COMMUNICATIONCIRCUITS Assoc. Prof. Zlatka Valkova‐Jarvis, PhD TECHNICAL UNIVERSITY OF SOFIA FACULTY OF TELECOMMUNICATIONS COMMUNICATION CIRCUITS Bachelor of Engineering Course Compulsory Subject, Code BTCEe25, Credits 6, Semester IV Lectures: 45 hours (15 three‐hour lectures); Laboratory Work: 15 hours (5 three‐hour labs) Elective Semester Project: Code BTCEe34, Credit 1, Semester V LECTURE CONTENT 1. Analogue linear one- and two-port circuits– basic parameters and descriptive equations, properties. Analysis of analogue circuits using generalized matrices. Basic elements used in analogue linear circuits 2. Representation of analogue immittance functions – analytical representation, pole-zero diagrams, frequency responses. The Hurwitz test. Frequency and impedance scaling 3. Passive linear one-port circuits – canonic schemes. LC and RC one-port circuits - basic properties, synthesis of Foster and Cauer canonic one-port schemes 4. Description of analogue linear two-port circuits in the s-domain – transfer function (TF), pole-zero diagrams. Description of analogue linear two-port circuits in the frequency domain – frequency responses, polar diagrams 5. The transfer function obtained from a given magnitude or phase response. Frequency responses conditioned by basic type poles and zeros. Pole-zero diagrams and transfer functions obtained from analogue frequency responses. Asymptotic Bode diagrams 6. Description of analogue linear circuits in the time-domain - impulse and step responses, analogue convolution. Analogue impulse and step responses due to different transfer function’s poles. Step responses of general bilinear and biquadratic analogue transfer functions LECTURE CONTENT 7. An introduction to two-port passive circuit synthesis - terms of realisability. Properties and limitations of structures with different elements and topology. Different types of two-port passive circuits. Synthesis of non-terminated L-shaped and lattice schemes 8. Amplitude correction – basic considerations, TF, cascade realisation, specifics of approximation. Passive and active amplitude-correction sections of first and second order. Attenuators 9. Phase correctors – basic principles of phase correction, transfer functions, cascade realisation. Delay circuits 10. Electrical filters- basic types, specifications. Approximation based on given requirements to the Magnitude Response. Frequency transformation for analogue filters. Basic filter transfer functions of n -th , first and second order 11. Polynomial (Butterworth, Chebyshev, Legendre, Bessel) and non-polynomial (Cauer, Inverse Chebyshev) classic approximations. Comparison of different approximations. Computer-based approximations – generalised equal-ripple and maximally-flat approximations 12. Synthesis of RC and single-terminated LC two-port ladder circuits with polynomial transfer functions. Analysis of ladder LC filter structures 13. Active circuits - basic principles of synthesis and design. Properties of cascade realisations. Low pole- quality factor (single-amplifier) first- and second-order polynomial and non-polynomial active filter sections – transfer functions and active models 14. Medium and large pole-quality factor active filter sections – basic models and realisations. Multi- amplifier universal biquadratic active filter sections. Switched-Capacitor active filters (SC-filters) 15. Direct active realisation methods by passive reactive schemes imitation. Gyrator models. Simulation of double-terminated LC-ladder circuits using frequency-dependent negative resistance and conductivity – Bruton transformations. Sensitivity – types and evaluation LECTURE CONTENT 16. Linear Time-Invariant (LTI) systems. Description of digital linear circuits in the time- and z-domains. The interdependence of the s-plane and z-plane 17. Description of digital linear circuits in the frequency domain. Аnalogue to digital transformation. Digital frequency response in relation to the basic type of poles and zeros. Pole-Quality Factor 18. Basic elements used in digital LTI systems. Recursive (IIR – Infinite Impulse Response) and non- recursive (FIR - Finite Impulse Response) digital circuits. Stability considerations 19. Analysis of digital LTI systems using Signal Flow-Graph (SFG) transforms and simple circuit descriptions in time- and z-domain 20. Non-recursive (FIR) digital circuit realisations. Direct, predictive direct, cascade, lattice, cascade lattice and linear-phase structures 21. Design of non-recursive digital systems using window functions. Mini-max design methods – Parks- McClellan algorithm. Design of FIR systems with Matlab 22. IIR digital circuit realisations. Direct, cascade, parallel, leader and lattice realisations. Realisations based on parallel all-pass structures 23. Design of IIR digital systems. Design based on the bilinear z-transformation, invariant Impulse Response method. Direct methods. Digital frequency transforms. Design of IIR systems with Matlab INFORMATION RESOURCES • Primary Sources 1. Lecture slides (available at - http://sopko-tu-sofia.bg). 2. LUTOVAC, M, D. TOSIC AND B. EVANS, Filter design for signal processing using Matlab and Mathematica, Prentice Hall, 2001. 3. SCHAUMANN, R AND M. E. VAN VALKENBURG, Design of analog filters, 2nd edition, The Oxford Series in Electrical and Computer Engineering, 2009. 4. MITRA, S., Digital signal processing. 4-rd Edition, McGraw-Hill, 2010. 5. ADYA ALKA, Circuits and Systems, Alpha Science Intl Ltd, 2010. 6. WING OMAR, Classical Circuit Theory, Springer, 2010. 7. COUCH L., Digital & Analog Communication Systems, 7th edition, Prentice Hall, 2006. • Supplementary Sources 8. MATLAB – R2010b + Toolboxes. MathWorks, 2010 and any other versions of MATLAB. 9. CHEN, W.K., The Circuits and filters handbook, CRC Press & IEEE Press, 2009. • Useful WWW addresses General materials on CC http://sopko.tu‐sofia.bg. Recorded seminars on MATLAB http://www.mathworks.com/company/events/archived_webinars.jsp Exchange of files for MATLAB http://www.mathworks.com/matlabcentral/fileexchange/loadCategory.do Interactive filter design http://scpd.stanford.edu/SCPD/js/brandingFrame/externalURL2.htm CC courses in GeorgiaTechU, USA http://csip.ece.gatech.edu/csipindex.html Materials on Circuits and DSP http://www.yov408.com/html/tutorials.php?&s=83 CC courses in Tampere, Finland http://www.tut.fi/public/oppaat/opas2004‐2005/kv/laitokset/InstituteofSignalProcessing/ On‐line courses and materials http://www.techonline.com/community/home Educational materials http://www.onesmartclick.com/engineering/digital‐signal‐processing.html MIT Open Courses http://ocw.mit.edu/OcwWeb/Electrical‐Engineering‐and‐Computer‐Science/index.htm METHOD OF ASSESSMENT Assessment Two-hour written exam with 12 problems to be solved. Unlimited use of books and sources. The final grade includes the assessments from the Lab works and from participation in the discussions during the lectures, with weighting coefficients as given in the table below. Standards of Assessment Excellent (6) – a very comprehensive knowledge of the entire course-material and of the main additional sources. A deep understanding, perfect reasoning and sound creativity in solving complex problems and applying the knowledge. Well-demonstrated original, deductive thinking, together with the capability to argue, define and defend a position connected with the subject. Very Good (5) – a comprehensive knowledge of the entire material with deep understanding, very good reasoning and manifest creativity in solving problems and applying the knowledge. Good (4) – more than basic knowledge with the ability to solve simple and more difficult problems but showing some lack of creativity in solving problems and applying the knowledge. Fair (3) – a basic knowledge and ability to solve simple problems. Final Assessment Components Component Weight 1 Written Exam 0,75 2 Lab Work Assessment 0.15 3 Participation in Lecture Discussions 0,10 LABORATORY WORK 1. Investigation of time-, frequency- and s-domain characteristics and descriptions of analogue circuits and transfer functions using MATLAB 2. Investigation of time-, frequency- and z-domain characteristics and descriptions of digital circuits and transfer functions using MATLAB 3. Magnitude approximation of given specifications. Realization and comparative studies of Butterworth, Chebyshev, Cauer, Inverse Chebyshev and Bessel approximations using MATLAB 4. Magnitude approximation of FIR and IIR digital with MATLAB. Design of non-recursive (FIR) digital filters using window functions. Mini-max methods of design. Design of IIR filters - Butterworth, Chebyshev, Cauer, Inverse Chebyshev and Bessel digital circuit approximations 5. Design of FIR and IIR filter structures with MATLAB. Basic digital circuit realizations - Direct, Cascade, Parallel, Ladder- lattice, and Lattice realizations. Analogue linear one- and two-port circuits– basic parameters and descriptive equations, properties. Analysis of analogue circuits using generalized matrices. Basic elements used in analogue linear circuits Linear circuits can be described in the complex s-domain (the Laplace domain), frequency domain (the Fourier domain), and time domain. Let's recap: The Laplace Transformis a powerful tool that is very useful in Electrical Engineering. The transform allows equations in the "time domain" to be transformed into an equivalent equation in the complex s-domain (s = σ + jω) and vice-versa. The Fourier Transformdescribes the system in the frequency domain and can be a specific case of the Laplace transform. The complex Laplace variable s can be separated into its real and imaginary parts: s = σ + j ω (σ is the real part of s, ωis the imaginary part of s, and j is the imaginary number). For σ = 0, i.e. s = j ω, the Laplace transform is the same as the Fourier transform if the signal is causal. The variable ωis known as the "radial frequency”, and is given the units rad/s (radians per second) 1. ANALOGUE LINEAR ONE-PORT CIRCUITS The descriptive parameter/function of a linear one- port circuit is: I mpedance: Complex Laplace Impedance Complex Fourier Impedance Z(s)=R + jX (s÷jω) Z(je)= R(jω) + jX(jω) R – resistance; X - reactance Admi ttance: Complex Laplace Admittance Complex Fourier Admittance Y(s) =1/Z(s) =G+ j X -1 (s÷jω) Y(je)= G(jω) + jS(jω) G =1/R– conductance; 1/X- susceptance Z(s) and Y(s) F(s) - I mmi tance functi on The descriptive equation of a one-port circuit is Ohm‘s law: ) ( ). ( ) ( ) ( ). ( ) ( s U s Y s I s I s Z s U = = U I F(s) 2. ANALOGUE LINEAR TWO-PORT CIRCUITS Two-por t passi ve ci r cui t par ameter s These consider the internal state of the circuit rather than its working behaviour and are also called static, primary, transducer matrices, internal. y-parameters - facilitate admittance matching calculations z-parameters - facilitate impedance matching calculations h-parameters (hybrid) f-parameters a-parameters (inverse transmission), providing electrical inputs b-parameters (transmission), providing electrical outputs - + U1 I1 U2 - + I2 ( ¸ ( ¸ ( ¸ ( ¸ = ( ¸ ( ¸ 2 1 22 21 12 11 2 1 U U y y y y I I ( ¸ ( ¸ ( ¸ ( ¸ = ( ¸ ( ¸ 2 1 22 21 12 11 2 1 I I z z z z U U ( ¸ ( ¸ ( ¸ ( ¸ = ( ¸ ( ¸ 2 1 22 21 12 11 2 1 U I h h h h I U ( ¸ ( ¸ ( ¸ ( ¸ = ( ¸ ( ¸ 2 1 22 21 12 11 2 1 I U f f f f U I ( ¸ ( ¸ ÷ ( ¸ ( ¸ = ( ¸ ( ¸ 2 2 22 21 12 11 1 1 I U a a a a I U ( ¸ ( ¸ ÷ ( ¸ ( ¸ = ( ¸ ( ¸ 1 1 22 21 12 11 2 2 I U b b b b I U Two-por t ci r cui t tr ansfer functi ons These consider the behaviour of the circuit and not the unnecessary internal detail and are also called secondary parameters, dynamic, external, working Tabl. 1-1 - + U1 I1 U2 - + I2 Dynamic parameter The Laplace domain The Fourier domain 1 Voltage transfer function 2 Currency transfer function 3 Transfer Impedance 4 Transfer Admittance 5 Input Impedance 6 Output Impedance ) ( ) ( ) ( 1 2 e e = e j U j U j T U ) ( ) ( ) ( 1 2 e e = e j I j I j T I ) ( ) ( ) ( 1 2 21 e e = e j I j U j Z ) ( ) ( ) ( 1 2 21 e e = e j U j I j Y ) ( ) ( ) ( 1 1 e e = e j I j U j Z in ) ( ) ( ) ( 2 2 e e = e j I j U j Z out ) ( ) ( ) ( 1 2 s U s U s T U = ) ( ) ( ) ( 1 2 s I s I s T I = ) ( ) ( ) ( 1 2 21 s I s U s Z = ) ( ) ( ) ( 1 2 21 s U s I s Y = ) ( ) ( ) ( 1 1 s I s U s Z in = ) ( ) ( ) ( 2 2 s I s U s Z out = An interaction between two-port passive circuit parameters and circuit transfer functions exists. Tabl. 1-2 y z h f a T U T 22 T 21 Z y 1 Z y + ÷ T 11 z T 21 Z z Z z + A T h 21 T 21 Z h Z h A + ÷ T 22 T 21 Z f Z f + T 11 12 T Z a a Z + T U / Z T = · 22 21 y y ÷ 11 21 z z h 21 h A ÷ 21 f 11 a 1 T I T y 11 21 Z y y A + T 22 21 Z z z + ÷ T 22 21 Z h 1 h + T 11 f 21 Z f f + A ÷ T 21 22 Z a a 1 + ÷ Z 21 T 11 y 21 Y y y + A ÷ T 22 21 Y z 1 z + T 22 21 Y h h + ÷ T f 11 21 Y f f A + T 22 21 Y a a 1 + Y 21 T 22 21 Z y 1 y + T 11 z 21 Z z z + A ÷ T h 11 21 Z h h A + T 22 21 Z f f + ÷ T 11 12 Z a a 1 + ÷ Z in T y 11 T 22 Z y Z y 1 A + + T 22 T 11 z Z z Z z + + A T 22 T h 11 Z h 1 Z h + A + T 11 f T 22 Z f Z f + A + T 21 22 T 11 12 Z a a Z a a + + Z out И y 22 И 11 Z y Z y 1 A + + И 11 И 22 z Z z Z z + + A И 