IT 2202PRINCIPLES OF COMMUNICATION 3104 UNIT I FUNDAMENTALS OF ANALOG COMMUNICATION 9 Principles of amplitude modulation, AM envelope, frequency spectrum and bandwidth, modulation index and percent modulation, AM Voltage distribution, AM power distribution, Angle modulation - FM and PM waveforms, phase deviation and modulation index, frequency deviation and percent modulation, Frequency analysis of angle modulated waves. Bandwidth requirements for Angle modulated waves. UNIT II DIGITAL COMMUNICATION 9 Introduction, Shannon limit for information capacity, digital amplitude modulation, frequency shift keying, FSK bit rate and baud, FSK transmitter, BW consideration of FSK, FSK receiver, phase shift keying – binary phase shift keying – QPSK, Quadrature Amplitude modulation, bandwidth efficiency, carrier recovery – squaring loop, Costas loop, DPSK. UNIT III DIGITAL TRANSMISSION 9 Introduction, Pulse modulation, PCM – PCM sampling, sampling rate, signal to quantization noise rate, companding – analog and digital – percentage error, delta modulation, adaptive delta modulation, differential pulse code modulation, pulse transmission – Intersymbol interference, eye patterns. UNIT IV SPREAD SPECTRUM AND MULTIPLE ACCESS TECHNIQUES 9 Introduction, Pseudo-noise sequence, DS spread spectrum with coherent binary PSK, processing gain, FH spread spectrum, multiple access techniques – wireless communication, TDMA and CDMA in wireless communication systems, source coding of speech for wireless communications. UNITV SATELLITE AND OPTICALCOMMUNICATION 9 Satellite Communication Systems-Keplers Law,LEO and GEO Orbits, footprint, Link modelOptical Communication Systems-Elements of Optical Fiber Transmission link, Types, Losses, Sources and Detectors. TUTORIAL: 15 TOTAL: 45 +15=60 TEXT BOOKS: 1. Wayne Tomasi, “Advanced Electronic Communication Systems”, 6/e, Pearson Education, 2007. 2. Simon Haykin, “Communication Systems”, 4th Edition, John Wiley & Sons., 2001. REFERENCES: 1. H.Taub,D L Schilling ,G Saha ,”Principles of Communication”3/e,2007. 2. B.P.Lathi,”Modern Analog And Digital Communication systems”, 3/e, Oxford University Press, 2007 3. Blake, “Electronic Communication Systems”, Thomson Delmar Publications, 2002. 4. Martin S.Roden, “Analog and Digital Communication System”, 3rd Edition, PHI, 2002. 5. B.Sklar,”Digital Communication Fundamentals and Applications”2/e Pearson Education 2007. IT 2202 1 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Question Bank SUBJECT: IT2202-PRNCIPLES OF COMMUNICATION Semester: III Department: IT UNIT I--Fundamentals Of Analog Communication PART-A What is modulation? Modulation is the process of changing any one parameter (amplitude, frequency or phase) of a relatively high frequency carrier signal in proportion with the instantaneous value of the modulating signal or message signal. 2. Define amplitude Modulation. Amplitude Modulation is the process of changing the amplitude of a relatively high frequency carrier signal in proportion with the instantaneous value of the modulating signal. Give the expression for AM modulated wave. Vam =Vc sin ωc t + m Vc cos (ωc - ωm ) t – m Vc cos (ωc + ωm) t ---------2 2 where, Vam - amplitude of modulated signal Vc - amplitude of carrier signal ωc = 2 π fc = carrier frequency ωm = 2π fm = modulating signal frequency. 3. Define Modulation index and percent modulation for an AM wave. Modulation index is a term used to describe the amount of amplitude change present in an AM waveform .It is also called as coefficient of modulation. Mathematically modulation index is m = Em/ Ec Where m = Modulation coefficient Em = Peak change in the amplitude of the output waveform voltage. Ec = Peak amplitude of the unmodulated carrier voltage. Percent modulation gives the percentage change in the amplitude of the output wave when the carrier is acted on by a modulating signal. 5. Give the bandwidth of AM? Bandwidth (B) of AM DSBFC is the difference between highest upper frequency and lowest lower side frequency. B= 2fm(max) fm(max) – maximum modulating signal frequency. 1. IT 2202 2 6. Draw the spectrum of AM signal. Vc mVc/2 fm mVc/2 fm fc-fm fc fc+fm frequency fLSB fUSB 7. Give the expression for modulation index in terms of Vmax and Vmin. m = Vmax – Vmin / Vmax + Vmin 8. Give the formula for AM power distribution. Ptotal = Pc [1 + m2 / 2] where, Ptotal – total power m- modulation index Pc – carrier power 9. Give the expression for total current. Itotal = Ic [1 + m2 / 2] 1/2 where, Itotal – total Current m- modulation index Ic – carrier current 10. Give the types of AM Modulation. DSBSC-Double sideband suppressed carrier. SSBSC- Single sideband suppressed carrier. DSBFC- Double sideband full carrier. VSBSC-Vestigial sideband suppressed carrier. 11. What are the disadvantages of conventional (or) double side band full carrier system? In conventional AM, carrier power constitutes two thirds or more of the total transmitted power. This is a major drawback because the carrier contains no information; the sidebands contain the information. Second, conventional AM systems utilize twice as much bandwidth as needed with single sideband systems. 12. Define Single sideband suppressed carrier AM. AM Single sideband suppressed carrier is a form of amplitude modulation in which the carrier is totally suppressed and one of the sidebands removed. 13. Define AM Vestigial sideband. AM vestigial sideband is a form of amplitude modulation in which the carrier and one complete sideband are transmitted, but only part of the second sideband is transmitted. 14. What are the advantages of single sideband transmission? The advantages of SSBSC are 1.Power conservation 2.Bandwidth conservation 3.Noise reduction IT 2202 3 21. IFRR is IFRR =(1+Q2ρ2)1/2 Where ρ= (fim/fRF)-(fRF/fim) Q – quality factor of preselector fim-image frequency fRF. 23. It requires less power to achieve a high percentage of modulation. if allowed to enter a receiver and mix with the local oscillator. Define image frequency rejection ratio. Heterodyne means to mix two frequencies together in a nonlinear device or to translate one frequency to another. 18. IT 2202 4 . For high level modulator class C amplifier is used. 19. will produce a cross product frequency that is equal to the intermediate frequency. Mathematically. 24.15. Define High-level Modulation. Define image frequency. In high-level modulators. In indirect frequency modulation. frequency of a constant amplitude carrier signal is directly proportional to the amplitude of the modulating signal at a rate equal to the frequency of the modulating signal. Complex receivers ii. What is the advantage of low-level modulation? An advantage of low-level modulation is that less modulating signal power is required to achieve a high percentage of modulation. Define direct frequency modulation. In direct frequency modulation. What is the advantage of low-level modulation? An advantage of low-level modulation is that less modulating signal power is required to achieve a high percentage of modulation. 17. Define instantaneous frequency deviation.RF frequency 22. Define Low-level Modulation. Define indirect frequency Modulation. An image frequency is any frequency other than the selected radio frequency carrier that. modulation takes place prior to the output element of the final stage of the transmitter. For low level AM modulator class A amplifier is used. What are the disadvantages of single side band transmission? i. 20. Tuning Difficulties 16. In low-level modulation. It requires a much higher amplitude modulating signal to achieve a reasonable percent modulation. the modulation takes place in the final element of the final stage where the carrier signal is at its maximum amplitude. The image frequency rejection ratio is the measure of the ability of preselector to reject the image frequency. Define Heterodyning. 25. using nonlinear mixing. phase of a constant amplitude carrier directly proportional to the amplitude of the modulating signal at a rate equal to the frequency of the modulating signal. 3. Hence transmitter power remains constant in FM 30. ii. The depth of modulation has But in FM the depth of modulation limitation in AM. The peak frequency deviation is simply the product of the deviation sensitivity and the peak modulating signal voltage and is expressed mathematically as Δf =K1Vm Hz. Carson rule states that the bandwidth required to transmit an angle modulated wave as twice the sum of the peak frequency deviation and the highest modulating signal frequency. Amplitude modulation Frequency modulation 1. State Carson rule. iii. Frequency deviation is typically given as a peak frequency shift in Hertz (Δf). Noise interference is more Noise interference is less 2. Simple circuits used in transmitter Uses more complex circuits in and receiver. Disadvantages: IT 2202 5 . Deviation ratio is the worst-case modulation index and is equal to the maximum peak frequency deviation divided by the maximum modulating signal frequency. 26. The peak-to-peak frequency deviation (2 Δf) is sometimes called carrier swing. Define frequency deviation. depth of modulation. 