z o for oc.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Z10c = Z1 + Z3 Z15c = Z1 + Z2Z3 Z2 + Z3 Z1 = Z10c = Z3 Z2 = Z20c − Z3 Z Z Z3 = Z2 + Z3 Z1 + Z3 − Z1 + 2 3 Z2 + Z 3 Z ( Z +Z ) −Z Z 3 2 3 = Z2 + Z3 3 2 Z2 + Z3 ( ) ( ) 2 Z3 = Z3Z2 + Z3 − Z2Z3 Z3 = Z3 ∴ Z3 = Z20c ( 210c − Z15c ) The input impedance of oc and SC lines are. VEL TECH eγl + e−γl ∴ 210c = Zo γl e + e−γl VEL TECH MULTI TECH TECH HIGHTECH VEL eγl + e−γl III ly Z13C = 20tan hγl= γl 20 −γl e +e since the line is symmetrical Z10c=Z20c ∴ Z3 = Z o Zo − Zo tan hrl tan h γl tan h γl Zo Zo − Zo tan2hγl = tan h γl tan h γl Zo = 1− tan2 h γl tan h γl Zo Zo 1 = sec2 hγl = × coshγl × tan h γl sin h γl cos h hγl Zo = sin h γl eγl + e−γl Z1 = Z2 = Z10c − Z3 = Zo γl e + e−γl = Z3 eγl + e−γl eγl + e−γl 2 Z 3 = Zo 2 eγl + e−γl ( ( ( ( )( ) ) ) 2 ) − Zo ( eγl + e−γl ) ( eγl + e−γl ) (e γl +e −γl )( e γl + e−γl ) Z3 = 2 Zo eγl + e−γl −1 eγl + e−γl = 2 Zo eγl + e−γl + 2 − e2γl − e−2γl − 2 eγl + e−γl 2 ( ( ) ) 4 2 z2 = z = o γl 2 o −γl γl −γl e −e e −e VEL TECH ( ) VEL TECH MULTI TECH TECH HIGHTECH 90 VEL . Section equivalent :- ZA = ZB = ZC = Z1Z2 + Z2Z3 + Z3Z1 Z (Z + Zc ) Z10c = A B Z2 Z A + Z B + Zc Z1Z2 + Z2Z3 + Z3Z1 Z (Z + Z B ) Z20c = c A Z3 Z A + Z B + Zc Z1Z2 + Z2Z3 + Z3Z1 ZA Z B Z15c = Z1 ZA + Z B Since the line is symmetrical VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 91 VEL .VEL TECH eγl + e−γl 2 Z10c − Z2 = Zo γl − γl −γl −γl e +e e +e γl e + e−γl − 2 = Z o γl e + e−γl 2 eγl − e−γl / 2 = Zo γl / 2 −γl / 2 γl / 2 −γl / 2 e −e e +e VEL TECH MULTI TECH TECH HIGHTECH VEL ( ) ( ( )( ) ) eγl / 2 − e−γl / 2 = Zo γl / 2 −γl / 2 e −e z1 = z2 = zo tan hγl/2 ∴ Zo tan hγl/2 Zo tan hγl/2 π . Z15c )(e γl / 2 2 − e−γl / 2 ) Z A = Zc = ZB = Zo tan hγl Z20c ( Z10c − Z15c ) 2 Zo Z2 sin hrl = = o = Zo sin hγl Zo / sin hγl Zo VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 92 VEL .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH z20C + z15c VEL ZA=Zc= Z20c − Z20c − Z20c ( Z10c − Z15c ) eγl + e−γl eγl + e−γl × Zo γl = Z o γl −γl e + e−γl e +e eγl + e−γl 2Z = Z o γl − γl o−γl −γl e −e e −e = 2 = Zo eγl + e−γl 2Z Zo γl − γl o−γl −γl e −e e −e = Zo eγl / 2 − e−γl / 2 ( ( eγl / 2 − e−γl / 2 ) Zo tan hγl Z 2oc. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL UNIT – III THE LINE AT RADIO FREQUENCIES PART – A 1. What is the nature and value of Zo for dissipation less line? For dissipation less line. Dissipation less there is used for transmission of power at high frequency in which losses are neglected completely. Zo is purely resistive and is given by L Z o = Ro = C 3. What are nodes and antinodes on a line? Nodes are points of zero voltage or current in a standing wave system and Antinodes are points of maximum voltage or current in a standing wave system. 4. 2. What is dissipation line? A line for which the effect of resistance ‘R’ is completely neglected is called dissipation less line. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 93 VEL . What is the value of ∝ and β for a dissipation less line? The value of attenuation constant ‘ ∝ ‘ for a dissipation less line is zero and the value of phase constant ‘ β ‘ for a dissipation less line is ω LC radian / m. What are the advantages of dissipation less line? The advantages of dissipation less line are • The line acts as a smooth line • No reflection takes place at receiving end • Standing waves are not produces. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 94 VEL .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 5. Draw the graph between standing wave ratio (s) and reflection co efficient (k) 6. Give the expression for skin depth. Skin depth or nominal depth of penetration is given by l S= meters ∏ f /m where ‘p’ is the resistivity of conductor in Ω /m ‘f’ is the frequency in HZ ‘m’ is absolute magnetic permeability of conductor in H/M 7. Calculate the reflection co efficient if VSWR of the line is 1. What are standing waves? When a transmission line is not terminated in its characteristic impedance. i. Give the relationship between VSWR and reflection coefficient for a transmission line The relationship between VSWR(S). Define standing wave ratio. Reflection coefficient is defined as ratio of reflected voltage or current to the incident voltage or current. The ratio of maximum and minimum magnitude of current or voltage on a line having standing waves is called standing wave ratio. Define reflection coefficient.2 9.5 =| K | +1.5 − 1. V I K = r (or ) K = − r Vi Ii The current ratio is negative because the reflected current suffer a 180° phase shift at VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 95 VEL . 11.5 =| K | 2.5 | K | 0.5. 1+ | K | 1− | K | 1+ | K | 1.eSWR = | Vmax | | I max | = | Vmin | | I miin | 10. It is denoted as ‘K’.5 ∴ | K |= 0. the traveling electromagnetic wave from generator at sending end is reflected completely or partially at the terminating end.5 | K | 0.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 8.5 = 1− | K | VSWR = 1+ | K |= 1. The combination of incident and reflected wave gives rise to standing waves of current and voltage with definite maxima and minima along the line. and reflection coefficient ’K’ is S= 1+ | K | 1− | K | 12. 15. The input impedance of short circuited line is 2π s Z sc = jRo tan λ where λ is wavelength s is length of the line Ro is characteristic impedance 17. The velocity of propagation for open wire dissipation less line is given by V = 3X108 m/sec 14.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL the receiving end while the reflected voltage does not. List out the features of power frequency line. Give the expression for input impedance of short circuited line. 16. Power transmission lines are operated at constant output voltage. The input impedance of dissipation less RF line is given by Z R + jRo tan β s Z in = Ro Ro + jZ R tan β s where Ro is characteristic impedance ZR is terminating impedance β is phase constant ‘s’ is length of the line. The input impedance of open circuited line is 2Π S Z oc = − jRo cot λ where λ is wave length s is length of the line VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 96 VEL . • • • Power transmission lines are electrically short in length. 13. Give the expression for input impedance of open circuited line. Give the velocity of propagation of open wire dissipation less line. with the length not λ exceeding 10 The power efficiency of power transmission line is very high as compared to other energy sources. Write the expression for input impedance of RF line. A 50 Ω line is terminated in load ZR = (90 + j60) Ω . 19. Define dissipation factor. A lossless line of 300 Ω characteristic impedance is terminated in a pure resistance of 200 Ω Find the value of SWR? Zo = 300 Ω VEL TECH . Name the device used for measuring standing wave. 21. Determine VSWR due to this load. What is the maximum resistive input impedance of a dissipation less line? The maximum resistive input impedance of a dissipation less line is Rmax = S Ro Where S is standing wave ratio and Ro is characteristic impedance. The derive used for measuring standing wave is directional coupler. ZR =200 Ω VEL TECH MULTI TECH TECH HIGHTECH 97 VEL .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Ro is characteristic impedance 18. 20. What is the minimum resistive input impedance of a dissipation less line? The minimum resistive input impedance of dissipation less line is R Rmin = s S where ‘S’ is standing wave ratio Ro is characteristic impedance 22. Dissipation factor is defined as the ratio of energy dissipated to energy stored in dielectric per cycle. S= 1+ | K | 1− | K | Z R − ZO 90 + j 60 − 50 = = Z R + ZO 90 + j 60 + 50 whereK = ∴s = 1+ 1− 23. 2 1. 4 terminated in its Ro value at both ends to prevent reflection.8 S = 1.VEL TECH S= 1+ | K | 1− | K | Z R − ZO 200 − 300 = Z R + ZO 200 + 300 100 =− 500 k = −0. 25. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 98 VEL . Sketch the standing waves on a line having open-or.short circuit termination. 26.2 0.2 = 1 − 0. Sketch the standing waves on a dissipation less line terminated in a load not equal to Ro. What is directional coupler? Directional coupler is a device which is used to measure standing waves. It consists of Coaxial transmission line having two small holes in the outer sheath 1 spaced by wavelength clamped over these holes is a small section of line.2 VEL TECH MULTI TECH TECH HIGHTECH VEL whereK = ∴S = 1 + 0.5 24. 31. The extreme left point on the real axis represents zero conductance while the extreme right point on the real axis represent infinite conductance. What do you mean by reflection loss? When there is mismatch b/w the line and load. the reflection takes place. It is used to couple a transmission line to a resistive load such as antenna. What are the uses of quarter wave line? The expression for input impedance of quarter wave line is given by Zs = RO 2 ZR This equation is similar to the equation for impedance matching using transformer. Hence the quarter wave line is considered as transformer to match impedance of ZR & ZS. the susceptance is inductive which is negative. 29. Because of this the energy delivered to the load by the line is less it composition with the power delivered to the load by a properly terminated line. Hence it is operated as one to one transformer. Explain how smith chart can be used as an admittance chart. The expression for the input impedance of the line is given as ZS = ZR. while’Xi’ axis becomes ‘bi’ axis. What is the practical value of SWR we can achieve by double stub matching? VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 99 VEL . If the smith chart is to be used for admittance. The main application of a half wave line is to connect a load to a source where both of them cannot be made adjacent. It is used as an impedance matching section. 30. Then above real axis. Thus the line represent repeats its terminating impedance. the ‘ri’ axis becomes ‘gi’ axis. State the use of half wave line.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 27. 28. This loss in power is called reflection loss. 1. This is called sub matching. Single stub matching 2. 38. 36. Give type of stub matching. Single stub matching 2. It is very useful device because d its simplicity. From mechanical stand point. 34. 37. 1. What do you mean by impedance circle diagram? VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 100 VEL . Double stub matching. What is stub matching? A section of transmission line is used as a matching by inserting then b/w load and source. 35.2 32.VEL TECH SWR = 1. Quarter wave transformer (impedance inverter) 2. What are the advantages of stub matching? 1. What are the advantage and disadvantages as quarter transformer? Advantages: 1. 2. Stub matching. 33. Double stub matching. adjustable susceptance are added in shunt with the line. It is so sensitive to change in freq. What is smith chart? VEL TECH MULTI TECH TECH HIGHTECH VEL Smith chart is an impedance (or) admittance chart which is used to calculate all the parameters of transmission line. It consists of two sets of circles. Disadvantages: 1. Name the impedance transformer that are used at higher frequencies? At higher frequency the impedance transformers consists as a section of transmission line in various arrangements as listed below. The length and characteristic impedance of the line remain unaltered. 1. Its behaviour can be easily calculated. 2. Advantages: 1. In circle diagram. Draw the family of constant S – Circle diagram. ‘S’ and ‘pl’ circles are not concentric making interpolation difficult. the locus of the i/p impedance. the resistance component * an impedance represented in rectangular form . Zin as electrical length pl is varied as a circle. 41. the resistive component ‘p’ and resistive component ’x’ and impedance are refines in circular form. This circle diagram is known as impedance circle diagram. In smith chart. 42. Disadvantages: 1. the movement in the clockwise corresponds to transverse from the load towards the generator and the movement in the anti clockwise direction corresponds to transverse from the generator towards the load. 2. What are the differences between circle diagram and smith chart? The basic difference between circle diagram and smith charts are: 1. What are the advantages and disadvantages * circle diagram. Only limited range impedance value can be contained in nc chart.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL If impedance are plotted in the form of R – X diagram it turns that for a loss less line terminated in some fixed impedance ZF. Explain the direction of movement towards generator or load in circle diagram. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 101 VEL . 39. 40. It is very useful in calculating line impedance and admittances. 2. In circle diagram. Determination of K in magnitude and direction 4. 45. 3. Any value of input impedance can be easily determined. Smith chart can also be used to determined load and impedance VEL TECH MULTI TECH TECH HIGHTECH 102 VEL VEL TECH . Determination of swr 5. Smith chart can be used as an admittance. Draw the family of constant x – circle in smith chart. Movement along the periphery of the chart. Give some applications of smith chart: 1. Normalizing impedance. 2.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 43. 4. 46. 2. 44. Draw the family of constant – R circles in smith chart. diagram Used for convey r + aj impedance into admittance. Plotting as an impedance 3. Give the properties or smith chart: The properties of smith chart are: 1. 47. The single stub matching system is useful only for fixed frequencies. Why short circuit stub is used in single stub matching? The short circuit is invariably used because. A loss less line has a characteristic impedance of 400 Ω . 48. 49. 2. It effective length may be varied by means of a shorting bar which normally takes the shapes of shorting plugs.0Ω Solution: K= Z O − Z R Z R − ZO Z O + Z R Z R + ZO ZR = 70 . It radiates ups power and 2.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 5. Determine the stands wave ratio with the following receiving and impedance Z1 = 70 + 0. Final adjustment of the sub has to the moved along the line shifts. What are the disadvantages of single stub marks? 1. 1. Zo = 400 k= 70 − 400 ⇒ −33 / 47 70 + 400 VEL TECH MULTI TECH TECH HIGHTECH 103 VEL VEL TECH . 50. Draw a diagram showing how a quarter wave transformer can be used for matching two lines. The input impedance and admittance of shot circuited line and open circuited line can be easily calculated. 31o K= 0. 55. Determine K of a line for with ZR=200u.8) ZR − Zo 200 − 692 −12 K= = = ZR + Zo 200 + 692 −12o 200 + ( 676 − j143.51 −9. Under this conditions. 1= λ tan −1 Z O Z 2Π λ 20 S= tan −1 2Π 2R 52. hence VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 104 . 53.4 −162.8 − j143. losses are absent.VEL TECH ∴| k |= 33 / 47 1+ | k | 1 + 33 / 47 s= = = 5.6 56. Give the formula to calculate the position and length of a short circuited stub. The position of the stub can be calculated using the formula. What is the need for stub matching in transmission lines? When line at high frequency is terminated into its characteristic impedance Ro. What is the nature of value of Zo for the dissipation less line? For the dissipation less line. What is dissipation less line? A line for which the effect of resistance R is completely neglected is called dissipation less line.91o K= 888.71 1− | k | 1 − 33 / 47 VEL TECH MULTI TECH TECH HIGHTECH VEL 51.8) L C ( ( ) ) 489. Zo is purely resistive of given by Zo = 1Zo = 54.55 −153. Zo=692 −12o o 200 − ( 676. then the line operates as smooth line. What is the range of values of standing wave ratio? The range of values of standing wave ratio is theoretically 1 to ∞. stub matching is used. So to provide impedance matching between line of its termination.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL maximum power is delivered with increased performance. 59. If using of line is 1.2. VSWR = 1+ 1K1 = 1. 60.21×10−7 log henrys / m 2∏ a a 12.51K1=1.5 then calculate its reflection co-efficient .5 1K1=0. Give the expression for L & C for coaxial line at high frequency L = 2 × 10−7 l n C= b henrys/m a 2∏∈ farads / m b ln a PART – B 1.51K1 2. If a line is not matched to its load then the energy delivered by the line of the load VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 105 VEL .5-1=. 57. Hence over open circuited stubs. Why are short circuited stubs preferred over open circuited stub? A high frequencies. Give the expression for L & C for open-wire line at high frequency? h0 d d l n henrys / m(or ) L = 9. But practically Ro of the line termination are not matching. 58. Write short notes on reflection losses on unmatched line. open circuited stubs radiated some energy which is not the case with short circuited stub. short circuited stubs are preferred.07 C= µµ f / m d ln a L= 61. What are the advantages of dissipation less line? i) ii) iii) The line acts as a smooth line No reflection takes place at the receiving and The standing waves are not produced.5-1.5 1− 1K1 1+1K1=1. (1) 2 Now this equation can be writes as Emax | =| Ei | + | Er |= I R Z R + RO (1+ | k | ............ In measurement of power and impedance on a Tx line we found that Emax = I R Z R + RO (1| k | .. One power Pi being transmitted in the incident wave and the other power Pr traveling back in the reflected wave.... Emax =| Ei | − | Er |= I R Z R + RO (1− | k | .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL is less than the energy delivered by the matched line to load.(4) Emin | Ei | − | Er | The total power transmitted along the line and delivered to the load is given as P= = | Emax | ....(2) 2 The minimum voltage is due to the difference of the incident and reflected waves and it is given as ... Due to this unmatched system reflected waves and standing waves are produced.. The ratio of the power ‘P’ delivered to the load to the power transmitted by the incident wave is P Pi − Pr | Ei |2 − | Er |2 E = = = 1− r 2 Pi Pi | Ei | Ei 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 106 VEL ...(5) Ro From the above expression we can recognize the transmitted power as the difference of two powers... | Emin | ( | Ei | + | Er |) (| Ei | − | Er |) = Ro Ro | Ei |2 − | Er |2 .. The voltage at a maximum voltage point is due to the in phase sum of the incident and reflected waves.....(3) 2 Hence the standing wave ratio is S= Emax | Ei | + | Er | = ... (6) ( S + 1) 2 Now the ratio power absorbed by the load to the power transmitted is plotted as a function of ‘S’ as shown in fig. =Π/4 λ 8 Z + jRO Z S = RO R → (3) RO + jZ R If the line is terminated in a pure resistance RR... then R + jRO Z S = RO R → (4) RO + jRR In equation (4) the numerator and denominator have identical magnitudes so equ ($) becomes Zs = Ro ……(5) β= VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 107 VEL ... 2 2. Eighth Wave Line: The input impedance of a line of length s = λ /8 is Z + jRO tan β s Z S = RO R → (1) RO + jZ R tan β s Z + jRO tan(Π / 4) Z S = RO R → (2) RO + jZ R tan(Π / 4 Whereβ = 2π / λ 2Π λ .....VEL TECH VEL TECH MULTI TECH TECH HIGHTECH = 1− | K |2 VEL S − 1 = 1− S + 1 4S = .. Explain Eighth wave line and half wave line... lowering the insulation resistance of the line.. since their effect would then be cumulative.(2) Thus a half wave length of line may be considered as a one to one transformer. Half – Wave Line: When a length of line having s = λ / 2 is used.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Thus an eighth – wave line can be used to transform any resistance to an impedance with a magnitude equal to Ro of the line. *It has greater utility in connecting a load to a source in cases where the load and source cannot be made adjacent.. Explain the principle and application of Quarter wave transformer for impedance matching (or) what are the features of a Quarter wave transformer? Quarter wave line Impedance Matching: The expression for the input impedance of dissipation less line is given as Z + jRO tan β s Z S = RO SR → (1) RO + jZ R tan β s Equ (1) is rearranged as ZR tan β s + jRO Z S = RO → (2) RO + jZ R tan β s For a Quarter wave line . (or) to obtain a magnitude match between a resistance of any value and a source of R0 internal resistance.. As a result insulators on a high frequency line should not be spaced at half wave intervals... 