22 h И 11 Z h Z h + A + И 11 И f 22 Z f 1 Z f + A + И 21 11 И 22 12 Z a a Z a a + + 3. ANALYSIS OF ANALOGUE CIRCUITS USING GENERALIZED MATRICES I. The method applies to analogue circuits having direct connection between the input and the output - this node is numbered as “zero”. The other nodes have the numbers from 1 to mstarting from the input and moving towards the output. II. A square matrix Y (m× m) is formed. Each element of the matrix Y is an admittance formed as follows: - in the main diagonal are the admittances with identical indexes Y ii - the sum of admittances connected to the node i in the circuit; - the remaining admittances Y ij are the sum of the admittances placed between node i and node j, with sign minus. III. The necessary adjugates Δ ij and the matrix’s determinant Δ are calculated and used to determine the required dynamic parameter via the formulae in Table 1-3 (the second column – for the matrix of admittances Y). Δ ij = (-1) i+j M ij , where M ij (referred to as the minor of Y) is the determinant of the (m− 1)×(m− 1) matrix that results from deleting row i and column j of Y. Dynamic parameter Connection with the matrix of: admittances impedances 1 2 U U T U = bb aa T aa ab Y , A + A A A + A A T bb ab Y 0 2 1 2 = = I U U T U aa ab A A bb ab A A 1 2 I I T I = A + A A ÷ T bb ab Z bb aa T aa ab Z , A + A A ÷ 1 2 21 I U Z = bb T ab Y A + A A aa T bb aa ab Y A + A A , 1 2 21 U I Y = aa T bb aa ab Z A + A A ÷ , bb T ab Z A + A A ÷ 1 1 I U Z in = A + A A + A T bb T aa bb aa Z Z , bb aa T aa T bb Z Z , A + A A + A 2 2 out I U Z = A + A A + A s aa s bb bb aa Z Z , bb aa s bb s aa Z Z , A + A A + A Z T is the impedance of the load (Y T – the admittance); Z s is the impedance of the source. Tabl. 1-3 NUMERICAL EXAMPLE 1: Determine the voltage transfer function, when Z T =∞, of the following analogue circuit: Soluti on: The admittances matrix [Y] is 3 × 3. The formulae for the voltage transfer function (Table 1-3) is: “a” denotes the number of the input node, i.e. a=1, while “b” denotes the number of the output node, i.e. b=3. For Z T = ∞, i.e. Y T =0, the voltage transfer function will be: We calculate the adjugates Δ 13 and Δ 11 : Hence, the voltage transfer function is: R L C sL Y sC Y R Y 1 ; ; 1 3 2 1 = = = | | ( ( ( ( ( ( ( ¸ ( ¸ ÷ ÷ | | . | \ | + + ÷ ÷ = ( ( ( ¸ ( ¸ ÷ ÷ + + ÷ ÷ = sL sL sL sL sC R R R R Y Y Y Y Y Y Y Y Y 1 1 0 1 1 1 1 0 1 1 0 0 Y 3 3 3 3 2 1 1 1 1 bb aa T aa ab U Y s T , ) ( A + A A = 11 13 ) ( A A = A A = aa ab U s T ( )( )( ) sRL sL R Y Y ab 1 1 1 1 3 1 3 1 13 = | | . | \ | ÷ | | . | \ | ÷ = ÷ ÷ ÷ = A = A + ( ) | |( ) ( ) sRL sRC sC R sL Y Y Y Y Y Y Y Y aa 1 1 1 1 2 1 3 1 1 2 3 3 2 1 3 11 + = | | . | \ | + = + = ÷ ÷ + + = A = A + 1 1 1 1 1 1 1 ) ( ) ( 2 1 3 3 1 + = + = | | . | \ | + = + = sRC sRL sRC sRL sC R sL sRL Y Y Y Y Y s T U 4. BASIC ELEMENTS USED IN ANALOGUE LINEAR CIRCUITS Active elements - the elements which are capable of delivering energy - Transistors, Triacs, Varistors, Vacuum Tubes, Relays, Solenoids and Piezo-Electric Devices; Passive elements - the elements which will receive the energy and dissipate or store energy - Resistor, Capacitor and Inductor (linear passive elements) ; Diodes, Switches and Spark Gaps (non- linear passive devices) Resi stor - R, O (or Conductance - G =1/R, S) Resistors are circuit elements that allow current to pass through them, but restrict the flow according to a specific ratio called "Resistance". Every resistor has a given resistance. Resistors are commonly used as heating elements. Potentiometers - resistors that have a var i able r esi stance as a function of position. Thermisters - resistors that have a variable resistance as a function of temperature. R Capacitor - C, F ; Z C (s)=1/sC Inductor - L, H ; Z L (s)=sL Ideal Transformer – n (U 2 =nU 1 ) coefficient of transformation Oper ati onal Ampli fi er s (Op Amps) - active circuit components; have 2 input terminals and 1 output terminal . U out = A(U 2 - U 1 ) –descriptive equation An Ideal Op Amp has: • i nfi ni te impedance, bandwidth, voltage gain • zer o: output impedance, offset (error) C L Sour ces Independent Sources - produce current/voltage at a particular rate that is dependent only on time. • Current sources - sources that output a specified amount of current. The voltage produced by the current source will be dependent on the current output, and the resistance of the load (Ohm's law). An ideal current source has R in = ·, O. I out = const in any kind of load A diagram of an ideal current source, I, driving a resistor, R, and creating a voltage V • Voltage sources - produce a specified amount of voltage. The amount of current that flows out of the source is dependent on the voltage and the resistance of the load (again, Ohm's law). An ideal voltage source has R in = 0, O. U out = const in any kind of load. A diagram of an ideal voltage source, V, driving a resistor, R, and creating a current I Dependent Sources - these are current or voltage sources whose output value is based on time or another value from the circuit. The following sources are possible: • Current-controlled current source - Z in = 0; Z out = · Descriptive parameter: o Descriptive equation: I 2 = o I 1 • Current-controlled voltage source - Z in = 0; Z out = 0 Descriptive parameter: z 11 Descriptive equation: U 2 = z 11 I 1 • Voltage-controlled current source – Z in = ·; Z out = · Descriptive parameter: y 21 = g m . Descriptive equation: I 2 = g m (U 2 - U 1 ) • Voltage-controlled voltage source - Z in = ·; Z out = 0 Descriptive parameter: µ Descriptive equation: U 2 = µ U 1 I mpedance Conver ter s • NI C-Negative I mpedance Converter Descriptive parameter: _ Descriptive equation: Z 1 =-_Z 2 One Op Amps realization: _=R 1 /R 2 Appli cati on: negative element circuits; active analogue circuits • GI C-Generalized I mpedance Converter Descriptive parameter: f(s) Descriptive equation: Appli cati on: active analogue circuits, active inductors, capacitor multipliers, FDNC, FDNR, etc. ) ( ) ( ) ( ) ( ) ( ) ( 4 2 3 1 s Z s Z s Z s Z s Z s Z T in = NIC Z 1 =- _ Z 2 GIC Z in =f(s)Z T Op Amp 1 Op Amp 2 Fr equency-dependent negati ve r esi stor s and conductor s •FDNC – Frequency Dependent Negative Conductivity – (D-element) •FDNR – Frequency Dependent Negative Resistance – (N-element) Descriptive parameter: D, F.s Descriptive equation: Realization: 1 GIC, 3 resistors and 2 capacitors. For example: ( ) C j j D j Y Ds s Y D D e e = e ÷ = e = 2 2 ) ( ; ) ( Descriptive parameter: N, H.s Descriptive equation: Realization: 1 GIC, 3 resistors and 2 capacitors. For example: ( ) L j j N j Z Ns s Z N N e e = e ÷ = e = 2 2 ) ( ; ) ( Application: Active analogue circuits based on Bruton transformations ( ) 2 2 3 1 4 2 4 2 3 1 1 1 ) ( Ns s R R R C C R sC sC R R s Z T T in = = = ( ) ( ) 2 2 2 4 2 3 1 4 2 3 1 1 1 1 ) ( Ds s Y Ds s R R C C R R R R sC sC s Z in T T in = ÷ = = = Gyr ator – a passive linear two-port entirely non-reciprocal device Descriptive parameter: R gy – gyration resistance Descriptive equation: Z 1 (s) Z 2 (s) =R gy =const (Z 1 (s) and Z 2 (s) are reverse impedances) An important property of a gyrator is that it i nver ts the cur r ent-voltage char acter i sti c of an electrical component or network. In the case of li near elements, the i mpedance i s also i nverted, i. e. a gyrator can make a capaci ti ve ci r cui t behave i nducti vely and vice versa Reali zati on: 1 NIC, 2 resistors and 1 D-element (FDNC): R gy R gy Z 1 = NIC Appli cati on of a gyr ator : It is primarily used in active filter design and miniaturisation. •to transform a load capacitance into an inductance •to reduce the size and cost of a system by removing the need for heavy and expensive inductors Representation of analogue immittance functions – analytical representation, pole-zero diagrams, frequency responses. The Hurwitz test. Frequency and impedance scaling 1. REPRESENTATION OF ANALOGUE IMMITTANCE FUNCTIONS 1.1. Analytical representation in s-domain A circuit transfer function is the ratio of two polynomials N(s) (Numerator) and D(s) (Denominator): impedance Z(s) admittanceY(s) Limitations of i mpedance Z(s) and admi ttance Y(s): The coefficients b i anda k must be positive real numbers. The highest (mand n) and the lowest powers of the numerator and denomi-nator polynomials (D(s) and N(s)) must not differ by more than one D(s) and N(s) are Hurwitz polynomials. For LC-one-port circuits they are modified Hur wi tz polynomi als. Hence, when H(s) is an immitance function (Z(s) or Y(s)), both N(s) and D(s) must be Hurwitz polynomials. ¿ ¿ = = ÷ ÷ ÷ ÷ = + + + + + + + + = = n j j j m i i i n n n n m m m m s a s b a s a s a s a b s b s b s b s D s N s H 0 0 0 1 1 1 0 1 1 1 ... ... ) ( ) ( ) ( . 1.2. Pole-zero Diagram , where •The poles and zer os can be only be either real or complex-conjugated. •The poles and zer os of the immittances are located in the left half-plane of the complex s-plane and/or on its boundary, the imaginary axis [ [ = = ÷ ÷ = ÷ ÷ ÷ ÷ ÷ ÷ = = n j j m i n n m s s s s H s s s s s s a s s s s s s b s D s N s H i m 1 1 0 2 1 0 0 0 ) ( ) ( ) )....( )( ( ) )...( )( ( ) ( ) ( ) ( 2 1 n m a b H = An analogue immittance function can be described in the frequency domain (the Fourier domain) by applying the substitution s⟶jω on the analytical expression in the complex s-domain (the Laplace domain): 1.3. One-port circuit representation in the frequency domain j Z j Z e e e e j Z e ·1 e ·2 j Z e ·1 e ·2 magnetostrictive resonator ( )( ) ( )( ) 2 2 4 2 2 2 2 2 3 2 2 1 5 5 3 3 1 4 4 2 2 0 ) ( ) ( e ÷ e e ÷ e e ÷ e e ÷ e e = e ÷ ÷ ÷ ÷ + + + + + + = e = j H j Z s b s b s b s a s a a s Z j s ) ( ) ( ) ( 2 2 2 2 2 1 e ÷ e e ÷ e e = e j H j Z Z(e) e =· 0 e ) ( ) ( ) ( 2 2 1 2 2 2 e ÷ e e ÷ e e = e H j j Z N S L 0 C L C 0 L C The frequency responses Z(jω)/j=Z(ω) of an inductor, capacitor, parallel and serial resonant one-port circuits are: piezoresonator Z(e) e=· 0 e 2. HURWITZ POLYNOMIAL 2.1. Strict Hurwitz Polynomial A polynomial P(s) is a strict Hurwitz polynomial whose • Coefficients are positive real numbers (a necessary but not sufficient condition) • Roots are located in the left half-plane of the complex s-plane. (a necessary and sufficient condition ) A strict Hurwitz polynomial has no any missing term of 's' A given polynomial can be tested to be Hurwitz or not by using the continued fraction expansion technique. The Hurwitz strict polynomial test - strict Hurwitz polynomial verification 1. The polynomial P(s) is split into a sum of two polynomials A(s) – even, and B(s) – odd, i.e. P(s)=A(s)+B(s). n n s a s a s a a s P + + + + = . . . ) ( 2 2 1 0 . . . ) ( 4 4 2 2 0 + + + = s a s a a s A . . . ) ( 5 5 3 3 1 + + + = s a s a s a s B 2. The ratio or (the polynomial in the numerator is the one with the higher degree). 3. +(s) expands in continued fraction using Euclid’s algorithm: 4. If a) the coefficients C i (i=1÷m), derived after the continued fraction expansion, are positive real numbers, and b) the total number of these coefficients is equal to the degree of the polynomial, i.e. m= n, the polynomial P(s) is a strict Hurwitz polynomial. . s C s C s C s C s m 1 . . . 1 1 1 ) ( 3 2 1 + + + + = + ) ( ) ( ) ( s B s A s = + ) ( ) ( ) ( s A s B s = + NUMERICAL EXAMPLE 1: Check if the polynomial is a strict Hurwitz polynomial Soluti on We determine the polynomials A(s) and B(s): - even polynomial of 4th order - odd polynomial of 3rd order Then we define the ratio: , and expand it in continued fraction: Clearly, a) all coefficients C 1 ÷C 4 , derived after the continued fraction expansion are positive real numbers, and b) the total number of the coefficients is 4, and the polynomial P(s) is of 4th degree as well. Therefore, P(s) is a strict Hurwitz polynomial 1 3 ) ( 2 3 4 + + + + = s s s s s P 1 3 ) ( 2 4 + + = s s s A s s s B + = 3 ) ( s s s s s B s A s + + + = = + 3 2 4 1 3 ) ( ) ( ) ( . 2.2. Modified Hurwitz Polynomial A polynomial P(s) is a modified Hurwitz polynomial whose roots are located in the left half-plane of the complex s-plane and/or on the imaginary axis which is its boundary (this is a necessary and sufficient condition ) A modified Hurwitz polynomial always has some missing terms of 's’. Only when the roots of the polynomial are located on the imaginary axis will the odd or even terms of 's' be missing. The Hurwitz modified polynomial test - modified Hurwitz polynomial verification A given polynomial can be determined to be modified Hurwitz or not by using the continued fraction expansion technique of the ratio: , where is the first derivative of the polynomial P(s) The same conditions as those of the strict Hurwitz polynomial test a) and b) must be met to conclude that the polynomial is a modified Hurwitz polynomial. ds s dP s P s ) ( ) ( ) ( = + ds s dP ) ( NUMERICAL EXAMPLE 2: Check if the polynomial is a modified Hurwitz polynomial Soluti on The odd terms of s are missing, so we apply the modified Hurwitz test: Evidently, a) all coefficients C 1 ÷C 4 , derived after the continued fraction expansion are positive real numbers, and b) the total number of the coefficients is 4, and the polynomial P(s) is of 4th degree as well. Therefore, P(s) is a modified Hurwitz polynomial 8 6 ) ( 2 4 + + = s s s P 3. FREQUENCY AND IMPEDANCE SCALING The main dimensions of the electronic elements are not chosen in the best possible way: Resistor - 1O; in practice - 1kO=1000O, 1MO=1000000O are used; Capacitor - 1F, in practice - 1nF=0,000000001F or 10pF=0,00000000001F; Inductor - 1Н, in practice - 1 mH(0,001H) or 1µН(0,000001Н). The frequency range used in communications is wide – from a few Hz to tens of GHz. This makes the description of the circuits more difficult. 