28. Phase of a constant amplitude carrier is varied directly proportional to the amplitude of the modulating signal at a rate equal to the frequency of the modulating signal. the deviation ratio is DR= Δf (max) / fm(max) 29. 27. 4. instantaneous value of the modulating signal. Power varies in AM depending on The amplitude of FM is constant. What are the advantages of angle modulation and also list its disadvantages. Frequency deviation is the change in frequency that occurs in the carrier when it is acted on by a modulating signal frequency. Mathematically. Amplitude Modulation is the process Frequency Modulation is the process of changing the amplitude of a of changing the frequency of a relatively high frequency carrier signal relatively high frequency carrier in proportion with the instantaneous signal in proportion with the value of the modulating signal. 31. Mathematically Carson’s rule is B=2(Δf +fm) Hz.The instantaneous frequency deviation is the instantaneous change in the frequency of the carrier and is defined as the first derivative of the instantaneous phase deviation. Write down the comparison of frequency and amplitude modulation. Advantages: i. more efficient use of power. Noise reduction. transmitter and receiver. Define Phase modulation. Improved system fidelity. 5. Define Deviation ratio. can be increased to any value by increasing the deviation. For an FM receiver with an input frequency deviation ∆f=4 kHz and a transfer ratio K= 0. Define deviation sensitivity for FM and PM and give its units. ii.01* 40 =0.01 V/k Hz. “Advanced Electronic Communication Systems” Fifth edition. FM: Change in output frequency occurs when amplitude changes in input signal.Vc = 20-5= 15V m= Vmax – Vmin / Vmax + Vmin =25-15 /25+15 = 0. Refer: Wayne Tomasi.150 3. fm =2000Hz and K1 =4 kHz /V Frequency deviation = δ = K1Vm = 4 kHz/v * 8V = 32kHz Modulation index = m = δ/ fm = 32 kHz/2000Hz = 16 37. 35. wider Bandwidth. Vm=8V. of significant sidebands for m found using Bessel’s table.4V PART –B 1. For an AM DSBFC transmitter with an unmodulated carrier power Pc=100w that is simultaneously modulated by 3 modulating signals with co-efficient of IT 2202 6 . “Advanced Electronic Communication Systems” Fifth edition. PM: Change in output phase occurs when amplitude changes in input signal. Vout = K * ∆f =0. determine Vout.151 2.25 36. An FM transmitter has a rest frequency fc =96MHz and a deviation sensitivity K1 = 4 KHz/V. uses more complex circuit in receiver and transmitter. 32.i. 34.(ii)Power in upper and lower sidebands(iii)Total transmitted power. What is Phase deviation ? The relative angular displacement (shift) of the carrier phase (rad) in respect to reference phase is called phase deviation(ΔӨ) 33. Refer: Wayne Tomasi. Give the expression for bandwidth of angle-modulated wave in terms of Bessel’s table. and m=1determine (i) Power in carrier. Derive the expression for total power in an AM DSBFC and draw the power spectrum. If a modulated wave with an average voltage of 20Vp changes in amplitude ±5V. determine the maximum and minimum envelope amplitudes and the modulation coefficients. Unit K1=(rad/s)/V. Vm = 20Vp Vc = 5 V m = Vmax – Vmin / Vmax + Vmin Vmax = Vm + Vc = 20+5= 25V Vmin = Vm . For an AM DSBFC wave with unmodulated carrier voltage of 10Vp and load resistance of 10 ohms. B= 2(n*fm) n=no. Unit K =(rad)/V. Page No. Determine the frequency deviation for a modulating signal Vm(t) = 8sin(2π 2000t). Determine the modulation index. Page No. Total transmitted power and then.286 8. Refer: Wayne Tomasi.145 7.146 5.5. Draw the output spectrum. phase or frequency) of the analog carrier is varied proportional to the information signal is called ad digital modulation. “Advanced Electronic Communication Systems” Fifth edition. Derive Eusf and Elsf in terms of Vmax and Vmin . determine. Refer: Wayne Tomasi. Compare PM and FM. Total coefficient of modulation.2. “Advanced Electronic Communication Systems” Fifth edition. Page No. Determine the side band frequencies of an angle-modulated wave. “Advanced Electronic Communication Systems” Fifth edition. IT 2202 7 .282 10.154 4.(v)Draw the output frequency spectrum.4 and m3=0.142 6.281 9. Refer: Wayne Tomasi. “Advanced Electronic Communication Systems” Fifth edition. Page No. Page No. “Advanced Electronic Communication Systems” Fifth edition. Refer: Wayne Tomasi.5Vp change in the amplitude of the envelope.(iv)Draw the output envelope. What is digital modulation? When the information signal is digital and any one of the parameters (amplitude.modulation m1=0. Refer: Wayne Tomasi. Refer: Wayne Tomasi. “Advanced Electronic Communication Systems” Fifth edition. Derive the expression for average power of an angle-modulated wave. m2=0. Derive the output expression for an AM DSBFC and also draw the AM spectrum. Page No. UNIT II--Digital Communication PART-A 1. The second input is a 10 kHz modulating signal whose amplitude is sufficient to produce a ± 7. Upper and lower sideband power. What is information capacity? 2. “Advanced Electronic Communication Systems” Fifth edition. One input to an AM DSBFC modulator is 500 kHz carrier with peak amplitude of 20Vp. (ii)Modulation co-efficient and percent modulation.(iii)Maximum and minimum amplitude of the envelope. Page No. Refer: Wayne Tomasi. Explain in detail about the Bandwidth requirements of angle-modulated wave. Refer: Wayne Tomasi. Total sideband power. Determine the following (i) Upper and lower side frequency. Page No. Page No.. “Advanced Electronic Communication Systems” Fifth edition. Page No.293. 8. It is the number of independent symbols that can be carried through a system in a given unit of time. Error probability less then Higher than QPSK AQSK What are Antipodal signals? In BPSK. there is no abrupt change in the amplitude and the modulated signal is continuous and smooth. 6. Quadrature phase Quadrature phase and modulation amplitude modulation 2. fb –channel capacity (bps) B-minimum Nyquist bandwidth (Hz) M. Minimum shift keying uses two orthogonal signal to transmit ‘0’ and ‘1’ in such a way the difference between these two frequencies is minimum. I= B log2 [1+ S/N] Where. 5. fb =2B log2 M Where. Such signals are called antipodal signals. Define minimum shift keying. Relatively complex 4. Hence. The two frequencies are Difference between two integer multiple of base frequencies minimum and band frequency and at the at the same time they are same time orthogonal. 4. What are the advantages of M-ary signaling scheme? IT 2202 8 .number of discrete level or voltage levels Compare QASK and QPSK. Has discontinuities when Phase discontinuities are phase changes from 0 to 1 removed by smooth phase or 1 to 0.All signal points placed on Signal points are replaced circumference of circle symmetrically about origin 3.Noise immunity better then Poor than QPSK. I= information capacity (bps) B= bandwidth S/N=signal to noise power ratio (unit less) Give the Nyquist formulation for channel capacity. FSK MSK 1. Give the expression for Shannon limit for information capacity. transition. Circuit is simple. Symbol ‘1’ => s1 (t) = √2P cos (2πf0 t) Symbol ‘0’ => s2 (t) = √2P cos (2πf0 t + π) Here observe that above two signals differ only in a relative phase shift of 1800. 9. orthogonal. 2. 7. the two symbols are transmitted with the help of following signals. Bandwidth (BW) = 4fb BW = fb/2 3. QPSK QASK 1. Give the difference between standard FSK and MSK.3. QASK 5. M-ary signaling schemes transmit bits at a time. Two bits form a symbol. 17. Quadrature amplitude modulation is a form of digital modulation where the digital information is contained in both the amplitude and phase of the transmitted carrier. Define peak frequency deviation for FSK. but the drawback is that error probability increases. where. compared to BPSK IT 2202 9 .coherent (envelope) detection: This type of detection does not need receiver carrier to be phase locked with transmitter carrier. What is a constellation diagram? It is also called as signal state-space diagram. 12. Bring out the difference between DPSK and BPSK. It does not need a carrier It needs a carrier at receiver at its receiver 2. Bandwidth requirement of M-ary signaling schemes is reduced. 4. 11. Define bit rate. Symbol duration = 2Tb. Peak frequency deviation (Δf) is the half the difference between either the mark and space frequency. Symbol duration = Tb. One bit forms a symbol. Coherent (synchronous) detection: In coherent detection. This is made physically possible by allowing ISI in the transmitted signal in controlled manner. BPSK QPSK 1. ts. 14. Define Baud rate. Bandwidth reduced More bandwidth. The detection is done by correlating received noisy signal and locally generated carrier. The advantage of such a system is that the system becomes simple. 16. 3. the rate of change at the input to the modulator is called the bit rate (fb) and has the unit of bits per second (bps). similar to phasor diagram where. 2. 15. 10. Two possible symbols Four possible symbols. 