3.. s = λ / 4 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 108 VEL .... *A group of capacitors may be placed in parallel by connecting them with sections of line n half waves in length. the input impedance is Z + jRO tan Π Z S = RO R → (1) Since tan Π = 0 RO + jZ R tan Π Z S = Z R . here it transforms a low impedance into a high impedance and vice versa... An application of the Quarter wave matching section is to couple a transmission line to a resistive load such as a antenna........... the input impedance of the line is equal to the square of Ro of the line divided by the load impedance..(5) VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 109 VEL . A Quarter wave line acts as a transformer to match a load of ZR ohms. Such a match can be obtained if the characteristic impedance Ro of the matching Quarter wave section of line is chosen as.(3) ZR Since tan Π /2 = infinity i..... R0 = Z S Z R .VEL TECH ∴ equ (2) becomes ZR tan Π / 2 + jRO Z S = RO RO + jZ R tan Π / 2 VEL TECH MULTI TECH TECH HIGHTECH VEL R 20 ZS = ..(4) A Quarter wave line may be considered as an impedance inverter.e...... impedance Ro of the Tx line The characteristic impedance Ro of the matching section should be Ro = RA RO .... The Quarter wave matching section is designed to have a characteristic impedance Ro chosen that the antenna resistance RA is transformed to a value equal to the characteristic..... It should then be connected between points corresponding to Imax or Emin at which places the transmission line has resistive impedances given by Ro /s or s/Ro.5 d= .563 Y1 = 3. A Quarter wave transformer may also be used if the load is not pure resistance.587 7.5Meters 40 The location of the stub nearest to the load is given by cos −1 | K | λ cos−1 0.587 The first voltage minimum occurs at y2 = λ / 2. = . 4. the matching transformer characteristic impedance should be. What type of single stub will be required to provide impedance when placed nearest to the load? Calculate its length and find its location. A lossless line having Ro = 300 ohms is terminated by a load resistance of 78 ohm.187 meters The susceptance of the line at the location of the stub will be positive. The frequency of operations is 40MHz. Π 4 Π 4 d = 0.587 Z R + ZO Z R + R 78 + 300 ∴| k |= 0.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The transformers also a single frequency or narrow band device. These lines are sometimes referred as copper insulator. For step down in impedances from the line value of Ro.563 meters Hence the stub is located at a distance of Y1 = 3. This application is illustrated is as shown below. where 300 λ= = 7. Solution : The reflection coefficient is given by K= Z R − ZO Z R − RO 78 − 300 = = k = −0. Hence a VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 110 VEL .75 – 0. The bandwidth may be increased by using two or more Quarter wave sections in series each accomplishing part of the total transformation.RO / S = RO / S Another application of the short circuited Quarter wave line is as an insulator to support an open wire line or the center conductor of a coaxial line.] RO = RO . 00 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 111 VEL .1832 ×100 Vload = 118. A 150 ohm transmission line is terminated by a load of 200 – j300. a.72meter 5. Determine the location and length of a short circuited matching stub.Vmax = 1. Solution: Given Zo = 150 ohm ZR = ZL = 200 – j 300 ohm The single stub is constructed as follows Zr = Z R 200 − 300 = ZO 150 a) Calculate Z r = 1.322OHMS . Solution a ) Z O = Z R .587) 7. b. The length of this short circuited stub is given by 1− | K |2 λ L= tan −1 2Π 2| K | L= 1 − (0.5 tan −1 tan (0.587 2Π L = 0.1832 c)Vmin = 100v.5 −1 7. What is the ‘S’ for the transformer? c. Find the load voltage. If the input voltage to the line is 100v.689) = 2Π 2 × 0.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL short circuited stub will be needed to provide the impedance match. Find ZO of the matching transformer.100 b) S = Z R / ZO = 140 /100 Z O = 118. 6. S = 1.RO = 14.32volts d) The location of Vmax on the N4 line is located at the load. A load of ZR = 140 ohms is to be connected to a line RO = 100 ohms by a quarter wave matching transformer.33 − j 2. 1.08λ = 12 7.126 λ f) Final step is the determination of the length of the stub.183 λ .23 +j 0.33 – 0.33 – j 200) d) Find the intersection of SWR circle and G = 1 circle mark it as point ‘D’ e) Extend the radial line from the centre to ‘D’ continue the line to the perimeter ‘E’ at 0..33 λ .8.25 ) λ = 0.33 λ i. Explain in details the constant ‘R’ circles and constant ’X’ circles in a smith chart.e. The susceptance at point ‘D’ is 0 = j 1.0057 λ The point ‘B’ is the normalized load admittance (0.0. So the susceptance contributed by the stub must be j 1.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL b) Mark this point as ‘A’ in the smith chart. Point ‘F’ corresponds to this for a short circuited line at 0. A modified form of a circle diagram for the dissipation less line is the smith chart developed by P. I1= 0.H smith.25 λ to 0.8. ‘R’ circles and 2. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 112 VEL . This chart consists of two circles.057 λ . The length of the matching stub is from 0.9 d)Extend the radial line from point ‘A’ through the center to the other side of the SWR circle and mark the point as ‘B’ and to the outer perimeter as point ‘C’ at 0. ‘X’ circles question) (For R circles & X circles diagram refer part A This chart is also know as circular chart. (0.36) equal to the reciprocal of the normalized load impedance at point A (1.183 λ The distance from load to the junction of the transmission line and the stub is I1 = 0. c) Draw the SWR circle through ‘A’ the ‘S’ value is 4. Equation (1) will yield a family of circle called R – circles and equation (2) will yield a family of circles called x – circles..... 1 − K r 2 − K x2 + 2 jK R + jx = (1 − K r ) 2 + K x2 Equating real and imaginary parts R= X= 2 1 − K r2 − K x ....T ...VEL TECH W .(2) (1 − K r ) 2 + K x2 When equation (1) and (2) are solved..K = K= Z R − Z O Z R / ZO − 1 = Z R − Z O Z R / ZO + 1 VEL TECH MULTI TECH TECH HIGHTECH VEL Zr −1 WhereZ r = Z R / Zo Zr + 1 Zr – Normalized terminating impedance 1+ K Hence Zr = 1− K Since ‘Zr” and ‘K’ are complete quantities we have R + jx = 1 + K r + jK x 1 − ( K r + jK x ) Rationalizing right hand side we get.....(1) (1 − K r ) 2 + K x2 2K x .....K .. we get two sets of circles.... The constant ‘R’ circles: Consider equ (1) and cross multiply R= 2 1 − K r2 − K x (1 − K r ) 2 + K x2 We get R (1 + K r2 − 2 K r + K x ) = 1 − Kr2 − Kx2 R + K r2 R − 2 RKr + RK x2 − 1 + Kr2 + Kr2 = 0 K r2 ( R + 1) + Kr2 ( R + 1) − 2 Kr R = 1 − R VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 113 VEL ..... ..VEL TECH 2 K r .(4) Equation (4) represents another family of circles called constant – X circles with centre. Constant X – circles: Consider the equation (2) and cross multiplying.R R2 (1 − R ) R2 + = + R + 1 (1 + R ) 2 (1 + R) (1 + R )2 2 2 K r2 + K x2 − R 1− R R K x2 + K r − + 1 + R + (1 + R) 2 1+ R R 1 K x2 + K r − − .. ‘X’ being the reactance can be positive or negative whenever ‘X’ is positive the circle lies above the horizontal line. ( K r − 1) 2 + ( K x − 1/ X )2 = (1/ X ... these circles are called constant –R circles having radius 1/1 + R and centre (R / 1 + R... These circles have their centers on the positive Kr axis and are contained in the region ‘0’ to ‘1’ as shown in the figure below.. O ).. This circle forms the periphery of the smith chart. When X=0 . we get 2K x (1 − K r2 ) + K 2 = X 2K x (1 − K r2 ) + K x2 = =0 X Adding (1/x2) on both sides in order to make the ‘Kx” terms a perfect square we get. R = o corresponds to a circle with center (0. (1....... the circle degenerates into a straight line because straight line is VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 114 VEL ..R ( R = 1) K r2 + K x2 − = 1− R ( R + 1) 2 K r .. (1 + R ) 2 2 K r ...0( on the plane K – plane. 1/x ) and radius (1/x on the k – plane as shown in figure below... All constant ‘R’ circles touch the point ( 1.... On the other hand when ‘X’ is negative the circle lies below the real axis Kx = 0..(3) 1 + R 1 + R 2 2 This equation represents the family of circles on the reflection coefficient plane..R 1 − R K r2 + K x2 − = R +1 1+ R Adding VEL TECH MULTI TECH TECH HIGHTECH VEL R2 on both sides to make it a perfect square we have ... 0 ) Including that R = ∞ which is the same as the point itself... ..VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL circle whose radius is infinity and for X = 0.. the radius 1/x will be infinity.... When we along the line from the load towards the source (generator). We know that yR is different from yo standing waves are set up..... All the circles touch the point (1. Consider a transmission line having a characteristic admittance y o terminated in a pure conductance yR as shown in the figure.... Since we connect stub in parallel with the main line it is easier to deal with the admittance as they can be added up. then up to that point matching has been achieved... Derive an expression for the position of attachment and length of short circuited stub will remove the standing wave on a large potion of a transmission line.(1) Z O + jZ R tan β s Convert impedance to admittance Y + jYO tan β s Yin = YO R YO + jYR tan β s Y Y + j tan β s Yin = O r . the input admittance will be varying for a maximum conductance through a parallel combination of conductance and inductance a minimum conductance and so on this cycle repeats for every λ / 2. The input impedance of a transmission line at any point is given as [ Z R + jZO tan β s ] Z in = ZO Z R + jZO tan β s . there will be a point at which the real part of the admittance is equal to the characteristic admittance. 0 ) 8.. obtained by using an appropriate length of a short circuited or open circuited line called stub is added in shunt at this point so as to obtain and 0 resonance with the susceptance already existing.. If a suitable susceptance. When the line is traversed from the point of maximum (or minimum) conductance to that of minimum (or maximum) conductance..(2) YO 1 + jYR tan β s YR Where Yr = (Normalized load admittance) Yo VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 115 VEL . ..s = tan −1 Z R / Z o λ λ ∴s = ........... Yr + (1 + tan 2 β s ) =1 1 + Yr tan 2 β s tan 2 β s (Yr − Yr2 ) = 1 − Yr Yr tan 2 β s = 1 tan 2 β s = 1/ Yr tan β s = Yo .......tan −1 Z R / Zo ..VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Yin = YIN (Normalized input admittance ………. the stub has too be located at a point where the real part is equal to unity.....(5) 2Π VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 116 VEL ..(3) Y0 Rationalizing equ (2) Yin = = Yr + j tan β s (1 − jYr tan β s) 1 + jYr tan β s (1 − jYr tan β s) Yr (1 + tan 2 β s ) + j (1 − Yr2 ) tan β s (1 + Yr2 tan 2 β s ) For on reflection Yin = 1 Thus..(4) YR This equation gives the location of the stub ‘S’ and can further simplified as β s = tan −1 Yo / YR 2Π ......... . 9.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL bs (1 − Yr 2 ) tan β s = ..(6) Yo (1 − Yr2 ) tan β s Substitute equation (4) in (6) 2 2 bs (1 − Yr / Yo ) Yo / YR = .. VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 117 ..Yo / YR = (1 − Yr2 / Yo2 ) Yo / YR = (1 − YR / Yo ) Yo / YR 1 + YR / Yo bs = (Yo − YR ) Yo Yo YR Advantages of stub matching: 1.. For final adjustment the stub has to be moved along the line slightly. 2.(7) Yo 1 + Yr2 / Yo2 ..... Single stub matching is used only for a fixed frequency because as the frequency changes... Its effective length may be varied means of a shorting the bars....... Determine the maximum value of conductance that can be matched by a double stub tuner with one stub at the load and the other stub at 3/8 back from the load. Disadvantages: 1.. 2..... the location of the stub will too be changed... This is possibly only in open wire line and on co – axial single stub matching may become inaccurate in practice.... It radiates less power. the new admittance value will be. bA = 1 + Yr2 tan 2 β S1 1 + Yr2 tan2 β S1 When a stub having a susceptance value is altered b1 is added at this point. The lengths of these stubs are adjustable but the positions are fixed. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 118 VEL . Let the first stub whose length is It 1 be located at the point ‘A’ at the distance of 'S1’ from load end then the normalized input admittance at that point will be Y Y + j tan β S1 YA = A = r YO 1 + jYr tan β S1 YA = = Yr + j tan β S1 1 − jyr tan β S1 × 1 + jyr tan β S1 1 − jyr s tan β S1 Yr (1 + tan 2 β S1 ) + j (1 − yr2 ) tan β S1 1 + yr2 tan 2 β S1 y A = s A + jbA Where.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL In order to overcome the two disadvantages of the single stub matching. y A = sA + jbA Since only the susceptance value is altered by the addition of the stub the conductance part remains unchanged The stub length at ‘B’ is adjusted such that the zero value yA is equal to ‘I’ How the location of the stub can be encountered in practice. Two short circuit stubs are connected. • The distance s1 can never be more than or equal to λ / 2. Yr Sec 2 β S1 (1 − Yr2 ) tan β S1 SA = . In common practice the distance s1 is of the order of 0. In order to avoid this loss. A single Stub matching is constructing by following the procedure as below. Step 1 : Calculate the normalized impedance of admittance.0 Ro S = 9 /16λ Load impedance may be calculated using the formula.1λ to 0. Solution The given values are. Give the method of constructing single stub matching using Smith Chart.25 and the value at ‘F’ point. Circle and mark this point as ‘B’ Step 5 : Extend the ‘AB’ radial line towards outer perimeter and mark it as ‘C’ Step 6 : Find the intersection of SWR circle and G = 1 circle mark it as ‘D’ Step 7 : Extend the radial line from the center to ‘D’ and continue the line to the perimeter and mark it as ‘E’ Step 8 : The difference between the points ‘C’ and ‘E’ gives the distance of the stub to load. Step 9 : Find the susceptance at the point ‘D’ to cancel this susceptance mark it in opposite direction and mark it as ‘F’. Step 2 : Mark this point as ‘A’ in the smith chart. Since matching is obtained between the point ‘B’ and generator. This is the procedure to construct a single stub matching using smith chart. This gives the length of the stub.5 + j 0. Find the load impedance normalized to R0 and also the standing wave ratio. Step 3 : Draw the SWR circle through ‘A’ Step 4 : Extend the radial line from point ‘A’ through the centre to the other side of the SWR . sometimes the first stub is located at the load itself. A 9/16λ long lossless line has Zs/Ro = 1.15λ 10. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 119 VEL . Zo = 1.9.VEL TECH • • VEL TECH MULTI TECH TECH HIGHTECH VEL The distance chosen will be either λ / v or 3 λ / 8.5 + j0. 11. Where Zs is the impedance and Ro is the characteristic resistance. we have reflection loss occurring to the right of ‘B’ due to mismatch. Step 10 : Find the difference between the values of 0. 4142) RO (0.9) = RO −0.tan = Z R / RO + j tan(9π / 8) RO 8 jZ (1.8198 + j1.4142 − (1.6272 + j 0.3757 VEL TECH MULTI TECH TECH HIGHTECH 120 VEL .3757 = 1− | K | 1 − 0.4142 − (1.4142) = Z R / RO + j (0.5 + j 0.544 1.6213 −0.5 + j 0. S = VEL TECH 1+ | K | 1 + 0.9)1 + 1 + R .4142) Z R j 0.5857 + 1 −0.9) + Z R / RO (0.6272 + j 0.6272 + j 0.5867 RO K= k= Reflection coefficient Z R − RO Z R / RO − 1 = Z R + RO Z R / RO + 1 0.6213 − 1) = j (0. R λ 16 9π Z R / RO + j tan 8 1.8198 + j1.5 + j 0.8198 + j1.37278 − j 0.5 − j 0.6213 = j 0.37569 + j 0.58 | k |= 0.VEL TECH Z O Z R + jRO tan βS = RO RO + jZR tan β s 2π 9 Z R / RO + j tan .4858 = −0.(0.9) −1.9 = Z 9π 1 + j R tan R 8 (1.3757 Standing wave ratio.8198 + j1.5 + j 0. ZS λ 16 = Z RO 2π 9 1 + j R tan .5 + j 0.5867 − 1 = 0.180 + j1.5 + j 0.5867 = 0.6213 ZR = 0.9)1 + VEL TECH MULTI TECH TECH HIGHTECH VEL jZ R 9π . 241λ The impedance of an open circuited line is given by 2π s Z oc = jRo cot λ Substitute the values.6 = 200 λ cot ( π − 2π s / λ ) = 17.S Substitute the values 2π s J 2π × 70 ×106 × 8 ×10 −6 = j 200 tan λ 2π s 15π × 70 tan = 200 λ 2π s = tan −1 17.75 × π = radians 180 S = 0.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL S = 2.6 s π − 2π s / λ = cos −1 17. Z ac = jR0 tan(2π / λ ). What length of the line will be required to obtain at the input of an inductance of 8 micro hertz at frequency of 70 mHz with far end short circuited? Repeat the calculation of open circuited received end. If a lossless line has Ro=200 ohms. Solution The input impedance of a short circuit 4d line is a pure reactance is given by.20 12. 2π s J 2π × 70 ×106 × 80 ×106 = − j 200 cot λ 2π −2π × 560 cot = −17.6 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 121 VEL .6 λ 86. . standing wave Ratio If the voltage magnitudes are measured along the length of a line terminated in a load other than Ro the plotted values will appear as Figure: Resistive load of value not equal to RO In the case of either OC(or)SC lines current magnitudes will be same except there will be a λ / ∆ shift of maxima and minima. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 122 VEL ...981.λ / 2 λ 180 ∴ s = 0... current mode occurs at a distance of o. For a open circuited like the voltage modes occur at a distance of λ / 4. The ratio of maximum to minimum magnitude of voltage or current in a line having standing waves is called Standing Wave Ratio. λ...e) are points of zero voltage (or)zero currents..5λ / 4. this modal pts get shifted by a distance of λ / 4 and voltage modes occur 0.4905λ VEL TECH MULTI TECH TECH HIGHTECH VEL 13) Explain the terms standing waves . Antinodes – maxima (i. λ ....25 × π /180 = −2 s 3. Nodes.3λ / 4... For a short circuited line.e) are points of maxima (voltage or current).5λ / 4.3λ / 4.. Nodes – Minimum (i.25 × π radians 180 π (1 − 2s / λ ) = 3. λ / 2. and current modes occurs at λ / 4.. λ / 2..VEL TECH 3. λ / 2. s = 0.25 = 1− . (or) Derive the I/Pimpedance of an dissipationless line. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 123 VEL . /Emin/=/Ei/-/Er/________(2) From S = | E max | Ei | + | Er | |= E min | Ei | − | Er | ÷ by Ei on both sides | Ei | Ei | + | Er | Ei | 1+ | k | = | Ei | Ei | − | Er | Ei | 1− | k | 1+ | k | S= (∴ k = Er | Ei ) _____(4) 1− | k | 1− | k | ( s ) = 1+ | k | S= 1k1 = S −1 S +1 = 1k1= E ma −1 E min E max + 1 (or ) E min Im ax −1 Im in Im ax +1 Im in ⇒ /K/= E max Im ax − Im in or E min Im ax + Im in 15) Draw the phaser diagram for the i/pimpedance on the line and concenent on it . / E max/ =| Ei | + | Er | _______(1) The voltage minimum occurs at pts. From the SWR it is clear that the points of voltage maxima occur at points where the incident and reflected waves are in phase with each other.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Scors = E max Im ax = E min Im in 14) Derive the interrelation between reflection co-efficient and standing wave Ratio(SWR). Where with each other. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 124 VEL .VEL TECH 2π s 2π s + jI R Ro sin E λ λ Zs = s = 2π s ER 2π s Is I R cos +j sin λ Ro λ ER cos ER cos β s + Zs = IR cos β + IR Ro sin β s ER ER j sin β s Ro I R j VEL TECH MULTI TECH TECH HIGHTECH VEL divide by c os β s 1+ ER Zs = I R 1 + divide by I R Ro janβ s ER ER j janβ s Ro IR j Ro tanβ s 1+ j ZR Zs = ZR ZR 1 + j janβ s Ro Z + jRo tan β s Z s = Ro R Ro + jzR tan β s __________(2) Another form of I/P Impedance Equ. ∴ i / p impedance is min if. we get a point of minimum impedance.e. s = 2β 1 + 1k1 zs (max)τ Zo = szo 1 − 1k1 zs (max) = szo If we travel a distance of λ / 4 from pt. where impedance is maximum. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 125 VEL .VEL TECH ER ( Z R + Z o ) β s (ej + ke = j β s) Es 2 zR zs = = Is zR + Zo β s I (ej − ke − j β s ) 2 Ro = R0 ej β s = / ej β s / tan −1 ej β s + ke − j β s ej β s − ke − j β s COS β s + j sin β s VEL TECH MULTI TECH TECH HIGHTECH VEL = COS 2 β s + sin 2 | β sτ | = 1 sin β s = tan −1 tan β sτβ cos β s φ angle of reflection coefficient 1 β s + 1k1 φ − β s ________(3) = Ro 1 β s − 1k1 φβ s 1 + 1k1 φ − 2 β s Ro 1 − 1k1 φ − 2 β s I/P Impedance is maximum at a distance of φ = 2 β s ________(1) φ i. so that jRo tan β s Ro ZSC = Ro ZSC = JRo tan β s As Z is purely reactive or imaginary then Let Z s = jλ s 2π s ∴ jxs = jRo tan λ xs 2π s = tan Ro λ VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 126 VEL .