1 10 . 2 10 . 4 10 10 . 2 1 1 1 ) ( 8 2 16 2 6 2 + + + = + + + = + + = ÷ ÷ ÷ s s s sRC LC s R sL sL R sC s Z s rad LC p / 10 . 5 , 0 1 8 = ~ e C L R 100O 2µH 200pF The main aim is to transform the value of the electronic elements and frequencies so that they are of the order of 1. In this way the coefficients in the transfer function will have values of the same order. I mpedance Scali ng When the impedance is divided by a coefficient k r , The nature of the impedance doesn’t change : sC sL R s Z 1 ) ( + + = ( ) ' ' ' ' 1 1 1 ) ( n n n r r r r r n sC sL R sCk k L s k R k sC sL R k s Z s Z + + = + + = + + = = ¦ ¦ ¦ ) ¦ ¦ ¦ ` ¹ = = = C k C k L L k R R r n r n r n ' ' ' ; ; I mpedance scali ng k r is chosen to be around the average value of the resistors. Fr equency Scali ng The frequency range is scaled by a coefficient k f so that the most important frequencies of the circuit become close to 1. To keep the analytic expression of the circuit functions unchanged a multiplication and a division of the coefficient k f is carried out. f n f n k f F f k = = e = O = e ; " " " 1 1 n n f f f f f n C j L j R C k k j L k k j R k j Z Z O + O + = e + e + = | | . | \ | e = ¦ ¦ ¦ ) ¦ ¦ ¦ ` ¹ O = e = = = = j k j k s S C k C L k L f f f n f n ; ; " " Fr equency scali ng The scaled impedance is easier to work with because it has better coefficient numbers than the unscaled impedance. Si multaneous Fr equency and I mpedance Scali ng f f f r n r f n r n k s S k C k k C k k L L k R R = e = O = = = ; ; ; ; NUMERI CAL EXAMPLE 3: Let us choose k r = R = 100 and k f = e p = 0,5.10 8 and scale the elements: . / 1 10 . 5 , 0 10 . 5 , 0 ; 1 10 . 2 . 10 . 5 , 0 . 10 ; 1 100 10 . 5 , 0 10 . 2 ; 1 100 100 8 8 10 8 2 8 6 s rad k F k Ck C H k k L L k R R f p f r n r f n r n = = e = O = = = = = = O = = = ÷ ÷ C L R 100O 2µH 200pF 1 1 ) ( 2 + + + = S S S S Z 1 10 . 2 10 . 4 10 10 . 2 ) ( 8 2 16 2 6 + + + = ÷ ÷ ÷ s s s s Z Hz F k f s rad k F k k C C H L k k L R k R f f f r n n f r n r , ; / , ; , ; , ; , = O = e = = O = Descali ng Passive linear one-port circuits – canonic schemes. LC and RC one-port circuits - basic properties, synthesis of Foster and Cauer canonic one-port schemes 1. CANONIC REACTIVE (LC) ONE-PORT CIRCUITS C 2 C 1 C 4 C 3 L 1 one DC circuit one circuit for frequency ω=∞ 5 inductors and 5 capacitors two DC circuits two circuits for frequency ω=∞ 3 inductors and 3 capacitors 1 ) 1 ( ) 1 ( 2 + = + = + = a d a c a a b aZ 1 Z 1 Z 2 bZ 1 cZ 2 dZ 1 1 ) 1 ( 1 2 2 2 + = + = + = a a d a a c a a b Z 1 Z 2 aZ 1 dZ 1 bZ 1 cZ 2 1 ; ) 1 )( 1 ( ) ( 1 ; ) 1 )( 1 ( ) ( 2 2 2 2 + = + + ÷ = + = + + ÷ = b b f a b b a d a a e b a b a c Z 1 Z 2 aZ 1 bZ 2 fZ 2 cZ 1 dZ 2 eZ 1 ab A B b a A b A B bB f b A B B A B e b A B bB d b B A B A B c 4 ; 1 2 2 ; 2 ) ( 2 2 ; 2 ) ( 2 ÷ = + + = + ÷ = + ÷ ÷ = ÷ + = ÷ + + = bZ 2 Z 1 Z 2 aZ 1 cZ 1 dZ 2 eZ 1 fZ 2 Table 3‐1 No equivalent one-port circuits conditions of equivalence 1. 2. 3. 4. 2. BASIC PROPERTIES OF PASSIVE LC ONE-PORT CIRCUITS EXAMPLE 1: 2 1 2 2 2 2 2 2 2 1 2 1 2 1 2 2 2 1 2 1 3 2 2 2 1 1 1 ) ( 1 ) ( ) ( ) ( 2 1 e + e + = + + + = + + + = + + = + = s s sH C L s C L L L L s sL C L s L L s C L L s C L s sL sL s Z s Z s Z C L L 1 2 0 2 2 1 2 1 0 2 1 2 1 2 2 2 2 1 1 ; ; 1 ; 1 3 , 2 3 , 2 e ± = ± = e ± = + ± = + = e = e j C L j s j C L L L L j s C L L L L C L PROPERTIES WHICH AFFECT CANONIC CIRCUITS 1. The difference between the number of inductors and capacitors is no more than one. 2. There is no more than one DC circuit and no more than one circuit for frequency ω=∞. PROPERTIES WHICH AFFECT RESONANCE 1. The number of resonances is one less than the number of elements. 2. The first resonance (in the lowest frequency) is of: • parallel type if there is a DC circuit; • serial type if there is no DC circuit. 3. When the frequency increases, the resonances alternate (parallel, then serial, then parallel again and so on). PROPERTIES RELATING TO POLES AND ZEROS 1. Each resonance frequency has a corresponding pair of “inner” poles or zeros located on the imaginary axis. 2. For s=0 and s=∞ there is always a pole or zero called “outer”. Characteristic row: s=· s=0 je 1 ‐je 1 je 2 je 3 ‐je 2 ‐je 3 s=-· ‐je 2v je 2v e 1 e 2 e 3........... e 2v s=· s=0 PROPERTIES RELATING TO Z(je) AND Z(s) 1. Analytical expression of an LC impedance consists of its numerator, the squares of the frequencies of the serial type resonances and the parallel frequencies in the denominator. ( )( ) ( )( ) 2 2 3 2 2 1 2 2 4 2 2 2 ) ( e ÷ e e ÷ e e ÷ e e ÷ e e = e H j j Z ‐ полюс в началото; ‐ полюс в началото; ‐ полюс в началото; + + + + + + = 5 5 3 3 1 4 4 2 2 0 ) ( s b s b s b s a s a a s Z ‐ pole for s=0 + + + + + + = 4 4 2 2 0 5 5 3 3 1 ) ( s b s b b s a s a s a s Z ‐ zero for s=0 PROPERTIES RELATING TO FREQUENCY RESPONSE 1. Since Z(jω)=jX(ω) and Y(jω)=jS(ω), the frequency responses of a reactance function are or . They have a sign and are always increasing functions located in 1 and 4 quadrants. j j Z X ) ( ) ( e = e j j Y S ) ( ) ( e = e ( )( ) ( )( ) 2 2 4 2 2 2 2 2 3 2 2 1 ) ( e ÷ e e ÷ e e ÷ e e ÷ e e = e j H j Z squares of the frequencies of the serial type resonances squares of the frequencies of the parallel type resonances 2. The multiplier H/je indicates the presence of a DC circuit and the first resonance in the frequency ω 1 is of serial type. In contrast, the multiplier jeHspecifies the absence of a DC circuit, and the first resonance ω 1 is of parallel type. 3. Analytical expressions of LC-impedances in Laplace domain Z(s) – reactance functions: squares of the frequencies of the serial type resonances squares of the frequencies of the parallel type resonances ‐ полюс в началото; EXAMPLE 2: ( )( ) ( )( ) 2 2 3 2 2 1 2 2 4 2 2 2 1 ) ( e ÷ e e ÷ e e ÷ e e ÷ e e = e L j j Z ( )( ) ( )( ) ( ) ( ) 2 3 2 1 2 3 2 1 4 4 2 4 2 2 2 4 2 2 3 5 1 2 3 2 2 1 2 2 4 2 2 2 2 1 ) ( e e + e + e + e e + e + e + = e + e + e + e + = s s s s s L s s s s sL s Z 2 3 2 3 1 2 2 1 1 ; 1 C L C L = e = e 3. SYNTHESIS OF LC-DRIVING-POINT FUNCTIONS 3.1. PARTIAL-FRACTION EXPANSION OF AN LC-DRIVING-POINT FUNCTION ) ( lim ); ( lim ; ) ( lim ) ( 2 2 0 0 2 2 0 2 2 s F s s A s sF A s s F A s s A s A s A s F l s l s s l l l l e + = = = e + + + = e ÷ ÷ ÷ · ÷ · · ¿ LC impedance describes a one-port circuit comprised of inductors and capacitors. - First LC-form of Foster - partial-fraction expansion of an LC impedance ¹ ¹ ¹ + + + ¹ ¹ ¹ + + ¹ ¹ ¹ + e + + = · · · ¿ ¿ l l l o l l l l l C L s s C s A s L s Z s Z s Z s s A s A s A s Z 1 1 1 1 ) ( ) ( ) ( ) ( 2 0 2 2 0 C 0 L · L l C l L 1 C 1 H A C L F A C F A C H A L l l l l l l l o , 1 ; , 1 , 1 ; , 2 2 0 e = e = = = = · · EXAMPLE 3: Synthesise the first form of Foster using the following normalised LC impedance: Apply impedance and frequency descaling by coefficients k r =10 3 and k f =10 6 respectively. Solution: ( )( ) ( ) 2 3 1 ) ( 2 2 2 + + + = s s s s s Z ( )( ) ( ) ( )( ) ( ) ( )( ) 2 1 3 1 lim ) ( lim 2 3 2 3 1 lim ) ( lim 1 2 3 1 lim ) ( lim 2 2 2 2 2 2 2 2 2 0 0 0 2 2 2 2 2 2 2 = + + = e + = = + + + = = = + + + = = ÷ ÷ e ÷ ÷ ÷ ÷ · ÷ · ÷ · s s s s Z s s A s s s s sZ A s s s s s s Z A s l s l s s s s l H A L F A C F A C H A L o 4 1 2 2 1 ; 2 2 1 1 1 3 2 2 3 1 1 ; 1 2 1 1 1 1 1 0 = = e = = = = = = = = = · · C 0 L · L 1 C 1 Descaling: . 2 10 2 10 10 2 ; 25 . 0 4 1 10 10 ; 3 2 10 3 2 10 10 3 2 ; 1 10 10 9 6 3 1 1 6 3 1 1 9 6 3 0 6 3 nF k k C C mH L k k L nF k k C C mH L k k L f r f r f r o f r = = = = = = = = = = = = = = H H H H · · - Second LC-form of Foster - partial-fraction expansion of an LC admittance ¹ ¹ ¹ + + + ¹ ¹ ¹ + + ¹ ¹ ¹ + e + + = ¿ ¿ ¿ · · · l l l l l l l l l C L s L s sL s C s Y s Y s Y s s A s A s A s Y 1 1 1 ) ( ) ( ) ( ) ( 2 0 0 2 2 0 C · L 0 C 1 C l L 1 L l F A L C H A L H A L F A C l l l l l l l o , 1 ; , 1 , 1 ; , 2 2 0 e = e = = = = · · EXAMPLE 4: Synthesise the second form of Foster using the following normalised LC admittance: Apply impedance and frequency descaling by coefficients k r =10 3 and k f =10 6 respectively. Solution: H A L F A C F A C H A L o 4 1 2 2 1 ; 2 2 1 1 1 ; 3 2 2 3 1 1 ; 1 2 1 1 1 1 1 0 = = e = = = = = = = = = · · The impedance and frequency descaling: . 6 1 10 6 1 10 10 6 1 ; 2 2 10 10 ; 2 1 10 2 1 10 10 2 1 ; 2 2 10 10 9 6 3 2 2 6 3 2 2 9 6 3 1 1 6 3 1 1 nF k k C C mH L k k L nF k k C C mH L k k L f r f r f r f r = = = = = = = = = = = = = = H H H H ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) 2 1 3 1 2 3 lim ) ( lim 2 1 3 1 2 1 lim ) ( lim 0 3 1 2 lim ) ( lim ; 0 3 1 2 lim ) ( lim 2 2 2 2 3 2 2 2 2 2 2 2 2 1 2 1 2 1 2 2 2 2 0 0 0 2 2 2 2 2 2 2 2 2 1 2 = + + + + = e + = = + + + + = e + = = + + + = = = + + + = = ÷ ÷ e ÷ ÷ ÷ ÷ e ÷ ÷ ÷ ÷ · ÷ · ÷ · s s s s s s s Y s s A s s s s s s s Y s s A s s s s s sY A s s s s s s s Y A s s s s s s s s C 1 C 2 L 1 L 2 ( ) ( )( ) 3 1 2 ) ( 2 2 2 + + + = s s s s s Y 3.2. CONTINUED-FRACTION EXPANSION OF AN LC-DRIVING-POINT FUNCTION Every LC-driving-point function can be represented as a sum of two functions: ) ( ) ( ) ( 2 1 s Z s Z s Z + = ) ( ) ( ) ( 2 1 s Y s Y s Y + = If Z(s) has a pole for s=∞, then Now, Z 1 (s) has a zero for s=∞, i.e. Y 1 (s)=1/Z 1 (s) has a pole for s=∞, which can be extracted: ) ( 1 ) ( 3 3 2 s Z sL s Y + = Here Y 2 (s) has zero for s=∞, but Z 2 (s)=1/Y 2 (s) has a pole for s=∞ and so on. ) ( ) ( 1 1 s Z sL s Z + = ) ( 1 ) ( 2 2 1 s Y sC s Z + = The input impedance is described by the following continued-fraction expansion: The obtained one-port circuit is the first (LP) form of Cauer: If Z(s) has a pole for s=0, that pole is extracted. The second (HP) form of Cauer is synthesised after the following continued-fraction expansion is done: Hence, when an input function (impedance or admittance) is expanded as a continued- fraction: The following two schemes are derived: As a rule: In order to synthesise the first (LP) form of Cauer, the input function (impedance Z(s) or admittance Y(s), the one which has pole for s=∞) is expanded as a continued-fraction as the polynomials N(s) and D(s) are presented in descending order of s. To synthesise the second (HP) form of Cauer, the input function having pole for s=0 (impedance Z(s) or admittance Y(s)) is expanded as a continued-fraction as the polynomials N(s) and D(s) are presented in ascending order of s. EXAMPLE 5: Synthesise the first and second form of Cauer using the following normalised LC impedance: Apply impedance and frequency descaling by coefficients k r =10 3 and k f =10 6 respectively. ( )( ) ( ) s s s s s s s s s Z 2 3 4 2 3 1 ) ( 3 2 4 2 2 2 + + + = + + + = Solution: The characteristic order of Z(s): a) First formof Cauer: Z(s) has pole for s=∞ and is expanded as a continued-fraction when the polynomials N(s) and D(s) are in descending order of s: . 6 1 10 6 1 10 10 6 1 ; 4 4 10 10 ; 2 1 10 2 1 10 10 2 1 ; 1 10 10 9 6 3 2 2 6 3 2 2 9 6 3 1 1 6 3 1 1 nF k k C C mH L k k L nF k k C C mH L k k L f r f r f r f r = = = = = = = = = = = = = = H H H H Descaling: b) Second formof Cauer: Z(s) has pole for s=0 and is expanded as a continued-fraction when the polynomials N(s) and D(s) are in ascending order of s: . 5 5 10 10 ; 25 2 10 25 2 10 10 25 2 ; 2 4 5 10 10 ; 3 2 10 3 2 10 10 3 2 6 3 2 4 9 6 3 2 3 6 3 1 2 9 6 3 1 1 mH L k k L nF k k C C mH L k k L nF k k C C f r f r f r f r = = = = = = = = = = = = = = H H H H Descaling: OUTER POLE AND ZERO COMBINATIONS CHARACTERISTIC ROW IMPEDANCE Z(jω) FREQUENCY CHARACTERISTIC X(e) IMPEDANCE Z(jω) FIRST FORM OF FOSTER CANONIC ONE‐PORT CIRCUIT (SERIAL TYPE) SECOND FORM OF FOSTER CANONIC ONE‐PORT CIRCUIT (PARALLEL TYPE) FIRST FORM OF CAUER CANONIC ONE‐PORT CIRCUIT (LOW‐PASS TYPE) SECOND FORM OF CAUER CANONIC ONE‐PORT CIRCUIT (HIGH‐PASS TYPE) 4. BASIC PROPERTIES OF PASSIVE RC ONE-PORT CIRCUITS 0 1 0 1 2 2 1 2 1 2 1 1 1 1 ) ( a s a b s b C sR R R C R sR sC R R s Z RC + + = + + + = + + = 2 2 2 2 2 2 2 2 1 2 2 2 1 1 ) ( ) ( C R C R R R R Z RC e + e + + = e 0 1 1 1 0 1 1 1 . . . . . . ) ( a s a s a s a b s b s b s b s F n n n n m m m m RC + + + + + + + + = ÷ ÷ ÷ ÷ PROPERTIES RELATING TO ANALYTICAL EXPRESSIONS Z RC (s) AND Y RC (s) 1. The degrees mand n of the polynomials N(s) and D(s) -for Z RC (s) ¬ m=n or m=n-1 (m>n is not possible); -for Y RC (s) ¬ m=n or m=n+1 (m<n is not possible). 2. The polynomials N(s) and D(s) must be strict Hurwitz polynomials. PROPERTIES RELATING TO POLES AND ZEROS 1. Poles and zeros are real numbers, located on the negative real semi-axis and alternate. 2. There is always a pole of impedance Z RC (s) (a zero of the admittanceY RC (s)) which is located next to or exactly in the coordinate home (s=0). 