13. Minimum bandwidth is Minimum bandwidth is equal twice of fb to fb. Baud= 1/ts. the relative position of peaks of phasors is shown. Correlative coding is implemented by duobinary signaling and modified duobinary signaling. What does correlative coding mean? Correlative coding allows the signaling rate of 2B0 in the channel of bandwidth B0. In digital modulation. ii. (Δf)=|fm-fs|/2. the local carrier generated at the receiver is phase locked with the carrier at the transmitter. 18. Define QAM. Differentiate coherent and noncoherent methods. Non. Compare binary PSK with QPSK. The coherent detection is a synchronous detection. The receiver knows this ISI.time of one signaling element (seconds). Hence effects of ISI are eliminated at the receiver.i. The rate of change at the output of the modulator is called baud rate. DPSK BPSK 1. Comparatively low.1) are represented by two different amplitudes of the carrier signal. S(t) = Acos2πft binary 1 0 binary 0 PART-B 1. 21. Minimum transmitted power. In ASK. What is an Offset QPSK? Offset QPSK (OQPSK) is a modified form of QPSK where the bit waveforms on the I and Q channels are offset or shifted in phase from each other by one-half of a bit time. It is not necessary to recover phase-coherent carrier. It is the process of extracting a phase-coherent reference carrier from a receiver signal. What is bandwidth efficiency? It is also called as information density or spectral efficiency. since it since it uses two bits for its uses only single bit reception 5.369 2. Page No. Maximum data rate ii.What do you mean by ASK? ASK (Amplitude Shift Keying) is a modulation technique which converts digital data to analog signal. Refer: Wayne Tomasi.3. Explain FSK bit rate. baud. Mention any four advantage of digital modulation over analog modulation. 24.368 IT 2202 10 . 20.Probability of error or bit Comparatively low error rate more than BPSK 4. Refer: Wayne Tomasi. i. It is also called as phase referencing 23. “Advanced Electronic Communication Systems” Fifth edition. v. is the ratio of the transmission bit rate to the minimum bandwidth required for particular modulation scheme. Minimum circuit complexity vi. iv. Error propagation more. Noise interference more Comparatively low 19. Minimum probability of symbol error iii. Minimum channel bandwidth. Page No. bandwidth and modulation index. What is DPSK? Differential phase-shift keying (DPSK) is an alternative form of digital modulation where the binary input information is contained in the difference between successive signaling elements rather than the absolute phase. “Advanced Electronic Communication Systems” Fifth edition. Explain on-off keying (OOK) or ASK. the two binary values(0. Maximum resistance to interfering signals 22. Define carrier recovery. The transmission of digitally encoded analog signals requires significantly more bandwidth than simply transmitting the original analog signal. it is completely determined by specifying its ordinates at a sequence of points spaced 1/2W seconds apart. iii. The binary number varies according to the amplitude of the analog signal. Refer: Wayne Tomasi.407 5. State the sampling theorem for band-limited signals of finite energy. Digital signals are better suited to processing and multiplexing than analog signals. Explain BPSK (transmitter and receiver) and also discuss about the bandwidth.381 4. Explain QPSK transmitter and receiver. Analog signal must be converted to digital codes prior to transmission and converted back to analog form at the receiver. “Advanced Electronic Communication Systems” Fifth edition. serial binary number for transmission. Digital pulses are less susceptible than analog signals to variations caused by noise. What are the advantages of digital transmission? i. IT 2202 11 . iv. ii.404 UNIT III --Digital Transmission PART-A 1. “Advanced Electronic Communication Systems” Fifth edition. Explain in detail about carrier recovery. Page No. Page No. Page No. Explain DPSK with an example. Discuss the operation of 16-QAM transmitters and receivers.3. 2. ii. Refer: Wayne Tomasi. Page No. Refer: Wayne Tomasi. What are the disadvantages of digital transmission? i. “Advanced Electronic Communication Systems” Fifth edition. Digital systems are better suited to evaluate error performance.398 7. analog signal is sampled and converted to fixed length. 4. Refer: Wayne Tomasi. 3. In pulse code modulation. Refer: Wayne Tomasi.376 6. If a finite energy signal g(t) contains no frequency higher than W Hz. Digital transmission systems are more noise resistant than the analog transmission systems. Define pulse code modulation. “Advanced Electronic Communication Systems” Fifth edition. thus necessitating additional encoding and decoding circuitry. “Advanced Electronic Communication Systems” Fifth edition. Page No. The advantage of digital transmission over analog transmission is noise immunity. Quantization is a process of approximation or rounding off. List the four predominant methods of pulse modulation. 8. 11. Pulse position modulation (PPM) iii. 13. low-amplitude signals are more likely to occur than large-amplitude signals. Assigning PCM codes to absolute magnitudes is called quantizing. This type of coding is called nonuniform or nonlinear encoding. Pulse duration modulation (PDM) 9. Define and state the causes of fold over distortion. The frequency that folds over is an alias of the input signal hence. What is the purpose of the sample and hold circuit? The sample and hold circuit periodically samples the analog input signal and converts those samples to a multilevel PAM signal. What is the Nyquist sampling rate? Nyquist sampling rate states that. The distortion is called aliasing or fold over distortion. The minimum sampling rate (fs) is equal to twice the highest audio input frequency (fa). dynamic range is DR= Vmax / Vmin 14. i. 7. distortion will result. What is the principle of pulse modulation? Pulse modulation consists essentially of sampling analog information signal and then converting those discrete pulses and transporting the pulses from a source to a destination over a physical transmission medium. The side frequencies from one harmonic fold over into the sideband of another harmonic. 10. Define overload distortion. overload distortion occurs. Therefore. If fs is less than two times fa. Pulse width modulation (PWM) ii. What is the necessity of companding? Companding is the process of compression and then expanding. Mathematically. Higher amplitude signals are compressed prior to transmission and then expanded in the IT 2202 12 . Define dynamic range. the names “aliasing” or “fold over distortion”. Pulse amplitude modulation (PAM) iv. 15. 12.5. If the magnitude of sample exceeds the highest quantization interval. which would increase quantization error. 6. Dynamic range is the ratio of the largest possible magnitude to the smallest possible magnitude. if more codes are used for lower amplitude. What is nonuniform or nonlinear encoding? With voice transmission. What is the advantage and disadvantage of midtread quantization? Advantage: less idle channel noise Disadvantage: largest possible magnitude for Qe 16. Define quantization. it would increase accuracy and fewer codes are used for higher amplitudes. What is codec? An integrated circuit that performs the PCM encoding and decoding functions is called a Codec (coder/decoder). the minimum sampling rate is equal to twice the highest audio input frequency. Timing inaccuracies ii. How is percentage error calculated? 12-bit encoded voltage-12-bit decoded voltage % Error = --------------------------------------------------------------X 100 12-bit decoded voltage 19. thermal noise. which indicates whether the sample is larger or smaller then the previous sample.size is 2. Companding is the means of improving dynamic range of communication systems. List the significance of eye pattern. IT 2202 13 . This interference is commonly called as intersymbol interference or ISI. Slope overload noise Granular noise 1. 20. What is an eye pattern? The performance of a digital transmission system can be measured by displaying the received signal on an oscilloscope and triggering the time base at data rate. Caused when step -size is small. The ringing tails of several pulses have overlapped. 18. The four primary causes of ISI are i. the only input to PAM sampler is random. Give the concept of delta modulation PCM. Such a display is called eye pattern or eye diagram. large. only single bit is transmitted. 2. The output is assigned a discrete value selected from a finite set of representation levels are reconstruction values that are aligned with the treads of the staircase. all waveform combinations are superimposed over adjacent signaling intervals. The peak-to-peak range of input sample values subdivided into a finite set of decision levels or decision thresholds 2. Compare slope overload and granular noise. Rather than transmit a coded representation of the sample.receiver. Insufficient bandwidth iii.What are the two fold effects of quantizing process? 1. thus interfering with major pulse lobe. 17. Phase distortion 22. 24. It discloses any noise or errors in line equalization. Amplitude distortion iv. It also gives the amount of ISI present. 