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL φ 2π + λ / 4 (∴λ = ) 2β β φ π s= + ) 2β β φ +π ∴s = 2β 1 + 1k1 −π 1 − 1k1 ∴ Z min = Z o = Zo 1 + 1k1 1 − 1k1 −π S= ∴ Zmin = zo s 16) Derive the I/P Impedance expression for a lossless line terminated one a)short circuit b)open circuit a)Short circuit I/P impedance of a dissipation less live is Z + jRo tan β s Z s = Ro R Ro + jZR tan β s for a short circuit live ZR = 0. =∞ 4 Zo λ X SJ . SJ Xs =∂ Zo λ Xs . =∞ 4 Zo X S J λ. s = ∂ Zo For an open-circuited live.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Xs Z x is called as normalized impedance. Z R = ∞ Ro 1 + j Z tan β S R Z oc = R∂ Ro + j tan β S ZR VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 127 VEL . The variation of sc = with length of Ro Ro Ro line S may be as for In this SJ . s = o 2 Zo 3λ X s SJ . .25. ∞. A load of admittance The normalized load admittance is given by YR = 1. 1/ R = 1. In a practical live there will 1o be a small resistance component of impedance indicating source power loss. shoot ckt. o 4 2 4 Zo Sc and Oc impedance is purely reactance value. = ∞.3 . λ λ λ Xs . 17. and P zero or infinite impedance are never achieved the actual values treading to minima and maxima.25+ j. However. S = 0.25 Go VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 128 VEL . the curves are for the ideal dissipatedness live.25+ j0. Find the length of location of single Go stub funec short circuited. For the first quarter wavelength.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 1 Z oc = Ro j tan β S −j 2π S Z oc = Ro = − jRo cot X tan β S Xs 2π S = − j cot Ro λ Similarly for open circuit. o. line acts as inductance and next wavelength it acts as capacitance. 25) = 1+ ( 1.0587λ 4π Length of the stub.25+ j.25 -.466o Calculating value of cos-1(1K1) Cos-1(1k1)=cos-1 (0.25 1/ R o Ro R o 2R − R o 2R = = Ro 2R 2R + R o 1− 2R 1− 1.34o = 0.1561-2.414 VEL TECH MULTI TECH TECH HIGHTECH VEL Calculation for length of location of stub:Case(1) S1 = φ + π − cos−1 ( 1k1) 2β φ + π − cos−1 ( 1k1) = 2π 2 λ -2.λI − 0.34o = 0.25+ ) j.26386.VEL TECH 1/ 2R = 1.414 π = .466+ -1.( 1.25+ j.3535-135o = 2.1561)=1.25-j.25 = -2.1561-141. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 129 VEL .25+ j0.25 0. 142 + 1. L= 1− 1k12 λ tan−1 2π −21k1 λ = tan−1 ( -1.013231) 2π L = 0.3535) 2 λ = tan−1 −2( 0.1469λ Case (2) S1 = = φ + π − cos−1 ( 1k1) .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 1− 1k12 λ tan−1 2π 21k1 VEL L= 2 λ −1 1− ( .466 + 3.353λ 18.3535) 2π = 0.3535) = tan 2( 0.3231) 2π 1− ( 0. Find the position of length of short circulated stub of same VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 130 VEL .414 .λ 4π −2.1662λ length of the stub. A loss less RF line has Zo of 600 Ω of is connected to a resistive load of 75Ω .3535) 2π λ = π − tan−1 ( 1.λ 4π S1 = 0. f.8918M The length of the stub is given by L= 1− 1k12 λ tan−1 2π 21k1 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 131 VEL .7777) π = .λ 4π π+ -cos-1 ( 0.2 4π = 0.777 75+ 600 675 K = 0.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL construction as line which would enable the main length of a line to be correctly terminated at 150MHz. K= K= ZR − Zo ZR − R o = Z R + Zo Z R + R o 75− 600 −525 = = −0.λ =c C 3× 102 λ= = = 2m f 150× 106 The reflection co-efficient is given by. f=150MHz Ro=600Ω ZR=75Ω Finding λ .7777 180o Case(1) S1 = φ + π − cos−1 ( 1k1) 2β 2π λ β= φ + π − cos−1 ( 1k1) S1 = .7777 πc = 0. Given . The stub must be located at a distance 0. Zin = 2s = 2 Ro 2R The source impedance Zs = 500 Ω Load impedance =2Ω = 200Ω VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 132 VEL . For a quarter wave transformer. Hence the stub location nearest to the load is calculated in case1.8918m from the load of the length of the stub required is 0.( 0.VEL TECH 1− ( 0.7777) π = . 19. the input impedance is given by.7777) 2 2 L= tan−1 2( 0. Design a quarter wave transformer to match a load of 200Ω to a source resistance of 500Ω .1222m.λ 4× π = 1.7777) 2 1− 1k12 λ λ −1 = tan−1 L= tan −2( 0.384) 2π π = 0.8777m Selecting a point located nearest to the load.4041) = ( π − 0.7777) 2π = 0.7777) 2π −21k1 2π λ 1 = tan−1 ( −0. Operating frequency is 200 MHz.1222m Case (2) S1 = φ + π + cos−1 ( 1k1) 2β VEL TECH MULTI TECH TECH HIGHTECH VEL π+ + cos-1 ( 0.108m The length of the stub is given by 1. Determine the reflection co-efficient at load. A loss less transmission line with Zo = 75Ω of electrical length l=0.5 = = 0.VEL TECH 2 Ro 200 VEL TECH MULTI TECH TECH HIGHTECH VEL 500 = 2 R o = 500× 200 2 R o = 100000 R o = 316.345m.375m 4 4 C-s=λ /4=.λ =c λ= C 3× 108 = = 1. input impedance of the line.3λ is terminated with load impedance of 2Ω =(40+j20)Ω .22Ω f = 200MHz wave length f.5m f 200× 106 ∴ The length of quarter wave line is given by S= λ 1. SWR of line. Solution: Given Zo=Ro=75Ω Zr=(40+J20)Ω Reflection co-efficient is given by VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 133 VEL . 20. 0548 Input impedance of the line is given by 2πs ZR + jR o tan λ Zin = Ro R o + j2R tan ( 2πs/ λ ) 2π× 0.3λ 75+ j( 40 + j20) tan λ =75 40 + j20 + j( −230.82) 75+ ( − j123.VEL TECH K= ZR − R o ZR + R o j20 ( 40+ ) − 75 ( 40+ j20) + 75 VEL TECH MULTI TECH TECH HIGHTECH VEL = -35+ j20 = 115+ j20 40.31129.7219.3453 = 2.345319.55) 40 − j210.74o = 116.82 = 75 136.1) + ( 61.3λ ( 40 + j20) + j75tan λ =75 2π× 0.1 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 134 VEL .86o = 0.3453 = 1− 0.55− j123.88o Standing wave ratio is S= 1+ 1k1 1− 1k1 1+ 0. 998× 108 m/ sec −6 −12 LC 2.2095/ m ∴ α= 0 β= 0.5 × 10-6 =15. A line with zero dissipation has R= 0.95Ω VEL TECH MULTI TECH TECH HIGHTECH VEL ( ) 21.ii)α iii) β iv) ν v) λ Given R=0. we can neglect R.5× 10 × 4.45 PF/m F=10MHz At f=10MHz. i) Characteristic impedance Zo = R o = L 2.167 −37.2093 rad/m iii) velocity of propagation 1 1 V= = = 2. WL=2π fL=2× π fL=2× π × 10× 10-6 × 2.5µ H/m d C=4.45× 10−12 UNIT – IV VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 135 VEL .03 = 75 1.5× 10−64.53Ω C 4.006Ω /m L=2.VEL TECH 214.7 − j52.25o = 75 o 183.84 −42.13m β 0.5× 10−6 = = 749.2095 ( ) 2.58 −79. If the line is operated at 10 MHz. L=2.006Ω /m.5 × 10-6 H/m C= 4.45PF/m.708Ω ∴ WL > > R at 10 MHz So according to standard assumption for the dissipation less line.45× 10−12 ii) Propagation constant ν = α + jβ = 0 + jw LC γ = α + jβ = 0 + j 2π× 10× 10−6 γ = α + jβ = 0 + j0.222o = 69. find i) Ro.45× 10 iv) wave length is given by 2π 2π γ= = = 29. [OR] The transverse electromagnetic (TEM) waves are waves in which both electric and magnetic fields are transverse entirely but have no components of E z and Hz. Mention the characteristic of TEM waves. is called the cut-off frequency of the wave guide. What is cut-off frequency? The frequency (fc) at which the wave motion ceases. The guide wavelength λc = 2π mπ ω µε − a 2 2 4. What is TEM wave or principal wave? TEM wave is a special type of TM wave in which an electric field E along the direction of propagation is also zero. 3. It is a special type of TM wave.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL PART – A 1. Examples: Parallel wire and transmission lines. Attenuation factor α = power lost /unit length 2 × power transmitted 7. The electromagnetic waves that are guided along or over conducting or dielectric surface are called guided waves. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 136 VEL . 2. Define attenuation factor. 5. What are guided waves? Give examples. Its cut-off frequency is zero. 6. It does not have either Ez or Hz component. Give the expression for guide wavelength when the wave transmitted in between two parallel plates. It velocity is independent of frequency. it is referred to as principal wave. Distinguish TE and TM waves. Derive the field components of the wave propagating between parallel plates. Electromagnetic waves along ordinary parallel wire 2. Give some examples of guided waves. It has z component of electric field Ez (direction of propagation) It has no z component of electric field Ez It has no component of magnetic field Hz. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 137 VEL . Define wave impedance. Wave in wave guides 3. (Ez = 0) (Hz= 0) 8. 1. Wave impedance is defined as the ratio of electric to magnetic field strength + Z xy = Ex Hy in the positive direction − Z xy = − Ex in the negative direction Hy 9. What are the characteristics of TEM waves? TEM wave is a special type of transverse magnetic wave in which the electric field E along the direction of propagation is also zero. Wave guided along the earth surface from a radio parameter to the receives. Consider an electromagnetic wave propagating between a pair of parallel perfectly conducting planes of infinite in the y and z directions as shown in Fig 2. It has z component of magnetic field Hz. 10. What do you mean by cutoff – frequency? Cut – off frequency can be defined as the frequency at which the propagation constant changes form being real to imaginary.1. transverse. 11.VEL TECH TE strength VEL TECH MULTI TECH TECH HIGHTECH TM strength VEL Electric field E is entire Magnetic field H is entirely transverse. fc = m 2a M ∑ PART – B 1. Maxwell’s equations for a non-conducting rectangular region and given as ∇ × H = − jωε E ∇ × E = − jωµ H ax ∂ ∇× H = ∂x Hx ay ∂ ∂y Hy az ∂ ∂z Hz ∂H x ∂H z − + ay ∂x ∂z ∂H y ∂H x + ax ∂x − ∂y ∂H z ∂H z = ax − ∂z ∂x = jωε a x Ex + a y E y + a z Ez Equating x .y and z components on both sides.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Fig Parallel conducting guides Maxwell’s equations will be solved to determine the electromagnetic field configurations in the rectangular region. ∂H z ∂H y − = jωε Ex ∂y ∂z ∂H x ∂H z − = jωε E y ∂z ∂x ∂H y ∂H x − = jωε Ez ∂z ∂y VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 138 VEL . (3) It is assumed that the propagation is in the z direction an the variation of field components in this z direction may be expressed in the form e-yz. Where VEL TECH γ is propagation constant γ =α +jβ VEL TECH MULTI TECH TECH HIGHTECH 139 VEL .. y and z components on both sides ∂Ez ∂E y − = jωµ H x ∂y ∂z ∂Ex ∂Ez − = jωµ H y ∂z ∂x ∂E y ∂Ex − = jωε H z ∂X ∂y The wave equation is given by ∇2 E = γ 2 E ∇2 H = γ 2 H Where γ 2 =(σ +jωε ) (jωµ ) For a non-conducting medium.VEL TECH ax ∂ Similarly. ∇ × H = ∂x Ex ∂e ∂E = ax z − Y ∂z ∂y ay ∂ ∂y Ey az ∂ ∂z Ez VEL TECH MULTI TECH TECH HIGHTECH VEL ∂Ex ∂Ez − + ay ∂x ∂z ∂E y ∂Ex + ax ∂x − ∂y = jωµ a x H x + a y H y + a z H z Equating x.... it becomes ∇ 2 E = −ω 2 µε E ∇ 2 H = −ω 2 µε H ∂ 2 E ∂2 E ∂2 E + 2 + 2 = −ω 2 µε E 2 ∂x ∂y ∂z 2 2 2 ∂ H ∂ H ∂ H 2 + 2 + 2 = −ω µε H ∂x 2 ∂y ∂z . the fields Hx.. β =0...e.. Ex and E To solve Hx’ VEL TECH y can be found out. VEL TECH MULTI TECH TECH HIGHTECH 140 VEL . γ H y = − jωε Ex ∂H z −γ H x − = jωε Ey ..(4) ∂x ∂H y = jωε Ez ∂x γ E y = − jωµ H x ∂Ez −γ Ex − = − jωε H y . (2) and (3).VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL If α =0..e... Let 0 = −γ H y e − yz = −γ H y ∂z ∂H x = −γ H x ∂z 0 E y = E y e − yz ∂E y Similarly = −γ E y ∂z ∂E x = −γ Ex ∂z There is no variation in the direction i.. wave propagates without attenuation. there is no wave motion but only an exponential decrease in amplitude.(5) ∂x ∂Ey = − jωε H z ∂x 2 ∂ E + γ 2 E = −ω 2 µε E 2 ∂x . Hy..(6) 2 ∂ H 2 2 + γ H = −ω µε H ∂x 2 2 ∂ E ∂2 H Where = γ 2 E and = γ 2H ∂x 2 ∂x2 Solving the equation (4) and (5) .... derivative of y is zero substituting the values of z derivatives and y derivatives in the equation (1). If α is real i. Let 0 H y = H y e − yz ∂H y Similarly. −γ E y Hx = jωµ Ey = 1 jωε ∂H z γ H x + ∂x Substituting the value of Ey in the above equation. 1 ∂H z − jωε γ H x + ∂x −γ ∂H z Hx = − γ H x + jωµ ∂x Hx = γ2 −γ ∂H H x = 1 + 2 = 2 z ω µε ω µε ∂x ∂H z H x ω 2 µε + γ 2 = −γ ∂x Hx = −γ ∂H Z 2 ω µε + γ ∂x 2 −γ 1 − jωµ jωε where h 2 = γ 2 + ω 2 µε To solve h y' ∂Ez = jωε E y ∂x γ H y = jωε Ex From the above equations. jωε H y − z γ γ ∂x VEL TECH MULTI TECH TECH HIGHTECH 141 VEL VEL TECH . jωε Hy = Ex γ γ Ex + Ex = 1 ∂Ez jωε H y − ∂x γ Substituting the value of Ex in the above equation.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL ∂H z = jωε Ey ∂x γ E y = − jωµ H x −γ Hx − From the above equation . Hy = jωε 1 ∂E . ∂E z = jωε H y ∂x jωε Hy = Ex γ Substituting the value of H y in the above equation.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Hy = −ω 2 µε jωε ∂E2 Hy − 2 2 γ γ ∂x ω 2 µε jωε ∂Ez H y 1 + 2 = 2 γ γ ∂x ∂E H y ( y 2 + ω 2 µε ) = jωε z ∂x Hy = ∂Ex − jωε 2 (γ + ω µε ) ∂x 2 h 2 = γ 2 + ω 2 µε Hy = To solve Ex. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 142 VEL −γ ∂Ez h 2 ∂x . − jωε ∂E2 h 2 ∂x γ Ex . γ Ex jωε ∂E z = jωε Ex ∂x γ −ωµε = Ex γ ω 2 µε ∂E γ Ex + Ex = z γ ∂x γ Ex ω 2 µε ∂E2 Ex γ + = − ∂x γ ∂E Ex [γ 2 + ω 2 µε ] = −γ 2 ∂x Ex = To solve Ey. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL γ Hx = ∂H z = − jωε Ey ∂x −γ 2 ∂H z Ey − jωε = − ∂x jωµ E y [γ 2 + ω 2 µε ] = jωµ jωµ ∂H z h 2 ∂x ∂H z ∂x Ey = h 2 = γ 2 + ω 2 µε Where The components of electric and magnetic field strength (E x,Ey,Hx and Hy) are expressed in terms of Ez and Hz. It is observed that there must be a z components of either e or H; otherwise all the components would be zero. Although in general case Ez and Hz may be present at the same time, it is convenient to divide the solutions into two sets. In the first case, there is a component of E in the direction of propagation (Ez), but no component of H in this direction . Such waves are called E or Transverse magnetic ™ waves. In the second case, there is a component of H in the direction of propagation (Hz), but no component of E in this direction. Such waves are called H waves or transverse Electric (TE) waves. 2. Derive the electro Magnetic field for TE waves. TRANSVERSE ELECTRIC WAVES: Transverse electric (TE) waves are waves in which the electric field strength E is entirely transverse. It has a magnetic field strength Hz in the direction of propagation and no component of electric field Ez in the same direction (Ez=0) Substituting the value of Ez =0 in the following equations. Ex = Then γ ∂E − jωε ∂E2 and Hy = 2 2 h ∂x h ∂x E x = 0 and H y = 0 The wave equation for the components Ey VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 143 VEL VEL TECH ∂2 E + γ 2 EY = −ω 2 µε EY 2 ∂x ∂2E = −ω 2 µε E y − γ 2 E y 2 ∂x =-(ω 2 µε + γ 2 ) E y But h 2 = y 2 + ω 2 µε ∂2 Ey ∂x 2 VEL TECH MULTI TECH TECH HIGHTECH VEL + h2 E y = 0 This is a differential equation of simple harmonic motion. The solution of his equation is given by Ey=C1 sin hx + C2 cos hx Where C1 and C2 are arbitrary constants. 0 − yz If Ey is expressed in time and direction ( E y = E y e ) then the solution becomes Ey=(C1 sin hx + C2 cos hx)e-yz The arbitrary constants C1 and C2 are determined from the boundary conditions. The tangential components of E is zero at the surface of conductors for all values of z. Ey=0 at x=0 Ey= 0 at x=a Applying the first boundary condition (x=0) 0=0+C2 C2=0 Then Ey=C1 sin hx e-yz Applying the second boundary condition (x=a) VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 144 VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH sinh a = 0 mπ h= a m=1,2,3........ mπ − yz Ey=C1Sin xe a ∂E y mπ mπ − yz = C1 cos xe ∂x a a VEL Where Therefore, ∴ Equations (5) are γ E y = − jωµ H x ∂E y ∂x = − jωε H z γ Ey jωµ Substituting the value of E y in the above equation: From the first equation , H x = − Hx = −γ mπ C1 sin jωµ a x e −γ x From the second equation , Hz = − 1 ∂E y jωε ∂x -mπ mπ C1Cos jωµ a a x e − yz x e− yz Substituting the value of Ey, in the above equation. = Hz = -mπ mπ C1Cos jωµ a a The field strengths for TE waves between parallel planes are mπ E y = C1Sin a −γ mπ − yz Hx = C1 Sin xe jωµ a −mπ mπ − yz Hz = C1Sin xe jωµ a a x e− yz .......(7) Each value of m specifies a particular field of configuration or mode and the wave associated with integer m is designated as TEm0.wave or TEm0 mode. The second subscript refers to another integer which varies with y. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 145 VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL If m =0 , then all the fields becomes zero, Ey=0,Hx=0, Hz=0. Therefore, the lowest value of m =1. The lowest order mode is TE10. This is called dominant mode in TE waves. The propagation constant γ =α +jβ . attenuation, α =0, only phase shift exists. γ =jβ Then the fields strengths for TE waves. mπ − j β z E y = C1Sin xe a −β mπ − j β z Hx = C1Sin xe jωµ a Hz = jmπ mπ C1Cos ωµ a a x e− jβ z If the wave propagates without The field distributions for TE10 mode between parallel planes are shown in fig. Fig. Electric and magnetic fields between parallel planes for the TE10. 3. Derive the Electromagnetic fields expression for TM waves. TRANSVERSE MAGNETIC WAVES: Transverse magnetic (TM) waves are in which the magnetic field strength H is VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 146 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL entirely traverse. It has an electric field strength Ez in the direction of propagation and no component of magnetic field Hz in the same direction (Hz=0) Substituting the value of Hz =0 in the following equations. −γ ∂H z jωµ ∂H z and E y = 2 2 h ∂x h ∂x Then Hx = 0 and E y = 0 Hx = [ Q Hz = 0] The wave equation for the component Hy ∂2 H y ∂x 2 + γ 2 H y = ω 2 µε H y ∂2H y ∂x 2 = −(ω 2 µε + γ 2 ) H y But h 2 = γ 2 + ω 2 µε ∂2 H y ∂x 2 + h2 H y = 0 This is also a differential equation of simple harmonic motion. The solution of this equation is H y = C3 sinh x +C4 cosh x Where C3 and C4 are arbitrary constants. If Hy is expressed in time and direction , then the solution becomes. H y = (C3 sinh x +C4 cosh x)e − yz The boundary conditions cannot be applied directly to Hy, to determine the arbitrary constants C3 and C4 because the tangential component of H I not zero at the surface of a conductor . However Ez can be obtained in terms of Hz. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 147 VEL .. γ H y = jωε Ex Ez = = x e − yz γ Hy jωε γ mπ C4 cos jωε a x e − yz The field strengths for TM waves between parallel planes are mπ − yz H y = C4 xe a γ mπ Ex = C4 cos jωε a Ez = jmπ mπ C4 sin ωε a a x e − yz x e − yz VEL TECH MULTI TECH TECH HIGHTECH 148 VEL VEL TECH ... mπ mπ − yz Therefore..(4)] ∂x 1 ∂H y Ez = jωε ∂x h = [C3 cos h x − C4 sinh x]e − yz jωε Applying the first boundary condition (E z =0 at x=0) C3 = 0 Then Ez = −h C4 sin hx e-yz jωε Applying the second boundary condition (Ez=0 at x=a) mπ a where m is a mode m=1. Ez = − C4 sin xe jωε a a jmπ mπ − yz =− C4 sin xe ωε a a h= mπ H y = C4 cos a But ..3...VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL ∂H Y = jωε Ez [eqn.2. Describe the field expression for TEM wave guided by parallel conducting plane.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The transverse magnetic wave associated with the integer m is designated as TMm0 wave or TMm0 mode. the propagation constant become γ =jβ .e. 4.. The field strengths for TM waves between parallel conducting planes are: mπ H y = C4 cos a Ex = x e− jβ z β mπ − j β z C4 cos xe ωε a jmπ mπ − j β z Ex = C4 cos xe ωε a The field distributions for TM10 wave between parallel planes shown in fig 2.3 Fig The TM10 wave between parallel planes. If m=0 all the fields will not be equal to zero i. The transverse electromagnetic (TEM) waves in which both electric and magnetic fields are transverse entirely VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 149 VEL . TRANSVERSE ELECTROMAGNETIC WAVES: It is a special type of transverse magnetic wave in which electric field E along the direction of propagation is also zero. Ex and Hy exist and only Ez=0. In the case of TM wave there is a possibility of m=0 If the wave propagates without attenuation (α =0). The field strength for TM waves are: mπ − j β z H y = C4 cos xe a β mπ − j β z Ex = C4 cos xe ωε a Ez = jmπ mπ − j β z C4 sin xe ωε A a for TEM waves E=0 and the minimum value of m=0 H y = C4 e − j β z β C4 e − j β z ωε Ez = 0 Ex = The fields are not only entirely transverse .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL but has not component of Ez and Hz. Wavelength λ = VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 150 VEL . Characteristics: For. but they are constant in amplitude between the parallel planes. the velocity of TEM wave independent of frequency and has the value c=3 x 108 m/sec. lowest value m=0 and dielectric is air. It is referred to as principal waves. Propagation constant γ = 0 − ω 2 µ 0ε 0 =jω µ 0ε 0 β = ω µ 0ε 0 Velocity v= ω 1 = =c β µ 0ε 0 2π c = β f Unlike TE and TM waves. VELOCITY OF PROPAGATION: The velocity with which the energy propagates along a guide is called group velocity dω dβ If the frequency spread of the group is small enough vg = dω dβ VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 151 VEL . 