3. For the frequency closest to infinity or infinity itself (s=·) the impedance Z RC (s) may have zero only, whereas the admittance Y RC (s) may have only a pole. 4. The first and the last special points (poles and zeros) are always of opposite type. PROPERTIES WHICH CONCERN FREQUENCY RESPONSES 1. Frequency responses are limited functions of the frequency (0<АЧХ<·). 2. Frequency responses Z RC (o) and Y RC (o) are purely theoretical in nature. 5. SYNTHESIS OF RC-DRIVING-POINT FUNCTIONS 5.1. PARTIAL-FRACTION EXPANSION OF AN LC-DRIVING-POINT FUNCTION ¿ o + + + = · l l l s A sC R s Z 0 1 ) ( C0 R - First RC-form of Foster ¿ o + + + = · l l l RC s sA sC R s Y 0 1 ) ( C · R - Second RC-formof Foster O o = = ¬ ) ` ¹ o + = = = o ÷÷ ÷ · ÷ · , ; F , 1 ) ( ). ( lim ); ( lim 1 ); ( lim l l s l 0 0 s l l l RC l RC s RC A R A C s Z s A s sZ C s Z R l F A C A R s Y s s A s s Y C s Y R l l l l l RC l s l RC s RC s l , ; , 1 ) ( . ) ( lim ; ) ( lim ); ( lim 1 0 0 o = O = ¬ ) ` ¹ o + = = = o ÷÷ · ÷ · ÷ EXAMPLE 6: Synthesise the first and second forms of Foster using the following normalised impedance: Solution: Check the realisability: First form of Foster: Second form of Foster: ( )( ) 3 1 2 ) ( + + + = s s s s Z ( )( ) ( ) ( )( ) ( )( ) ( )( ) O = = o = = = ¬ ) ` ¹ = + + + + = o + = O = = o = = = ¬ ) ` ¹ = + + + + = o + = = + + + = = = + + + = = ÷÷ o ÷÷ ÷÷ o ÷÷ ÷ ÷ · ÷ · ÷ · , 2 1 1 5 . 0 ; F 2 1 2 1 3 1 2 ) 3 ( lim ) ( ). ( lim ; , 2 1 1 5 . 0 ; F 2 1 2 1 3 1 2 ) 1 ( lim ) ( ). ( lim ; 0 3 1 2 lim ) ( lim 1 ; 0 3 1 2 lim ) ( lim 3 3 3 3 3 s 3 s 3 1 1 1 1 1 s 1 s 1 0 0 0 s s 3 1 l RC l RC s RC s RC A R A C s s s s s Z s A A R A C s s s s s Z s A s s s s s sZ C s s s s Z R ( ) ( ) ( )( ) F C R R F C s s s s s s z Z s Y , 4 1 ; , 2 ; , 3 2 ; , 1 2 5 . 0 2 3 2 3 1 1 2 2 0 = O = O = = + + + = + + + = = · Only Z RC (s) may have a pole for s=0. In this case, the impedance is expanded as a continued-fraction provided that the polynomials N(s) and D(s) are presented in ascending order of s. As a result the realization called second (HP) RC-form of Cauer is derived, consisting of alternating longitudinal capacitors and transversal resistors, starting with a capacitor. The same canonic realization will be obtained in case of Y RC (s) expansion when its first pole is next to s=0. The first element derived during the synthesis will be a transversal resistor. 5.2. CONTINUED-FRACTION EXPANSION OF AN RC-DRIVING-POINT FUNCTION Only Y RC (s) may have a pole for s=∞ and if this pole exists, we expand the RC admittance in a continued-fraction as the polynomials N(s) and D(s) are presented in descending order of s. Thus transversal capacitors and longitudinal resistors are derived. Each capacitor corresponds to a pole for s=∞. This leads to a chain-based realisation called first (LP) RC-form of Cauer. Another way to synthesize the LP form of Cauer is to expand the RC impedance in descending order of s, in this case the last pole of Z RC (s) is real and is located next to s=∞. The first obtained element now is a longitudinal resistor followed by a transversal capacitor, etc. The rules for synthesis of the canonic first and second RC-forms of Cauer are summarised in Table 3-2: the last pole for s=∞ the first pole for s=0 the last pole next to s=∞ the first pole next to s=0 Z RC not possible Z RC – in ascending order of s HP form of Cauer not possible Y RC – in ascending order of s HP form of Cauer Y RC Y RC – in descending order of s LP form of Cauer not possible Z RC – in descending order of s LP form of Cauer not possible o j e s1= · o1 =0 o j e o j e s 1 = o1 o o1 j e Table 3‐2 EXAMPLE 7: Synthesise the first and second form of Cauer using the following normalised impedance: Solution: Check the realisability: First (LP) form of Cauer: Since Z RC (s) has a zero for s=∞, then Y RC (s) has a pole for s=∞. Hence, Y RC (s) will be expanded in a continued-fraction when the polynomials N(s) and D(s) are in descending order of s: ( )( ) 3 1 2 ) ( + + + = s s s s Z ( )( ) ( ) 2 3 4 3 4 2 3 1 2 ) ( 2 2 + + + = ¬ ¬ + + + = + + + = s s s s Y s s s s s s s Z Second (HP) form of Cauer: The first pole of Z RC (s) is next to s=0, therefore Y RC (s) must be expanded as a continued-fraction to derive the HP RC-form of Cauer. The polynomials N(s) and D(s) are in ascending order of s: ( )( ) ( ) 2 3 4 3 4 2 3 1 2 ) ( 2 2 + + + = ¬ + + + = + + + = s s s s Y s s s s s s s Z Description of analogue linear two-port circuits in the s-domain – transfer function (TF), pole-zero diagrams. Description of analogue linear two-port circuits in the frequency domain –frequency responses, polar diagrams 1. DESCRIPTION OF ANALOGUE LINEAR CIRCUITS FOURIER TRANSFORM } } · · ÷ e ÷ ÷ · e ÷ e e t = e = ¬ = = e ¬ d e j H j H F t h F dt e t h t h F j H F t j t j ) ( 2 1 )] ( [ ) ( ) ( )] ( [ ) ( 1 1 0 LAPLACE TRANSFORM } } · · ÷ ÷ ÷ · · ÷ ÷ t = = ¬ = = ¬ ds e s H j s H L t h L dt e t h t h L s H L st st ) ( 2 1 )] ( [ ) ( ) ( )] ( [ ) ( 1 1 ( ) ( ) ( ) e ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ e = j H s H t h j s Laplas Ti me domai n Complex Laplace domai n (s-domai n) Complex Four i er domai n h(t) – Impulse Response; H(s) – Schematic Function; H(jω) – Schematic Function in frequency domain. A linear analogue system (one-, two- or many- port circuit) is fully described if the output signal y(t) (in time domain) or Y(s) (in s-domain), can be calculated for any given input signal x(t) (in the time domain) or X(s) (in the s-domain) ANALOGUE LINEAR SYSTEM h(t), H(s), H(jω) x(t) y(t) X(s) Y(s) If we have a circuit with impulse response h(t) in the time domain, with input x(t) and output y(t), we can find the Schemati c Functi on H(s) of the circuit, in the Laplace domain, by transforming all three elements: Therefore, the Schematic Function denoted with H(s), can be defined as either the Laplace- transformed representation of the impulse response, ) ( ). ( ) ( ) ( ) ( ) ( s H s X s Y t h t x t y L = ÷÷ ÷ - = )] ( [ ) ( ) ( t h L dt e t h s H st = = } · · ÷ ÷ or the ratio of the circuit output to its input in the Laplace domain: Input Output s X s Y s H = = ) ( ) ( ) ( •In case of one-port circuits H(s) is an impedance Z(s) or admittance Y(s); • The Schematic Function H(s) of a two-port circuit is a “transfer” (a ratio of the circuit input over the circuit output) function Т(s) which can be: a voltage transfer function Т U (s), a currency transfer function Т I (s), a transfer impedance Z 21 (s), and a transfer admittance Y 21 (s). The transfer function can be obtained by one of two methods: Transform the impulse response Transform the circuit, and solve. When the Laplace transform is applied on the differential equation of the analogue system the transfer function is obtained. H(s) is the ratio of two polynomials N(s) and D(s) having the same coefficients as those in the differential equation below. N-Numerator; D-Denominator Limitations of tr ansfer functi ons T(s) The coefficients a k must be positive real numbers, while b i can also be negative; The degree of the numerator m and denominator n may differ arbitrarily but must always be m≤ n; D(s) must be a Hurwitz polynomial, but this is not essential for N(s). ¿ ¿ = = ÷ ÷ ÷ ÷ = + + + + + + + + = = n j j j m i i i n n n n m m m m s a s b a s a s a s a b s b s b s b s D s N s T 0 0 0 1 1 1 0 1 1 1 ... ... ) ( ) ( ) ( ¿ ¿ = = = m i i i i j j n j j dt t x d b dt t y d a 0 0 ) ( ) ( 2. DESCRIPTION OF ANALOGUE LINEAR TWO-PORT CIRCUITS 2.1. TRANSFER FUNCTION – ANALYTICAL EXPRESSION Polynomial TF 0 1 1 1 ... ) ( ) ( ) ( ) ( a s a s a s a s b s D s b s D s N s T n n n n m m m m + + + + = = = ÷ ÷ Non-polynomial TF 0 1 1 1 2 2 2 2 2 2 1 2 ... ) .( . . ) )( ( ) ( ) ( ) ( a s a s a s a s s s Hs s D s N s T n n n n l m + + + + e + e + e + = = ÷ ÷ · · · Reaction Action ) ( ) ( ) ( 1 ) ( = = = s N s D s T s G ‐ Transmission Constant Transfer functions are powerful tools for analysing circuits. If we know the transfer function of a circuit, we have all the information we need to understand the circuit, and we have it in a form that is easy to work with. When the transfer function is obtained, we can say that the circuit has been "solved" completely. There are two types of Transfer Functions T(s): The power of the denominator n is the order of the Transfer Function T(s). Mi ni mum phase TF – the zer os are on the left half-plane only Non-mi ni mum phase TF – some of the zer os are on the right half-plane Maxi mum phase TF– zeros are positioned on the right half-plane •The poles and zer os can only be either real or complex-conjugated. •The poles of the transfer functions are located in the left half-plane of the complex s-plane, while zer os can be anywhere in the s-plane , ) ( ) ( ) )....( )( ( ) )...( )( ( ) ( ) ( ) ( 1 1 0 2 1 0 0 0 2 1 [ [ = = ÷ ÷ = ÷ ÷ ÷ ÷ ÷ ÷ = = n j j m i n n m s s s s H s s s s s s a s s s s s s b s D s N s T i m n m a b H = where 2.2. POLE-ZERO DIAGRAM 2.3. FREQUENCY RESPONSES A system function can be represented in both the Laplace s-domain and the frequency domain: ) ( ) ( ) ( ) ( ) ( ) ( 0 0 e e = e ÷ ÷ ÷ ÷ = = e = = = ¿ ¿ j D j N j T s a s b s D s N s T j s n k k k m i i i ( ) ( ) ) ( ) ( ) ( ) ( ... ... ... ) ( 2 ' 1 2 1 1 2 1 5 4 3 2 1 4 4 2 2 0 4 4 3 3 2 2 1 0 e e + e = = e + e = = ÷ e + e ÷ e + ÷ e + e ÷ = = + e + e ÷ e ÷ e + = e B j A jB A b b b j b b b b b j b b j b j N ( ) ( ) ) ( ) ( ) ( ) ( ... ... ... ) ( 2 ' 2 2 2 2 2 2 5 4 3 2 1 4 4 2 2 0 4 4 3 3 2 2 1 0 e e + e = = e + e = = ÷ e + e ÷ e + ÷ e + e ÷ = = + e + e ÷ e ÷ e + = e B j A jB A a a a j a a a a a j a a j a j D Then, ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 1 2 ' 2 2 ' 2 2 2 2 ' 2 1 2 ' 1 2 ' 2 2 2 2 ' 2 ' 1 2 2 1 ' 2 2 ' 2 2 ' 2 2 ' 1 1 ' 2 2 ' 1 1 e + e = e e + e = e + ÷ e + e + e + = = e ÷ e ÷ · e + e + = e + e + = e e = e jT T B j A B A B A A B j B A B B A A B j A B j A B j A B j A B j A B j A j D j N j T even function odd function The numerator and denominator polynomials are respectively: ( ) 2 ' 2 2 2 2 2 ' 1 2 2 1 2 2 2 1 ) ( ) ( ) ( B A B A T T j T T e + e + = e + e = e = e Magnitude Response the module of T(jω) is an even frequency function: ) ( 2 1 ) ( ) ( ) ( ) ( ) ( e ¢ e = e = e + e = e ÷ ÷ ÷ ÷ j j s e T jT T j T s T Hence, ) ( ) ( ln ) ( ln ) ( ln ) ( e ¢ + e = e = e e ¢ j T e T j T j dB T T Np T T dB Np ), ( lg 20 ) ( ), ( ln ) ( e = e e = e dB Np Np dB T T T T ) ( 115 . 0 ) ( ) ( 686 . 8 ) ( e = e e = e ( ) dB T T a dB dB ), ( lg 20 ) ( e ÷ = e ÷ = e Logarithmic Frequency Responses (for Transfer Functions Т(s) only) Attenuation 0 100 200 300 400 500 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 M a g n i t u d e R e s p o n s e Frequency ANALOGUE BAND‐PASS FILTER ‐ CAUER 0 100 200 300 400 500 ‐60 ‐50 ‐40 ‐30 ‐20 ‐10 0 M a g n i t u d e R e s p o n s e i n d B Frequency ANALOGUE BAND‐PASS FILTER ‐ CAUER 0 50 100 150 200 250 300 350 400 450 500 0 10 20 30 40 50 60 A t t e n u a t i o n i n d B Frequency ANALOGUE BAND‐PASS FILTER ‐ CAUER ( ) ' 2 ' 1 2 2 1 ' 2 1 2 ' 1 1 2 ) ( ) ( B B A A B A A B arctg T T arctg e + ÷ e = e e = e ¢ . s d d t delay , ) ( ) ( e e ¢ ÷ = e ( ) ) (e ¢ ÷ = e b Group Delay Phase Constant Phase Response the argument of T(jω) is an odd frequency function: 2.4. POLAR DIAGRAM MATLAB EXAMPLE: Analogue 6 order BP filter has transfer function: Draw the pole-zero diagram and the frequency responses of the filter. Solution: Ns=[11.8835 6.0943e‐13 788874.3449 2.7785e‐8 4753403994.8018 2.0515]; Ds=[1 58.702 69709.9822 2661613.6619 1394.199644.e+6 234.81 e+8 799.9e+10]; pzmap(Ns,Ds); Zero‐pole plot 10 8 2 3 5 4 3 5 6 7 2 8 3 3 4 13 5 10 . 9 , 799 10 . 81 , 234 19 , 1394 10 . 6 , 26 10 . 7 , 69 702 , 58 0515 , 2 10 . 34 , 475 10 . 778 , 2 10 . 87 , 788 10 . 