21. What is idle channel noise? When there is no analog input signal. It used to determine the effects of degradations introduced into pulses as they travel to the regenerator. Slope of analog signal is Original input signal has greater than delta modulator relatively constant amplitude and can maintain the reconstructed signal has variation the were not present in the original signal. Thus. Caused when step. This noise is called idle channel noise. 23. What is ISI and give its causes. 455 What is the advantage of DPCM? With a neat block diagram explain transmitter and receiver. 4. Page No. 28. “Advanced Electronic Communication Systems” Fifth edition.425 2. PART-B 1. “Advanced Electronic Communication Systems” Fifth edition. 6.458 Write notes on ISI and eye pattern. “Advanced Electronic Communication Systems” Fifth edition. Write a note on aliasing.What is PAM? PAM is the pulse amplitude modulation. 5. Refer: Wayne Tomasi. Page No.Define quantization error? Quantization is the value of which equals the difference between the output and input values of quantizer. Caused when the step size is small. Explain PCM with a neat block diagram. Page No. “Advanced Electronic Communication Systems” Fifth edition.427 What is companding? Explain in detail Analog and digital companding. What do you mean by slope overload distortion in delta modulation? Slope of analog signal is greater than delta modulator can maintain. Refer: Wayne Tomasi. 26. 3.What is nyquist rate? The minimum sampling rate of 2W sample per second for a signal bandwidth of W hertz is called the nyquist rate. “Advanced Electronic Communication Systems” Fifth edition. Refer: Wayne Tomasi.460 IT 2202 14 . 27. Explain PCM sampling with necessary diagrams and circuits . “Advanced Electronic Communication Systems” Fifth edition. In pulse amplitude modulation. Page No. Page No.25. How slope over and granular noise can be minimized. the amplitude of a carrier consisting of a periodic train of rectangular pulses is varied in proportion to sample values of a message signal.442 With a neat block diagram explain Delta modulation. Refer: Wayne Tomasi. Refer: Wayne Tomasi. Page No. Refer: Wayne Tomasi. but it is generated by a well-defined logic. the carrier frequency randomly changes among different slots. Let the data signal be b (t) and pseudo-noise signal be c (t). Previously this antijam capability was used in military application. Hence it is called pseudorandom (or pseudo-noise) sequence. 6. The pseudo noise sequence can be generated by a feedback shift register and combinational logic. Then the modulated signal is given as. The spectrum spreading is accomplished before transmission through the use of code that is independent of data sequence. It looks like pulses. These frequency slots are called hops. i) Unwanted interference is rejected. iii. This signal is binary in nature.In this case. m (t) = b (t) c(t). Define pseudo noise sequence. What are the applications of Spread spectrum modulation? i. ii. Some commercial applications also use spread spectrum because of its antijam capability.UNIT IV -SPREAD SPECTRUM AND MULTIPLE ACCESS TECHNIQUES PART A 1. The data is transmitted in these hops. 3. IT 2202 15 . The same logic is used at transmitter and receiver. Low probability of intercept is an application of spread spectrum in military . It has following important advantages. ii) Protection against antijamming signals is also provided. What do you mean by direct sequence spread technique? The data sequence directly modulates the pseudo noise sequence. iii) Multipath interference rejection. What is frequency hop spreading? In frequency hop spread spectrum. Because of wide spectrum only small portion of the signals is in fade. The spread spectrum has the ability to resist the effect of intentional jamming. This is because the spread spectrum signal has the ability to resist the effects of multipath fading. 4. Spread spectrum is defined in two parts i. The same code is used in receiver to despread the signal so that the original data may be recovered. 2. ii. Hence the sequence is rather ‘pseudo’ random. 5. Spread spectrum is used in mobile communications. The sequence is not completely random. Define spread spectrum. The pseudo noise sequence is a noise like high frequency signal. the signal spectral density is kept small such that the presence of the signal is not detected easily. Data of interest occupies a bandwidth in excess of the minimum bandwidth necessary to send the date. What are the advantages of spread spectrum modulation? Spread spectrum modulation spreads the message signal over wide bandwidth with the help of special code (key). They have relatively short acquisition time. 2. The length of the run is equal to the length of the subsequence. 3.More than one symbols are More than one frequency hops are transmitted per frequency hop. These systems bandwidth are very large ii. It requires wideband channel with small phase distortion ii. Define fast frequency hopping When several frequency hops take place to transmit one symbol. Chip rate is equal to hop rate. i. then it is called fast frequency hopping. These systems need complex frequency synthesizers ii. 15. List the disadvantages of frequency hopping systems i. Correlation property: The auto correlation function of maximum length sequence is periodic and it is binary valued. The distance effect is less 10. Symbol rate higher than hop rate. iv. Unrecognized receivers find it most difficult to detect direct sequence signals iii. Time division multiple access (TDMA) ii.e. Compare slow and fast frequency hopping. 1’s or 0’s within one period of the sequence. The pseudo-noise generator should generate sequence at high rates iv. 13. List the advantages of direct sequence systems i. They need error correction. Slow Frequency Hopping Fast Frequency Hopping 1. 11. It has long acquisition time iii. then it is called slow frequency hopping. Frequency division multiple access (FDMA) iii. It has best discrimination against multipath signals 8. required to transmit one symbol. They are not useful for range and range –rate measurement iii. 16. Define slow frequency hopping When several symbols are transmitted in one frequency hop (slot). 12. Code division multiple access (CDMA). What is processing gain? Processing gain is defined as the ratio of the bandwidth of spreaded signal to the bandwidth of the unspreaded signal 14. List the disadvantages of direct sequence systems i. They can be programmed to avoid some portions of the spectrum iii. This means the symbol rate is less than hop rate. List the advantages of frequency hopping systems i. Mention the classification of multiple access protocols.Chip rate is equal to symbol rate. IT 2202 16 . What are the properties of maximum length sequence? Balance property: The number of 1’s is always one more than the number of zeros in each period of a maximum length sequence Run property: Run means subsequence of identical symbols i. Hop rate higher than symbol rate. This system is distance relative 9.7. This means the symbol rate is higher than hop rate. This system has best noise and antijam performance ii. 1104 2. It does not require external synchronization networks. 19. Code-excited LPC 22. using spread spectrum is related by processing gain (PG) PG=Tb / Tc Where. unique word and correlation spike.17. Complete bandwidth of the channel is available to the user in given time slot. of users is increased But it is easy to add new user to the system. Spread Jammer over the entire measure of the spectrum of transmitted signal. Explain in detail the notion of spread spectrum. What does CEPT stand for? Conference of European Postal and Telecommunication Administrations (CEPT) is a commonly used TDMA frame format for digital satellite systems. Retuning the Jamming signal over the frequency band of transmitted signal. “Advanced Electronic Communication Systems” Fifth edition. Describe what a reference burst is for TDMA and explain the following terms: preamble. CDMA offers gradual degradation in performance when the no. Refer: Wayne Tomasi. Refer: Simon Haykin. Give the expression for probability of gain. 3. 23. Page No. carrier recovery sequence. Refer: Wayne Tomasi. 2. 2. IT 2202 17 .B 1. TDMA is in satellite communication.What are the features of code Division multiple Accesses? 1. What are the two different techniques used in speech coding for wireless communication? i. 18. the resultant would be zero cross correlation between chip codes. Page No. ii.1106 3. Page No. bit timing recovery. “Advanced Electronic Communication Systems” Fifth edition. 21. If offers an external interference rejection capability. Such a code is called orthogonal codes. “Communication Systems” Fourth edition. The bandwidth of signal before and after encoding. the time of the channel is shared by multiple users. Multi-pulse excited Linear Predictive Coding (LPC). Tb – bit duration Tc – chip duration 20.What are the two function of fast frequency hopping? 