5. all frequencies down to zero can propagate along the guide.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Fig. Describe the different of various of propagation between 2 plates and prove that Vg Vp=C2. The TEM wave between parallel planes The cut-off frequency for TEM wave is zero fc = m =0 2a µε (m=0) This means that for TEM wave. the ratio of E to H between the parallel planes for a traveling wave is E = H µ0 ε0 The fields distribution are shown. These losses will modify the field strength by the introduction of multiplying α factor e. It is always less then the free space velocity c. The attenuation factor α that is caused by losses in the walls of the VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 152 . In actual wave guides. The field strengths between parallel conducting planes for TE.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL may be considered to be a constant through the group. It is denoted by dω vp = dβ It is always greater then the free space velocity c. there are some losses. Phase velocity is defined as the velocity of propagation of equiphase surfaces along a guide.z. Obtain the attenuation factor in parallel plane. The phase-shift is given by mπ β = ω µε − a 2 2 mπ Squaring on both sidesβ 2 = ω 2 µε − a Differentiate 2β d β = 2ω d ωµε − 0 dω β 1 = − d β ωµε ω β µε vg = 2 v2 vp dω vg = group velocity dβ ω phase velocity Where v p = β 1 v= free space velocity µε The product of group velocity and phase velocity is the square of free space velocity vg p v = 2 v = 2 c vg v p = c 2 6. TM and TEM waves have been obtained without any loss. ∴ Power lost/unit length 2α Wav = = 2α power transmitted Wav The attenuation factor α is α= Power lost/unit length 2 × power transmitted This is the attenuation factor for more general case of guided wave transmission. Derive an expression for attenuation Factor for TE Waves.VEL TECH guide is determined as follows. 7. The electric and magnetic field strengths between perfectly conduction parallel planes for TE waves are VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 153 VEL . TM and TEM waves. The α can be determined for TE. VEL TECH MULTI TECH TECH HIGHTECH VEL The voltage and current phasors in waveguide are V = V0 e −α z e− j β z I = I0 e −α z e− j β z The average power transmitted is Wav = 1 Real part of [ VI*] 2 1 = Real part of V0 e −α z e− j β z I 0 e−α z e j β z 2 1 * = Real part of V0 I 0 e −2α z 2 Where I* is the complex conjugate of I * I * = I 0 e −α z e j β z The rate of decrease of transmitted power −∂Wav 1 = 2α Re al V0 I* e −2α z 0 ∂z 2 = 2α Wav This is the power lost per unit length or power dissipated per unit length. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH mπ − j β z E y = C1 sin xe α Hx = Hx = −β mπ C1 sin ωµ α x e− jβ z x e− jβ z VEL jmπ mπ C1 cos ωµα α The amplitude of linear current density in the conduction planes will be equal to the tangential component of H at x=0 and x = a. Rs = ωµ 2α The power loss in two conducting (upper and lower) planes m2π 2C12 ωµ / 2α 1 2 2 × J zy Rs = 2 ω 2 µ 2α 2 The power transmitted in the z direction is 1 Re al E × H * 2 1 Power transmitted/unit area = = − E y H x 2 β c12 mπ = sin 2 x 2ωµ a = Power transmitted in z direction for a guide 1 metre wide with a spacing between conductors of metres is VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 154 VEL . J zy = H z = mπ C1 ωµα 1 2 J zy Rs 2 The power loss in each conduction plane is Where Rs is the surface resistance. VEL TECH x=a VEL TECH MULTI TECH TECH HIGHTECH VEL β c12 2 mπ ∫0 2ωµ sin a x= β c2 mπ x dx = 1 sin2 2ωµ a x dx a mπ sin 2 x 2 β c1 x a − = mπ 2ωµ 2 4 a 0 2 βc a = 1 4ωµ Power transmitted/unit length = = β C12 a 4ωµ α= The attenuation factor power lost/unit length 2 × power transmitted m 2π 2C 2 ωµ / 2σ ω 2 µ 2a2 = β C 2a 2× 1 4ωµ 2m 2π 2 ωµ / 2σ β ωµ a 3 2 2 α= mπ Substitute the value of β = ω µε − a α= 2m 2π 2 ωµ / 2σ mπ ωµ a 3 ω 2 µε − a 2 The attenuation factor α decreases from infinity at cut-off frequency to a low value at higher frequency. mπ 2 a 2 2 α= ωµ 2σ 2 mπ ωµ a ω µε − a VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 155 VEL . The Expressions for E and H for the Transverse magnetic waves between perfectly conducting parallel planes are VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 156 VEL . 2 2 α= 2ω c2 µε Rs f ω µα µε 1 − c f 2 2 = ω 2 Rs c ω 2 f µ a 1− c ε f f 2 Rs c f 2 2 α= f a η 1− c f 2 Q η = µ ε 8. Derive on Expression for Attenuation Factor for TM Waves. ω µε − a 2 2 2 mπ 2 2 ω µε − = ω µε − ω c µε a ω = ω µε 1 − c ω f = ω µε 1 − c f Substituting this value.off frequency.VEL TECH mπ 2 Rs a mπ ωµ a ω 2 µε − a 2 2 2 VEL TECH MULTI TECH TECH HIGHTECH VEL = where Rs = ωµ 2σ mπ At cut. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL mπ − j β z H y =C4 cos xe a β mπ − j β z Ey = C4 cos xe ωε a Ez =mπ mπ C4 sin jωε a a x e− jβ z The amplitude of the current density in each plane is J = C4 The power loss per unit length in each conducting plane is 1 2 1 2 j Rs = C4 Rs 2 2 ω µm Rs = 2σ m The power transmitted down the guide per unit are 1 Re al (E × H*) 2 1 = (E x H y ) 2 1 βC mπ mπ = 4 cos x C4 cos x 2 ωε a a 2 1 β C4 mπ = cos 2 x 2 ωε a = Power transmitted in the z direction for a guide 1 metre wide with a spacing between conductors of ‘s’ metre is 2 1 β C4 1 β C42 2 mπ cos x dx = a 2 x∫0 ωε 4 ωε a = Power lost/ unit lengh Attenuation factor (α ) = 2 × power transmitted a C2 4 α= 2 ω µm 2σ m β C42 a 4ωε VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 157 VEL . VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 2ωε = VEL ω µm 2σ m aβ 2 But mπ β = ω µε − a 2 The attenuation factor for TM waves is αTM = 2ω ωµ m / 2σ m mπ a ω 2 µε − a 2 The attenuation factor for TE waves is 2m 2π 2 αTE = ωµ a Dividing α TE 3 ωµ m 2σ m 2 mπ ω µε − a 2 by α TM 2 α TE α TM m 2π 2 mπ ωµ a 3 a = = 2 ωε ω µε a mπ a = 2 ( 2π f ) µε 2 But m 1 2a µε = 2 f m 1 fc = . 2a µε f c2 α TE = α TM f 2 2 αTE VEL TECH f = c αTM f VEL TECH MULTI TECH TECH HIGHTECH 158 2 VEL . VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The attenuation factor for TM waves is αTM = 2 ωε R s mπ a ω 2 µε − a 2 mπ 2 substituting the value of = ω c µε a 2 αTM = = 2 ωε R s a ω 2 µε − ω c2 µε 2 ωε R s ω aω µε 1 − c ω = 2R s f µ a 1− c ε f 2R s f a η 1− c f Rs = 2 2 2 = µ ∴η = ε Where = 2 ∴αTM = π f µm σm π fc µm σm f fc f fc f 1− c f 2 π f µm σm aη 2 or α TM = π fc µm σm 2 f c fc a η 1 − f f VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 159 VEL . x2 = 3 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 160 VEL . to zero. 2 2 3 dα K 3 x ( x − 1) − x (2 x) = 1 2 dx ( x 2 − 1) x3 2 2 2 x −1 K x 2 − 1 3x 4 − 3 x 2 − 2 x4 = 2 x 3 ( x 2 − 1) 2 dα =0 Equating dx K x 2 − 1 x 4 − 3 x2 =0 2 x 3 ( x 2 − 1) 2 4 2 x − 3x = 0 ( x 2 − 1) 2 x4 = 3 x2 Dividing by x2. fc TM VEL TECH MULTI TECH TECH HIGHTECH VEL is determined by equating derivative of α with respect to Let π fc µm σm K= aη f x= fc 2 α =K x 1 1− x x3 x2 −1 1 2 =K x3 2 =K 2 x −1 Differentiating with respect to ‘x’.VEL TECH The minimum value of α f . After substituting this value.VEL TECH ∴ But x= ± 3 f x= fc VEL TECH MULTI TECH TECH HIGHTECH VEL f =± 3 fc Take only positive value of frequency f = 3 fc The attenuation α TM reaches a minimum value at a frequency of 3 times the cut-off frequency and then increases with frequency. 3 2 = 2. Z xy + = Ex . α min = 2 Rs 1 η a 1. TEM and TM waves. Zxz + = − x Hy Hz The wave impedances for the negative directions of the coordinates are VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 161 VEL . Wave impedances are defined by the following ratios of electric to magnetic field strengths for the positive directions of the coordinates. Z zy + = − E Ez . However. Define Wave Impedance. and only one impedance constant is involved. Hy Ey Hx Z yz + = Ey Hz . Obtain the wave impedance expressions for TE.5R s ηa 9. in the three dimensional wave propagation power may be transmitted along three axes of the coordinate system and consequently three impedance constants must be defined. Zzx + = Ez Hx Z yx + = − . WAVE IMPEDANCES In transmission-line theory power is propagated along one axis only. the wave impedance is given by + Z yx = − Ey Hx = jωµ jβ ωµ = β + Z yx = ωµ mπ ω µε − a 2 2 = ωµ mπ ω µε 1 − / ϖ µε a 2 = µ ε 1 f 1− c f 2 mπ Q ω c = a µε + Z yx = η f 1− c f 2 mπ + At cut-off frequency ω 2 µε = . Hy − Zxz = Ex Hz For TE waves.VEL TECH Ex . Z = = + yx 2 ωµ ω 2 µε 2 mπ 2 Q ω µε >> a µ µ = =η ε µε VEL TECH MULTI TECH TECH HIGHTECH 162 VEL VEL TECH . wave impedance Z yx becomes infinity. Ey Hz VEL TECH MULTI TECH TECH HIGHTECH − Z yz = − VEL − Z xy = − . At a very high frequency (greater than cut-off frequency) wave impedance becomes. − Zzx = − Ez Hx − Z yx = Ey 1+ x − Z zy = Ez . Hy . VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL At ω >> ω c. the wave impedance becomes zero. the wave impedance is given by mπ ω 2 µε − Ex β a + Z xy = = = H y ωε ωε 2 ω µε = mπ 1− / ω µε a ωε ω 1− c ω 2 2 2 µ = ε + xy f Z = η 1− c f 2 2 mπ At cut-off frequency ω µε = . the wave impedance is equal to the intrinsic impedance. a At very high frequency (greater than cut-off frequency). Z+yx =η For TM waves. the wave impedance is + Z xy = Ex Hy β ωε µo = εo = =η VEL TECH o intrinsic impedance VEL TECH MULTI TECH TECH HIGHTECH 163 VEL . the wave impedance becomes ω 2 µε ωε µ = =η ε + Z xy = η + Z xy = For TEM wave. Does the propagation take place in each case. Given: TE10 mode : m = 1. n = 0 a = 10 cm = 0. 10 GHz For free space µ = µ o and ε = ε Free space velocity c = o 1 = 3 × 108 m / sec µ oε o Propagation constant is given by mπ 2 γ= − ω µε a π 2π f = − a c 1 2f =π − a c VEL TECH 2 2 2 2 [Q ω = 2π f ] 2 VEL TECH MULTI TECH TECH HIGHTECH 164 VEL .1 m f = 100 MHz. TM and TEM waves between parallel planes are shown as functions of frequency in figure. Wave impedance versus frequency characteristics of waves between parallel conducting plane 10.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL + Z xy = ηo The wave impedances for TE. A parallel plane wave guide consists of two sheets of good conductor separated by 10 cm. when the wave guide is operated in TE10 mode. Find the propagation constant at frequencies of 100 MHz and 10 GHz. For f = 10 GHz 2 × 10 ×109 1 γ =π − (0.346 Nepers / metre Here propagation constant γ has real value. 11.. A pair of perfectly conducting planes are separated 8 cm in air.e..95 fc (iv) phase shift (v) phase velocity and group velocity (vi) wavelength measured along the guiding walls Given : TM1 mode : m = 1 1 εo = F /m a = 8 cm = 0.e. γ = jβ .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL For f = 100 MHz 2 × 100 ×106 1 =π − (0. i.1) 2 3 × 108 2 = π 100 − 3 = 31.07 2 2 2 2 Here γ has imaginary value i. γ = α Hence no propagation takes place at 100 MHz. (i) cut-off frequency (ii) characteristic impedance (iii) attenuation constant for f = 0. For frequency of 5000 MHz with the TM1 mode excited find the following.1) 2 3 ×108 200 = π 100 − 3 = j 207. Hence propagation takes place at 10 GHz.08 m 36π ×109 µ o = 4π ×10−7 H / m f = 5000 MHz (i) Cut-off frequency: VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 165 VEL . e.875 GHz (ii) Characteristic impedance: z= µo = 120π εo η = 120π ohms or 377 ohms (iii) Propagation constant becomes attenuation constant (i.08 = 18.95 f c 2 2 2 Cut-off wave length λc = v 2a = fc m VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 166 VEL . ω = 2π f ] µε fc = m v 2a But v = c = 3 × 108 m/sec 1 × 3 × 108 fc = 2 × 0.95 fc mπ 2 γ =α = − ω µε a mπ 2π f = − a v f = 0. f = 0.. a 2 2 VEL TECH MULTI TECH TECH HIGHTECH VEL 2 1 µε [Q v = 1 . real value) if the operating frequency is less than the cut-off frequency.VEL TECH mπ 2 γ= − ω µε a mπ ω c µε = a mπ ωc = .75 × 108 Hz = 1. 08 (v) Phase velocity: vp = 2 2 ω β 2π f = β 2π × 5000 ×106 = 97.08 = 3.25 3 = 97.27 0.95 a a = mπ 1 − (0.236 ×108 m / sec or VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 167 VEL .26 Nepers / m (iv) Phase shift: mπ β = ω 2 µε − a f = 5000 MHz 2π f mπ β= − v a 2 2 2 2 × 5000 ×106 1 =π − 3 ×108 0.08 100 =π − 156.0975 = 12.95 2a a mπ mπ = − 0.95) 2 a 2 2 2 2 VEL TECH MULTI TECH TECH HIGHTECH VEL = 39.VEL TECH m mπ α= − 2π 0. a = 0.875 × 109 1− 6 5000 × 10 2 = 3.VEL TECH vp = c f 1− c f 2 VEL TECH MULTI TECH TECH HIGHTECH VEL = 3 ×108 1. find the following for the TE1 mode.236 ×108 = 2. (vi) Wave guide wavelength: λ λg = 2 λ 1− λc But λ = 3 × 108 = 0.78 m / sec.06 m 50 ×108 12. (i) cut-off frequency (ii) angle of incidence of the planes (iii) phase velocity and group velocity is it possible to propagate TE3 mode? Given : TE1 mode : m = 1. For a frequency of 6000 MHz and plane separation = 7 cm.236 ×108 m / sec c2 Group velocity vg = vp (3 ×108 ) 2 = 3.07 m f = 6000 MHz (i) Cut-off: VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 168 VEL . 92° VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 169 VEL .4286 ×108 1− 6 6000 ×10 = 3. = (iii) Angle of incidence c cos θ c 3 ×108 cos θ = = v p 3.934) = 20.4286 ×108 Hz (ii) Phase velocity vp = c f 1− c f 2 = 3 × 108 21.2118 × 108 m / sec Group velocity vg = c2 vp (3 ×108 ) 2 3.802 m / sec. 2a µε = 2 VEL TECH MULTI TECH TECH HIGHTECH VEL m v 2a 1 = × 3 ×108 2 × 0.2118 ×108 vp = θ = cos −1 (0.VEL TECH mπ ω c 2 µε = a m 1 fc = .07 f c = 21.2118 × 108 = 2. VEL TECH mπ 2 γ= − ω µε a mπ 2π f = − a v m 2f =π − a v 2 2 2 2 2 2 VEL TECH MULTI TECH TECH HIGHTECH VEL 6 3 2 × 6000 ×10 =π − 3 ×108 0.337 Nepers / m Propagation constant has real value is v = α . a = 0. Consider a parallel plate wave guide with plate separation 20 cm with the TE10 mode excited at 1 GHz.735 = 48.735 − 16000 = π 236.07 2 = π 1836.2 m f = 1 GHz Propagation constant mπ 2 γ= − ω µε a 9 1 2 × 10 × 2 =π − 8 0.2 3 × 10 2 2 2 Q v = = j 39. 13. Propagation is not possible for TE3 mode. Find the propagation constant.3 radians / m Cut-off Frequency 1 = µε 1 µ oε o ε r VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 170 VEL . the cut-off frequency and guide wavelength assuming ε r = 4 for medium of propagation in the guide. Given: TE10 mode : m = 1. n = 0. Assuming the dominant mode.3 m 2 λg = 0. n = 0 f = 1 GHz a = 0. calculate: (i) cut-off wavelength (ii) guide wavelength (iii) group velocity (iv) phase velocity and (v) wave impedance Given : TE10 mode : m = 1.4 m 3 × 108 λ= 1×109 = 0.VEL TECH fc = = = m 2a 1 µε 1 c = µε εr VEL TECH MULTI TECH TECH HIGHTECH VEL m c 2a 4 3 ×108 2 × 0.16m 14.3 0.3 1− 0.05 m b = 0.2 × 2 = 375 MHz Guide wavelength λg = λ λ 1− λc 2a λc = = 2 × 0.5 × 5 cm. A 4 GHz signal is propagated in a rectangular wave guide with internal dimensions of 2.4 2 = 0.2 m = 0.025 m (i) Cut-off wave length VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 171 VEL . 10 m (ii) Guide wave length : λg = λ λ 1− λc 2 c f 3 × 108 = = 0.075 λg = = 0.984 × 108 m / sec.075 1− 0.075 m 4 × 109 0.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL λc = 2a m = 2 × 0.1134 m 2 0.1134 = 1.05 = 0.1 λ= (iii) Group velocity λ vg = c λg 0. (iv) Phase velocity VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 172 VEL .075 8 = 3 × 10 0. An uniform plane wave at 2.1 f c = 30 × 108 Hz vp = 3 × 108 30 ×108 1− 9 4 ×10 2 = 4. find the time average power flow per unit area. 15.5356 ×108 m / sec.45 GHz is transmitted through a medium having σ =2.17 s/m ∑=47∑ o µ =µ o.075 = 4. (iv) Wave impedance z0 zTE = 2 λ 1− λc Intrinsic impedance z0 = 120π ohms zTE = 120 π 2 0.075 1− 0. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 173 VEL .c 2a 1 × 3 ×108 0.5356 × 108 m / sec. Find the complex propagation constant. phase velocity and the wave impedance of the medium.1134 8 = × 3 × 10 0. or λg vp = c λ 0.1 = 570 ohms. If the electric field mag is 10V/m.VEL TECH vp = c f 1− c f mv = 2a 2 VEL TECH MULTI TECH TECH HIGHTECH VEL fc = = m . 85+j356.63o γ =58.45× 109Hz VEL TECH MULTI TECH TECH HIGHTECH VEL W=2π f=2π × 2. What are the characteristics of TE and TM waves? (1) Propagation constant VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 174 VEL .65 α =58.45GHz=2.24Ω jωµ o σ + jω∑ Time average power flow / unit area 1 Pav= EH 2 1 = × 10× 192=96w/m2 2 Intrinsic impedance η = 16.854× 10-12=416.17s/m ∑=47∑o=47× 8.14× 10-12 F/m µ =µ o=4π × 10-7 H/m γ = jωµ ( σ + jω∑ ) = jωµσ-ω2µ ∑ = -123922.85N/m β =356.51+ j41977.580.VEL TECH Given : F=2.65 rad/m Vp=ω /β =4316× 108m/s ωµ wave impedance Z = β = 54.45× 109=15.394× 109 σ =2.94 = 361. < ω µ ∑. 2 mπ . ω µ ∑ is < becomes real with value of α . without any propagation. For f<fc a (ii) Cut – off frequency:at f= fc 2 2 mπ 2 o= −ω µ∑ a mπ ω2µ ∑ = a ωc µ ∑ = mπ a mπ a 2 2 2πfc µ ∑ = fc = m 2a µ ∑ VEL TECH MULTI TECH TECH HIGHTECH 175 VEL VEL TECH . mπ 2 2 0 For f<fc . γ = β α = a 2 mπ . 2 At higher frequencies value of ω µ ∑ becomes greater than making a γ . ∴ there a is only attenuation. ω 2 µ ∑.∴ γ at lower freq.imaginary. and β =0. ω µ ∑.VEL TECH h2 = γ 2 + ω2µ 2 ∑ γ 2 = h2 − ω2µ 2 ∑ r = h2 − ω2µ 2 ∑ mπ sub h= a mπ 2 r= −ω µ∑ a γ = α + jβ 2 VEL TECH MULTI TECH TECH HIGHTECH VEL mπ . γ = α β= a mπ 2 2 0 For f<fc . VEL TECH (iii) Phase – constant :mπ sub ωc2µ ∑ = in eqn of γ = a γ = ωc2µ ∑ −ωc2µ ∑ γ = µ ∑ − ωc2 − ω2 γ = j µ∑ 2 VEL TECH MULTI TECH TECH HIGHTECH mπ − ω2µ ∑ a 2 VEL ( ) ω2 − ωc2 γ = j µ ∑ f 2 − fc2 = jβ β = 2π µ ∑ f 2 − fc2 (iv) cut-off wave length λc = fc = V fc m 1 mv = but v= 2a µ ∑ µ ∑ 2a 2a µ ∑ j1 × m µ∑ 2a m λc = λc = (v) Group wavelength:- VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 176 VEL . VEL TECH v= λ= ω = fλ β / ω 2πf = / β f βf 2π mπ ω2µ ∑ − a 2 2 VEL TECH MULTI TECH TECH HIGHTECH VEL λg = mπ 2 sub = ωc µ ∑ a λg = 2π ω µ ∑ − 1− = 2π ωc2 ω2 ωc2 2πf 1− 2 ω V f λg = ωc2 1− 2 ω vi) Velocity of propagation:ω ω V= = mπ 2 β ω2µ ∑ − a mπ 2 2 sub = ωc µ ∑ a V= ω ω µ ∑ 1− V= Vo f 1− c f 2 −ωc2 ω2 UNIT – V VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 177 VEL . Write down the Maxwell’s equation for the loss-free region within the guide? ∂Hz + γ Hv = jω ε Ex. Write down the propagation constant for a rectangular guide for TM waves. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 178 VEL . = − jw µ H z ∂x ∂y ∂x ∂y 2. In waves between parallel plates what are the classification by field configurations? 1. and y = b Ev = Ez = 0 at x = 0 and x = a. For Rectangular guide shown in figure what is the boundary condition? Ex = Ez = 0 At y = 0. ∂y ∂x ∂H y −∂H x ∂E y −∂E x = jw ∈ Ez . 5. What is the wave equation for E2 and H2? 2 ∂ 2 E z ∂ 2 Ez + + γ E z = −ω 2 µε Ez 2 2 ∂x ∂y 2 ∂ 2H z ∂ 2 Hz + + γ H z = −ω 2 µε Hz 2 2 ∂x ∂y 3.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL PART – A 1. Transverse Magnetic Waves (TM) 2. ∂Ez + γ EV = − jw µ Hz ∂y ∂Ez ∂H z + γ E x = jw µ Hv + γ H x = − jw ∈ EV . Transverse Electric waves (TE) 4. what is cut-off frequency? Cut-off frequency is that frequency for which the corresponding half wavelength is equal to the width of the guide. 7. is fc = 1 2π µε mπ nπ = + a b 2 2 the frequency below which wave 9. that is the cutoff wavelength which wave propagation will not occur is 2 λ0 = 2 2 m n a +b 10. For TE10 wave. propagation constant for a perfectly conducting wall? α = 0 such that ω > ω β mπ nπ = ω µε − − a b 2 2 2 c 1 The value of ω c = µε mπ nπ = + a b is 2 2 8. The Cut-off frequency that propagation will not occur. The cur-off wavelength. What is the propagation constant for an ordinary transmission line? γ =α + jβ α → Attenuation constant β → Phase constant.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL ν = G 2 − ω 2 µε = A 2 + B 2 − ω 2 µε 2 2 mπ nπ 2 = + b − ω µε a 6. 11. What is the expression of attenuation constant. What is dominant wave? The wave which has the lowest cut-off frequency is called the dominant wave. the cut-off frequency is independent of the VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 179 VEL . 17. Write down the maxwell’s equation for non-conducting medium. z = η 1 − f VEL TECH 2 VEL TECH MULTI TECH TECH HIGHTECH 180 VEL . ∇xH = jωξ E ax ay ∂ ∂ ∇xH = ∂x ∂y Hx Hy [∴ σ =0] az ∂ ∂z Hz = jω ε ( a xEx+ a yEy+ a zEz] 15. Waves that are guided along the surface (walls) of the tube is called waveguide. 13. Hence the TE10 mode is the dominant mode of a rectangular waveguide. n=1). This wave has the lowest cut-off frequency. 