094 , 6 883 , 11 ) ( + + + + + + + + + + + = ÷ ÷ s s s s s s s s s s s s T -18 -16 -14 -12 -10 -8 -6 -4 -2 0 -300 -200 -100 0 100 200 300 Pole-Zero Map Real Axis (seconds -1 ) I m a g i n a r y A x i s ( s e c o n d s - 1 ) Magnitude Response in relative units [T,w]=freqs(Ns,Ds,9000); m=abs(T); figure(2); plot(w,m); ylabel('Magnitude Response'); xlabel('Frequency'); title('ANALOGUE BAND-PASS FILTER - CAUER'); axis([0 500 0 1]); grid Magnitude Response Logarithmic Magnitude Response m1=20*log10(m); figure(3); plot(w,m1); ylabel('Magnitude Response in dB'); xlabel('Frequency'); title('ANALOGUE BAND-PASS FILTER - CAUER'); axis([0 500 -60 0]); grid; Magnitude Response in dB 0 100 200 300 400 500 ‐60 ‐50 ‐40 ‐30 ‐20 ‐10 0 M a g n i t u d e R e s p o n s e i n d B Frequency ANALOGUE BAND‐PASS FILTER ‐ CAUER 0 100 200 300 400 500 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 M a g n i t u d e R e s p o n s e Frequency ANALOGUE BAND‐PASS FILTER ‐ CAUER Logarithmic Attenuation a=-20*log10(m); figure(4); plot(w,a); ylabel('Attenuation in dB'); xlabel('Frequency'); title('ANALOGUE BAND-PASS FILTER - CAUER') axis([0 500 0 60]); grid; Attenuation in dB Phase Response fi=angle(T); figure(5); subplot(211), plot(w,fi*180/pi); ylabel('Phase Response in deg'); xlabel('Frequency'); title('ANALOGUE BAND-PASS FILTER - CAUER'); axis([0 500 -180 180]); grid; subplot(212), plot(w, unwrap(fi)*180/pi); grid; axis([0 500 -300 300]); ylabel('Phase Response in deg'); xlabel('Frequency'); Phase Response 0 50 100 150 200 250 300 350 400 450 500 0 10 20 30 40 50 60 A t t e n u a t i o n i n d B Frequency ANALOGUE BAND‐PASS FILTER ‐ CAUER 0 100 200 300 400 500 ‐100 0 100 ANALOGUE BAND‐PASS FILTER ‐ CAUER P h a s e R e s p o n s e i n d e g Frequency 0 100 200 300 400 500 ‐200 0 200 P h a s e R e s p o n s e i n d e g Frequency 1. THE TRANSFER FUNCTION OBTAINED FROM A GIVEN MAGNITUDE RESPONSE Very often the Magnitude Response T(e) is known and the Transfer Function T(s) is needed in order to synthesise the circuit. To obtain T(s) from a given magnitude response T(e) we apply the substi- tution e→s/j in the Magnitude Response power of two T 2 (e)=T(je) 2 . Let’s recall from mathematics that: A complex function can be represented: T 2 (e)=T(je) 2 =T(je)T * (je) where “*” denotes complex-conjugated; T * (je)=T(-je) for real coefficient functions as all transfer functions are. Then ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( * 2 s R s T s T j T j T j T j T j T s j = ÷ ÷ ÷ ÷ ÷ e ÷ e = e e = e = e The transfer function obtained from a given magnitude or phase response. Frequency responses conditioned by basic type poles and zeros. Pole-zero diagrams and transfer functions obtained from analogue frequency responses. Asymptotic Bode diagrams How to dissociate T(s) from R(s)? It is known that , then . D(s) must be a strict Hurwitz polynomial, i.e. the poles of the transfer function are located on the left half-plane of the complex s-plane. Therefore, the roots of R(s), which are placed there, are roots of D(s), while those on the right half-plane of the complex s-plane are roots of D(-s). In the case of a minimum phase TF derivation, the roots of the numerator of the R(s) in the left half-plane are roots of N(s), while those on the right half-plane are roots of N(-s). When the derived TF is a non-minimum phase TF, more than one solution for N(s) and T(s) is possible, providing the same Magnitude Response but different Phase Responses. ) ( ) ( ) ( s D s N s T = ) ( ) ( ) ( ) ( ) ( ) ( s D s N s D s N s T s T ÷ ÷ = ÷ EXAMPLE 1: Derive a minimum phase and a non-minimum phase TF having the following Magnitude Response: Solution: The substitution jω⟶s leads to: In the square of the Magnitude Response analytical expression we substitute e 2 with -s 2 and e 6 with ‐s 6 : Then calculate the roots of the numerator and the denominator of R(s): ) ( ) ( ) ( ) ( ) ( 1 4 1 4 ) ( 6 2 6 2 2 2 2 s R s D s N s D s N s s j T s = ÷ ÷ = ÷ ÷ ÷ ÷ ÷ ÷ ÷ e + e + = e ÷ = e 6 2 1 4 ) ( ) ( e + e + = e = e T j T ... ; ; ; 4 4 3 3 2 2 s j s s j s ÷ e ÷ e ÷ ÷ e ÷ e ) 1 )( 1 )( 1 )( 1 ( ) 2 )( 2 ( ) 1 )( 1 ( ) 2 )( 2 ( 1 4 ) ( 2 2 3 3 6 2 s s s s s s s s s s s s s s s R + ÷ + + + ÷ + ÷ = + ÷ + ÷ = ÷ ÷ = The poles in the left half-plane set up the polynomial D(s)=(s+1)(s 2 +s+1). A minimum phase TF will have the zero s 01 placed on the left side of the imaginary axis: The only possible non-minimum phase TF having the zero s 02 on the right half-plane is: ) 1 )( 1 )( 1 )( 1 ( ) 2 )( 2 ( ) 1 )( 1 ( ) 2 )( 2 ( 1 4 ) ( 2 2 3 3 6 2 s s s s s s s s s s s s s s s R + ÷ + + + ÷ + ÷ = + ÷ + ÷ = ÷ ÷ = 75 , 0 5 , 0 3 , 2 j s ± ÷ = 1 1 ÷ = s 2 01 ÷ = s 75 , 0 5 , 0 6 , 5 j s ± = 1 4 = s 2 02 = s 1 2 2 2 ) 1 )( 1 ( 2 ) ( 2 3 2 + + + + = + + + + = s s s s s s s s s H MF 1 2 2 2 ) 1 )( 1 ( 2 ) ( 2 3 2 + + + ÷ = + + + ÷ = s s s s s s s s s H NMF 2. THE TRANSFER FUNCTION OBTAINED FROM A GIVEN PHASE RESPONSE Applying the substitution ω=s/j in the analytical expression of the Phase Response: we obtain: The polynomials L(s) and M(s) are related as follows: L(s)+M(s)=N(s)D(-s) Hence, the procedure for obtaining a TF T(s) from a given Phase Response φ(ω) is as follows: 1. Substitute ω=s/j in the analytical expression of φ(ω) and derive the polynomials L(s) and М(s); 2. Calculate the roots of L(s) + М(s); 3. For a minimum phase TF T(s): the roots of L(s) + M(s) placed in the left half-space are roots of N(s) and those in the right half-space, roots of D(-s) (their mirror images with respect to the jω axis are the roots of D(s)); 4. When it comes to a non-minimumTF T(s), there are many polynomials N(s) and D(s) with a different distribution of the poles of L(s)+M(s). ( ) ' 2 ' 1 2 2 1 ' 2 1 2 ' 1 1 2 ) ( ) ( B B A A B A A B arctg T T arctg e + ÷ e = e e = e ¢ ( ) ) ( ) ( / s jM s L arctg j s ÷ ÷÷ ÷ e ¢ = e EXAMPLE 2: Derive a minimum phase and a non-minimum phase TF having the following Phase Response: Solution: In the analytical expression of the Phase Response we substitute e by s/j, e 2 by -s 2 , e 3 by –s 3 /j, etc.: The polynomials L(s), M(s) and L(s)+M(s) =N(s)D(-s) are: 4 4 6 ) ( 2 3 ÷ e e ÷ e = e ¢ arctg ) ( ) ( ) 4 4 ( 6 4 4 6 ) ( 2 3 2 3 s M j s L arctg s j s s arctg arctg j s = ÷ ÷ + ÷÷ ÷ ÷ e e ÷ e = e ¢ ÷ e ) 2 2 )( 2 ( 4 6 4 ) ( ). ( ) ( ) ( 2 2 3 + ÷ ÷ = ÷ + ÷ = ÷ = + s s s s s s s D s N s M s L s’ 3 s 1 s 2 s’ 2 s’ 1 s 3 o je 1 -j j -1 -2 2 The roots of L(s)+M(s) are: s 1 =1+j; s 2 =2; s 3 =1‐j. The minimum phase and non-minimum phase TF are, respectively: ; 4 6 4 1 ) 2 2 )( 2 ( 1 ) ( ) ( ) ( 2 3 2 + + + = + + + = = s s s s s s s D s N s T MP 2 2 2 ) ( ) ( ) ( 2 + + ÷ = = s s s s D s N s T NMP ¦ ) ¦ ` ¹ ÷ ÷ = + = 4 4 ) ( 6 ) ( 2 3 s s M s s s L 3. MAGNITUDE AND PHASE RESPONSE DERIVATION FROM POLE-ZERO PLOT n m n k k m i j s n k k m i n n m a b H s j s j H s s s s H s s s s s s a s s s s s s b s T i i m = ÷ e ÷ e ÷ ÷ ÷ ÷ ÷ ÷ = ÷ ÷ ÷ ÷ ÷ ÷ = [ [ [ [ = = e = = = , ) ( ) ( ) ( ) ( ) )....( )( ( ) )...( )( ( ) ( 1 1 0 1 1 0 2 1 0 0 0 2 1 ( ) ( ) . ; ) ( ) ( 0 0 0 e ¢ e ¢ ÷ e = ÷ e ÷ e = ÷ e k i i i j k k j e s j s j e s j s j ( ¸ ( ¸ e ¢ ÷ e ¢ = = = e ¢ = e ¢ ¿ = ¿ = [ [ [ [ ÷ e ÷ e = ÷ e ÷ e = e n k k m i i i k i i j n k k m i n k j k m i j e s j s j H e s j e s j H j T 1 1 0 0 ) ( ) ( 1 1 0 1 ) ( 1 ) ( 0 ) ( Magnitude Response Phase Response ¿ e ¢ ÷ ¿ e ¢ = e ¢ = = n k k m i i 1 1 0 ) ( ) ( ) ( Phase Response: dB s j s j H T k n k m i i , lg 20 lg 20 lg 20 ) ( 1 0 1 dB ÷ e ÷ ÷ e + = e ¿ ¿ = = Magnitude Response: EXAMPLE 3: 1 01 ) ( o + o + = s s s Т 1 01 ) ( o + e o + e = e j j j Т 01 01 ) ( o e = e ¢ arctg 1 1 ) ( o e = e ¢ arctg ( ) 2 2 2 2 1 01 1 01 ) ( e + o e + o = o + e o + e = e = e j j j Т T 1 01 ) ( o e ÷ o e = e ¢ arctg arctg 2 2 1 1 e + o = e + o j 2 2 01 01 e + o = e + o j - pole - zero 4.FREQUENCY RESPONSES CONDITIONED BY BASIC TYPE POLES AND ZEROS 4.1. FREQUENCY RESPONSES CONDITIONED BY A NEGATIVE REAL POLE 1 1 1 1 1 ) ( 1 ) ( o + e = e ÷ ÷ ÷ ÷ o + = e = j j Т s s Т j s Magnitude Logarithmic Magnitude Response Response 2 2 1 1 1 ) ( e + o = e Т dB T , lg 20 ) ( 2 2 1 1 e + o ÷ = e dB T T , 6 2 lg 20 lg 20 2 lg 20 lg 20 lg 20 ) ( ) ( 2 1 1 1 2 1 1 2 1 1 2 ÷ = ÷ = = e + e ÷ = = e + e ÷ = e ÷ e ¬ e = e dB T T T , lg 20 ) 0 ( ) ( 0 1 0 1 1 o ÷ = = = e ¬ = e dB T T , 3 2 lg 20 lg 20 2 lg 20 ) ( 0 1 1 1 1 1 ÷ = ÷ o ÷ = o ÷ = o ¬ o = e dB T , lg 20 ) ( 1 1 e ÷ ~ e ¬ o >> e 1. 2. 3. To derive the graphic of T(e) dB we calculate it in several frequencies: = straight line with slope –6 dВ/oct 3 dB error Phase Response: We calculate φ 1 (e) in several frequencies: 1 1 ) ( o e ÷ = e ¢ arctg error deg 0 , 0 0 ) 0 ( 0 1 1 = = o ÷ = ¢ ¬ = e rad arctg 1. 2. 3. deg 45 , 4 1 ) ( 1 1 1 1 1 ÷ = t ÷ = ÷ = o o ÷ = o ¢ ¬ o = e rad arctg arctg deg 90 , 2 ) ( 1 1 1 ÷ = t ÷ ~ o e ÷ ~ e ¢ o >> e rad arctg deg 3 . 84 47 . 1 10 10 ) 10 ( 10 deg 7 . 5 1 . 0 1 . 0 1 . 0 ) 1 . 0 ( 1 . 0 1 1 1 1 1 1 1 1 1 1 ÷ = ÷ = ÷ = o o ÷ = o ¢ ¬ o ÷ = ÷ = ÷ = o o ÷ = o ¢ ¬ o rad arctg arctg rad arctg arctg 4.2. FREQUENCY RESPONSES CONDITIONED BY REAL ZERO Real negative zero e + o = e ÷÷ ÷ o + = o ÷ = e = j j T s s T s j s 0 10 0 10 0 0 ) ( ) ( e + o ÷ = e ÷÷ ÷ o ÷ = o + = e = j j T s s T s j s 0 10 0 10 0 0 ) ( ) ( ; Real positive zero dB T , lg 20 ) ( 2 2 0 0 1 e + o + = e 0 10 ) ( o ÷ e = e ¢ arctg Logarithmic Magnitude Response Phase Response 0 10 ) ( o e = e ¢ arctg dB T , lg 20 ) ( 2 2 0 0 1 e + o + = e Logarithmic Magnitude Response Phase Response 4.3. FREQUENCY RESPONSES CONDITIONED BY COMPLEX CONJUGATED POLES 2 2 2 , 1 e ± o ÷ = j s ( )( ) ( )( ) 2 2 2 2 0 2 1 0 1 2 2 2 2 2 2 1 2 2 1 1 1 ) ( e + o = o = ¬ + + = e + o + e ÷ o + = ÷ ÷ = a a a s a s j s j s s s s s s T 2 2 2 2 0 e + o = = e a c The denominator is a Hurwitz polynomial. Frequency of the complex-conjugated poles Quality factor of the poles 1 0 2 2 2 2 2 2 2 2 2 1 2 1 2 a a Q П = o e + = o e + o = Then the TF can be re-written as follows: 2 2 0 0 0 1 2 0 1 2 2 1 1 1 ) ( C П C s Q s a s a a a s a s a s s T e + e + = + + = + + = · s s П Q 5 . 0 Small: Q П <3÷5 Medium: Q П =5÷20 Large: Q П >20 Extra Large: Q П >100 2 2 2 1 0 2 4 1 1 lg 20 ) 0 ( ) ( 2 2 1 1 П П Q Q T T a a Q П m C m ÷ + = e ¬ ÷ = ÷ e = e e ÷ ÷ e ÷÷ ÷ e + e ÷ ÷ = e · ÷ e lg 40 ) ( ) ( lg 20 ) ( 2 2 1 2 2 2 0 2 Т a a Т Q П determines the type of the Magnitude and Phase Responses Always m C e > e 2 m b e = e 1. – t гр (ω) – monotone decreasing curve 2. – t гр (ω) – the graphic has a maximum for the frequency 2 2 2 0 1 2 ) ( e ÷ e e e ÷ = e ÷ e ÷ = e ¢ C П C Q arctg a a arctg ( ) s Q Q d d t П C C C П C гр , ) ( ) ( 2 2 2 2 2 2 2 2 2 e e + e ÷ e e + e e = e e ¢ ÷ = e 577 . 0 3 1 = s П Q 577 . 0 3 1 = > П Q 2 max 4 1 1 2 1 П Q C ÷ + ÷ e = e Q п Т 2 (e) max t ГР uneven curve 4.4.FREQUENCY RESPONSES CONDITIONED BY COMPLEX CONJUGATED ZEROS Complex conjugated zeros Frequency of the complex-conjugated zeros Quality factor of the zeros 1 0 0 2 0 2 0 0 2 b b Q = o e + o = 2 0 2 0 0 0 e + o = = e b c ( )( ) ( )( ) 2 0 2 0 1 2 0 0 0 0 0 0 2 0 0 2 1 0 ) ( C C s Q s b s b s j s j s s s s s s T e + e + = = + + = = e + o + e ÷ o + = = ÷ ÷ = 0 0 0 2 , 1 e ± o ÷ = j s ( )( ) ( )( ) 2 0 2 0 1 2 0 0 0 0 0 0 2 0 0 2 1 0 ) ( C C s Q s b s b s j s j s s s s s s T e + e ÷ = = + ÷ = = e + o ÷ e ÷ o ÷ = = ' ÷ ' ÷ = 0 0 0 2 , 1 e ± o + = ' j s Complex conjugated zeros on the imaginary axis 4.5. FREQUENCY RESPONSES CONDITIONED BY A ZERO AT S=0 ( )( ) ( )( ) 2 0 2 0 0 0 0 20 2 1 ) ( e + = e + e ÷ = ÷ ÷ = s j s j s s s s s s T 0 0 2 , 1 e ± = j s 2 ) ( ; ) ( t e = e = e = j j j T s s T . 2 ) ( ; lg 20 ) ( ; ) ( const dB T T dB = t = e ¢ e = e e = e 0 0 = s 4.6. FREQUENCY RESPONSES CONDITIONED BY A POLE AT S=0 CONCLUSIONS 1. Each zero of the transfer function, when the frequency is high enough, increases the steepness of the Logarithmic Magnitude Response by 6 dB/oct (20 dB/dec) and raises the Phase Response by π/2 rad (90 deg). 2. Each pole of the transfer function, when the frequency is high enough, decreases the steepness of the Logarithmic Magnitude Response by -6 dB/oct (-20 dB/dec) and reduces the Phase Response by -π/2 rad (-90 deg). 3. When ω→∞ λ= –6(n-m), dB/oct; φ(ω)→[(m-n) π]/2, rad . 2 ) ( ; , lg 20 ) ( ; ) ( const dB T T dB = t ÷ = e ¢ e = e e = e 2 1 1 ) ( ; / 1 ) ( t ÷ e = e = e = j e j j T s s T 0 1 = s FREQUENCY RESPONSES CONDITIONED BY REAL POLES AND ZEROS NEGATIVE REAL POLE o 1 eb = o1 T( e ), dB T 0 T 0 -3 e -6dB/oct eb = o1 ¢ ( e ) -45 ° -90 ° lg e -45 ° /dec 0,1 eb 10 eb NEGATIVE REAL ZERO o 01 ea = o01 T( e ) dB T 01 e +6dB/oct ea = o01 ¢ ( e ) 90 ° 45 ° lg e -45 ° /dec 0,1 ea 10 ea A POLE at s=0 ¢ ( e ) lg e -90 ° T( e ), dB e -6dB/oct 1 A ZERO at s=0 T( e ), dB e +6dB/oct 1 ¢ ( e ) lg e 90 ° 20 dB/dec = 6 dB/oct The jumps in the curve are not approximated PROBLEM: From a given TF or pole- zero diagram, plot the Logarithmic Magnitude Response and Phase Response via graphical summations. ( )( ) ( )( ) 0 1 2 2 2 1 2 2 1 2 2 ) ( a s a s s s s s s s s s s s T + + = o + o + = = ÷ ÷ = 5. ASYMPTOTIC BODE DIAGRAMS Plainly, s 1 = –σ 1 = –ω b1 s 2 = –σ 2 = –ω b2 Only His unknown. Using the expression: we obtain: ( )( ) ( )( ) 0 1 2 2 1 2 1 2 ) ( a s a s H s s H s s s s H s T + + = o + o + = ÷ ÷ = dB s j s j H T j n j m i dB i , lg 20 lg 20 lg 20 ) ( 1 0 1 ÷ e ÷ ÷ e + = e ¿ ¿ = = dB T H b b , lg 20 lg 20 lg 20 2 1 0 e + e + = INVERSE PROBLEM: From a given Logarithmic Magnitude Response, determine: 1. Poles and zeros of the TF 2.Analytical expression of the TF 3.The graph of the Phase Response 6. Asymptotic Bode diagrams – using Matlab Band-pass Transfer Function High-pass Transfer Function 3 4 ) ( ) ( ) ( 2 + + = = s s s s D s N s T Ns1=[0 1 0]; Ds1=[1 4 3]; bode(Ns1, Ds1); grid 3 4 ) ( ) ( ) ( 2 2 + + = = s s s s D s N s T Ns2=[1 0 0]; Ds2=[1 4 3]; figure(2); bode(Ns2, Ds2); grid Low-pass Transfer Function 3 4 1 ) ( ) ( ) ( 2 + + = = s s s D s N s T Ns3=[0 0 1]; Ds3=[1 4 3]; figure(3); bode(Ns3, Ds3); grid Description of analogue linear circuits in the time domain - impulse and step responses, analogue convolution. Analogue impulse and step responses due to different transfer function poles. Step responses of general bilinear and biquadratic analogue transfer functions 1. DESCRIPTION OF ANALOGUE LINEAR CIRCUITS IN THE TIME DOMAIN DIFFERENTIAL EQUATION When the coefficients a j and b i are known, it is possible to calculate y(t) for a given x(t): ¿ ¿ = = = m i i i i j j n j j dt t x d b dt t y d a 0 0 ) ( ) ( The coefficients a j and b i depend on the parameters of the circuit elements R, L, C, etc.; n (the order