1. What is TDMA? In time division multiple access. PART . What are orthogonal codes? If half the bits within a code were made the same and half were exactly the opposite.488. Describe the operation of CEPT primary multiplex frame. Page No. 8. List the satellite orbits. These paths are called orbits of the satellite. With the help of transmitter and receiver diagram explain frequency hopped spread spectrum and discuss about slow –frequency hop and Fast-frequency hop.482. Inclined orbits are virtually all orbits except those that travel directly above the equator or directly over the north and south poles. Explain the generation of pseudo-noise sequence with an example. Explain direct sequence-spread spectrum with appropriate waveforms and expressions. and a user network of earth stations. 2. The transmitter-receiver combination in the satellite is known as a transponder. Refer: Wayne Tomasi. Satellite orbits about the earth are either circular or elliptical. “Communication Systems” Fourth edition. “Communication Systems” Fourth edition. Page No. IT 2202 18 .490. Page No. UNIT-V --SATELLITE AND OPTICAL COMMUNICATION PART-A 1. List the properties of maximum length sequence with examples. What is a satellite system? A satellite system consists of one or more satellite space vehicles. State the basic function of satellite transponder. Page No. Refer: Simon Haykin. Refer: Simon Haykin.480 7.499. The satellite orbits are: • Inclined orbit • Polar orbit • Equatorial orbit 4. The basic function of a transponder is amplification and frequency translation. Refer: Simon Haykin.1108 5.4. 3. Define inclined orbit. “Communication Systems” Fourth edition. “Advanced Electronic Communication Systems” Fifth edition. The satellite can be rotated around the earth through various paths. Refer: Simon Haykin. Page No. 6. Define orbit. These orbits are used to cover the specific application areas. Explain CDMA and also give the orthogonal condition of the signals in CDMA. a ground – based station to control the operation of the system. “Communication Systems” Fourth edition. Angle of elevation: It is the vertical angle formed between the direction of travel of an electromagnetic wave radiated from an earth station antenna pointing directly toward a satellite and the horizontal plane. 9. Define geosynchronous orbit. When the inclination of the orbit is not zero and eccentricity is not zero. All the points in this orbit are at equal distance from earth surface. 13. Therefore this orbit is called geostationary orbit. 14. Define polar orbit. 7.66 GHz iii. Define perigee and apogee. It is defined as the horizontal pointing angle of an earth station antenna. What is station keeping? The process of maneuvering a satellite within a pre assigned window is called station keeping. IT 2202 19 . How are satellites classified based on elevation? i. Medium Earth orbit (MEO): 1. Define Azimuth angle. The point in the orbit where the satellite is closest to the earth is called the perigee.5.2GHz-1. Define a transponder. Angle of inclination: It is the angle between the earth’s equatorial plane and the orbital plane of a satellite measured counter clockwise at the point in the orbit where it crosses the equatorial plane traveling from south to north. 11. The circular equatorial orbit is exactly in the plane of equator on earth. 10. It is an RF-RF repeater. 8. Define angle of inclination and angle of elevation. Satellite orbits with inclinations of 90°are called polar orbit. The period of geosynchronous orbit is equal to the period of revolution of earth with itself. it is called as geosynchronous orbit.5 GHz ii. What is a footprint? The geographical representation of a satellite antenna’s radiation pattern is called a footprint or footprint map. and (3) the square of the time of revolution of a planet divided by the cube of its mean distance form the sun gives a number that is the same for all planets. Define geostationary orbit. What is its basic function? A satellite radio repeater is called a transponder. State the laws of planetary motion. The point in the orbit where the satellite is farthest from the earth is called the apogee.(2) the line joining the sun and a planet sweeps out equal areas in equal intervals of time. Geosynchronous earth orbit (GEO): 2 GHz -18 GHz 12. and a satellite in this orbit appears to be stationary to the point of earth. Low earth orbit (LEO): 1 GHz -2. 6. Kepler’s law may be simply stated as (1) the planets move in ellipses with the sun at one focus. Polar orbits are used for special applications like navigational satellites. 15. List the optical detectors. The optical sources used for optical fiber communication are • Light Emitting Diodes and • Solid-state lasers. What is an optical communication system? It uses light as the carrier of information. 17.16. The optical detectors used for optical fiber communication are IT 2202 20 . What is the basic transponder configuration? There are three basic transponder configurations used in communication systems. State the frequency range of optical fiber communication. What are the major subsystems in a communication satellite? The major subsystems in a communication satellite are • Communications subsystems • Power subsystems • Telemetry tracking and control (TTC) subsystems • Propulsion subsystems • Attitude stabilization subsystems • Antenna subsystems 19. What are the techniques for increasing channel capacity? Two of the techniques for increasing channel capacity are: • Frequency reuse • Spatial isolation 18. 21. They are • Single conversion transponder • Double conversion transponder • Regenerative transponders 20. State the uplink frequency and downlink frequency. It uses glass or plastic fiber cables to contain the light waves and guide them 23. State the major subsystems in a satellite earth station. List the applications of a satellite Some of the applications of a satellite are: • Surveillance or observation • Navigation • TV broadcast • Satellite telephones 22.1x1014 to 4x1014 24. List the optical sources used for optical fiber communication. Light frequencies range . 25. The major subsystems in a satellite earth station are: • Transmit subsystems • Receive subsystems • Power subsystems • Antenna subsystems • Telemetry tracking and control (TTC) subsystems • Ground control equipment (GCE) subsystems. A typical uplink frequency is 6 GHz and a common downlink frequency is 4 GHz. Closed circuit TV systems used in buildings for security.• Photodiodes and • Avalanche photodiodes. Aircraft communication and Aircraft controls k. 27. 26. Interference immunity: fiber optic cables do not radiate signals as some electrical cables do and cause interference to other cables. Define critical angle. When the light ray strikes the interface between the air and the glass. 2. c. List the advantages of optical fiber systems. They are also insulators so are not susceptible to lightning strikes as electrical cables are. Wider bandwidth: they have higher information carrying capability. eliminating microwave radio link. Greater safety: fiber optic cables don’t carry electricity. 3. 1. 7. Strength: fiber optic cables are stronger than electrical cables and can support more weight. Lightweight: glass or plastic cables are much lighter than copper cables and offer benefits in those areas where low weight is critical (i.velocity of light in given material (meters per second) 29. Local and long distance telephone systems b. n-refractive index(unit less) c-velocity of light in free space (3*10^8 meters per second) v. n=c/v where. State the applications of optical fibers. d. there is no shock hazard. Nuclear plant instrumentation. Computer networks. They are also immune to pickup of interference from other sources. 5.. wide area and local area.e. therefore more can be contained in a smaller space. aircraft). 28. g. Small size: practical fiber optic cables are much smaller than electrical cables in diameter. TV studio-to-transmitter interconnection. 30. 6. f. Security: fiber optic cables cannot be “tapped” as easily as electrical cables. there is less signal attenuation over long distances. Define total internal reflection. Shipboard communications. Refractive index or index of refraction (n) is ratio of the velocity of propagation of a light ray in free space to the velocity of propagation of a light in a given material. they don’t radiate signals that can be picked up for eavesdropping purposes. 8. Lower loss: with fiber optic cables. at an angle greater than the critical angle. 4. the light ray does not pass through the interface IT 2202 21 . a. Therefore. The angle of incidence at which the refracted angle becomes 90°to the normal is said to be the critical angle. Define refractive index. This action is known as total internal reflection. in which the refractive index of the core is not constant. Mode refers to the various paths that the light rays can take in passing through the cable. There are two classifications: • Single mode and • Multimode.into the glass. The cladding is also made of glass or plastic of low refractive index. in which there is a sharply defined step in the index of refraction where the fiber core and cladding interface. A plastic jacket is put over the cladding for insulation. reaching a peak at the center and then declining as the outer edge of the core is reached. The main reasons for the losses in optical fiber cable are • Light absorption • Scattering and • Dispersion. Graded index fiber refers to the fiber. 