14. What are the wave impedance for different modes? fc For TM. What is Dominant mode? The lowest mode for TE wave is TE10 (m = 1. n = 0) whereas the lowest mode for TM wave is TM11 (m = 1. Zyx = y . Write down the wave equation for rectangular waveguide. ∂ 2 E z ∂ 2 Ez + + γ 2E z = −ω 2 µε Ez 2 2 ∂x ∂y 16.VEL TECH dimension b. What is a waveguide? VEL TECH MULTI TECH TECH HIGHTECH VEL A hallow conducting metallic tube of uniform section (rectangular or circular) is used for propagating electromagnetic waves. What is wave impedance? The wave impedances defined as the ratio of electric field intensity to magnetic field intensity are E E E + + + Z xy = x . Zzx = z Hy Hx Hx 18. 12. What do you mean by propagation of waveguide? Propagation of waveguide can be considered as a phenomenon in which waves are reflected from wall to wall and hence pass down the waveguide in a zig-zag fashion. I) = I 20. What is attenuation constant in terms of power? The attenuation constant is α= Power lost 2 x power transmitted. What is characteristic impedance? For transmission lines the integrated characteristic impedance Zo can be defined as in terms of the voltage current or interms of the power transmitted for a given voltage or given current. What is attenuation constant in propagation? It is given by Power lost per unit length α= 2 x power transmitted. V i.e. How will you calculate power? The power transmitted is obtained by integrating the axial component of the pointing vector through the cross-section of the guide the pointing vector Pz is given by Pt = ½ |Etrans||Htrans| 23. What are the sources of attenuation in wave guide? Attenuation for propagating modes results when there are losses in the dielectric and in the imperfectly conducting guide walls. 22.. For TEM. z = η = µ ε 19. 24. What is attenuation constant for TM11 mode? VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 181 VEL . Zo (V. 21.VEL TECH z= VEL TECH MULTI TECH TECH HIGHTECH VEL η fc 1− f 2 For TE. Write the field expression for rectangular TM waves. 2 f a b 2 VEL TECH MULTI TECH TECH HIGHTECH VEL α= 25. Ex0 = Ey0 Hx0 Hy0 − j βc B cos Bx sin Ay h2 − j βc = A sin Bx cos Ay h2 jωε c = 2 A sin Bx cos Ay h − jωε c = B cos Bx sin Ay h2 27. Draw the variation in attenuation with frequency due to wall losses in a rectangular waveguide? TM11 TE10F 26. Write the field expression of TE wave guide. Hx0 = Hy0 jβ CB sin Bx cos Ay h2 jβ = 2 CA cos Bx sin Ay h Hz0 = C cos Ay cos Bx jωµ CA cos Bx sin Ay h2 − jωµ Ey0 = CB sin Bx cos Ay h2 28.VEL TECH a b 2Rs 2 + 2 b a fc 1 1 ηab 1 − 2 . What are the types of waveguides? Ex0 = 1. Circular waveguide VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 182 VEL α in (dB| m) . Rectangular waveguide 2. m = 1.28 6 cm = 2.016 cms determine the cut off frequencies for TE11 and TM11 modes? a = 2. Why are rectangular waveguides preferred over circular waveguides? Rectangular waveguides are preferred over circular waveguides because of the following reasons. The waveguides are employed for transmission of energy at every high frequencies where the attenuation caused by waveguide is smaller.286 x 10-2 m b = 1. For an air filled copper x-band waveguide with dimension a = 2. The wave cannot be propagated. 33. Y = α .286 cms and b = 1. 1. This non-propagating mode is known as evanescent mode. fc = = c m n + 2 a b 2 2 1 2π µε mπ nπ a + b 2 2 3 × 108 102 102 = + 2 2.e. It does not maintain its polarization through the circular waveguide. Elliptical waveguide VEL TECH MULTI TECH TECH HIGHTECH VEL 29. Waveguide are used in microwave transmission circular waveguides used as attenuation and phase shifters.VEL TECH 3. The cut of frequency for TE11 mode is same as that of TM11. n = 1 Cut-off frequency. 2.286 1. What is an evanescent mode? When the operating frequency is lower than the cut-off frequency the propagation constant becomes real i. 31.016 cm = 1. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 183 VEL . Rectangular waveguide is smaller in size than a circular waveguide of the same operating frequency. 30. 32. Mention the application of waveguides.. Why is rectangular or circular form used as waveguides? Waveguides usually take the form of rectangular or circular cylinders because of its simpler form in use and less expensive to manufacture.156 GHz.016 x 10-2 m For TE11 mode.016 2 2 FC = 16. λ c Wave impedance. What is the wave impedance of TEM waves in waveguide? VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 184 VEL . λ c = 22 C Cut-off frequency. fc = 22 38. fc = 2 µε 2 1 1 a +b 2 2 ν 1 1 = + 2 a b 39.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 34. Region of number TE fc 37. What is the cut-off wavelength and cut-off frequency of the TE10 mode in a rectangular waveguide? Propagation TM Cut-off wavelength. Hy Ey and Ez 36. η = 1 1 1 a +b 2 2 2 Cut-off frequency. Which are the non-zero field components for the TE10 mode in a rectangular waveguide? Hx. Hz and Ey 35. Which are the non-zero field components for the TM11 mode in a rectangular waveguide? Ex. What is the cut-off wavelength and cut-off frequency of the TM 11 mode in a rectangular waveguide? 2 Cut-off wavelength. Draw a neat sketch showing the variation of wave impedance with frequency for TE and TM waves in a waveguide. I ) = II * VV * Zo ( w . Define character is the impedance in a waveguide. Loss due to attenuation of signals below cut-off frequency. Write down the expression for phase velocity in a waveguide? ν = ω mπ nπ ω µε − − a b 2 2 2 43. I ) = 2W Where V and I are peak phasors. 2.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Wave impedance of TEM becomes the intrinsic impedance of the medium Z =η = µ ε η 40. What are the types of power loss in waveguides? There are two types of power loss occurs in waveguides 1. V i. I) = I 2w Zo ( w .e. 44. W is the power transmitted. 45. The larger dimension of the cross section of a rectangular waveguide is VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 185 VEL . For transmission lines the integrated characteristic impedance can be defined as in terms of the voltage current ratio or in terms power transmitted for a given voltage or a given current.. Write down the expression for wave impedance of TM mode? f Z = η 1− c f 2 42. * indicates complex conjugate. Loss due to dissipation with in the waveguide walls and the dielectric with in the waveguide. Zo (V. Write down the expression for wave impedance of TE mode? Z= 2 f 1− c f 41. 02 = 7. it cannot be There fore for a displacement current the guide requires an axial component of E. Explain why TEM waves are not possible in rectangular waveguide.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 2cm.02 = 0. (May 2006) fc = 2 2 For the modes TM01 and TM10.04mts 46. The cut-off frequency of rectangular waveguides is v m n vm fc = a + a = 2 a Since n = 0 2 3 × 108 1 1. C m π nπ + 2π a b 48. We know the dominant TE mode is TE10 mode. A rectangular wave guide has the following dimensions l=2.5GHz 2 2π Cutoff wavelength λ c 2 2 = mπ nπ a − b 2 2 = 2a m since n = 0 λc = 2 ( 0. Explain why TM01 and TM10 modes in a rectangular wave guide do not exist.02 ) 1 = 0.) Since TEM wave does not have actual component of a E or H.127 cm.54cm b=1. 47. 2004. The given data are a = 2 cm. Fe > f Where f → frequency of the wave to be propagated ∴ The dominant mode in TM mode VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 186 VEL . (Apr. thus m = 1 and n = 0. Calculate the cut off frequency for TE11 mode.27cm.5 × 108 9 = 0. Find the cut off frequency and wavelength for the domain and TE mode. which is not present in TEM waves. propagated with in a single conductor wave guide.5 × 10 = 7. wave guide thickness = 0. 52. η = . Give the attenuation factor for TM waves wµ 2σ α= Power lost / length 2 × Power transmitted/length where the power loss in the 4 walls of the guide is the sum of losses in x=0 and y=0. ∴Z = Ey Hx = Ex Hy 50. Define the difference between the wave impedance and the intrinsic impedance. Show the excitation method of TE11 and TM11 modes is a rectangular wave guide. Which mode is the dominant mode in a rectangular wave? The dominant modes are. The surface impedance is defined by the conductivity. ε The wave impedance is the radio between the electrostatic energy and the magnetic field energy.VEL TECH is TM11 VEL TECH MULTI TECH TECH HIGHTECH VEL 49. The intrinsic impedance η is given by the ratio between the permeability and the permittivity. TE10 and TM11 But the lowest mode is TE10 53. Define surface impedance. M For free space. (Nov 2005) VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 187 VEL . 377Ω for cu. Rs = 51. (Apr. the dominant mode TE10. f 56. For a rectangular waveguide. What is a guided wavelength? λ 1− λc c where λ = and λc = cut off wave length. Circular waveguides are used as attenuators and phase shifters.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Fig. Mention the applications of circular waveguides. Mention the dominant modes in rectangular and circular waveguides. 2005) 2Rs = ηab 1 a2 Rs = a b a 2 + b2 1 fc + 2 1− b c wµ 2σ 2 α TM where 55. Give the attenuation factor for TM wave is a rectangular wave guide. For a circular waveguide VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 188 λg = λ 2 . Which mode in a circular waveguides has alternation effect decreasing with increasing in frequency? TE01 58. 57. (a) TE11 (b) TM11 54. fc = hnm 2π µε where hnm = ( ha ) nm a 60. Determine the cut-off frequency of a circular waveguide with a diameter of 2.566 x 102 fc = h11 = ( ha ) 11 [ (ha)11 = 3. VEL TECH MULTI TECH TECH HIGHTECH VEL 59.566 × 102 × 3 × 108 2π = 12.VEL TECH the dominant mode TE11. in TE11 mode. ( ha ) nm hnm = a ( ha ) 11 = a 3. fc = hnm 2π µ m . Write down the expression for cut-off frequency in a circular waveguide. Calculate the cut off frequency of copper tube with 3 cm diameter inside with air filled.85 × 2 × 102 = 2. 2 Dominant mode is TE11. a= 3 × 10 −2 m 2 a 3. 2.263 x 102 a= fc = hnm 2π µε 1 8 Q µε = 3 × 10 VEL TECH MULTI TECH TECH HIGHTECH 189 3.36 = 3.36 × 10−2 m.85 × 2 = 3 × 10−2 = 2.85] 2.36 cms operating in the dominant mode.25 GHz 1 Q µε = c 61.263 × 102 = × 3 × 108 2π VEL TECH VEL . since it requires a smaller diameter for the same cut-off wavelength. Define Q of a waveguide. VEL TECH MULTI TECH TECH HIGHTECH VEL 62.VEL TECH fc = 15. 67. Give the relation b/w quality factor and attenuation factor of a waveguide? Q= ω 2Vg 2 65.58 GHz. Which of the following wave guide is easier to manufacture? Circular ⇒ Rectangular ⇒ Elliptical ⇒ none Ans (a) circular ⇒ 66. 69. Which of the following waveguide / transmission line would offer the maximum attenuation? a) Coaxial b) Rectangular waveguide c) Circular waveguide. Draw the variation of attenuation as a function of frequency for different VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 190 VEL . 63. Why is TM01 mode prepared to the TE01 mode in a circular waveguide? TM01 mode is preferred to the TE01 mode. 64. Quality factor Q is given by Q = ω = energy stored/unit length energy lost / unit length / second. 68. A circular waveguide will behave like a) Low pass filter b) Band pass filter c) High pass filter d) Non of the above Ans: (c) High pass filter. Ans: (a) coaxial. What are the uses of circular wave guides? Circular waveguides are used as attenuators and phase shifters. What the cut-off freq or critical frequency below which the transmission of a wave will not occur? hnm fc = 2π µε where hnm = ( ha ) nm a 73. 1 ρ ∞ r 2 P = C1 ∑ ( −1) r ! 2 r =0 ( ) 2r This series is convergent for all values of p1 either real complex. What is the expression for TM waves in circular guides? H20 = Cn Jn (hp) cos n φ VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 191 VEL . It is called Bessel function of first kind of order zero d is denoted by J0 (P). 72. What is Bessel function? In solving for the electromagnetic fields within guides of circular cross section. VEL TECH MULTI TECH TECH HIGHTECH VEL Attenuator TE11 TM01 TE11 70.VEL TECH modes. a differential equation known as Bessel’s Equation is encountered. What is the wave impedance of a circular waveguide? The wave impedance of a circular waveguide is the ratio of the resultant Frequency transverse electric fields to the transverse magnetic field. For TM waves E 2TM = H 71. VEL TECH − j βCn Jn (hp) cos n φ h jn βCn Hφ 0 = Jn (hp) sin n φ h2 p ωµ 0 Fp0 = H β p −ωµ 0 0 Eφ = Hp β Hp0 = VEL TECH MULTI TECH TECH HIGHTECH VEL 74. What are the boundry conditions for TM waves in circular guides? The boundry conditions to be met for TM waves are that Eφ = q at P = a. 75. What is the wave impedances at a point? + Z xy = Ex Hy + Zy z = Ey Hz + Zzx = Ez Hx + Z yx = −E y Hx + Z zy = − Ez Hy + Zxz = −E x Hz 76. What is the work impedence for waves guided by transmission lines / wave guides? 2 2 E x + Ey Etrans = Z2 = 2 2 Htrans H x + Hy Zz = ZPφ = -Zφ p = Zz ( TM ) = β ωε µ w 2 1− c 2 ε w ω 2 = η 1- c 2 ω β = DE 77. Draw the characteristic between wave impedance and frequency. TE Wave impedance TM VEL TECH fc VEL TECH MULTI TECH TECH HIGHTECH 192 f VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 78. Define wave impedance. The wave impedances are defined by the following rations of electric to magnetic field strengths for the positive direction as well as negative direction of the co-ordinates. Z+ = xy Z+ = yz Z+ = zx Ex Hy Ey Hx Ez Hx Z -yx = Z -zy = Z -xz = Ey Hx Ez Hy Ex Hz 79. Write condition for minimum attenuation for TM waves. dα = 0 f d fc α TM = 2wμ wε . σ βa = 2wμ . σ 2πfε mπ ωμε- .a a 2 simplifying we get, f= 3fc 80. Write the wave impedances for TE, TM Q TEM at cut off frequency. * For TE waves at fc 1 ZTE =α For TM waves at fc 1 ZTM = 0 For TEM waves at fc 1 ZTEM = η 81. Relationship between α TM VEL TECH and α TE: VEL VEL TECH MULTI TECH TECH HIGHTECH 193 VEL TECH 2 αTE fc = α TM 2 f VEL TECH MULTI TECH TECH HIGHTECH VEL The ratio between the attenuation factors of TM and TE waves is given by the ratio between the cutoff frequency and the wave frequency. 82. Give the equation for the power loss at the magnetic field. Power loss = I2R = (JYZ)2 . Rs = (Hy)2 . Rs (Hy) = C4 ωμ 2σ \ Power loss= 2 C4 PART – B 1. Write the expression for transverse magnetic waves in rectangular wave guides. The wave equations are partial differential equations that can be solved by the usual technique of assuming a product, solution. This procedure leads to two ordinary differential the solutions of which are known. Nothing that Ez (x, y, z) = Ez0 (x, y) e-y2 Ez0 = xy Where X is a function of x alone, and y is a function of y alone. y 2 d 2x d 2y + x 2 + y xy = −ω 2 µε xy 2 dx dy Putting h2 = y + ω VEL TECH 2 2 µε as before VEL TECH MULTI TECH TECH HIGHTECH 194 VEL VEL TECH d 2x dy + x 2 + h 2 xy = 0 2 dx dy Divide by XY 1 d 2x 1 d2y + h2 = − X dx 2 y dy 2 VEL TECH MULTI TECH TECH HIGHTECH VEL y The above equation equates a function of x alone to a function of y alone. The only way in which such a relation can hold for all values of x and y is due to have each of these function equal to some constant. 1 d 2x + h 2 = A2 2 X dx 1 d 2y = − A2 2 y dy The solution of equation is X = c1 cos Bx + c2 sin Bx B2 = h2 – A2 The above solution is y = c3 cos Ay + c4 sin Ay This gives E20 = xy = c1c3 cos Bx cos Ay+c1c4 cos Bx sin Ay + c2c3 sin Bx cos Ay + c2 c4 sin Bx sin Ay. The constants c1, c2, c3, c4, A and B must now be selected to fit the boundary condition. E2 0 = 0 when x = 0, x = a, y = 0 y = b. If X = 0 the general expression. E20 = c1 c3 cos Ay + c1 c4 sin Ay E20 = c2 c3 sin Bx cos Ay + c2 c4 sin Bx sin Ay When y = 0 reduces to E20 = c2 c3 sin Bx For this to be zero for all values of x if it possible to have either c 2 (or) c3 equal VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 195 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL to zero (assuming B ≠ 0) putting c2 =0 would make E20 identically zero. So instead c3 will be put equal to zero. E20 = c2 c3 sin Bx sin Ay In addition to the amplitude constant c = c2 c4 If x = a E20 = c sin Ba sin Ay. In order for this to vanish for all values of y (and assuming A ≠ 0) (Because A = 0 would make E20 identically zero) The constant B must have B= mπ when m = 1, 2, 3 a mπ x sin Ab a Again if y = b; E20 = c sin A= nπ where n = 1, 2, 3. b Therefore the final expression for E20 is = c sin mπ nπ y x sin a b − j βc B cos Bx sin Ay h2 − j βc = A sin Bx cos Ay h2 jωε c A sin Bx cos Ay h2 − jωε c B cos Bx sin Ay h2 y = jβ Ex0 = Ey0 0 Hz = 0 Hy = B= mπ nπ and A = a b VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 196 VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL These expressions show how each of the components of electric and magnetic field strengths varies with x and y. that is M the z direction. ωc = 1 µε 1 mπ nπ a + b 2 2 2 fc = 2π µε 2 m π nπ a + b 2 λc = m n a +b 2 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 197 VEL . This propagation constant met with in ordinary transmission-line theory is a complex number. For low frequencies where ω 2 µ ε is small y will be a real number. is shown by putting back intro each of these expressions the factor ejwt. The variation with time and along the axis of the guide. That is y = a + j β where a is the mathematics constant and β is the phase shift constant.xz and then taking the real port. If y is real. In the division of the fields it was found necessary to restrict the constants A and B to the values given by expressions. The attenuation constant a is zero for all frequencies such that w > wc. β must be zero and there can be no phase shift along the tube mπ nπ β = ω µε − − a b 2 2 2 2 2 This means there can be no wave motion along the tube for low frequencies y = jβ . A2 + B2 = h2 h2 = y 2 + ω µ ε y = h 2 − ω 2 µε = A2 + B 2 − ω 2 µε mπ nπ 2 = + b − ω µε a The above equation defines the propagation constant for a rectangular guide for TM waves. Derive the Expression for Transverse Electric waves in Rectangular waveguides. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 198 VEL . Y → function of y only. y) = Hz0 (x. y) e-vz Hz0 (x. The wave equation. 2 f it will be longer L= 2π 2 2 2. The lowest cut off of frequency will occur for m = n = 1 This frequency TM waves which can be propagated through the guide. 2 Hz (x. This particular wave is called the TM11 wave for obvious reasons. Therefore the lowest possible values for the either m or n is unity. Substituting the value of H2 in wave equation. Since the wave length in the guide is given is given by L = v than the corresponding free space wave length. mπ nπ ω µε − − a b In the above expressions the only restriction on m and n is that they be integers.VEL TECH fcλ c = v0 v= VEL TECH MULTI TECH TECH HIGHTECH VEL ω = β ω mπ nπ ω µε − − a b 2 2 2 This last expressions indicates than the velocity of propagation of the wave in the guide is greater than the phase velocity in free space. However from the above equations it is seen that is either m or n is zero fields will all be identically zero. High order waves require higher frequencies in order to be propagated along a guide of given dimensions. in a rectangular waveguide is given by ∂ H z ∂ 2 Hz + + γ 2H z = −ω 2 µε Hz 2 2 ∂x ∂y The solution of the equation is. y) = xy Let Where x → function of x only. VEL TECH d 2x d 2y + x 2 + γ 2 xy = −ω 2 µε xy dx 2 dy d 2x d 2y + x 2 + h 2 xy = 0 dx 2 dy where G2 = γ + ω Dividing by xy 1 d 2x 1 d2y + + h2 = 0 2 2 X dx y dy 1 d 2x 1 d2y 2 +h = − X dx 2 y dy 2 2 VEL TECH MULTI TECH TECH HIGHTECH VEL y y µ ε The equation relates a function of x alone to a function of y alone and this can be equated to a constant. X = C1 cos Bx + C2 sin Bx Similarly − 1 d 2y = A2 2 y dy 1 d 2y + A2 = 0 2 y dy The solution of this equation is y = C3 cos Ay + C4 sin Ay But H20 = xy = (c1 cos Bx + c2 sin Bx) (c3 cos Ay + c4 sin Ay) VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 199 VEL . 1 d 2x + h 2 = A2 X dx 2 1 d 2x + h 2 − A2 = 0 2 X dx let 1 d 2x + B2 = o 2 X dx The solution of this equation is. Ex = − γ 2 ∂E2 jωµ ∂H z − 2 h 2 ∂x h ∂y For TE waves Ez = 0 − jωµ ∂H z h 2 ∂y − jωµ = [ -c1 c3 A sin Ay cos Bx – c2 c3 A sin Ay sin Bx + c1 c4 A cos Bx cos Ay + c2 h2 c4 A cos Ay sin Bx] Applying Boundary conditions. b nπ . so it is bitter to select A = The general solution is Ex0 = jωµ [ c1 c3 Ay cos Bx + c2 c3 A sin Ay + Sin Bx] h2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 200 VEL . it is possible either B = 0 or A = It B = 0. the above solution is identically new. when y = 0. y = b If y = 0. the general solution is Ex = − jωµ (c1 c4 A cos Bx + c2 c4 A sin Bx ] = 0 h2 For Ex = 0.c1 c3 A sin Ay sin Bx – c2 c3 A sin Ay Sin Bx] h2 Ex = 0 nπ . C4 = 0 ( c4 is common) Then the general solution is Ex = − jωµ [ .