of the derivative of the output signal y(t)) is the order of the differential equation and of the analogue system itself. Descriptive parameters: the coefficients a j and b i. Problem: differential equations of high order are difficult to solve. STEADY-SPACE DESCRIPTION Space equati on Output equati on IMPULSE AND STEP RESPONSES The impulse and step responses are related as follows: The impulse response can be used to determine the output from the input through the convolution operation: } · · ÷ t t ÷ t = - = d t h x t h t x t y ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( t t dt t x d Bu Ax + = ) ( ) ( ) ( t t t y Du Cx + = x(t) - state vector, u(t) - input signal vector, y(t) - output signal vector, A, B, C and D - descriptive matrices ¹ ´ ¦ > < = 0 1 0 0 ) ( t за t за t u u(t ) t 1 Linear system h(t), g(t), T(s) x(t) y(t) The step response g(t) is the output that the circuit will produce when an unit step function u(t) is the input: ( ) dt t dg t h = ) ( ( ) ; ) ( dt t h t g } = Linear system h(t), g(t), T(s) x(t) y(t) o(t) t ¹ ´ ¦ = = = o 0 1 0 0 ) ( t за t за t The i mpulse response h(t) is the output that the circuit will produce when an ideal impulse function (Dirac impulse) o(t) is the input: 2. ANALOGUE CIRCUITS – TIME-DOMAIN CONSIDERATIONS LAPLACE TRANSFORM INVERSE LAPLACE TRANSFORM ( ¸ ( ¸ = = ÷ ÷ s s T L t g s T L t h ) ( ) ( )] ( [ ) ( 1 1 )] ( [ ) ( )] ( [ ) ( t g sL s T t h L s T = = LAPLACE TRANSFORM OF SOME OFTEN-USED SIGNALS 0 ), ( > t t x ( ) | | ) (t x L s X = 0 ), ( > t t x ( ) | | ) (t x L s X = 1 s 1 t 0 sine 2 0 2 0 e + e s ) (t o 1 t 0 cose 2 0 2 e + s s t e o ÷ o + s 1 t e t 0 sine o ÷ ( ) 2 0 2 0 e + o + e s t 2 1 s t e t 0 cose o ÷ ( ) 2 0 2 e + o + o + s s t te o ÷ 2 2 2 1 o + o + s s t sh 0 e 2 0 2 0 e ÷ e s 3. NATURAL AND UNNATURAL PARTS OF THE OUTPUT SIGNAL 3.1 . NATURAL AND UNNATURAL PARTS OF AN OUTPUT SIGNAL Transfer Function: Input Signal: ) ( ) ( ) ( s D s N s T = ) ( ) ( ) ( s D s N s X x x = The natural part of Y(s) is determined only by the TF T(s) , while the unnatural part of Y(s) is set by the input signal X(s). Output Signal Unnatural part Natural part [ [ [ [ = = = = ÷ ÷ ÷ ÷ = = = n j j x m i x n j j m i x x s s s s s s s s H s D s N s D s N s X s T s Y i i 1 1 0 1 1 0 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ). ( ) ( 3.2. AN ALTERNATIVE DERIVATION OF THE TIME-DOMAIN CHARACTERISTICS Y(s) expands into a sum of partial fractions: Apply the Inverse Laplace Transform: ¿ ¿ = = ÷ + ÷ = x n j j x j x n j j j s s A s s A s Y 1 1 ) ( ) ( ). ( lim s T s s A j s s j j ÷ = ÷ ) ( ). ( lim s X s s A xj s s xj xj ÷ = ÷ ) ( ) ( ) ( ) ( 1 1 t y t y e A e A t y s Y unnat n j nat t s xj n j t s j ILT x xj j ¿ ¿ = = + = + = ÷÷ ÷ When x(t)= δ(t), then y(t) has no unnatural part and y(t)=h(t). Therefore h(t) can be derived from the TF T(s), when only the natural part is considered. 4. IMPULSE RESPONSE CONDITIONED BY DIFFERENT TYPES OF POLES Practical (in the left half-plane) and purely theoretical (in the right half-plane) poles are considered: Y(s) expands into a sum of partial fractions: where The Inverse Laplace Transform is applied: Grouping the complex-conjugated pairs А 3 and А * 3 , А 4 and А * 4 , А 5 and А * 5 , corresponding to the complex conjugated poles according to the rule “the sum of a complex function and its complex-conjugated pair is equal to twice the real part of the complex function“: ( ) ( )( )( )( )( )( )( )( ) 0 1 2 0 1 2 2 3 2 4 4 4 4 2 2 2 2 3 3 5 1 ) ( a s a s a s a s s j s j s j s j s j s j s s s s s N s Y + ÷ + + e + e + o ÷ e ÷ o ÷ e + o + e ÷ o + e + e ÷ o + o + = 4 4 5 4 4 5 2 2 4 2 2 4 3 3 3 3 5 2 1 1 0 ) ( e + o ÷ + e ÷ o ÷ + e + o + + e ÷ o + + e + + e ÷ + o ÷ + o + + = - - - j s A j s A j s A j s A j s A j s A s A s A s A s Y ( ) ( ) s Y s s A j s s j j ÷ = ÷ lim t j t t j t t j t t j t t j t j t t e e A e e A e e A e e A e A e A e A e A A t y 4 4 4 4 2 2 2 2 3 3 5 1 5 5 4 4 3 3 2 1 0 ) ( e ÷ o - e o e ÷ o ÷ - e o ÷ e ÷ - e o ÷ o ÷ + + + + + + + + = | | | | | | t j t t j t t j t t e e A e e A e A e A e A A t y 4 4 2 2 3 5 1 5 4 3 2 1 0 Re 2 Re 2 Re 2 ) ( e o e o ÷ e o o ÷ + + + + + = The complex residiuums are presented as follows: then Hence, , ; ; 5 4 3 5 5 4 4 3 3 u - - u - - u - - = = = j j j e e e A A A A A A ( ) | | ( ) | | ( ) | | 5 4 4 4 2 2 3 3 5 1 Re 2 Re 2 Re 2 ) ( 5 4 3 2 1 0 u + e o - u + e o ÷ - u + e - o o ÷ + + + + + = t j t t j t t j t t e e e e e e A e A A t y A A A | | ( ) ( ) ( ) 5 4 4 2 3 3 2 1 0 cos 2 cos 2 cos 2 ) ( cos Re 4 2 5 1 5 4 3 u + e + u + e + u + e + + + = ¬ ¬ ¢ = o - o ÷ - - o o ÷ ¢ t e t e t e A e A A t y e t t t t j A A A 5. STEP RESPONSE OF A FIRST ORDER TRANSFER FUNCTION 5.1. FIRST ORDER TRANSFER FUNCTION (BILINEAR TF): Step Response is derived where Then: For t=0, t=· и t=1/a 0 =1/σ 1 , the Step Response is: i.е. after a time period of t=1/a 0, 63%of the step process is complete. 0 0 1 1 0 1 ) ( a s b s b s s H s T + + = o + o + = 0 1 0 ) ( 1 ) ( ). ( ) ( a s A s A s T s s T s X s Y + + = = = Natural part Unnatural part ) 0 ( ) ( . lim 0 0 0 0 T a b s Y s A s = = = ÷ 0 0 1 0 0 0 1 0 1 ) ( ). ( lim 0 a b b a b a b s Y a s A a s ÷ = ÷ + ÷ = + = ÷ ÷ 0 0 0 1 0 0 0 1 0 ) ( a s a b b s a b a s A s A s Y + ÷ + = + + = | | 0 , ) ( ) ( 0 0 0 1 0 0 1 > | | . | \ | ÷ + = = ÷ ÷ t e a b b a b s Y L t g t a | | | | ) 0 ( ) ( 63 . 0 ) 0 ( 37 . 0 ) ( ) 0 ( ) ( 1 1 ; ) ( ; ) 0 ( 0 1 0 0 1 0 0 0 0 0 0 1 g g g g g g e a b b a b a g a t a b g t b g t ÷ · + = · ÷ + · = | | . | \ | ÷ + = | | . | \ | ¬ = = · ¬ · = = ¬ = ÷ τ =1/a 0 - time-constant of the Step Response Represented by τ the Step Response will be: The transition process concludes after a time of 5τ, i.e. g(5τ) ~ g((·). Important Result: It was shown that: From the other side: Therefore, g(0)= T(∞) and g(∞)=T(0), i.e. if the transfer function Т(s) is known, two values of the Step response g(t) are known as well: g(0) and g(∞). This applies to every TF and Step Response. | | t ÷ · ÷ + + · = / ) ( ) 0 ( ) ( ) ( t e g g g t g ; ) ( ; ) 0 ( 0 0 1 a b g b g = · = 1 0 0 0 0 1 ) ( ; ) 0 ( ; ) ( b T a b T a s b s b s T = · = + + = g(t) g(·) g(τ) g(0) 0 t 5t t 0.63[g(·)-g(0)] g(0)<g(·), т.е. b 1 <b 0 /a 0 g(t) g(0) g(τ) g(·) 0 t 5t t 0.63[g(0)-g(·)] g(0)>g(·), т.е. b 1 >b 0 /a 0 EXAMPLE 1: Draw the Step Response of a two‐port circuit having TF . Calculate g(0), g(∞), the time τ, of the transition process; g(τ), the time for the transition process to complete. Solution: Then: g(0)=b 1 =1 g(∞)=b 0 /a 0 =1/2 τ=1/a 0 =1/2 –time‐constant of Step Response The transition process completes in 5τ=5/2=2.5s The Step Response is: | | ( ) 685 . 0 1 2 1 63 . 0 1 ) 0 ( ) ( 63 . 0 ) 0 ( ) ( = ÷ + = = ÷ · + = t g g g g ( ) 2 1 + + = s s s T ( ) 2 1 1 2 1 0 0 1 0 0 1 = = = ¬ + + = + + = a b b a s b s b s s s T g(t) g(0)=1 g(τ)=0.685 g(·)=0.5 0 t=0.5 5t=2.5 t 0.63[g(0)-g(·)] g(0)>g(·), т.е. b 1 >b 0 /a 0 5.2 STEP RESPONSE OF A SECOND-ORDER TRANSFER FUNCTION(BIQUADRATIC TF) 2 2 2 0 0 0 2 0 1 2 2 0 1 2 2 ) ( c c s Q s s Q s H a s a s a b s b s b s T e + e + e + e + = + + + + = H H a b T a b T a b H c = = · e e = = = 2 2 2 2 0 0 0 2 2 ) ( ; ) 0 ( ; ) 0 ( ) ( ); ( ) 0 ( 2 1 0 2 T b g T b g = o o = · · = = | | . | \ | o ÷ + o ÷ o ÷ o ÷ + o ÷ o o ÷ o + o o = ( ¸ ( ¸ = o ÷ o ÷ ÷ t t e b b b e b b b b s s T L t g 2 1 2 0 2 1 2 2 2 1 0 1 1 2 1 2 1 2 2 1 0 1 1 ) ( ) ( Real poles: ) )( ( ) ( 2 1 0 1 2 2 o + o + + + = s s b s b s b s T g(t) g(·) g(τ) g(0) 0 t 5t t g(t) g(0) g(τ) g(·) 0 t 5t t Q О – zeros quality factor Q П – poles quality factor ω О – natural frequency of zeros ω C – natural frequency of poles Complex conjugated poles (s 1,2 =-o 2 ± je 2 ): 2 2 2 2 0 1 2 2 2 2 2 2 0 1 2 2 ) ( ) )( ( ) ( e + o + + + = e + o + e ÷ o + + + = s b s b s b j s j s b s b s b s T t e b b b t e b b b s s T L t g t t 2 2 2 2 2 2 2 2 0 1 2 2 2 2 2 2 0 2 2 2 2 2 0 1 sin 1 cos ) ( ) ( 2 2 e | | . | \ | o ÷ e + o o ÷ + e + e | | . | \ | e + o ÷ + e + o = ( ¸ ( ¸ = o ÷ o ÷ ÷ For t=0 and t=· the Step Response is: ) 0 ( ) ( ); ( ) 0 ( 2 0 2 2 2 2 0 2 2 2 2 2 0 2 2 2 2 2 0 T b b g T b b b b g c = e = e + o = · · = = | | . | \ | e + o ÷ + e + o = + When Т(s) is of I order, g(t) increases or decreases fluently from g(0) to g(∞). When Т(s) is of II order: А) for two real poles – as for the TF Т(s) of I order; B) for complex-conjugated poles - g(t) increases or decreases fromg(0) to g(∞), but with an additional harmonic process in frequency ω 2 . t e 2 o ÷ 2 e EXAMPLE 2: Draw the Step Response of a two‐port circuit having TF . Calculate g(0) and g(∞). Solution: Then: g(0)=b 2 =1 g(∞)=b 0 /a 0 =1/2 We calculate the discriminant of the denominator polynomial a 2 s 2 +a 1 s+a 0 =0. The TF has a pair of complex‐conjugated poles, i.e. the Step Response is: ( ) 2 4 . 0 1 2 2 + + + = s s s s T ( ) 2 1 1 2 4 . 0 1 0 0 2 0 1 2 2 0 1 2 2 2 2 = = = ¬ + + + + = + + + = a b b a s a s a b s b s b s s s s T ¬ < ÷ = ÷ = 0 2 . 1 . 4 4 . 0 4 2 0 2 2 1 a a a D g(t) g(0)=1 g(·)=0.5 0 t 6. IMPULSE AND STEP RESPONSE OF AN LP FILTER τ И h(t) 1 0.5 0 τ З t t H 0 τ З τ У t 0.01 g(t) 1 0.9 0.5 0.1 ¸ Delay time τ з – the time taken for h(t) to reach maximum value or the time for g(t) to increases to one‐half of its set value in the filter output; Width of the Impulse Response τ И – the width of h(t) at 0.5 of the maximum value; Rise time of the Impulse Response τ Н – the time that g(t) increases from 0.1 to 0.9 of its set value; The maximum rebound of the Step Response ‐ γ Settling time of the transition process τ У – the time for rebounds of the Step Response to reduce to 1% of its set value. Elmer formula ( ( ¸ ( ¸ | . | \ | ÷ + ÷ t ~ t ÷ ~ t 0 2 0 2 2 0 2 1 2 0 2 1 0 1 0 1 3 2 2 ; a a b b b b a a b b a a H The link between the bandwidth BW of the filter at the 3dB level and the width of the Impulse Response: τ И f Г ≥0.318, f Г ‐ cut‐off frequency in Hz. When the rebounds of the Step Response of an LP filter are γ< 5%, then: τ Н ω Г ≈ 2.2, respectively τ Н f Г ≈ 0.35, where ω Г = 2tf Г , rad/s. An LP filter should have a cut‐off frequency f Г not less than 0.35τ, in order to pass an impulse which is τ wide. 7. STEP RESPONSE OF SECOND-ORDER SECTION TF Real poles Complex-conjugated poles EXAMPLE 3: An analogue BP filter of 6th order has the following transfer function: Draw the time responses of the filter. Solution: Ns=[0.034089 7.74e‐18 0.063161 ‐6.019e‐18 0.012909 ‐3.6517e‐19]; Ds=[1 0.27987 2.0005 ‐0.36567 1.2311 ‐0.10599 0.23305]; 23305 , 0 10599 , 0 2311 , 1 36567 , 0 0005 , 2 27987 , 0 10 . 6517 , 3 012909 , 0 10 . 019 , 6 063161 , 0 10 . 748 , 7 034089 , 0 ) ( 2 3 4 5 6 19 2 18 3 4 18 5 + + + + + + ÷ + ÷ + + = ÷ ÷ ÷ s s s s s s s s s s s s T Impulse Response impulse(Ns,Ds); grid; ylabel('amplitude (volts)');xlabel('time'); title('IMPULSE RESPONSE OF AN ANALOGUE BP FILTER'); Step Response figure(2); step(Ns,Ds); grid; ylabel('amplitude (volts)');xlabel('time'); title('STEP RESPONSE OF AN ANALOGUE BP FILTER'); 0 0.2 0.4 0.6 0.8 1 ‐0.2 ‐0.15 ‐0.1 ‐0.05 0 0.05 0.1 0.15 0.2 STEP RESPONSE OF AN ANALOGUE BP FILTER time (seconds) a m p l i t u d e ( v o l t s ) 0 0.2 0.4 0.6 0.8 1 ‐30 ‐20 ‐10 0 10 20 30 IMPULSE RESPONSE OF AN ANALOGUE BP FILTER time (seconds) a m p l i t u d e ( v o l t s ) An introduction to two-port passive circuit synthesis - terms of realisability. Properties and limitations of structures with different elements and topology. Different types of two-port passive circuits. Synthesis of non-terminated L-shaped and lattice schemes The two‐port circuit synthesis performs the following procedures and in the order specified: 1. Check the terms of realisability – to assess whether it is possible to synthesise the scheme using known methods provided that the requirements of the scheme to synthesise are set. 2. Approximation – specified requirements are submitted by functions describing the time and frequency characteristics. For example, to derive the transfer function from given requirements for Magnitude Response. 3. Synthesis – scheme derivation and dimensioning. Checking the terms of realisability is the first procedure ‐ this is very important because it verifies the realisability of the scheme synthesis. 