37. the cladding adds strength. Describe the construction of optical fiber cable. 32. the light follows a single path through the core and a multimode fiber is one in which. Instead it varies smoothly and continuously over the diameter of the core. What is meant by step index fiber and graded index fiber? Step index fiber refers to the fiber. 35. What is single mode fiber and multimode fiber? A single mode fiber is one in which. 33. 31. This ensures that proper interface is achieved so that the light waves remain within the core. Define mode. which is called the core. When this occurs the angle of reflection is equal to the angle of incidence as if a real mirror were used. is usually surrounded by a protective cladding. What is meant by modal dispersion? The stretching of the pulse due to the attenuation of the light in the cable and increase in duration of arrival times of various light rays is referred to as modal dispersion. The portion of a fiber optic cable that carries the light is made from either glass (silica) or plastic. IT 2202 22 . It means the core has one constant refractive index N1 while the cladding has another constant refractive index N2. In addition to protecting the fiber. List the reasons for the losses in optical fiber cable. Define absorption. 34. Absorption refers to how the light waves are actually “soaked up” in the core material. 36. the light takes many paths through the core. The fiber. Page No. less intense and are good only for short distances whereas.594 3. Scattering refers to the light lost because of light waves entering at the wrong angle and being lost in the cladding due to refraction. the Lasers are monochromatic. 43. Explain in detail loses in optical fibers. Refer: Wayne Tomasi. 41. Name the multiplexing scheme used for fiber optic communication. Classify and explain the Fiber optic cable based on index of refraction and mode. Refer: Wayne Tomasi.580 5. about light wave propagation. Whereas multimode step index fibers are the easiest to make and the least expensive. What are the advantages of single mode step index fiber over multimode step index fiber? Single mode step index fibers are of extremely small size and therefore difficult to make and very expensive. “Advanced Electronic Communication Systems” Fifth edition. “Advanced Electronic Communication Systems” Fifth edition. and highly intense and can be used over long distances. Fiber optic connectors are the optical equivalent of electrical plugs and sockets.38. Connectors are special mechanical assemblies that allow fiber optic cables to be connected to one another. Refer: Wayne Tomasi. What are the advantages of lasers over LED’s? LED’s covers a narrow spectrum of frequencies.583 2. Explain optical sources used in fiber optic communication. Page No. wavelength division multiplexing. “Advanced Electronic Communication Systems” Fifth edition. The attenuation of a fiber optic cable is expressed in decibels per unit of length. Connectors are used at the repeater units and at the end of the cable applied to the light source or photo detector. Refer: Wayne Tomasi. Draw the basic block diagram of a Fiber optic communication system and explain the function of each block. PART . Fiber optic connectors can be spliced by gluing. 42.B 1.597 4. Write short notes on cable splicing and connectors. Explain. Define scattering. Page No. in detail. The standard decibel formula used is 40. Page No. State the expression for attenuation in optical fiber cable. IT 2202 23 . The multiplexing scheme used for fiber optic communication is WDM. “Advanced Electronic Communication Systems” Fifth edition. 39. coherent. “Advanced Electronic Communication Systems” Fifth edition.1076 9. Refer: Wayne Tomasi. “Advanced Electronic Communication Systems” Fifth edition. “Advanced Electronic Communication Systems” Fifth edition. Page No. Page No.1058 8. DEGREE EXAMNIATION.613 7. NOV/DEC 2005 Third Semester IT 2202 24 . Page No.607 6. Page No. Page No. Explain the satellite system with a neat block diagram. “Advanced Electronic Communication Systems” Fifth edition.E. “Advanced Electronic Communication Systems” Fifth edition. What is a geosynchronous satellite? Discuss its advantages and disadvantages. Explain optical sources used in fiber optic communication. Refer: Wayne Tomasi. Page No. Refer: Wayne Tomasi. Write notes on satellite elevation categories and satellite orbital patterns. TECH. State and explain the laws of planetary motion. Refer: Wayne Tomasi.1056 10. “Advanced Electronic Communication Systems” Fifth edition. Refer: Wayne Tomasi.1062 B./B.Refer: Wayne Tomasi. 3. 3.5 = [1 + m2 / 2] ½ =1. Itotal = Ic [1 + m2 / 2] ½ => Itotal/ Ic = [1 + m2 / 2] 1/2 14/12..33% 2.Calculate the percentage of modulation employed. Noise interference is more Noise interference is less 2.2544 =1+m2/ 2 m=0. AM uses a Super heterodyne receiver to mix two frequencies together in a nonlinear device or to translate one frequency to another. The depth of modulation has But in FM the depth of limitation in AM. Amplitude Modulation is the process Frequency Modulation is the process of of changing the amplitude of a changing the frequency of a relatively relatively high frequency carrier signal high frequency carrier signal in in proportion with the instantaneous proportion with the instantaneous value value of the modulating signal. transmitter and receiver. If the rms value of the aerial current before modulation is 12. of the modulating signal. circuits in IT 2202 25 . Draw the block diagram of a super heterodyne AM receiver and explain why the usual AM radio receiver uses a super heterodyne system.7133 M= 71. 4. assuming no distortion. Write down the comparison of frequency and amplitude modulation.5A and during modulation is 14A. Amplitude modulation Frequency modulation 1. Simple circuits used in transmitter Uses more complex and receiver. modulation can be increased to any value by increasing the deviation. using nonlinear mixing.Information Technology IT1202-PRINCIPLES OF COMMUNICATION Time: Three hours Answer all Questions Maximum: 100 marks PART A – (10 *2 =20marks) 1. Draw the signal constellation of QPSK and give comments on QPSK IT 2202 26 . (a)PSK (b) ASK ( c) FSK 6. depth of modulation.find m(t). FSK. An angle modulated signal is described by Xc(t)=10 cos[2π (106) t + 0.] m(t)=Vc sin[ωc t + KpVm cos ωm t] m(t)=10 cos[2π (106) t + 1 sin (103) π t]. PSK for NRZ coded binary sequence and represent also each case mathematically. Power varies in AM depending on The amplitude of FM is constant. Sketch the waveform representation of ASK.1 sin (103) π t] considering Xc(t) as PM Signal with Kp=10. Hence transmitter power remains constant in FM 4.5. 5. Draw Eye pattern and interpret it properly. Sketch the structure of a CDMA system and give the orthogonal condition for the signals in CDMA. +1350. 8. DA – distortion at sampling time.QPSK is an M-ary encoding scheme with N=2 and M=4. Time Division Multiple access (TDMA) ii. If a finite energy signal g(t) contains no frequency higher than W Hz. Define sampling theorem. Code Division Multiple access (CDMA) iii. i.Margin over noise JT – distortion at zero crossings ST – time interval over which the wave can be sampled. -450 and -1350. Frequency Division Multiple access (FDMA) 10. 7. 9. Mention the classification of multiple access protocols. With two bits there are four possible conditions and four possible output phases +450 . MN . Balanced Modulator PSK Modulator RF Modulator High-power amplifier & BPF N-bit code word generator IF Carrier RF Carrier CDMA Encoder Data in X Orthogonal code IT 2202 Logic 1=+1 1 1 –1 1 –1 -1 27 Logic 0 = -1 -1 –1 1 –1 1 1 . it is completely determined by specifying its ordinates at a sequence of points spaced 1/2W seconds apart. m=3 DR= Δf (max) / fm(max) Δf= 3KHZ (1)fm is doubled . B=10 KHZ (2) fm is by one half.Compute the maximum frequency deviation(∆f) of the output of the FM transmitter and the carrier frequency fc is f1=200 kHz. ηmax=33. ∆f1=25Hz. (i) s(t) Product Modulator Local Oscilltor v(t)Vo(t) LPF (ii) Psb = ma 2Ec 2 / 4 Pt = Ec2/2[1+ma 2/2] η = Psb / Pt at μ =1 .Assume FM and fm=1kHz.calculate the modulation index and find the bandwidth when (1)fm is doubled (2) fm is by one half Carson’s rule is B=2(Δf +fm) Hz. B= 7KHz (b) (i)A block diagram of an indirect (Armstrong) FM transmitter is shown in fir. (i) Verify that the message signal m(t) is recovered from a modulated DSB signal by first multiplying it by a local sinusoidal carrier and then passing the resultant signal through a low pass filter(1) in the time domain(2) in the frequency domain.8MHz. (a) Consider an angle modulated signal Xc(t) = 10 cos (ωc(t) + 3 sin ωm(t)). ma = 1 substituting . flo=10.flo =2MHz IT 2202 28 .(ii) Derive the efficiency η of ordinary AM and show that for a single tone AM.8 MHz.3% 12. n1=64 and n2=48 out Frequency Frequency X m(t) Narrow multiplier multiplier band FM f1 X n1 f3 X n2 ∆f1 f2=n1f1 fc ~ ∆f3 ∆f2 ∆f flo f2=n1f1 = 64*200 kHz=12.3% at μ =1. ηmax = 33.