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL = c1 c3 cos Ay cos Bx + c2 c3 cos Ay sin Bx + c1 c4 cos Bx sin Ay + c2 c4 s in Ay sin Bx It is known that. For Ex = 0. Ex = \ E2 = 0. b If y = b. either A = 0 or B = a Ey0 = Since A = 0. will make Ey identically zero. it is better to take B = Ey0 = Ex0 − jωµ [ c1 c3 B sin Bx cos Ay + c1 c4 B sin Bx Sin Ay] h2 jωµ = 2 [c1 c3 A sin Ay cos Bx + c2 c3 A sin Ay Sin Bx] h mπ a Substituting the value of c2 = c4 = 0 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 201 VEL .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Similarly for Ey Ey = − γ 2 ∂E2 jωµ ∂H z + 2 h 2 ∂y h ∂x = jωµ ∂H z h 2 ∂x jωµ [ -c1 c3 B cos Ay Sin Bx + c2 c3 cos Ay cos Bx -c1 c4 B sin Bx sin Ay + c2 c4 B sin B h2 sin Ay cos Bx] = Applying boundry conditions. Ey = 0. then Ey0 = 0 jωµ Bc1 sin Bx [c3 cos Ay + c4 Sin Ay] h2 mπ For Ey0 = 0. c2 = 0 Then the general solution is Ey0 = jωµ [. Ey0 = x = 0 and x = a jωµ [c2 c3 B cos Ay+c2 c3 B cos Ay + c2 c4 B Sin Ay] h2 For Ey0 =0.c1 c3 B cos Ay sin Bx – c1 c4 B sin Bx Sin Ay] h2 If x = a. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 202 VEL .VEL TECH jωµ c1 c3 A sin A cos Bx h2 − jωµ Ey0 = c1 c3 B sin Bx cos Ay h2 Let C = c1 c2 Ex0 = Ex0 = jωµ CA sin A cos Bx h2 − jωµ CB sin Bx cos Ay h2 nπ mπ . o Hx = but Ey = − j β ∂H z h 2 ∂x jωµ ∂H z h 2 ∂x ∂H z h2 = ⋅ Ey ∂x j ∂ωµ Substituting the value of o Hx = − j β h2 ⋅ Eo 2 h jωµ y ∂H z in the above Hx0 equation. ∂x = −β 0 E ωµ y Substituting the value of Ey0 in the above Hx0 equation.B= b a VEL TECH MULTI TECH TECH HIGHTECH VEL Ey0 = Where A = Similarly for Hx0 o Hx = −γ h2 −γ = 2 h ∂H z jωε ∂Ez + 2 ∂x h ∂y ∂H z ∂x for propagation γ = jβ . Hy0 = β jωµ CA sin Ay cos Bx] ωµ h 2 jβ CA sin Ay cos Bx h2 H20 = xy = c1 c3 B cos Ay cos Bx + c2 c3 cos Ay sin Bx +c1 c4 cos Bx sin Ay + c2 c4 sin Ay Hy0 = VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 203 VEL . ∂y o Hy = = β 0 E ωµ x Substituting the value of Ex in the above equation Hy0.VEL TECH = − β − jωµ CB sin Bx cos Ay] ωµ h 2 VEL TECH MULTI TECH TECH HIGHTECH VEL Similarly for Hyo o Hy = − γ ∂H z jωε ∂Ez − 2 h 2 ∂y h ∂x −γ ∂H z = 2 h ∂y [∴ E2 = 0] For propagation γ = jβ 0 Hy = − j β ∂H z h 2 ∂y − jωµ ∂H z h 2 ∂x But Ex = ∂H z −h2 = ⋅ Ex ∂y jωµ Substituting this value of − j β ( −h2 ) o ⋅ Ex h 2 jωµ ∂H z in the above Hy0 equation. jβ CB sin Bx cos Ay h2 jβ Hy0 = 2 CA cos Bx sin Ay h 0 Hz = C cos Ay cos Bx jωµ Ex0 = 2 CA cos Bx sin Ay h − jωµ Ey0 = CB sin Bx cos Ay h2 nπ mπ Where A = . fc. mπ nπ β = ω µε − − a b 2 2 2 c iv and λ are found to be identical to ωc = 1 µε 1 mπ nπ a + b 2 2 2 fc = 2π µε mπ nπ a + b 2 The corresponding cut off wavelength is 2 λc = 2 2 m n a +b The velocity of propagation. λ those of TM waves. But c2 = c4 = 0 Hz0 = c1 c3 cos Ay cos Bx C = c1 c3 Hz0 = C cos Ay cos Bx VEL TECH MULTI TECH TECH HIGHTECH VEL The field equation of TE waves are as follows.B= b a Hx0 = For TE waves the equation for β .VEL TECH sin Bx. ω v= β VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 204 VEL . n = 0 mπ nπ h= − a b π = +0 a =π a The field expressions are Hx0 = jβ CB sin Bx cos Ay h2 2 2 2 H 0 x = π = ( a) jβ 2 C π a sin π a x ( ) ( ) Hy0 = 0 j β ac sin π a x π ( ) Hz0 = C cos π a x Ex0 = 0 ( ) VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 205 VEL . n =0) whereas the lowest mode for TM waves is TM11 (m = 1. This wave has the lowest cut-off frequency. Hence the TE10 mode is the dominant mode of a rectangular wave guide. 2 λ= 2π 2 2 The lowest mode for TE wave is TE10 (m = 1. For TE10 mode m = 1. n = 1). Because the TE 10 mode has the lowest attenuation of all modes in a rectangular wave guide and its electric field is definitely polarized in one direction every where. Explain about dominant mode in rectangular waveguide.VEL TECH = VEL TECH MULTI TECH TECH HIGHTECH VEL ω mπ nπ ω µε − − a b 2 2 2 mπ nπ ω µε − − a b 3. π fc = =c 1 µε a 1 π a 2 π 2π µε a 2a 1 . y.z) and the taking the real part of the product. t) = Ex0 = Hy0 = 0 For TE10 mode m = 1. y. y. t) = C cos π a x cos (wt .β z) Ey0 (x. 2 1 a VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 206 VEL .VEL TECH − jωµ 2 VEL TECH MULTI TECH TECH HIGHTECH VEL E 0 y = π ( a) C π a sin π a x ( ) ( ( ) = − j µwac sin π a x π ) The instantaneous field expressions for the dominant TE10 mode are obtained by the β phasor expressions in above equations with ej(wt .β z) π ( ) ( ) ωc = = C.β z) π ( ) Hz0 (x. z. z.3 x 108 m/sec where C = µ0ε 0 2 λc = λc = 2a. z. n = 0 w µa C sin π a x sin (wt . t) = − β ac sin π a x sin (wt . Hx0 (x. The surface current density on surface waveguide walls is given by Js = an x H At t = 0 When x = 0 Js = −ay Hz VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 207 VEL .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL For TE10 mode the cutoff wave length is equal to twice the width of the guide. TE10 Wave Electric and magnetic field configuration for the lower order mode in a rectangular wave guide. Its cut off frequency is independent of the dimension ‘b’ the field configurations for the lower order TE waves in rectangular wave guider. Zzx = z Hy Hx Hx VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 208 VEL . Explain Wave Impedance. 4. In a Cartesian coordinate s/m three wave impedances (impedance constants) must be defined.β z) When x = a J s = ay H z Js (x = a) = ay C cos β When y = 0 Js = ax Hz −ay Hx z β ac Js (y = 0) = ax C cos π a x cos β z .az sin π a x sin β z π When y = b Js (y = 0) = J3 (y = b) The surface currents on side walls at x = 0 and at y = b are selected the below figure. The wave impedances defines as the ratio of electric field intensity to magnetic field intensity are E E E + + + Z xy = x . ( ) ( ) Surface currents on wave guide walls for TE10 mode in rectangular wave guide.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Js (x = 0) = −ay C cos (0 . Zyx = y . wc2 µ ε h2 fc2 = 2 ( 2π ) µε = h2 The cutoff frequency is h fc = 2π µε For propagation λ must be imaginary λ = jβ λ = h2 − w 2 µε β = w 2 µε − h2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 209 VEL . ZTM = E x −E y β = = Hy Hx ωε It is known that. λ = 0 2 h 2 − ω c µε = 0 h2 – wc2 µ ε 2 wc = h 2 µε = 0. Zyx = − y . (i. the wave impedance which is seen in the direction of propagation z is given by Zz = zxy = Zyz Zz = 2 2 E x + Ey 2 2 H x + Hy For TM waves in a rectangular waveguide.) E E E − − − Z xy = − x .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The wave impedance in the opposite directions are the negative of those given above.e. the propagation constant is y = h 2 − ω 2 µε At cut-off frequency fc. Zzx = − z Hy Hx Hx For waves guided by transmission liens or wave guides. W < wc λ =a = h2 − w 2 µε 2 2 = w c µε − w c µε =w c w µε 1 − wc 2 2 (or) f a =h 1 − fc VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 210 VEL . When the operating frequency is lower than the cut-off frequency. The propagation constant is real.VEL TECH 2 = w 2 µε − w c µε VEL TECH MULTI TECH TECH HIGHTECH VEL wc =w µε 1 − w 2 f = w µε 1 − c f The wave impedance of TM waves Z+ M = p 2 ωε 2 f w µε 1 − c f = ωε = µ ZTM f 1− c ε f 2 f = η 1− c f 2 Where η is a characteristic impedance η= µ ε The wave impedance of propagating TM modes in a waveguide with a loss less dielectric is purely resistive and is always less than the intrinsic impedance of the dielectric medium. The wave impedance in a non propagation mode is ZTM = λ jωε 2 f h 1− fc = jωε f − jh = 1− ωε fc for f < fc fc 2 wave impedance is purely reactive. indicating that there is no power flow for f < For TE waves in a rectangular wave guide ZTE = = E x −E y = Hy Hx ωµ β For propagation Y must be imaginary y = j β y = j β = h2 − w 2 µε β = w 2 µε − h2 2 = w 2 µε − w c µε wc =w µε 1 − w f = w µε 1 − c f 2 2 The wave impedance of TE waves VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 211 VEL . waves are attenuated as e-y2 = e-az with z. Therefore a waveguide exhibits the property of a high pass filter.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL For a given mode the waves with frequencies lower than the cutoff frequency cannot be propagated i.e. It propagates if the operating frequency is greater than the cutoff frequency. indicating that there is no power flow for f < fc. propagation constant becomes real. When the operating frequency is lower than the cut-off frequency. f y = h 1− fc 2 The wave impedance in a non-propagating mode ZTE = = jωµ λ jωµ 2 f h 1− fc f<fc Wave impedance is purely reactive. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 212 VEL . For TEM waves in a rectangular wave guide.e.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL µ ZTE = ω β = ωµ f w µε 1 − c f 2 µ = ε 2 f 1− c f η ZTE = 2 f 1− c f Where η is the characteristic impedance η= µ ε The wave impedance of propagating TM modes in a waveguide with a loss less dielectric is purely resistive and is always larger than the intrinsic impedance of the dielectric medium. The propagation codes does not take place. i. The wave impedance for different are given below. Mod e TM TE Wave Impedance f Z = η = 1− c f η 2 Z= fc 1− f z =η = 2 TEM µ ε VEL VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 213 . Wave impedance versus frequency characteristics of waves between parallel conducting plane.VEL TECH Wave impedance ZTEM = −E x jωµ = Hy λ VEL TECH MULTI TECH TECH HIGHTECH VEL ( or ) ZTEM = λ jωε The propagation constant y = a + jβ = jβ = jω µε Substituting in wave impedance equation ZTEM = jω µε = jωε jωµ jω µε ( or ) ZTEM = µ ε µ ε = Wave impedances of TEM waves is the characteristic impedance of any medium ZTEM = η The variation of wave impedance with frequency is below. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL The phase velocity in a waveguide is given by γ = ω = β ω f w µε 1 − c f 2 f where β =w µε 1 − c f 1 r = 2 fc µε 1 − f V = Vo f 1− c f 1 2 2 Where Vo = µε The wave length in the waveguide is V λ= = f Vo f 2 f 1− c f 2 λ f Since c = o f λc ho h= 2 λo 1− λc 2 λ= or λo λc 2 λc2 − λ0 λ2 = λo2λc2 2 λc2 − λ0 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 214 VEL . It is possible for several modes to exist simultaneously in waveguides. Explain the Excitation methods for various modes? In order to launch a particular mode. In figure the probe is parallel to that y – axis and so produces lines of E in the y direction and lines of H which the in the xz plane. a type of probe is chosen which will produce lines of E and H that are roughly parallel to the lines of E and H for that mode possible methods for feeding rectangular waveguides are shown. In figure the probe parallel to the z-axis produce magnetic field lines in the xy plane.VEL TECH 2 2 λ 2 ( λc2 − λo ) = λo λc2 2 2 λ 2λc2 − λo λc2 = λo λc2 2 2 λ 2λc2 = λo λ 2 + λo λc2 VEL TECH MULTI TECH TECH HIGHTECH VEL λo2 ( λ 2 + λc2 ) = λ 2 λc2 λ02 = λ0 = λ 2 λc2 λ 2 + λc2 λλc λ + λc 5. This is the perfect field configuration for the TE10 mode. if the VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 215 VEL . In Figure the parallel produces fed with opposite phase tend to set up the TE20 mode in figure the probes which are parallel to the z-axis produces electric field liens in the xy plane for TE11 mode. This is the perfect field configuration for the TM11 mode. = ωµ m v 2f 2c 2 m2a m2 b + 2σ m 8η 2fc4 b a Power transmitted. due to this the is some less occurs in the waveguide. For the practical consideration we assumed that the waveguides has infinite conductivity that the waveguides has infinite conductivity thus there is the no loss. Rs = Surface impedance = b ωµ m a 2 2 ∴P = ∫ H x dx + ∫ Hy dy 2σ m 0 0 ωµm 2σ m Substitute the value of Hx and Hy integrate and simplify we get. Attenuation due to waveguide walls can be defines as α= 1 Power lost in guide walls 2 Power transmitted TM Waves The current induced in the waveguide alls depend on the magnitude of the Hx and Hy at the surfaces these the power lost and be written as b a 2 1 2 2 P = Js Rs = Rs ∫ Hx dx + ∫ Hy dy 2 0 0 2 2 $ Where Js2 = H. but some high value. However the waveguide dimensions are often chosen so that only the dominant mode can exist.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL frequency is above cut-off for these particular modes. 6. n = H x i + Hy j = linear current density per meter length per conducting wall. Derive the equation for Alternation in waveguides. which is known as “waveguide attenuation”. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 216 VEL . but impractical the conductivity is not infinite. for y 2 plane. 2 2 h a 2 Rs aA2 β 2 mπ 2 a = + 4 A .VEL TECH Pρ = 1 E x H y + Ey H x 2 VEL TECH MULTI TECH TECH HIGHTECH VEL a b 1 mπ c mπ x 2 nπ = 2 ωεβ ∫ ∫ cos sin a y dxdy 2 aω c µε a 0 0 1 nπ c mπ x 2 nπ y + ωεβ ∫ ∫ sin2 cos a dxdy 2 2 bω c µε a 0 0 a b abc 2 f PT = 8η fc 2 f 1− c f 2 watt ωµ m v 2f 2c 2 n2 a m 2 b + 2 4 a 1 2σ m 8η fc b α= 2 2 abc 2 f fc 1− 8η fc f n m + 2 ωµ m v f b a N /m α= 2 2σ m 2η fc2 f 1− c f for TE waves 1 2 1 2 The power lost = ∫ Jsx Rs dx = ∫ Jsz Rs dz 2 2 R = s 2 = Rs 2 2 b a 2 β 2 mπ 2 2 mπ x 2 mπ x ∫ A cos dx + ∫ h 4 a A sin a dx a 0 0 2 2 π A2 β 2 mπ 2 a + 4 A . for y 2 plane 2 2 2 h a 2 2 Rs A2 β 2 mπ nπ + Total power lost = = ( a + b ) + 4 2 h a a Power transmitted PT = ½ [Ex Hz + Ey Hx] VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 217 VEL . VEL TECH 1 A2ωµβ PJ = 2 h4 VEL TECH MULTI TECH TECH HIGHTECH VEL nπ 2 a b 2 mπ x 2 nπ ∫ ∫ cos a sin a y dxdy b 0 0 2 a b mπ mπ x nπ y + sin2 cos2 dxdy ∫∫ a a a 0 0 ab 2 ωµβ abA2 f = A 2 = ωµω µε 1 + c 8 ω c µε 8 f 2 ab µ 2 fc f = A 1+ c 8 ε f f 1 Power lost α− 2 Power transmitted R A2 = s 2 2 2 β 2 mπ nπ + ( a + b ) + 4 h a a 2 2 After simplifications. The propagation characteristics of TE & TM waves are obtained as follows. Explain the characteristics of TE & TM waves. From the above analysis we got mπ nπ h = p + ω µω = A + B = a b 2 2 2 2 2 2 2 mπ nπ 2 or P = b − ω µω a 2 2 2 or mπ nπ P= − ωµΣ a b 2 2 We know ‘P’ is Q complex number I e VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 218 VEL . πµ m α= δm 2a fc 2 1 + 2 b f nepers/m fc aη 1 − c f 7. 854 × 10 The cutoff wavelength is VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 219 −7 −12 = 1 VEL . mπ nπ 2 Hence α = a + b −ω Σ mπ nπ At high frequencies ω µΣ >> a b 2 2 2 2 2 Thus ‘p’ becomes purely imaginary [i. α = 0] hence the wave propagates mπ 2 nπ 2 j β = j ω µΣ − + a b 2 At the transition ‘p’ becomes zero. is defined as cut-off frequency’ At F = fc: P = 0 Hence the equation (1) becomes 2 2 mπ nπ 0= − ω 2 µΣ a b mπ nπ 2 or ω c µΣ = a b 2 2 1 mπ nπ or ωc2 = µΣ a b or fc = f= mπ nπ 2π µΣ a b 1 v m n + 2 a b 2 2 2 2 2 2 or = 3 × 108 m / sec Where v = 1 µΣ 4π × 10 × 8.e.e. p = α . the wave is hence the wave cannot propagate. the frequency at which ‘p’ becomes just zero.VEL TECH P = α + jβ mπ nπ 2 ∴P = α + jβ b − ω µΣ a 2 2 VEL TECH MULTI TECH TECH HIGHTECH VEL mπ nπ At low frequencies ω µΣ << + a b 2 2 2 Thus ‘P’ becomes real i. W. λg = 2π = β 2π mπ 2 ~ π 2 ω 2 µΣ − + a b Practically the guide wavelength is different from free space wavelength. λα & v = fc f µΣ ∴ λc = λ λ 1+ λc 2 squaring on both sides we get.T. λcα 1 1 . mπ nπ ω µΣ = + a b 2π λg = = 2 ω 2 µΣ − ω c µΣ 2 c 2 2 2π Thus ω ω µΣ 1 − c ω 2π v v λ = = = 2 2 2 fc fc λ ω 1.T.K. It is denoted by.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH v m 2 n 2 a b = 2ab VEL λc = v = fc v 2 ( mb ) 2 + ( na ) 2 Guide wavelength: It can be defined as the distance traveled by the wave in order to undergo a phase shift of 2π radians.K. f 1+ 1− f f λc 1 W. λ λ = 1− λ λc g 1 1 1 = 2+ 2 2 λ λc λg VEL TECH 2 2 VEL TECH MULTI TECH TECH HIGHTECH 220 VEL . VEL TECH Phase Velocity:- VEL TECH MULTI TECH TECH HIGHTECH VEL W. The wave propagates in the waveguide Wavelength λg is greater than the free space wavelength λ . thus the velocity of propagation is defined as the rate at which the wavelength changes its phase in terms of λg v p = λg f = = 2π f . mπ 2 nπ 2 β = ω µΣ − + a b = ω 2 µΣ − ω 2 µΣ = µΣ(ω 2 − ω 2 ) c Differentiate the above equation w. ‘w’ we get. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 221 VEL .t. Group velocity It can be defined as vg 2 = dω dP W.T.λg 2π = 2π f ω = (2π / λg ) β = ω mπ 2 nπ ω µΣ − a b 2 ω 2 ω µΣ − ωc µΣ 2 mπ nπ ω µΣ = + Since at cut-off frequency a b 2 c 2 2 vp = Then vp = ω ωc ω µΣ 1 − ω v fc 1− c 2 = v fc 2 1 − f = velocity of light The phase velocity is the velocity of TE & TM waves.r.T.K.K. TE11.016cm F = 10GHZ length = 1m TE10 .286cm.56 GHZ 2 2 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 222 VEL . What is the alteration for metre length of the guide when operating at the frequency of 10GHZ? Given a =2. For an a/r filled copper x –brand wave side with dimensions a=2. n=0 c fc = 2a 3 × 108 = 2 × 2.VEL TECH dβ 2ωµΣ = = 2 dω 2 (ω 2 − ωc )µΣ VEL TECH MULTI TECH TECH HIGHTECH VEL µΣ ωc 1− ω 2 fc 1− 2 dω fc f 1− or = =v c dβ µΣ or λ dω vg = = v 1− λ dβ y 2 2 8.TE2 m π nπ ωc 2µ ∈= + a b 1 2 2 2 m π nπ ωc = + M∈ a b c m n fc = + 2 a b for TE10 mode: m=1.286 × 10 −2 = 6. b=1.TE01.286cm. b=1.016cm determine the cut-off frequency of the first four propagating modes. 156 GHZ for TE02 mode: m = 0.016 × 10−2 for TE11 mode m=1. n = 1 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 223 VEL . For TE01..016 = 16.76 GHZ 2 × 1. m π nπ 2 γ= + b − ω M∈ a m n 2f = + a b c 2 2 2 2 2 If the operating frequency is less than the cut-off frequency alternation takes place i.016 × 10−2 Pr opagation constant.. propagation constant γ = α .VEL TECH for TE01 mode : m = 0.e. TE11. For TE11 mode. n = 2 fc = c2 2b 2 2 c 3 × 108 = = = 29. propagation does not take place. m =1.53 GH2 b 1.e. n = 1 fc = c 2b VEL TECH MULTI TECH TECH HIGHTECH VEL 3 × 108 = = 14.286 1. n=1 c 1 1 fc= + 2a b 2 2 3 × 108 102 102 = + 2 2. TE02 modes propagation will not take place i. 8 3 × 10 2 VEL ν = α = 265.5 Nepers / m Alternation = αl = 227. A rectangular waveguide has cross-section dimensions a = 7cm & b = 4cm.016 2 2 VEL TECH MULTI TECH TECH HIGHTECH 2 × 10 × 109 .88 × 1 = 581. m=0.88 Nepers 9.016 3 × 10 α = 227.016 3 × 10 α = 581. n=1 102 2 × 10 × 109 γ=π − 8 1.4 Nepers for TE01 mode.77 Nepers in Alternation = αl = 337.5Nepers 2 × 102 2 × 10 × 109 γ=π − 8 1.4 × 1 = 337.286 1. λo = c f 3 × 108 = = 5cm 6 × 109 2 2 2 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 224 VEL .88 Nepers / m Alternation = αl = 581.VEL TECH 102 102 ν=π + 2. Determine all the modes which will propagate through the waveguide at frequency of 6 GHZ Given a = 7cm = 7 x 10-2m b = 4cm = 4 x 10-2m f = 6GHZ The cut-off wavelength 2 λc = 2 2 m n + a b I m ethod.5 × 1 =227. λ c. TE01.n = 1 λ c = 2b = 2 × 4=8cm λ c > λo . TE11. m=1.. for TE11 mode.228 rad / m propagation is possible for TE11 mode for TE zo mode m = 2. for TE01 mode : m = 0. since λ c.e. should be greater then the minimum wavelength (λ o) for propagation. n=1 102 102 γ = π + − 1600 7 4 = 87. n = 0 2 × 102 γ=π − 1600 7 = j87.VEL TECH for TE10 mode : m = 1.propagation is possible for TE01 mode. TE20 modes will propagate VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 225 VEL 2 2 .λ o propagation is possible for TE10 mode.95rad / m Propagation is possible for TE20 mode For TE02 mode: 2 × 102 γ=π − 1600 = 94.n = 0 λ c = 2a = 2 × 7=14cm VEL TECH MULTI TECH TECH HIGHTECH VEL If λ c>λ o then the propagation takes place the cut off wavelength . γ = α (∴ β =0) Propagation will not take place in this made ∴ TE10.24 4 γ Becomes real value i. 145 × 10−2 m Characteristic wave impedance VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 226 VEL . A rectangular waveguide measures 3 x 5 cm in equally of ahs a 10GHz signal propagated in it.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 10. Given a = 5 cm = 5 x 10-2 cm b = 3 cm = 3 x 10-2 cm f = 10 GHz.145cm / 2. n = 0 Cut – off wavelength 2a λC = m =2 × 5 × 10 −2 = 10 × 10-2m Cut – off frequency c fc = λC 3 × 108 = 10 × 10 −2 = 3GHz Guide wavelength λ0 λg = 2 fe 1= f But λ 0 = c f 8 3 × 10 = 10 × 109 = 3cm or 3 × 10-2m λg = 3 3 1− 10 2 = 3. TE10. Calculate the cut off wavelength the guide wavelength of the characteristic wave impedance for the TE10 mode. m = 1. B) TE30 & TE12 wave the same cut off frequency.5 GHz the guide wave length for TE10 mode is 90% of the cut off wave length. Given f = 7.VEL TECH 2z = η f 1− c f 120π 2 2 VEL TECH MULTI TECH TECH HIGHTECH VEL ( Q η=120πΩ ) = 3 1− 10 = 395.81λ0 = 1.24 cm VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 227 VEL 2 2 .5GHz λg = 0.81× 4 = 7.9 λe λ0 = c f 3 × 108 = = 4cm 7.5 × 109 λ0 λ 1+ 0 λC 2 2 λg = λg = λc 1 2 λ 1− 0 λc cross multiplying λs λg − =1 λ 0 λc 2 λg λg = 1+ λ0 λc = 1+(0.19Ω 11. Design a rectangular waveguide with the following specifications a) At a 7.81 λg = 1.9) 2 = 1. 