1. TERMS OF REALISABILITY Checking the terms of realisability is the first procedure; this is very important as it verifies the realisability of the circuit. The most commonly‐used terms of realisability are as follows: I. The principle of causality – system reaction can not precede the action that causes that reaction. Mathematically, the principle of causality is described by the Paley‐Wiener theorem, where Т(e) is the Magnitude Response: II. Condition for stability – when the denominator polynomial D(s) is a Hurwitz polynomial whose roots (poles of the TF) are located in the left half‐plane of the complex s‐plane, the TF offers a stable circuit. Otherwise, the synthesis will result in an unstable scheme without actual practical applicability. III. Requirement for the coefficients (b i and a k ) and the degree of the polynomials (m and n): • The coefficients a k must be positive real numbers (D(s) must be a strict Hurwitz polynomial), while b i can also be negative (N(s) may not be a Hurwitz polynomial). • The degree of the numerator m and denominator n may differ arbitrarily but must always be m ≤ n. · < e e + e } · d T 0 2 1 ) ( ln ¿ ¿ = = ÷ ÷ ÷ ÷ = + + + + + + + + = = n k k k m i i i n n n n m m m m s a s b a s a s a s a b s b s b s b s D s N s T 0 0 0 1 1 1 0 1 1 1 ... ... ) ( ) ( ) ( 2. BASIC PASSIVE STRUCTURES 2.1. SYMMETRIC AND ASYMMETRIC STRUCTURES Parameters measured at the input terminals coincide with those measured at the output terminals. Obviously, the symmetry is at the transverse axis of the two‐port circuit. symmetric asymmetric 2.2. BALANCED AND UNBALANCED STRUCTURES The symmetry is at the longitudinal axis of the structure. balanced unbalanced 2.3. GROUNDED CIRCUITS Often unbalanced structures have a direct connection between the input and output terminals, thus transforming the two‐port (four‐terminal) circuit into a three‐terminal circuit (grounded circuit). When balanced and grounded circuits are incorrectly connected bypassing of elements is possible, such as the Z a (s) in the scheme below (between the grounded terminals of the first and the third circuits there is a short circuit). In cases like this, an optocoupler or decoupling transformer (ideal transformer with a 1:1 ratio) is used tо connect balanced and unbalanced circuits. 3. PROPERTIES AND LIMITATIONS OF STRUCTURES WITH DIFFERENT ELEMENTS AND TOPOLOGY 4. DIFFERENT TYPES OF TWO‐PORT PASSIVE CIRCUIT REALISATIONS 4.1. DIRECT REALISATIONS The given TF is subjected to synthesis procedures, leading directly to a circuit realisation. I. Direct Lattice Realisation ТF of idling mode TF in active load mode R Tн =1O: II. Direct T‐shaped realisation with bilateral coherent loads Т‐shaped realisation is used in the synthesis of amplitude and phase correctors. When Z 1 (s)Z 2 (s)=R 2 =const, the characteristic impedances will be active and identical. The TF will be as follows: . . 2 1 2 1 const R Z Z Z Z Z Т c c = = = = = ) ( ) ( ) ( ) ( 2 2 1 s Z R s Z s Z R R s T + = + = const R Z Z Z Z Z a b M c c = = = = = 0 2 1 . a b a b ПХ U Z Z Z Z z z s T + ÷ = = 11 21 ) ( H U b a b a T T U s Т Y Y Y Y y y R y R y s Т H H H H H H ) ( 2 1 1 ) ( 22 21 22 21 = + + ÷ = + ÷ ¬ + ÷ = 0 1 1 1 2 2 2 2 2 2 1 2 ... ) .( . . ) )( ( ) ( ) ( ) ( a s a s a s a s s s Hs s D s N s T n n n n l m + + + + e + e + e + = = ÷ ÷ · · · III. Direct Ladder Realization Ż i can be either a single element (inductor or capacitor) or a parallel/serial oscillating circuit. Polynomial Realisation (Ż i is a single element) • Lowpass (LP) TF: • Highpass (HP) TF: • Bandpass (BP) TF: Non‐polynomial Realisation (Ż i is a parallel or serial oscillating circuit) • Lowpass (LP) TF – m=0; 0≤2l<n general limitation: m+2l≤n • Highpass (HP) TF – m+2l=n; m>0 • Bandpass (BP) TF – 0<m+2l<n; m>0 • Bandstop (BS) TF – 0 1 1 1 ... ) ( ) ( ) ( ) ( a s a s a s a s b s D s b s D s N s T n n n n m m m m + + + + = = = ÷ ÷ 0 1 1 1 0 ... ) ( a s a s a s a b s T n n n n + + + + = ÷ ÷ 0 1 1 1 ... ) ( a s a s a s a s b s T n n n n n n + + + + = ÷ ÷ 0 1 1 1 ... ) ( a s a s a s a s b s T n n n n m m + + + + = ÷ ÷ 0 2 2 2 2 1 . . . ; 2 ; 0 a n l m l = e e e = = · · · 4.2. CASCADE REALISATIONS The TF factorises into first and second order multipliers (factors): Each factor is synthesised and the derived sections are connected in a cascade. I. Cascade‐coherent implementation All k sections have the same input and output impedances R 0 , and the last section is loaded coherently R Т = R 0 . II. Cascade‐untied implementation “Untying” means to eliminate the interference between two or more circuits. Each section in a Cascade‐untied implementation is the load for the next one and accomplishes its own TF, Magnitude and Phase Response as they would have been realized if the other section were missing. The aim is to provide extreme load: short circuit (Z Т ≈0) or an idling mode (Z Т = ∞) ) ( . . . ) ( ) ( ) ( ) ( ) ( 2 1 s T s T s HT s D s N s T k = = buffer stage high impedance input to output low resistance – each section in the cascade realisation operates in idling mode. This mode is often used, since the active schemes with Op Amps easily achieve Z in → ∞ and Z out ≈ 0. low resistance input to output high impedance – short circuit at the output In order to satisfy the untying conditions, usually “buffer stages” are used. They can be resistor groups, dependent sources or active schemes. ) ( ) ( ); ( ) ( 2 1 e >> e e >> e j Z j Z j Z j Z out in out in ) ( ) ( ); ( ) ( 2 1 e << e e << e j Z j Z j Z j Z out in out in 5. SYNTHESIS OF NON‐TERMINATED L‐SHAPED AND LATTICE SCHEMES 5.1. SYNTHESIS OF NON‐TERMINATED L‐SHAPED REALISATIONS Non‐terminated schemes usually work in idling mode. For the TF: the mathematically correct solution: is an electrotechnically‐absurd solution since Z 1 (s) and Z 2 (s) must be fractional‐rational functions. This introduces auxiliary polynomial L(s), such that both impedances are of RC‐type: , ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 1 2 1 2 s D s N s Z s Z s Z s U s U s T ПХ = + = = ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 1 2 1 2 s N s D s Z s D s Z s D s Z s Z s N s Z ÷ = ÷ = ¬ = + = ) ( ) ( ) ( ) ( ) ( ) ( ) ( 1 2 s L s N s D s Z s L s N s Z ÷ = = EXAMPLE 1: Synthesise a non‐terminated L‐shaped two‐port circuit having the following TF: Solution: Checking the terms of realisability shows that the zeros and poles of the TF T(s) are located correctly for RC ladder schemes: zeros: s 01 =0, s 02 = ∞, poles: p 1 =‐0.36, p 2 =‐1.39. Then: ) ( ) ( 2 7 4 ) ( 2 s D s N s s s s T = + + = ) ( ) 5 , 0 )( 1 ( 4 ) ( 2 6 4 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 1 2 s L s s s L s s s L s N s D s Z s L s s L s N s Z + + = + + = ÷ = = = -1 je o -0.5 je o 0 Z 2 (s)¬ Z 1 (s)¬ -0.7 je o 0 Z 2 (s)¬ -1 je o -0.5 Z 1 (s)¬ -0.7 ) 7 , 0 ( ) 5 , 0 )( 1 ( 4 ) ( ) 5 , 0 )( 1 ( 4 ) ( 7 , 0 1 ) 7 , 0 ( ) ( ) ( 1 2 + + + = + + = + = + / / = = s s s s s L s s s Z s s s s s L s s Z Z 1 (s) is synthesised by first form of Foster: ¿ = o + + + = + + + = 1 ' ' 0 ' 1 1 ) 7 , 0 ( ) 5 , 0 )( 1 ( 4 ) ( l l l s A sC R s s s s s Z ; 35 , 0 86 , 2 1 86 , 2 ) 7 , 0 ( 5 , 0 )( 1 ( 4 lim ) ( . lim 1 ; 4 ) 7 , 0 ( ) 5 , 0 )( 1 ( 4 lim ) ( lim ' 0 0 1 0 ' 0 s 1 s ' F C s s s s s s Z s C s s s s s Z R s s = = ¬ = + / + + / = = O = + + + = = ÷ ÷ · ÷ · ÷ O = = = = ¬ ¦ ) ¦ ` ¹ = + + = = + = ÷ ÷ ÷ ÷ 49 , 0 7 , 0 34 , 0 94 , 2 34 , 0 1 34 , 0 ) 5 , 0 )( 1 ( 4 lim ) ( ) 7 , 0 ( lim ' 1 ' 1 0,7 s 1 0,7 s ' 1 R F C s s s s Z s A The one‐port circuit Z 1 (s) is: R’=4O C 0 ’ =0.35F C 1 ’ =2.94F R1 ’ =0.49O Z 2 (s) is synthesised by first form of Foster as well: ¿ = o + + + = + = 1 " " 0 " 2 1 7 , 0 1 ) ( l l l s A sC R s s Z ; 0 7 , 0 lim ) ( . lim 1 ; 0 7 , 0 1 lim ) ( lim 0 2 0 " 0 s 2 s " = + = = = + = = ÷ ÷ · ÷ · ÷ s s s Z s C s s Z R s s O = = = ¬ ¦ ) ¦ ` ¹ = + + = = + = ÷ ÷ ÷ ÷ 43 , 1 7 , 0 1 ; 1 1 7 , 0 7 , 0 lim ) ( ) 7 , 0 ( lim " 1 " 1 0,7 s 2 0,7 s " 1 R F C s s s Z s A C 1 ” =1F R 1 ” =1,43O The one‐port RC‐circuit Z 2 (s) is: Thus, the synthesised L‐shaped RC two‐port circuit having TF T(s) will be: R’=4O C 0 ’ =0,35F C 1 ’ =2,94F R 1 ’ =0,49O C 1 ” =1F R 1 ” 1,43O Optimal synthesis with cuts of zeros and poles: ) 1 ( ) ( + = s s s L -1 je o 0 Z 2 (s)¬ -1 je o -0.5 Z 1 (s)¬ s s s s s s s L s s s Z s s s s s L s s Z ) 5 , 0 ( 4 ) 1 ( ) 5 , 0 )( 1 ( 4 ) ( ) 5 , 0 )( 1 ( 4 ) ( 1 1 ) 1 ( ) ( ) ( 1 2 + = + + + = + + = + = + / / = = R’=4O C 0 ’ =0.5F C 1 ” =1F R 1 ” 1O 5.2. SYNTHESIS OF NON‐TERMINATED LATTICE SCHEMES The transfer function idling expressed by z‐parameters and impedances Z a (s) and Z b (s) is : 1. An auxiliary polynomial L(s) is introduced. It divides the numerator N(s) and denominator D(s) polynomials : 2. Analytical expressions of Z a (s), Z b (s), z 11 and z 21 are determined: If T(s) is eligible for RC‐realization, which is the most common case, auxiliary polynomial L(s) is chosen so that Z a (s), Z b (s) and z 11 are RC‐impedances. 3. Z a (s) and Z b (s) are synthesised by any of the methods for the synthesis of RC‐ impedances. ) ( ) ( ) ( 11 21 пх s D s N Z Z Z Z z z s T a b a b U = + ÷ = = a b a b U Z Z Z Z s L s D s L s N s T + ÷ = = ) ( ) ( ) ( ) ( ) ( пх . ) ( ) ( ; ) ( ) ( ) ( ) ( ; ) ( ) ( ; ) ( ) ( ) ( ) ( 21 11 s L s N z s L s N s D s Z s L s D z s L s N s D s Z b a = + = = ÷ = EXAMPLE 2: Synthesise a non‐terminated lattice two‐port circuit having the following idling transfer function: Solution: Checking the terms of realisability shows that the zeros and poles of the TF T(s) are located in places appropriate for RC‐lattice schemes. zeros: s 01,02 = 1± j (in the right half‐plane), poles: p 1 =‐0.32, p 2 =‐4.68 (on the negative real semi‐axis). Then, Z a (s), Z b (s) and z 11 will be as follows: 3 10 2 2 2 ) ( 2 2 пх + + + ÷ = s s s s s T 9 , 11 ; 1 , 0 zeros ) ( ) 9 , 11 )( 1 , 0 ( ) ( 1 12 ) ( ) ( ) ( ' 0 ' 0 2 2 1 ÷ = ÷ = ¬ + + = + + = ÷ = s s s L s s s L s s s L s N s D Z a 67 . 1 ; 1 zeros ) ( ) 67 , 1 )( 1 ( 3 ) ( 5 8 3 ) ( ) ( ) ( ' ' 0 ' ' 0 2 2 1 ÷ = ÷ = ¬ + + = + + = + = s s s L s s s L s s s L s N s D Z b 68 , 4 ; 32 , 0 zeros ) ( ) 68 , 4 )( 32 , 0 ( ) ( 3 10 2 ) ( ) ( ' ' ' 0 ' ' ' 0 2 11 2 1 ÷ = ÷ = ¬ + + = + + = = s s s L s s s L s s s L s D z L(s) can be of second- or third-order. We choose the lowest possible order, i.e. second. The pole‐zero diagrams help to choose L(s) correctly. ) ( ) 68 , 4 )( 32 , 0 ( 11 s L s s z + + = ) ( ) 67 , 1 )( 1 ( 3 s L s s Z b + + = ) ( ) 9 , 11 )( 1 , 0 ( s L s s Z a + + = -1.67 ‐1 je o je o je o -1 -1 -0.32 -0.1 Z a Z b z 11 -4.68 -11.9 A possible choice is: , . Then: ) 5 . 1 ( ) )( ( ) ( 2 1 + = o ÷ o ÷ = s s s s s L ) 5 . 1 ( 5 8 3 ) ( ; ) 5 . 1 ( 1 12 ) ( 2 2 + + + = + + + = s s s s s Z s s s s s Z b a The following scheme is derived after the synthesis on the first form of Foster. Clearly, there are too many elements for second order realisation. Supposing the roots of the polynomial L(s) are selected so that some of the zeros of Z a (s) or Z b (s) are cancelled out, the scheme realisation will be simplified. When L(s)=s(s+1), the zero s 01 = –1 of Z b (s) and the root ‐1 of L(s) will cancel each other. Z a (s) and Z b (s) are synthesized via first form of Foster : For Z a (s) we derive: . ) 67 , 1 ( 3 ) 1 ( ) 67 , 1 )( 1 ( 3 5 8 3 ; ) 1 ( ) 9 , 11 )( 1 , 0 ( 1 12 2 2 2 2 s s s s s s s s s s Z s s s s s s s s Z b a + = + + + = + + + = + + + = + + + = ; 84 , 0 19 , 1 ) 1 ( ) 9 , 11 )( 1 , 0 ( lim ) ( . lim 1 ; 1 ) 1 ( ) 9 , 11 )( 1 , 0 ( lim ) ( lim 0 0 0 0 s s F C s s s s s s Z s C s s s s s Z R s a s a = ¬ = + + + = = O = + + + = = ÷ ÷ · ÷ · ÷ O = = = = ¬ ¦ ) ¦ ` ¹ = + + = = + = ÷ ÷ ÷ ÷ 81 , 9 1 81 , 9 1 , 0 81 , 9 81 , 9 ) 9 , 11 )( 1 , 0 ( lim ) ( ) 1 ( lim 1 1 1 s 1 s 1 R F C s s s s Z s A a 1 For Z b (s) we derive: ; 3 ) 67 , 1 ( 3 lim ) ( lim s s ' O = + = = · ÷ · ÷ s s s Z R b . 2 , 0 01 , 5 ) 67 , 1 ( 3 lim ) ( . lim 1 ' 0 0 0 ' 0 F C s s Z s C s b s = ¬ = + = = ÷ ÷ C 0 =0,84F R 1 =9,81O R’ 1 =3O C 1 =0,1F R=1O C’ 0 =0,2F The scheme of the non‐ terminated lattice two‐port realisation will be as follows : Amplitude correction – basic considerations, TF, cascade realisation, specifics of approximation. Passive and active amplitude-correction sections of first- and second- order. Attenuators 1. AMPLITUDE CORRECTION – BASIC CONSIDERATIONS On passing through electrical circuits, signals suffer from linear distortions consisting of spectral component ratio changes, called amplitude‐frequency distortions. When the Magnitude Response is a frequency‐independent constant Т(ω)=T 0 =const.; a(ω)=a 0 =const such distortions are missing. This condition of undistorted transmission is usually limited to a certain range of frequencies – the one in which the spectrum of the transmitted signal is located. Fig. 8‐1 shows the correction of a band‐pass‐type circuit with attenuation а b (ω). To make this attenuation a 0 =const., it is necessary for each frequency to а b (ω) to add attenuation equal to the difference between the desired a 0 and the present а b (ω). The two‐port circuit realises that attenuation: а k (ω)=a 0 ‐а b (ω) is called an amplitude corrector . circuit amplitude corrector Fig. 8‐1 ω a(ω) a k (ω) a b (ω) a E =a b +a k =a0 2. AMPLITUDE‐CORRECTION ANALOGUE TRANSFER FUNCTIONS [ [ e + e + e + e + + + = j j j j j j j j j i i i s Q s s Q h s H b s c s H s T i 2 0 0 2 2 0 0 2 0 0 ) ( Amplitude correctors can be synthesised as direct or cascade structures. In practice, cascade‐coherent implementation is the most often used. For this purpose, the amplitude corrector TF, which is always of polynomial type, factorises into first‐ and second‐order transfer functions, which are synthesised as first‐ and second‐order sections connected cascade‐coherently. 