------------------------Product code Recovered chip code --------------------------Correlation ---------------------------1 1 -1 1 -1 -1 1 1 -1 –1 1 1 ---------------------------1 1 1 -1 -1 -1 ------------------------------1 –1 1 -1 1 1 1 1 -1 -1 1 1 ------------------------------1 -1 -1 1 1 1 PART B—(5*16=80 marks) 11.(1). f3 = f2 . /B.E. Refer: Wayne Tomasi. Illustrate the generation of DPSK signal with an example sequence. Page No. A broadcast radio transmitter radiates 5 KW power when the modulation percentage is 60% .How much is the carrier power? Pt = 5kW. unique word and correlation spike.24 kW 2. “Communication Systems” Fourth edition.460 15. NOV/DEC 2006 Third Semester Information Technology IT1202-PRINCIPLES OF COMMUNICATION Time: Three hours Maximum: 100 marks Answer all Questions PART A – (10 *2 =20marks) 1.m=0.6) 2 / 2] = 4.6 or 60% Ptotal = Pc [1 + m2 / 2] Pc = Ptotal / [1 + m2 / 2] = 5* 102 / [1+ (. bit timing recovery. Page No. TECH. Phase modulation is the first integral of FM. Refer: Wayne Tomasi. ii. DEGREE EXAMNIATION. Page No. Refer: Simon Haykin. Refer: Wayne Tomasi. (ii) Briefly describe the operation of CEPT primary multiplex frame. 3. Page No. Ө (t) – instantaneous phase deviation IT 2202 29 . (b)Write short notes on intersymbol interference. What are the two major limitations of the standard form of amplitude modulation? i.1104 B. Ө (t) = ∫ Ө’ (t) dt where. Illustrate the relationship between frequency modulation (FM) and phase modulation.490 (b) (i) Describe what a reference burst is for TDMA and explain the following terms: preamble. AM less efficient because most of power is transmitted in carrier. “Advanced Electronic Communication Systems” Fifth edition. (a) With neat transmitter and receiver block diagram of Direct Sequence spread coherent phase shift keying. The effect of noise is more because of amplitude variations. (a) With the help of a block diagram (transmitter and receiver) explain the principle of a differential phase shift keying. explain its principle of operation.407 14. carrier recovery sequence. “Advanced Electronic Communication Systems” Fifth edition.fc =n2 f3 = 48*2 =96 MHz ∆f = n2∆f3 = 76.8kHz 13. “Advanced Electronic Communication Systems” Fifth edition. fm=2kHz ∆f = 5 kHz BW= 2(∆f + fm ) =2(5 *103 + 2 * 103)=14 KHz Compare QASK and QPSK. The width of the eye opening defines the time interval over which the received wave can be sampled without error from intersymbol interference. Symbol ‘1’ => s1 (t) = √2P cos (2πf0 t) Symbol ‘0’ => s2 (t) = √2P cos (2πf0 t + π) Here observe that above two signals differ only in a relative phase shift of 1800. QASK. Relatively complex 4.Noise immunity better then Poor than QPSK. Quadrature phase modulation Quadrature phase and amplitude modulation 2. Circuit is simple. MN . 5. the two symbols are transmitted with the help of following signals. it is completely determined by specifying its ordinates at a sequence of points spaced 1/2W seconds apart. 7. 8. Such signals are called antipodal signals. QPSK QASK 1. Error probability less then AQSK Higher than QPSK What are Antipodal signals? In BPSK. Ө’ (t) –instantaneous frequency deviation A carrier is frequency modulated with a sinusoidal signal of 2 kHz resulting in a maximum frequency deviation of 5 kHz.All signal points placed on Signal points are replaced circumference of circle symmetrically about origin 3.Margin over noise JT – distortion at zero crossings 30 IT 2202 . If a finite energy signal g(t) contains no frequency higher than W Hz. 6. 5. It is apparent that the preferred time for sampling is the instant time at which the eye is open widest. Find the bandwidth of the modulated signal. State sampling theorem for band-limited signals of finite energy. How is eye pattern used to measure intersymbol interference in a data transmission system? i.4. (a) (i) Explain BPSK transmitter and receiver with the help of block diagram. ST – time interval over which the wave can be sampled.More than one symbols are transmitted per frequency hop. When the effect of intersymbol interference (ISI) is severe traces from the upper portion of the eye pattern cross traces from the lower portion with the result that the eye is completely closed. Slow Frequency Hopping 1. Symbol rate higher than hop rate.282. 3. “Advanced Electronic Communication Systems” Fifth edition. Training mode b. IT 2202 31 . Determine the modulation index and the approximate bandwidth of the FM signal if the frequency of the modulating signal is 100 kHz. Illustrate the two modes of an adaptive equalizer. 13. List out the comparison between slow and fast frequency hopping. with the help of neat block diagram and frequency spectrum. (a)Determine the spectrum of single tone FM wave for an arbitrary value of the modulation index β. Page No. Decision Directed mode Decision device x[n] Adaptive y[n] a ^[n] a[n] Training equalizer [wk] sequence generator e[n] ∑ 10. m= ∆f/ fm =1 BW= 2(∆f + fm ) = 400kHz. Hop rate higher than symbol rate. a. (b)(i) A 20 MHz carrier is frequency modulated by a sinusoidal signal such that the maximum frequency deviation is 100 kHz. PART-B (5 X 16=80 MARKS) 11. Chip rate is equal to hop rate.9. 2. (a) (i)Explain the process of demodulating DSBSC signal by using synchronous demodulator. Refer Nov/Dec 2005 –11 (i) 12. Fast Frequency Hopping More than one frequency hops are required to transmit one symbol.Chip rate is equal to symbol rate. Refer: Wayne Tomasi. DA – distortion at sampling time ii. explain the principle of operation frequency hopped (FH) spread spectrum system Refer: Simon Haykin. (a) (i) Draw the functional block diagram of a direct sequence spread spectrum modulator and demodulator and explain. April/May 2008 Third Semester Information Technology IT1202-PRINCIPLES OF COMMUNICATION Time: Three hours Maximum: 100 marks Answer all Questions PART A – (10 *2 =20marks) 1. Page No. Refer: Simon Haykin. “Communication Systems” Fourth edition. TECH. Page No. Page No.480. Page No./B.5kHz IT 2202 32 . Determine the encoded sequence and transmitted phase sequence Input 1 0 1 1 1 0 0 0 1 1 XNOR 0 0 1 1 1 1 0 1 0 0 0 Output Output 180 180 0 0 0 0 180 0 180 180 180 phase 14. “Communication Systems” Fourth edition.376.499. DEGREE EXAMNIATION. “Advanced Electronic Communication Systems” Fifth edition. Vc mVc/2 fm mVc/2 fm fc-fm fc fc+fm frequency fLSB fUSB 2. Draw the spectrum of AM signal.E. (b) With the help of transmitter and receiver block diagram. fm=2.490. (b) (ii)The bit stream 1011100011 is to be transmitted using DPSK. (ii) What are the applications of spread spectrum techniques? Refer: Simon Haykin. B. Calculate the BW of FM signal whose frequency deviation is 75KHz and signal frequency is 2.5KHz.Refer: Wayne Tomasi. “Communication Systems” Fourth edition. to BPSK 3. Symbol rate higher than hop rate. 8.∆f = 75 kHz BW= 2(∆f + fm ) =2(75 *103 + 2. Minimum transmitted power.it does not need a carrier at its It needs a carrier at receiver receiver 2.Chip rate is equal to symbol rate. since it it uses two bits for its reception uses only single bit 5. IT 2202 33 . required to transmit one symbol.e.5 * 103)=155 KHz 3. 3. 7. iv. Minimum channel bandwidth 6.Bandwidth reduced compared More bandwidth.error propagation more. Define source-coding theorem. 1’s or 0’s within one period of the sequence. What are the properties of pseudo noise sequence? Balance property: The number of 1’s is always one more than the number of zeros in each period of a maximum length sequence Run property: Run means subsequence of identical symbols i. Write the condition required to avoid the slope over load distortion in delta modulation. Mention any four advantages of Digital Modulation over analog modulation. Chip rate is equal to hop rate. Correlation property: The auto correlation function of maximum length sequence is periodic and it is binary valued. since Comparatively low. Define phase modulation. Phase of a constant amplitude carrier is varied directly proportional to the amplitude of the modulating signal at a rate equal to the frequency of the modulating signal. 10. Minimum probability of symbol error iii.Probability of error or bit Comparatively low error rate more than BPSK 4. Bring out the difference between DPSK and BPSK. i. Hop rate higher than symbol rate. The length of the run is equal to the length of the subsequence. but smaller than entropy. Maximum data rate ii.More than one symbols are More than one frequency hops are transmitted per frequency hop. What is PAM? The amplitude of constant width. constant-position pulse is varied according to the amplitude of the sample of the analog signal. 4. 2.noise interference more Comparatively low 5. 