04) ( 12.04 a = 2 a = 4.00687 = 0.24cm VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 228 VEL . n = 2 m n + a b = 0.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL But TE10 : λg = 0.29 = = 8.067cm 4 m n + = 2 a b λc 2 2 2 for TE 30 : for TE12 : m = 1.9) 2 = 1.9 λc λ λc = g 0.055 A b2 = 0.0619 − 0.24cm =7.04 cm 0.81 λg = 1.067 cm λs λg = 1+ λo λo 2 2 λc = 2 2 ( 2/b ) 2 = 4 1 − 2 (8.9 2a λc = m λc =2a λ a = c 2 8.069 ) 2 = 1+(0.055 b = 8. λc = 2a m λm a= c 2 a = 12.02 cm.9 7.5cm a = 12.81 > 0 = 1081× 4 =7. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 229 VEL .04cm But 0.067 cm.9 λc 7. 2 m n a b 2 2 2 2 For TE12: λc = m n 2 + = 4/λ a b 4 1 2 − = 2 2 b (8.04 2 a = 4.0 ~) (12.067) = 0.067cm.5 cm a =12.00687 = 0. n=2. M=1.055 b2 = 4 0.02cm a= For TE30: λc = 2a m > xm a= 2 a = 12.24 = = 8.055 b = 8.0619-0.9 2a λc = m For TE10: λc = 2a a = λc / 2 8.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL λ9 = 0. How many TE modes will this waveguide transmit at frequencies below 4 GHz? How these mode are designated & What are their cut-off frequencies? Given a = 0.08 2 Propagation constant.1 0. An air filled hollow rectangular conducting waveguide has cross section dimensions of 8x10 cm.08m. mπ nπ 2 ν= + − ω µΣ a b mπ nπ 2π f = + − a b c =π m n 2f + − a b c 2 2 2 2 2 2 2 2 f 4 ×109 40 = = c 3 ×108 3 m n 2 × 40 γ =π + + 0. f = 4 GHz λ = c/ f = 3 ×108 = 0. b = 0.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 12.08 3 Let m = 1.075m. 4 ×109 The cut-off wavelength λc = 2 m n a b 2 2 2 2 = m n 0. n = 0 2 2 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 230 VEL .1m.1 0. m = 2.4 rad/m m=0. n=2 2 2 1 2 80 γ =π + − 1 0.8 3 VEL TECH 2 2 2 VEL TECH MULTI TECH TECH HIGHTECH 231 VEL .08 3 = j74rad/m m=1. n = 0.08 3 =j64 rad/m.66rad/metre 2 2 VEL TECH MULTI TECH TECH HIGHTECH VEL 1 80 m=0. n=1.1 0. n=1 γ = π 0+ − 0. 1 80 γ =π 0 + + 0.08 3 =j39. 2 2 2 2 80 γ =π +0− 0.1 3 = j 55. n=2 2 2 2 80 γ =π 0+ + 0. 2 2 2 2 1 1 80 γ =π + − 0.1 3 = j77.08 3 = j74 rad/m m=1.1 rad/m m=1. n=1.VEL TECH 1 80 γ =π +0− 0. 08 = 1. For TE10.1 = 3 GHz TE02: 3 ×108 2 fc = 0+ 2 0. TE02. TE10 : c m n fc = + 2 a b 2 2 2 3 × 108 1 fc = +0 2 1 = 1. TE20. TE11. n=2 VEL TECH MULTI TECH TECH HIGHTECH VEL 2 2 80 γ =π + − −1 0. γ is imaginary. These are the nodes will be nodes will be propagated. TE01.457 GHz 2 13. A rectangular air filled copper waveguide with dimension 2cm x 1cm VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 232 VEL . TE01: fc = 3 ×108 1 0+ 2 0.5 GHz fc = 3 ×108 2 +0 2 0.VEL TECH = 11. propagation will take place.08 3 2 2 It propagation constant γ is imaginary. The corresponding wavelength for each mode is given by.1 0.08 = 3.08 2 2 TE21: 3 ×108 2 1 fc = 2 0.1 0. TE21 modes.875 GHz 2 2 2 TE11: TE20: 3 × 108 1 1 = 2 0.08 = 2.5 GHz.708 nepers/m m=2. phase velocity.02 ×10−2 m VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 233 VEL .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL cross-section & 300m length is operation at a cuttz with a dominant mode find cut-off frequency.8 × 107 for copper . guide wavelength. characteristic impedance and atomization.5GHz Guide wavelength λg = λ fc 1− f 2 = 3.33 × 10−2 m 9 ×109 The dominant mode is TE10. Assume σ = 5. Given a = 2cm = 2 ×10−2 m b = 1cm = 1× 10−2 m f = 9 ×109 Hz l = 30cm = 30 ×10−2 m c λ= f = 3 ×108 = 3.5 1− a 2 Phase velocity VP = c fc 1− f 2 λ = 6.33 ×10 −2 7. Cut-off frequency c c fc = = λ 2a 3 ×108 = 2 × 2 × 10−2 f c = 7. 5 2.VEL TECH 3 ×108 7.5 1− 9 = 682ohms.47 ×10-2 × 1× 2 × × 2 9 = 7.5 1− a VEL TECH MULTI TECH TECH HIGHTECH = VEL VP = 5. What are the reasons for impossibilities of TEM mode in a rectangular VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 234 VEL . = 120π 2 ZTE Surface resistance Rs = = π fµ σ π × 9 ×109 × 4π ×10−7 5 : 8 × 10−7 Rs = 2.43 ×108 m / sec Characteristic impedance 7 ZTE = 2 fc 1− f 7. Alteration constant 2b fc 2 Rs 1 + a c α= 2 fc µ b o 1− Σo f 2 1 7.02 Nepeirs/m −2 2 Total Alteration α l = 0.5 1×10 ×120π 1 − 9 = 0.30 = 0. 14.475 × 10−2 ohms.02 × 0.006nepers. Calculate the cut-off frequencies of the following modes. Ey. Ey. If use apply amperes circuit law tot his magnetic field.t x & y. which is not the case in a rectangular waveguide.286 cm and b=1. 2) In the expression for the field components. if this so. Hxy & Hy have to be constant w. TE20. then this will violate the boundary conditions E tan = 0< H normal =o.016 cm. then this wave cannot be a TEM wave. 3) Let us assume that a TEM wave exist inside a rectangular waveguide which is a single conductor system. Also find our which of the modes will propagate along the waveguide and which of them will evince when the signal frequency is 10 GHz? The cut-off frequency for a rectangular waveguide equation. then there should be a centre conductor to provide return path. If this is to be satisfied than the transferees field components Ex. 15. Existence of TEM wave means the magnetic field must the entirely in the transverse plane. There no TEM wave can exist inside a single conductor waveguide. div H = 0 that is the magnetic field lines must form closed loops in the x-y transverse plane inside the waveguide. the lines of this magnetic field aloud these closed path us must be equal to the current enclosed in the axial direction current or a displacement orient in the 2-direction existence of such displacement current will require an axial component of electric field E2 B~A if E2 is present then this wave cannot a TEM wave. For a magnetic field. Further. TEM wave means that there is no variation of electric & magnetic fields in the transverse plane.r. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 235 VEL . if we put E 2=H2=0 then all other field components Ex. A x-band waveguide which is over filled has inner dimensions of a =2. TE10. X 11 = kx 2 X y11 = ky 2 y ∴ There is no wave propagation if both E2 & H2=0 since ∂ ∂ & are not zero for a ∂x ∂y rectangular waveguide. if the boundary conditions are to be satisfied for a rectangular waveguide. Hx & Hy will also be zero as could be seen from equation.VEL TECH waveguide? VEL TECH MULTI TECH TECH HIGHTECH VEL TEM mode cannot propagate in a rectangular waveguide can be proved by the following argument: 1) By definition. Tm11. if instead of displacement current conduction current exist. TM21 and TM12. Hence no TEM wave propagation can take for a rectangular waveguide. This argument holds good for any ingle conductor waveguide. 016)2 1/ 2 1/ 2 3 ×1010 2.32 For the TM11 mode = 16.286 For the TE01 mode put m= 0 and n = 1.765 + 0.016 1/ 2 1/ 2 1/ 2 3 ×1010 (2.5 = 2 2.16 GHz For the TM21 mode 3 ×1010 4 1 f c / TM 21 = + 2 a 2 b2 4 1 = 1.16GHz f c / TM 11 will be same as above Fc /TM11 = 16.286)2 (1.n = 2π µε a b For the TE10 mode put m=1 and n=0 in the above equation c π2 c 3 1010 fc / TEVo = × = = × = 6. c f c / TE11 = 2π c a 2 + b2 = 2 2 2 a b π π2 2 + 2 b a 2 1/ 2 = 1 2b µε = 3 ×1010 2 × 1.28GHz 10 1/ 2 1/ 2 For the TM12 mode VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 236 VEL .5 ×1010 [0.887]1/ 2 = 19. π 2 f c / TE01 = 2 2π mε b 1 For the TE20 mode c 4π c 2π c f c / TE20 = a z = 2π a = a 2π 3 ×1010 = = 13.56GHz 2π 92 2a 2 2.12GHz 2 × 2.286)2 + (1.286) (1.5 ×10 + 2 2 (0.016)2 = 2 (2.VEL TECH 1 VEL TECH MULTI TECH TECH HIGHTECH 1/ 2 VEL mπ 2 nπ 2 fcm.286 For the TE11 mode.06) =1. fc= Vg fc = 1− C c 2 ( 0.286) (1. Given: Vg C = 0. 1/ 2 Since the signal frequency is 10 GHz only the modes with cutoff frequencies less than 10 GHz will propagate and the others will evanesce. Find the characteristic impedance also.875)1/ 2 = 30.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 1/ 2 VEL 3 ×1010 1 4 f c / TM 12 = + 2 a 2 b2 10 1 4 = 1.016) = 1.9 f = 10GHZ a=? Z TE = ? ( i) fc Vg = C 1 − c V 2a 2 2 sub.9 ) fc = 1− c 2 ∴ fc = 4.24 GHz.5 ×1010 (0. TM21 and TM12.5 ×10 + 2 2 (2. The waves that will propagate is only TE10 mode. TE11. The modes that will evanesce are: TE 01. TE20.358GHZ for f= 10 GHZ VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 237 VEL . then what be the breath of the waveguide.191 + 3. A 10GHZ signal is propagated is a dominant mode in a rectangular wave guide if vg is to be 90% of the free space velocity of the light. 16. 5 GHZ the guide λ g for TE10 is 90% of λ c (b) TE30 & TE12 have same fc.8λ 2 2 c λg = 1.0344m and Z TE = η fc 1− b 2 Z TE = 418.9 ) λ 2 λg = 1. Given: f = 7.9 λc λg = λ λ 1− λc 2 2 λg = λ 2 1 λ 1− λc 2 2 2 λg λg λ − = 1 λ λ λc λg 2 = 1 + ( 0.868Ω 17.358GHZ 2a ∴ a = 0. Design a rectangular wave guide with following specifications (a) at .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL ( ii) fc= V = 4.5GHZ λg = 0.8 f λg = 0.053m VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 238 VEL . we get. b=2.9 λg 0.08m 3V = 5.08x109 Hz 2a 1 2π µε m π nπ a + b 2 2 fc = fc = fc ( 30 ) = fc ( 12 ) at TE12 ⇒ λ c ( 12) = 2π m π nπ a + b 2 2 sub a=0.08m.058=2a ∴ a = 0.029 m ( b) TE30 = fc= 3V = fc 2a 3V 2a ∴ λc = 2a 3 TE10 & TE30 have sane 0.063m VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 239 VEL .VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL λc = = 0.059 ∴a = 3 λc 2 a = 0.058 WKT λc = 2a 0.024 fc & λc λ c ( 30 ) = 0. ... These same functions can be expelled to appear in any two dimensional problem in which these is circular symmetry. One solution to this equation can be obtained by assuming a power series solution... either real or complex... VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 240 r α 2r . 2 ⋅4 2 ⋅4 ⋅6 2 (1/ 2e) = C1 ∑ ( −1) (r1 ) T =0 this series is convergent for all values of p. The solution of the equation dents to Bessel functions.. The differential equation involved in there problems these the form.. These functions will be considered briefly in this section in propagation for the following section on circular wave-guides.. Substitution of this assumed solution back into and equating the coefficients of like power leads to a series solution for the direrential eqauation.... d2p 1 dp n2 + + 1− p = 0 dp2 e de p2 Where n is any integer. The propagation of waves within a circular cylinder . iT is called bissels funtion of the first kind of order zero and is denoted by the symbol... Derive the Equation for Beset function: Beset Function : In solving for the electromagnetic fields with in guides or circular cross section a differential equation known as Bessel’s equation is encountered. and the electromagnetic field distribution about an in finitely long wire.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 18. p = a0 + a1e + a2p2 + . The folowing series is obtained... p = p1 = C1 2 2 2 ( 2!) ( 3!) 2 4 2 P P −P = C1 1 − 2 + 2 2 2 2 2 + ... for example in me special case when n=0 d2p 1 dp + +p = 0 dp2 p de When the power series is interasted in and the sums of the coefficients of each power of p are equated to zero.. Examples of such problems are the vibrations of a circular membrane.. 4 6 1 1 e e 2 1 − e + 2 − 2 + . In the zero order of this solution of the secend kind the following series is obtained 2 P N0 (p) = ln + r Jo(p) π R 2 α 1 r ( 1/ 2p ) 1 1 = − ∑ ( −1) 1 + 2 + 3 + ..VEL TECH J0 (p) VEL TECH MULTI TECH TECH HIGHTECH VEL The zero order refers to the fact that if is the solution case of n=0..3…etc are designated. Because all the neumarn function become infinte at p=0.2.. thermust be two linearly independent solutions for each value of n . The second soulution is known as bessel’s function of the seond kind. or neuman’s function and is designated by the symbol Nn (p) Where again n indicates the order of the funtion. The coresponded solution for n=1. The second solution may be obtained ina manner somewhat similar to thaeir used for the first but starting with a slightly different senes that is suitably manipulated to yield a solution. these second solutions cannot be used for any physical problem in which the origin is included us for example the hellow waveguide problem VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 241 VEL . J0 (p). J2 (p). + r 2 π r =1 ( r!) 2r p = AJ0 (p) + Bw 0 (p) A plot of Jo (p) and No (e) is shown is shown in figure. J3 (p) where the subscript n donotes the order of the bessel function since second order differential equation. in order to simplify the applications of the boundary conditions. Derive the solution of the field equations of cylindrical co-ordinates. The method of solution of the electromagnetic equations for guided of-circular cross section is similar to that followed for rectangular guides however.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL In is apperent the (except near the origin for No (p) ) these curves beat a marked similary to J E ( e) → No( P ) → 2 cos( P − π / 4) πe 2 sin( P − π / 4) πe 19. it is expedient to express the field equations & the wave equations in the cylindrical co-ordinate system. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 242 VEL . VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL In cylindrical co-ordinates in a non-conducting region mass ell’s equations are. ∂Η 2 + γ Ηϕ = jωΣE ρ ρ∂ϕ ∂E2 + γ Eφ = − jωΣEφ ρ∂φ −γ H ρ − ∂E2 = jωµ Eφ ∂ρ γ Eρ − ∂E2 = jwµ H φ ∂ρ 1 ∂ ( ρ H φ ) ∂H ρ − = jωΣE2 ρ ∂ρ ∂φ 1 ∂ ( ρ Eφ ) ∂E ρ − = − jωµ Hz ρ ∂ρ ∂φ These equations can be combined to give h2 H ρ = j ωΣ ∂E2 ∂Η 2 −γ ρ ∂φ ∂ρ ∂E2 γ ∂Η 2 − − ∂ρ ρ ∂φ ∂E jωµ ∂H 2 h 2 E ρ = −υ 2 − ∂ρ ρ ∂φ h 2 H φ = − jωΣ H2E ϕ = γ ∂Σ 2 ∂H 2 + jωµ l ∂φ ∂ρ h 2 = γ 2 + ω 2 µΣ The wave equation in cylindrical co-ordinates for E2 is ∂ 2 f 2 1 ∂ 2 E2 ∂ 2 E2 1 ∂E2 + 2 + + = ω 2 µΣΕ 2 2 2 2 ∂ρ ρ ∂φ ∂2 ρ ∂ρ proceeding in a manner similar to that followed in the rectangular case. let E2 = P(P) e (φ ) e VEL TECH γ2 c = E2 e γz VEL TECH MULTI TECH TECH HIGHTECH 243 VEL . VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 244 VEL . Q = ( An cos φ + b ∩ sin nϕ ) Through by h2. gives P(eh)=Jn(eh) Where Jn (ph) is Bessel’s function of the first kind of order n sub the solution of (3) & (5) in (2 E2 = S-7 ( ( ρ h)( An cos nϕ )e The solution of Hz will have exactly the save form as fel E2 & can ∴ be written H 2 = Jn( Ph)(cn cos nϕ + Dn sin nϕ )e V2 V2 for Tm waves the remaining beld components can be obtained by inserting (b) into equation (b) for 7E waves (7) must be inserted into the set corresponding to(b). Derive the equation for Tm & TE waves in circular waveguides. The structure of above equation is.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH alone. 1 d 2P 1 1 d 2Q + + + + h2 = 0 2 2 2 P dP ρ P QP dQ 2 d 2 P QdP d 2Q + + P 2 + PQγ + ω 2 µΣPQ = 0 2 dP Pdl dQ As before equation above can be broken up in to two ordinary different equations d 2Q = − n 2Q dQ 2 d 2 P 1 dP 2 −n2 + +h dP 2 P dP P 2 P = 0 Where n is a constant. 20. VEL Sub the Where P(p) is a function of P alone & Q(φ ) is a function of φ expression for E2 in the wave equation gives. equation (32) is founded in to d 2P 1 dp n2 + + 1 + p=0 d ( ph) 2 Ph d ( ph) ( p 2)2 This is a standard form of Bessel’s equation is term of (eh) using only the solution that is finite at (eh)=0. Q by PQ. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL As in the case of rectangular guides. This in turn means that only the first few roots of (8) will be of practical interest the first few roots are. There is an infinite number of possible TM 2 waves corresponding to the infinitive number of roots of (8) As before h2= γ + ω 2 µΣ & in the case of rectangular guides h2 must be less than ω 2 µ 2 for transmission to occur. The Bonn dry conditions require the E2 must vanish at the surface of the guide ∴ from (6) Jn(ha) = 0 (8) Where a is the radius of the guide.405 (ha)11 = 3. The various Tm waves will be referred to as Tm0.25 – (ha)12 = 7. (ha)01= 2. There extract high frequencies will be required.02 The first subscript refers to the value of n & the second refers to the roots n their order of magnitude. The expression for Tm waves in circular guides are E20=An Jn (h ρ )cos n ϕ 1 ρ ° = = − jAn2 ω Σn Jn((h) sin nϕ ) H h ρ VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 245 VEL . & Tm12 etc.85 (ha)02= 5. it is convenient to divide the possible solution for circular guides into transverse magnetic & transverse electric waves for the TM waves H2 is identically zero and the wave equation for E2 is used. Since γ = h 2 − ω 2 µΣ this gives for β β mn = ω 2 µΣ − h2 nm the cut-off or critical frequency below which transmission of a wave will not occur is hnm 2π mΣ Where fc = (ha)nm a The phase velocity is hnm= V= ω ω 2 µΣ − h2 nm from equation (b) the various components of TM waves can be computed of Tm waves can be computed in terms of E2. 33 The corresponding TE waves are referred to as TE01. For transverse electric waves E2 is identically zero & H2 is given by equation (6).VEL TECH − jAn ωΣ jn( ρ h) cos nϕ h β El ° = Hϕ ° ωΣ β Eϕ ° = H ρ° ωΣ Hϕ ° = VEL TECH MULTI TECH TECH HIGHTECH VEL The variation of each of there field components with time & in the 2 direction are shown by multiplying each of the expressions of (3) by the factor ei (ω t − β z ) & taking the real part.83 (l a)102 = 7. of course that the roots of equation (7) are to be used in connection with TE waves only. β . The first few of these roots are (ha)101= 3.84 (ha)112 = 3. The equations shows that the wave having the lowest cut-off frequency is the VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 246 VEL . The relative amplitudes of An & Bn determine the orientation of the field in the guide. λ .e. In the original expression (6) for E 2 1 the arbitrary constant bn has been part equal to zero. TE11 & so on The equations for fc. & for a circular guide & any particular value of n1 the ϕ =0. &γ are identical to those for the Tm waves. can always i. It is understood. the remaining for Tm waves in circular guides are. H 2 ° = CnJn(hl ) cos nϕ − j β cn Jn '( h ρ ) cos nϕ h φ jo β cn Hϕ ° = 2 Jn(h ρ )sin nϕ h ρ −ωµ Eρ° = H ρ° β Hl ° = The boundary conditions to be met for Tm waves are that Eφ at ρ = a1 from (b) Eφ is proportional to ∂Hz / ∂ρ & ∴ to Jn’ (h ρ )’ (h ρ ) where the prime denote the derivative write (h ρ ) ∴ for Tm waves the boundary conditions require that Jn −1 (ha) = o & its the roots of (47) which must be determined.02 (ha)111= 1. by substituting (6) into (5). oriented to make either An or Bn equal to new. interest centers on the wave impedance which is seen when looking in the direction of propagation that is along the =axis. 21. + − − Hy = −Ey Hx = Ex 2 + Ey 2 Hx + Hy 2 2 = β ωΣ ∴ 2xy = 2yx = β = 2z ωΣ The wave impedances looking in the 2-direction are equal & may be put equal to 2 z. Derive the expression for wave impedance & characteristic impedance. For waves guided by transmission lines 1 wave guider. A similar inspection of egn (b) for in waves in circular grids shows that for them also 2z=2pφ = -2φ ι = β ωΣ If is seen that for in waves in rectangular a circle girder a indeed in cylindrical grids of any cress-section the wave impedance in the direction of propagation U Constant over the cross section of the guide.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL TE11 wave. The wave impedances at a point have been defined by equation already. ω c2 ΜΣ = h 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 247 VEL . β = ω 2 µ − h2 & that the cut-off angular frequency ω c has been defined is that frequency that wakes. & is the same for girder of different shapes reaching that. The wave having the next lowest cut-off frequency is the Tm01. where E 2z = trans = H trans Σx 2 +Σy 2 Hx 2 + Hy 2 is the ratio of the total transverse electric field transverse electric field straits to the total transverse magnetic field strength. γ= uo ω 1 1 = = = 2 2 β µΣ 1 − (ω c / ω ) 1 − (ω c2 / ω 2 ) Where Vo = 1 µΣ . d the wave impedance reducer to.