9. Differentiate between Fast FH and Slow FH with respect to symbol rate and hop rate. The entropy represents a fundamental limit on the average number of bits pr source symbol necessary to represent a discrete memoryless source in that it can be made as small as. DPSK BPSK 1. Slow Frequency Hopping Fast Frequency Hopping 1. “Advanced Electronic Communication Systems” Fifth edition. PART – B 11. “Advanced Electronic Communication Systems” Fifth edition. (a) (ii) Derive the waveform of power relation between carrier power and total transmitter power of AM signal. Refer: Wayne Tomasi. (a)(ii)What is meant by Intersymbol interference? (ISI). (b)(ii)Define MSK and how is it different from FSK. Suggest a suitable signal processing that has to be done to the baseband signal to overcome ISI. Refer: Wayne Tomasi.a. (b) (i) Distinguish between FM and AM. Refer: Wayne Tomasi.319. Refer: Wayne Tomasi. Increasing the clock frequency reduces the probability of slope overload noise. Draw the QPSK waveform for a digital data 100110. b. under modulation and 100% modulation. Page No. increase the magnitude of the minimum step size. Page No. (a) With block diagram explain QPSK transmitter and receiver. Page No.149 (b) (ii) Draw the waveform of AM signal for over modulation.381.375 14. Page No. (a) Under modulation (b) 100% modulation (c) over modulation 12. “Advanced Electronic Communication Systems” Fifth edition. “Advanced Electronic Communication Systems” Fifth edition. IT 2202 34 . 13. S(t) = Acos2 binary 1 ft 0 binary 0 4. Page No.479. Refer: Wayne Tomasi. the two binary values(0. Refer: Simon Haykin.1) are represented by two different amplitudes of the carrier signal. “Advanced Electronic Communication Systems” Fifth edition. “Advanced Electronic Communication Systems” Fifth edition.Refer: Wayne Tomasi. What is the difference between QASK and QPSK ? QPSK QASK 1.1104 & 1108 (ii) Draw DS spread spectrum modulation with block diagram.Tech Information Technology IT 2202-Principles Of Communication Question Paper Answer all questions PART-A Regulation 2008 1. In ASK. Mathematically modulation index is m = Em/ Ec Where m = Modulation coefficient Em = Peak change in the amplitude of the output waveform voltage.490. Quadrature phase Quadrature phase and modulation amplitude modulation IT 2202 35 . Why is ASK called as ON-OFF keying ? ASK (Amplitude Shift Keying) is a modulation technique which converts digital data to analog signal.460 15. Derive an expression for processing gain PG. Ec = Peak amplitude of the unmodulated carrier voltage. 2009 Anna University B. “Communication Systems” Fourth edition. (a)(i) What is spread spectrum modulation? Describe the following features of spread spectrum modulation: (1) Antijamming (2) Ranging (3) Multiple accessing (4) Message security. Define amplitude modulation. 2. Refer: Simon Haykin. Page No. Amplitude Modulation is the process of changing the amplitude of a relatively high frequency carrier signal in proportion with the instantaneous value of the modulating signal. (b) (i) Distinguish CMDA and TDMA. Page No. “Communication Systems” Fourth edition.It is also called as coefficient of modulation. (ii) Draw a PN sequence generator to generate 7 bit sequence. What is modulation index? Modulation index is a term used to describe the amount of amplitude change present in an AM waveform . Page No. 3. scattering losses Chromatic (or) wavelength. Hence it is called pseudo-random (or pseudo-noise) sequence. 10. Time division multiple access (TDMA) iv. 7. Error probability less then Higher than QPSK AQSK 5.All signal points placed on Signal points are replaced circumference of circle symmetrically about origin 3. Define Kepler's three laws of planetary motion. Code division multiple access (CDMA). 6. and (3) the square of the time of revolution of a planet divided by the cube of its mean distance form the sun gives a number that is the same for all planets. Define Pseudonoise sequence. What are the different types of multiple access techniques ? iii. Define sampling rate. (4) (2) AM power distribution. It looks like pulses.Noise immunity better then Poor than QPSK. The pseudo noise sequence is a noise like high frequency signal. dispersion Radiation losses Model dispersion Coupling losses PART-B 11.2. (a) (i)Write short notes on (1) AM voltage distribution. What are the losses encountered by optical fiber ? Absorption loss Material or Rayleigh. The sequence is not completely random. The pseudo noise sequence can be generated by a feedback shift register and combinational logic. Hence the sequence is rather ‘pseudo’ random.(2) the line joining the sun and a planet sweeps out equal areas in equal intervals of time. 9. 8. Kepler’s law may be simply stated as (1) the planets move in ellipses with the sun at one focus. Relatively complex 4. thus interfering with major pulse lobe. Circuit is simple. the minimum sampling rate is equal to twice the highest audio input frequency. This signal is binary in nature. The same logic is used at transmitter and receiver. sampling rate states that. but it is generated by a well-defined logic. (4) IT 2202 36 . This interference is commonly called as intersymbol interference or ISI. QASK 5. Frequency division multiple access (FDMA) v. What is ISI ? The ringing tails of several pulses have overlapped. “Advanced Electronic Communication Systems” Fifth edition. IT 2202 37 . Frequency modulation Noise interference is less Frequency Modulation is the process of changing the frequency of a relatively high frequency carrier signal in proportion with the instantaneous value of the modulating signal. Noise interference is more 2.369-372 (OR) (b) (i)Compare the various types of digital modulation techniques. (6) Refer: Wayne Tomasi. (ii) The phase deviation constant in a phase modulation system is K=0. (8) Refer: Wayne Tomasi. (2) 2fm =500*2=1000Hz (4)Total power delivered to the load of 600O. Amplitude Modulation is the process of changing the amplitude of a relatively high frequency carrier signal in proportion with the instantaneous value of the modulating signal. (4) 12.364-375 (ii)Write short note on spectrum and bandwidth of FSK. “Advanced Electronic Communication Systems” Fifth edition.145-153 (ii) An audio frequency signal 10sin2p*500t is used to amplitude modulate a carrier of 50 sin2p*10^5t.Calculate the maximum phase deviation when the modulating signal of 10V is applied.Refer: Wayne Tomasi. (2) Ptotal = Pc [1 + m2 / 2] (OR) (b) (i)Compare FM and AM. Page No. (2) Vm/vc=10/50=0. 3. (2) USB= fc+fm=500+100000=100500 Hz LSB=fc-fm=90500 Hz (3)BW required. Page No.01rad/v. (a) (i)Explain the principle of FSK transmitter and receiver (10) Refer: Wayne Tomasi. Page No. (1) Modulation index.Calculate. “Advanced Electronic Communication Systems” Fifth edition.(12) Amplitude modulation 1.2 (2)Side band frequencies. The depth of modulation has limitation in AM. “Advanced Electronic Communication Systems” Fifth edition. But in FM the depth of modulation can be increased to any value by increasing the deviation. (ii)Explain the eye pattern in base band digital transmission with neat diagram. (a) (i)Explain the elements of PCM system with a neat block diagram. Page No. Companding is the means of improving dynamic range of communication systems.499. (10) Refer: Simon Haykin. Higher amplitude signals are compressed prior to transmission and then expanded in the receiver.425-427 (ii)What is companding ? (4) Companding is the process of compression and then expanding.462-463 13.In this case. “Advanced Electronic Communication Systems” Fifth edition. The information signal operates at 100Hz. v. “Communication Systems” Fourth edition. (a) (i)Write short notes on Frequency hop spread spectrum. “Advanced Electronic Communication Systems” Fifth edition. (12) Refer: Wayne Tomasi. the signal spectral density is kept small such that the presence of the signal is not detected easily. Previously this antijam capability was used in military application. “Advanced Electronic Communication Systems” Fifth edition. (8) Refer: Wayne Tomasi. (4) (ii) Describe the operations of DPCM system with relevant diagram. This is because the spread spectrum signal has the ability to resist the effects of multipath fading.455-456 14. (OR) (b) (i)Find the signal amplitude for minimum quantization error in a delta modulation system if step size is 1 volt having repetition period 1 ms. Because of wide spectrum only small portion of the signals is in fade. Low probability of intercept is an application of spread spectrum in military . Page No. (12) Refer: Wayne Tomasi. (6) The spread spectrum has the ability to resist the effect of intentional jamming. Page No. (ii)Explain the applications of spread spectrum techniques. Page No. Some commercial applications also use spread spectrum because of its antijam capability. iv. Spread spectrum is used in mobile communications. IT 2202 38 . “Advanced Electronic Communication Systems” Fifth edition.(OR) (b) (i)Give a detailed account of multiple access techniques. “Advanced Electronic Communication Systems” Fifth edition.1076 (ii)Write short notes on LEO and GEO orbits. (10) i. (8) Refer: Wayne Tomasi. Page No. Page No. “Advanced Electronic Communication Systems” Fifth edition.1104 & 1108 15. (6) Refer: Wayne Tomasi.580 IT 2202 39 .Frequency division multiple access (FDMA) iii. (8) Refer: Wayne Tomasi. Page No. “Advanced Electronic Communication Systems” Fifth edition. Page No.Time division multiple access (TDMA) ii. (ii)Compare TDMA and CDMA. (a) (i)Discuss briefly the basic satellite communication system .Code division multiple access (CDMA). (16) Refer: Wayne Tomasi.1062 (OR) (b) Enumerate the elements of an optical fiber transmission link.