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL it follows that β = ω M Σ 1 − (ω c 2 / ω 2 ) Then from (3) a (+) the wave impedance is the 2-direction for Tm waves is 2 z (Tm) = m 1 − (ω c 2 / ω 2 ) Σ = 7 1-(ω c 2 / ω 2 ) Thus for any cylindrical guide the wave impedance for Tm waves is dependent only on the intrinsic impedance of the dielectric & the ration of the frequency to the cut-off frequency For TE waves the same conclusion can be recharged. found that. The wavelength in the guide measured in the direction of propagation. d µ & Σ are the constants of the dielectric. Zz (TEM) =7 The dependence of β on the ratic of frequency to cut-off frequency as shown by (3) effects the phase velocity & the wavelength in a corresponding manner thus the phase or wave velocity in a cylindrical guide of any cross section is given by. is V 2π 1 λ= = ρ β f µΣ 1 − (ω c 2 / ω 2 ) 1-(ω c 2 / ω 2 ) Where λ o is the wavelength of a TEM wave of frequency f in a dielectric having the constants µ & Σ . Since ω 2 / ω 2 = λ o 2 / λ c2 it follows that = >o VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 248 VEL . Z z (TE ) = However for TE waves it is ωµ 7 β 1 − (ω c2 / ω 2 ) for TEM waves between parallel planes or on ordinary parallel wire or co-trig transmission lines the cut-off frequency is zero. the characteristic impendence for the TE 10 wave in a rectangular guide is found to be. 20 can be defined in terms of the voltage-current ratio or in terms of the power transmitted for a given voltage or a given current.I ) = π bGwm π b 2 = = 2a β 2a π b7 2a 1 − ( fc 2 / f 2 ) Terms of the second definition.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL λ= λo λc λc2 − λo2 λo = λλc A quantity of great usefulness in correction with ordinary two-conductor transmission liner is the characteristic impeding. That is for or infinitely long line. I ) 8a x Terms of the third definition the integrated characteristic impedance. For ordinary transmission fives these definitions are equivalent but for wave guides they lead to three values that depend upon the guide dimensions in the same way but which differ by a constant. Un=-Hx= − j β ac πx sin π a The total longitudinal current in the lower face is − j 2a 2 β c I = ∫ J 2 dx = π2 b a Then the integrated characteristic impedance by the first definition 20(V . The voltage will be taken as the maximum voltage from the lower face of the grids to the upper face this warms at x=9/2 & has a value. I 2ω ν v∗ z o = ∗ . Z o (w. For example consider the three definitions given by (6) for the case of the TE 10 mode in a rectangular grids. 20= π 2ω Where V & I are peak phasors. I) = π 2b π Z z = Z o (u . v Zo = . 20 of the line for such lines . VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 249 VEL . 22. At the function will increase the eventually decrease the power transmission. If a device is used in a circular waveguide in such a wary that it excites only a 2 – component of electric field. A common method of excitation of TM modes in a circular waveguide by co-axial line is shown. or the other hand if a device is placed in a circular waveguide in such a way that it exists only the 2-component of magnetic field. I ) a π VEL TECH MULTI TECH TECH HIGHTECH VEL Zo(W. the wave propagative through the guide will be in the TM mode. V) = Explain the excitations of modes in circular waveguides:TE modes here no 2 component of an Electric field .VEL TECH 2b 4 Z z Z o (u . & TM modes have no 2 – component of magnetic field. 1) Cut-off wavelength 2) Group velocity 3) Grid wavelength 4) Phase velocity VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 250 VEL . The methods o excitation for various modes in circular waveguides are shown. Given a circular waveguide used for a signal at a frequency of 11GHz propagated in the TE11 mode & the internal diameter is 4-5 cm. when the guide is connected to the source by a co-axial a discontinuity problem. At the end of the co-axial line a large magnetic field exists in the direction of propagation the magnetic field from the co-axial line will excide the TM modes in the guide however. Calculate. it is after necessary to place a turning device around the function in order suppress the reflection. the traveling wave will be in the TE mode. 029 × 108 m / sec. Phase velocity : λ = g c λ VP 0.84 λ = c/f 3 ×108 = 11× 109 = 0.0768 2 iii) = 0. A λ 1− λc 2 ii) Guide wavelength: λg = = 0.02727 iv) = 0.029 8 = 3 ×10 0.25 1. Group velocity: VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 251 VEL .02727 1.02727 0. 0.25 for TE01.029 m.02727 m. i) cut – off wavelength: 2π a λc = (ha )11 = 2 π × 2.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 5) Characteristic impedance Given f = 11 GHz d = 4. a = 2. (ha)11 = 1.5 cm.84 = 7.68 cm / 0.0768 m. 80u ×108 m / sec v) Characteristic impedance: = 20 λ 1− λc 120π 2 Z = 0.068. a=A/2 =2cm For TE11: (ha)11 = 1.02727 8 = 3 ×10 0.3m Xc 1.0333m.2727 1− 0.029 = 2. 23. Given: F = 9 GHZ D = 4cm. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 252 VEL .8 β cm / 0.VEL TECH λu = c λ g Vg VEL TECH MULTI TECH TECH HIGHTECH VEL 0.33cm / 0.029 2 = 1108 ~ . 9 × 109 Guide wavelength.8~ x = c/ f = = 3 ×102 = 3. Calculate the cut-off wavelength the guide wavelength & the characteristic wave impedance diameter is ucm for a g GHZ signal propagated in it is the TE11 mode.8~ cut-off wavelength 2π a = ( ha ) 11 2π a = 6. 83 Zz =435. 6.8~1 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 253 VEL .33 1. 6.33 Σ. Determine the cut-of frequencies of the first two propagating modes of circular waveguide with a=0.25 if the guide is 50cm in length operating at f=13 GHz determine the attenuation Given : A=0. 24. µ = µ 0 l = 50cm.83 2 = 3.81cm/ 0. = 0.25.5cm = 0.5x10-2m Σr = 2.0381 m.832 × 3 × 108 for = 2π × 0.35 1.5 cm & Σ1 = 2.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL λ9 = = λ 1 − (λ / λ c ) 2 3.1 = 3.6 GHz for TE11 mode: (ha)11=1. Diacharacteristic wave: m Polence = = η 1 − ( > / > c) 2 120π 2 3.5 × 102 =36.832 3.5m f = 13∠Hz Cut-off frequency: (ha )11c Fc = 2π a Fx TE01 mode: (ha) 0.7 ohms. 5 x 0.8 ~ 1× 3 × 108 2π × 0.5 = 324.26 Nepers/m 25. Propagation constant γ becomes real value i.e.8~1 A = 5 x 10-2m (a) Cut –off frequency: (ha )11 c = 2π a 1.5~ than the Alternation α l = 648.832 2π × 13 × 10 × 2.5 ×10 2 α = 648.8 ~ 1× 3 × 103 2π × 5 ×10−2 b) The phase constant in the guide is . A TE11 mode is propagating through a circular waveguide.5 Nepers / m.5 ×10 −2 VEL TECH MULTI TECH TECH HIGHTECH Fc= VEL Propagator constant γ = hnm − ω 2 µΣ (ha) nm 2 = + ω µΣ a (or) TE01 2 2 1 c = ∴ µΣ Σγ 2 (ha )01 2π f Σγ γ= − c a 2 9 3. The radius of the guide is 1cm and the guide content an air dielectric a) Determine in the out-off frequency b) Determine the wavelength in the guide for an operating frequency of 3GHz.25 = − 2 3 ×108 0. = VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 254 VEL .VEL TECH 1. v= α If the length of the waveguide is 0. c) Determine the wave impedance in the guide. For TE11 mode (ha)11 = 1. 8 ~ 1 β = (2π × 3 ×109 ) 2 4π × 10−7 × 8. An air filled circular loan guide having an inner radius of icon is excited in dominant mode at 10 GHz.3 cm.9 λ = 12.9 2TE = 465 ohms. mode (ha)11 = 1.854−12 − −12 5 ×10 the wavelength in the guide is 2π β 2π λ= 50.. Find the cut-off frequency: guide wavelength and wave impedance. 26. Find the bandwidth. For TE.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL β = ω 2 µ − h2 f = 3 × 109 µ = 4π ×10−7 Σ = 8. For operation in dominant mode only.854 ×10−12 ( ha)11 h= a 1.84) F = 10x10-9H2 A = 1x10-2m a) unit off frequency: VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 255 VEL . Find the a) cut off frequency of dominant mode at 10GHz . The dominant mode is TE. λ= c) the wave impedance is 2TE = = ωM β 2π × 3 × 109 × 4π × 10−7 50. 795 = 2.795 1− 10 = 792 Bandwidth = cut-off frequency of TM01-cut-off frequency of TE.405 fc of TM 01 = 1.VEL TECH fc = (ha ).3 ×10−2 m c) wave impedance: zTE = η fc 1− f 120π 2 8.49 . c 2π a 1.405 × 3 ×108 2 × π ×1×10 −3 Band width = 11..841× 3 × 108 = 2 ×1× 10−2 VEL TECH MULTI TECH TECH HIGHTECH VEL b) Guide wavelength: λg = λ 2 fc 1− 2 1 λ = c/ f 3 ×102 = 10 ×10−9 = 3 × 10−2 m λg = 3 × 10−2 8.795 1− 10 2 = 6.49 GHz RECTANGULAR WAVEGUIDES VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 256 VEL .8. ( ha ) 01 c fc of TM 01 = 2π a (ha) 01 = 2.695 GHz = 2 =11. Calling that constant as kx2. that is Ez ( x. y ) = 0 2 ∂x ∂y 0 If we use the method of separation of variables. Consider the shape of the rectangular waveguides above with dimensions a and b (assume a > b) and the parameters e and m. we get the following equation.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 27. °2 N xy Ez0 + h 2 Ez0 = 0 0 − gz Since Ez (x.z) = Ez ( x. Derive the field configuration. Y ( y ) we get 2 d 2 X ( x) 1 d Y ( y) − = + h2 X ( x ) dx 2 Y ( y ) dy2 1 Since the right side contains x terms only and the left side contains y terms only. should be solved from equation for TM mode. For TM waves. y ) = X ( x ) . we get. y ) e . they are both equal to a constant.y. cut-off frequency and velocity of propagation for TM waves in rectangular wave guides. ∂2 0 ∂2 + 2 + h 2 E x ( x. d 2 X ( x) + k x2 X ( x ) = 0 2 dx 2 d Y ( y) 2 + k yY ( y ) = 0 2 dy 2 2 2 Where k y = h + kx VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 257 VEL . Hz =0 and E. we have H x0 = jW ε ∂Ez0 h 2 ∂y jW ε ∂Ez0 h 2 ∂x 0 Hy = − Y ∂Ez0 h 2 ∂x γ ∂Ez0 0 Ey = 2 h ∂y 0 Ex = − From these equations. y ) = E0 sin α 2 2 2 From k y = h − kz . x where kx = mp/a. m= 1. we conclude that X(x) is in the form of sin k.3… Y (y) is in the form of sin kyy. y ) = 0 Ez0 ( a. b ) = 0 From all these.2. n=1. we get VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 258 VEL . 0 ) = 0 Ez0 ( x. y ) = 0 Ez0 ( x.3…. 0 So the solution of Ez ( x. Also we have the boundary conditions of.2. we have nπ x sin y b V / m mπ h = a 2 nπ + b 2 For TM waves. where ky = np/b. we should solve for X and Y from the preceding equations. y ) is mπ Ez0 ( x. Ez0 ( 0.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Now. the cut-off wave number is mπ nπ kc = + a b 2 2 and therefore. Therefore. y ) = − Where mπ nπ γ = j β = j w µε − − a b 2 2 2 Here. y ) = − 0 H x ( x. y ) = − VEL TECH MULTI TECH TECH HIGHTECH VEL γ mπ mπ nπ x sin y E0 cos 2 h a a b γ nπ h2 b jwε h2 jwε h2 mπ Eo sin a nπ x cos y b nπ x cos y b nπ x sin y b 0 E y ( x. When we observe the above equations we see that for TM modes in rectangular waveguides neither m nor n can be zero. 1 f = 2 εµ VEL TECH m n + a b 2 2 ( Hz ) VEL TECH MULTI TECH TECH HIGHTECH 259 VEL . Therefore. the lowest mode for rectangular waveguide TM mode is TM11 Here. M denotes the number of half cycle variations of the fields in the x-direction and n denotes the number of half cycle variations of the fields in the y-direction. This is because of the fact that the field expressions are identically zero if either m or n is zero. y ) = − nπ b mπ b mπ Eo sin a mπ Eo cos a 0 H y ( x.VEL TECH 0 E x ( x. β = k 2 − kc2 The cut-off frequency is at the point where g vanishes. m and n represent possible modes and it is designated as the TM mn mode. the wavelength of a plane wave in the filling medium. Such modes are called cut-ff or evanescent modes. we get from the expressions for Ex and Hy (see the equations above).VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL Since I= u/f. we have the cut-off wavelength λc = 2 m n + a b 2 2 ( m) At a given operating frequency f. the dominant frequency is ( 1 f c ) 11 = 2 εµ 1 1 + a b 2 2 ( Hz ) The wave impedance is defined as the ratio of the transverse electric and magnetic fields. The attenuation constant due to the losses in the dielectric can be found as follows: VEL TECH VEL TECH MULTI TECH VEL TECH HIGHTECH 260 . The mode with the lowest cut-off frequency is called the dominant mode. only those frequencies . Since TM modes for rectangular waveguides start from TM11 mode. ZTM = Ex Y jβ jβ βη = = = ⇒ ZTM = H y jW ε jwε jwε k The guide wavelength is defined as the distance between two equal phase planes along the waveguide and it is equal to λ= 2π 2π > =λ β k Which is thus greater than 1. Therefore. The modes with f < fc will lead to an imaginary b which means that the field components will decay exponentially and will not propagate. The phase velocity is up = w w > = β k 1 µε Which is greater than the speed of light (plane wave) in the filling material Attenuation for propagating modes results when there are losses in the dielectric and in the imperfectly conducting guide walls. which have fc < f will propagate. − 2 d 2 X ( x) 1 d Y ( y) = + h2 2 2 X ( x ) dx Y ( y ) dy 1 Since the right side contains x terms only and the left side contains y terms only.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 2 2 2 VEL f f f σ γ = j β = j k − k = jk 1 − c = jw 1 − c = jw µ ε + 1− c jw f f f 2 2 c 28. Derive the field configuration cut-off frequency propagation for TE waves in rectangular wave guide. 0 − gz Since Hz (x. they are both equal to a constant. ∂2 ∂2 + 2 + h 2 H z0 ( x. we get the following equation.y) =X(x). we get. that is Hz0(x. Y(y) we get. y ) = 0 2 ∂x ∂y If we use the method of separation of variables. d 2 X ( x) + k z2 X ( x ) = 0 2 dx VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 261 VEL . TE Modes and velocity of Consider again the rectangular waveguide below the dimensions a and b (assume a>b) and the parameters e and m.z) = H z ( x. For TE waves Ez =0 and Hz should be solved from equation of TE mode. Calling that constant as kx2. y ) e .y. y ) = H 0 cos a 2 2 2 From k y = h − kz . we must solve for X and Y from the preceding equations. we have. nπ x cos y b ( A / m ) mπ nπ h = + a b 2 2 2 For TE waves. we get mπ H z0 ( x. Aslo we have the following boundary conditions: ∂H z0 = 0 ( Ey = 0) at x=0 dx ∂H z0 = 0 ( Ey = 0) at x=a dx 0 ∂H z = 0 ( Ex = 0 ) at y=0 dy ∂H z0 = 0 ( Ex = 0 ) at y = b dx From al these. we obtain VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 262 VEL .VEL TECH d 2Y ( y ) 2 + k yY ( y ) = 0 dy 2 2 2 2 Where k y = h − kx VEL TECH MULTI TECH TECH HIGHTECH VEL Here. we have γ H =− 2 h γ 0 Hy = − 2 h 0 x 0 Ex = ∂H z0 ∂x ∂H z0 ∂y jwµ ∂H z0 h 2 ∂y jW µ ∂H z0 h 2 ∂x 0 Ey = From these equations. m and n represent possible modes and it is shown as the TE mn mode. Here. the cut-off wave number is mπ nπ kc = + a b 2 2 And therefore. fc = 1 2 εµ m n + ( Hz ) a b 2 2 Since I = u/f. Therefore. y ) = γ mπ h2 a γ nπ h2 b mπ H 0 sin a mπ H 0 cos a nπ x cos y b nπ x sin y b 0 H y ( x. y ) = Where mπ nπ γ = jβ = j w µ s − − a b 2 2 2 As explained before. we have the cut-off wavelength VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 263 VEL . y ) = 0 H x ( x. β = k 2 − kc2 The cut-off frequency is at the point where g vanishes. y ) = VEL TECH MULTI TECH TECH HIGHTECH jwµ nπ mπ nπ x sin y H 0 cos 2 h b a b jwµ mπ h2 b mπ H 0 sin a nπ x cos y b VEL 0 E y ( x.m denotes the number of half cycle variations of the fields in the x-direction and n denotes the number of half cycle variations of the fields in the y-direction. .VEL TECH 0 E x ( x. it is the dominant mode of a rectangular waveguide with a>b and so the dominant frequency is ( f c ) 10 = 1 2a µε ( Hz ) The wave impedance is defined as the ratio of the tranverse electric and magnetic fields. only those frequencies. the wavelength of a plane wave in the filling medium. we get from the expressions for E x and Hy (see the equations above). Since TE 10 mode is the minimum possible mode that gives nonzero field expressions for rectangular waveguides. The modes with f<fc will not propagate The mode with the lowest cut-off frequency is called the dominant mode.VEL TECH 2 m n + a b 2 2 VEL TECH MULTI TECH TECH HIGHTECH VEL λc = ( m) At a given operating frequency f. which have f>fc will propagate. The phase velocity is up = w w > = β k 1 µε Which is greater than the speed of the plane wave in the filling material The attenuation constant due to the losses in the dielectric is obtained as follows: f f f σ γ = j β = j k − k = jk 1 − c = jw 1 − c = jw µ ε + 1− c jw f f f 2 2 c 2 2 2 VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 264 VEL . Therefore. ZTE = Ex jwµ jwµ kη = = ⇒ ZTE = Hy γ jβ β The guide wavelength is defined as the distance between two equal phase planes along the waveguide and it is equal to λg = 2π 2π > =λ β k Which is thus greater than 1. (PART – TIME) FOURTH SEMESTER REGULATION 2005) PART . VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 265 VEL . NOVEMBER/DECEMBER 2008 FIFTH SEMESTER ELECTRONICS AND COMMUNICATION ENGINEERING EC 1305 – TRANSMISSION LINES AND WAVEGUIDES (COMMON TO B.016 cm. b= 1. DEGREE EXAMINATION. with dimensions a=2.286 cm.E.VEL TECH After some manipulation. we get VEL TECH MULTI TECH TECH HIGHTECH VEL αd = cn f 2 1− c f 2 = k 2 tan δ 2β Example: Consider a length of air-filled copper x-band waveguide. Briefly discuss the difference between wavelength and period of a sine wave.A 1./B.TECH. find the cut-off frequencies of the first four propagating modes. Solution: From the formula for the cut-off frequency 1 fc = 2 εµ c m n m n air − filled → + + 2 a b a b 2 2 2 2 ( Hz ) B.E. (a) Derive the general transmission line equations for voltage and current any point on a line. the attenuation constant is 74. Distinguish between wave guides and cavity resonator.048× 10-4 ∠ 88. 6. Or (b)(i) Write a brief notes on frequency and phase distortions.2oΩ . L. 9. PART – B 11. G and C per meter and the phase velocity on the line. Calculate the line parameters R. (ii) The characteristic impedance of a 805m-long transmission line 94∠ 23. Give the minimum and maximum value of SWR and reflection coefficient. Find the attenuation and phase shift constant of a wave propagating along the line whose propagation constant is 1. 3. 7. VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 266 VEL . Calculate the cut-off frequency of a rectangular wave guide whose dimensions are ‘a’=2.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 2. How is the TE10 mode launched or initiated in rectangular wave guide using an open ended coaxial cable? 8.5cm operating at TE10 mode.5cm and ‘b’=1. Enumerate the properties of TEM waves between parallel planes of perfect conductors. Why is the quarter wave line called as copper insulator? 5. Why is the Bessel’s function of the second kind (neumann’s function applicable for the field analysis inside the circular wave guide? 10. 4.8o.5× 10-6 Np/m and the phase shift constant is 174× 10-6 rad / m at 5KHz. Plot the frequency – versus – attenuation characteristic curve of TM and TE waves guided between parallel conducting plates. Use the SMITH chart to find. And also derive the expressions for the cut off frequency and phase velocity from the propagation constant. (2)VSWR. Find the cut-off frequency and Cut-off wave length. cut of frequency and velocity of propagation for TE waves in rectangular wave guide. (a) Derive the field configuration.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL 12.3λ long and (4)Input admittance of the line. (ii) Describe the Velocity of propagation of wave between a pair of perfectly conducting plates. Or (b) (i) Discuss on the characteristics of TE. Or (b) A TE10 wave at 10 GHz propagates in a X-band copper rectangular wave VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 267 VEL . (3) Input impedance of the line. (ii) A 50Ω loss less transmission line is terminated in a load impedance of ZL=(25+j50)Ω . TM and TEM waves between parallel conducting planes. 14. Find the characteristic impedance of the quarter wave transformer. (ii) A parallel perfectly conducting plates are separated by 5cmin air and carries a signal with frequency of 10 GHz in TM11 mode. (a) (i) A 75Ω loss less transmission line is to be matched to a resistive load impedance of ZL=100Ω via a quarter-wave section. Or (b) A 50Ω loss less feeder line is to be matched to an antenna with ZL(75-j20)Ω at 100MHz using SINGLE shorted stub. given that the line is 3. (a)(i) Derive the components of Electric and Magnetic field strength between a pair of parallel perfectly conducting planes of infinite extent in the ‘Y’ and ‘Z’ directions. 13. Calculate the stub length and distance between the antenna and stub using smith chart. The planes are separated in X direction by “a” meter. (1)Voltage reflection coefficient. (ii) Write a brief note on excitation of modes in circular wave guides. velocity of propagation.VEL TECH VEL TECH MULTI TECH TECH HIGHTECH VEL guide whose inner dimensions are ‘a’=2. 15. (a)(i) Derive the expression for TM wave components in circular wave guides using Bessel function.1. Guide wave length and Wave impedance.3cm and ‘b’ =1cm. Phase velocity. Or (b) Derive the equation for Q – factor of a rectangular cavity resonator for TE 101 mode. µ r=1. Calculate the cut-off frequency. which is filled with Teflon ε r=2. Phase constant. *************** VEL TECH VEL TECH MULTI TECH TECH HIGHTECH 268 VEL .