2.1. FIRST‐ORDER AMPLITUDE‐CORRECTION TRANSFER FUNCTION 0 0 c b > 0 0 b c > 0 0 ) ( b s c s H s T I + + = a ∞ a(ω) a ∞ /2 H=b 0 /c 0 ω ω 0 ω T(ω) c 0 /b 0 1 σ jω ‐b0 ‐c0 HP type LP type σ jω ‐b0 ‐c0 ω 1 T(ω) c 0 /b 0 ω a 0 a(ω) a 0 /2 H=1 ω 0 2 0 0 2 2 0 0 2 2 0 0 2 2 0 0 2 ) ( e + e + e + e + = e + e + e + e + = s Q s s Q h s H bs s hbs s H s T II Second‐order TF of amplitude correctors are typical biquadratic functions as follows: 2.2. SECOND‐ORDER AMPLITUDE‐CORRECTION TRANSFER FUNCTION h>1–band‐pass type; The Magnitude Response T(ω) has a maximum Hh for ω=ω 0 . Two cases are possible depending on the value of the coefficient h: h<1 –band‐stop type; The Magnitude Response T(ω) has a minimum Hh for ω=ω 0 . h T(ω) 1 ω ω 0 ω ω 0 a(ω),Np a 0 /2 a 0 ω’ ω ” h>1 H=1 h>1 H=1/h h 1 T(ω) a 0 /2 a(ω),Np a 0 h<1 H=1 h<1 H=1 ω ’ ω 0 ω ” ω ω 0 ω 3. AMPLITUDE‐CORRECTION CIRCUIT APPROXIMATION Since the design of the amplitude correctors uses the known attenuation curve a в (ω) or Magnitude response T в (ω) of the corrected circuit, approximating the amplitude corrector doesn’t pose a serious mathematical problem. The curve of the corrector a к (ω) or T к (ω), can be approximated by standard functions. In practice, it is simply the aggregation of the attenuation curves of the first- and second-order sections. By summation of the four curves, given on slides 2 and 3, arbitrarily complicated curves for the amplitude corrector’s attenuation can be obtained. Thus a cascade realisation transfer function is derived and the sections in the cascade are synthesised. Amplitude correctors and their constituent sections must have constant characteristic impedance, in order to be connected coherently to each other and to the corrected circuit. The lattice realisation (fig. 8‐2) has these properties when Ż а and Ż b are reciprocal impedances, i.e. describe reverse one‐port circuits. Lattice realisation scaled TF is: 1 1 1 1 ) ( + ÷ = + ÷ = н н н н b b a a н U Z Z Z Z s Т Fig. 8‐2 4. PASSIVE AMPLITUDE‐CORRECTION CIRCUITS Т‐shaped realisation (fig. 8‐3) is very often used as an amplitude‐correction section. When the impedances Z 1 (jω) and Z 2 (jω) describe reverse to R one‐port circuits, the T‐shaped circuit TF is: For a given TF (first‐ or second‐order) which is a section of a cascade realisation amplitude‐correction TF, and when R L =R S =R (L‐Load, S‐Supply), synthesising cascade section circuits comes down to determining the impedances Z 1 (s) and Z 2 (s). These impedances usually result in LC or RLC one‐port circuits: L‐shaped realisation (fig. 8‐4) is also sometimes used. Ż 1 and Ż 2 again must be reverse impedances . . ) ( ) ( ) ( ) ( 2 2 1 s Z R s Z s Z R R s T + = + = . ) ( 1 ) ( ) ( ) ( ; ) ( ) ( 1 ) ( 1 2 2 1 R s T s T s Z R s Z s T s T R s Z ÷ = = ÷ = Fig. 8‐3 Fig. 8‐4 R R Ż 1 Ż 2 R R R R Ż 1 Ż 2 R R 4.1. PASSIVE AMPLITUDE‐CORRECTION FIRST‐ORDER SECTIONS 0 0 0 0 0 0 ) ( b c b s c s c b s T LP > + + = . , ; , ; , 1 1 ; , 1 0 0 0 0 2 0 0 0 2 0 0 1 0 0 1 F c Rb b c C b c Rb R H c b R L b c R R ÷ = O ÷ = | | . | \ | ÷ = O | | . | \ | ÷ = ( ) . , ; , ; , ; , 1 0 0 0 2 0 0 2 0 0 0 1 0 0 1 O ÷ = ÷ = O ÷ = ÷ = c b Rc R H c b R L c c b R R F c b R C Т‐shaped realisations may have their zeros anywhere in the complex s‐plane except for the real axis. ; ) ( 0 0 0 0 b c b s c s s T HP < + + = Passive Amplitude‐correction section of first‐order and of HP type Passive Amplitude‐correction section of first‐order and of LP type R R R 1 R 2 L 2 C 1 R R R 1 R 2 C 1 L 1 0 0 b c < 0 0 b c > 4.2. PASSIVE AMPLITUDE‐CORRECTION SECOND‐ORDER SECTIONS Q b h bs s hbs s s Q s s Q h s s T BS 1 , 1 ) ( 2 0 0 2 2 0 0 2 2 0 0 2 2 0 0 2 = < e + e + e + e + = e + e + e + e + = ( ) ( ) H h Rb L F h Rb C h h R R , 1 ; , 1 1 ; , 1 0 1 0 1 1 e ÷ = ÷ e = O | . | \ | ÷ = ( ) ( ) H h b R L F R h b C h Rh R , 1 ; , 1 ; , 1 0 2 0 2 2 ÷ e = e ÷ = O ÷ = Passive amplitude‐correction section of second‐order and of BP type: Q b h bs s hbs s h s Q s s Q h s h s T BP 1 , 1 1 1 ) ( 2 0 0 2 2 0 0 2 2 0 0 2 2 0 0 2 = > e + e + e + e + = e + e + e + e + = ( ) ( ) ( ) H bh h R L F h R hb C h R R , 1 ; , 1 ; , 1 0 1 0 1 1 e ÷ = ÷ e = O ÷ = ( ) H h Rhb L F Rhb h C h R R , 1 ; , 1 ; , 1 0 2 0 2 2 ÷ e = e ÷ = O ÷ = Passive amplitude‐correction section of second‐order and of BS type: R R R 1 R 2 L 2 C 1 L 1 C 2 1 1 = < H h h H h 1 1 = > R R R 1 R 2 L 2 C 1 C 2 L 1 5. ACTIVE AMPLITUDE‐CORRECTION SECTIONS 5.1. ACTIVE AMPLITUDE‐CORRECTION FIRST‐ORDER SECTIONS The individual АRC‐sections connect on the principle „high impedance input to low resistance output“, while only the input and the output of the complicated cascade realisation ensure coherent connection. The easiest way to provide coherent connection is the use of buffer stages before the first and after the last section in the cascade realisation. First‐order TFs have poles and zeros on the negative real axis and can be synthesised as RC L‐shaped structure after which a buffer stage with a variable gain control OpAmp (Operational Amplifier) is connected. 0 0 0 0 ) ( b c b s c s s T HP < + + = . 1 , 1 , 2 0 0 1 0 0 0 1 = ÷ = ÷ = R c b C c c b R . 1 , , 0 0 2 0 2 0 0 1 c b C b R b c R = = ÷ = 0 0 0 0 0 0 ) ( b c b s c s c b s T LP > + + = Fig. 8‐5 Fig. 8‐6 5.2. ACTIVE AMPLITUDE‐CORRECTION SECOND‐ORDER SECTIONS Amplitude‐correction TF of band‐pass and band‐stop type can be synthesised via the same biquadratic active scheme (fig. 8‐7). The type (BP or BS) is determined by the value of the coefficient h. The TF of the section in fig. 8‐7 is scaled by frequency so that ω 0 =1. 2 1 5 4 2 1 6 1 5 1 4 2 1 3 2 2 1 5 4 3 6 2 5 4 1 1 2 1 3 2 2 6 2 0 0 2 2 0 0 2 1 1 1 1 1 ) ( C C G G C C G G G C G C C G s s C C G G G G G G G C G C C G s s G G bs s hbs s H s T + + ( ¸ ( ¸ + ÷ + | | . | \ | + + + + ( ¸ ( ¸ + + · ÷ | | . | \ | + + | | . | \ | + = e + e + e + e + = When we equalise the coefficients of the identical powers of s of the biquadratic TF and the section TF, the elements of the scheme will be: ( ) p p p p p n n n p p n p n p hbCG G C bCG G C G G H H G G H G G G G C G H G G G G G hbCG C G ÷ + ÷ + = ÷ = ÷ = = = = ÷ = 2 2 2 2 6 4 2 3 5 2 2 1 2 2 ; 1 ; ; ; ; 2 where C 1 +C 2 =C and G p are arbitrarily chosen. When an independent adjustment of the TF parameters is needed, schemes with 2 or 4 OpAmps are used. Fig. 8‐7 C 1 R 2 R 1 C 2 R 3 R 4 R 6 R 5 6. ATTENUATORS Attenuators, as well as amplitude‐correction circuits, impact on the amplitudes of the frequency components of the signal. Attenuators have permanent attenuation for all frequencies and permanent input and output impedances, usually identical to each other. Symmetric attenuators are called extenders because they cause the working effect of longer circuit (circuit with higher attenuation). Depending on the type of elements, the attenuators are reactive and resistive. In telecommunications, resistive attenuators are most often used. The most simple attenuator is an L‐shaped voltage divider (fig. 8‐8а). It can be realised as a potentiometer (Fig. 8‐8b) whose input and output impedances shift as the attenuation changes. sha R R a th R R T T = = 2 1 ; 2 2 2 2 ; 2 2 1 a cth R R sha R R П П = = T‐ and П‐shaped schemes are more often used. They are depicted in fig. 8‐9а and b respectively, together with the formulae for their synthesis (а, Np and the input impedances R T or R П are given). For higher attenuations (over 5‐10 Np) resistors have very small values. Hence attenuators with high attenuation are synthesised as cascade schemes. Fig. 8‐8 (а) (b) Fig. 8‐9 (а) (b) R 1 /2 R 2 Ż T Ż T R 1 /2 Ż H R 1 2R 2 2R 2 Ż H R 1 R 2 Ż П Ż T R 1 R 2 R ( ) . 1 ; 1 2 1 ÷ = ÷ = a a e R R R e R The circuits in Fig. 8‐9а and b cannot be adjustable, because changing the value of one of the elements to change the attenuation will also change the input impedances. Keeping input impedances unchanged is possible only if all three resistors are changed at once, which is difficult. A good solution for an adjustable attenuator with unchangeable input impedances is depicted in fig. 8‐10. Fig. 8‐10 For given input resistor R and attenuation а : the following analytical expressions for synthesis are obtained (when R 1 R 2 =R 2 =const., i. e. Z c1 =Z c2 =R=const): Np R R R R a , 1 ln 1 ln 2 1 | | . | \ | + = | . | \ | + = R R R 1 R 2 R R Phase correctors – basic principles of phase correction, transfer functions, cascade realisation. Delay circuits 1. PRINCIPLES OF PHASE CORRECTION Phase‐frequency distortions change the initial phase ratios of the spectral components of processed signals, which can be compensated for by so‐called phase‐frequency correctors. These correctors must influence the initial phases of the spectral components signals but keep the amplitudes of these components unchanged. This implies that the Magnitude Response (or Attenuation) of a phase‐frequency corrector must be a frequency‐independent constant within the frequency range of the transmitted signal, i. e. . ) ( ; ) ( ) ( 0 0 const a a const T T j T = = e = = e = e For this reason, phase‐frequency correctors are also called all‐pass circuits. A phase‐frequency corrector is cascaded to the corrected circuit adding to its Group Delay (GD) t ГР.b value t ГР.k such that the total GD t ГР.E is permanent (fig. 9‐1а). In regard to the phase constant b E it should be a linear frequency function (fig. 9‐1b). (а) (b) Fig. 9‐ 1 Fig. 9‐2 For instance, the GD correction of a BP fifth‐order Butterworth filter (fig. 9‐2) consists of following: 1. From the total GD, which is a horizontal line, the GD of the corrected circuit is extracted thus delivering the GD of the corrector: t гр.E ‐ t гр.НЧФ = t гр.k 2. The obtained curve (the expected GD of the corrector t гр.k ) is approximated by different GD curves of the first‐ and second‐order phase‐correction sections, thus specifying the exact order of the all‐pass TF. 3. The curves of each section are optimised so that the overall error is minimal. Total GD: t гр.E Filter GD: t гр.LPF GD of the phase corrector:t гр.k t гр . This described methodology requires cascade realisation of the phase‐frequency corrector. In order to prevent additional distortion due to incoherent load, the phase‐frequency corrector should have permanent input and output resistances. The phase‐frequency corrector must be symmetric so as to be able to be connected at any point of the corrected circuit. 2. ANALOGUE ALL‐PASS (PHASE) TF The polynomial in the denominator is a strict Hurwitz polynomial: ) ( . . . ) ( 1 2 2 1 0 [ = ÷ = + + + + = n i i n n n s s a s a s a s a a s D In order for the Magnitude Response to be constant at all frequencies either N(s)=D(s), which would make the circuit meaningless, or N(s) and D(s) must be conjugated polynomials: ) ( ) (‐ ) ( 1 3 3 2 2 1 0 [ = + ÷ = + ÷ + ÷ = = n i i n s s a s a s a s a a s D s N Therefore, an all‐pass TF will be: ) ( ) ( ) 1 ( ) ( ) (‐ ) ( ) ( ) ( 1 1 3 3 2 2 1 0 3 3 2 2 1 0 [ [ = = ÷ + ÷ = + + + + + ÷ + ÷ = = = n i i n i i s s s s s a s a s a a s a s a s a a s D s D s D s N s T All‐pass TF are typical non‐polynomial and non‐minimum phase TF – the all zeroes are located in the right‐half of the s‐plane. Since N(s) and D(s) are conjugated polynomials, the zeroes are mirror‐symmetrical to the poles. Cascade realisation is the most often used and an all‐pass cascade TF will be: [ [ | + o + | + o ÷ + o ÷ o = = l l l l l k k k s s s s s s s T s T 2 2 ) ( ) ( The following all‐pass first‐ and second‐order sections exist: 2.1. ALL‐PASS FIRST‐ORDER TF s s s T + o ÷ o = ) ( I Phase constant ( ) ( ) o e = e ¢ ÷ = e arctg b 2 I I ( ) ( ) dB T a 0 I I = e ÷ = e Attenuation ( ) ( ) 2 2 I гр.I 2 e + o o = e e = e d db t Group Delay σ jω ‐σ σ b 1 e e=o t/2 t t ГР.1 e 2/o 1/o e=o 2 0 0 2 2 0 0 2 2 2 II ) ( e + e + e + e ÷ = | + o + | + o ÷ = s Q s s Q s s s s s s T Phase constant ( ) 2 2 0 0 2 II 2 2 ) ( e ÷ e ee = e ÷ | oe = e ¢ ÷ = e Q arctg arctg b II ( ) ( ) dB H T a , lg 20 II II ÷ = e ÷ = e Attenuation Group Delay ( ) ( ) ( ) ( ) | . | \ | + ÷ ÷ e = e o + e ÷ | e + | o = e e = e 1 4 1 4 2 2 2 2 2 0 2 2 2 2 II гр.II Q Q Q Q d db t · | | . | \ | ÷ ÷ e = e 1 1 4 2 Q m 2.1. ALL‐PASS SECOND‐ORDER TF b 2 ω e=o t 2t tГР2. ω 4Q/ω0 ω 0 2Q/ω0 tmax ω m The GD curve t ГР.2 has a maximum when Q > 5. In practice, however, smaller values for Q are used. When Q < 0.5 the two poles of T II (s) become real and the TF can be presented as a product of two TF each of first‐order. SYNTHESIS OF LATTICE SCHEMES WITH PERMANENT INPUT RESISTANCE These realisations are usually LC or RLC coherently loaded by , where Ż a (s) and Ż b (s) are reverse impedances. When Ż a and Ż b are scaled by coefficient k r =R T =R 0 , so that R Tн = R 0н =1 O, then: Thus, for a given TF: The two sections are synthesised and the elements are descaled by the same coefficient k r =R T =R 0 . On the output the descaled resistor R T is plugged in. b a T Z Z R R = = 0 H H H H H H H H a b a b a b Y Y R Y Y R Z Z / 1 1 / 1 1 2 2 0 0 = ¬ = = ¬ = = 1 1 1 1 1 1 1 1 ) 1 ( 1 2 ) ( 2 2 + ÷ = + ÷ = + ÷ = + ÷ = + ÷ = + + ÷ = H H H H H H H H H H H H H H b b a a b b a a a a b a b a H U Z Z Z Z Y Y Y Y Y Y Y Y Y Y s Т Н U Н U a b Н U Н U a s T s T Y s Y s T s T s Y Н Н Н ) ( 1 ) ( 1 1 ) ( , ) ( 1 ) ( 1 ) ( + ÷ = = ÷ + = 3. PASSIVE ALL‐PASS SECTIONS OF FIRST‐ AND SECOND‐ORDER The synthesis of the all‐pass first‐ and second‐order sections follows the synthesis procedure for synthesis of coherently loaded lattice structures (fig. 9‐3). Fig. 9‐3 Passive all‐pass I order sections: scaled by R; o = o = / 1 ; / 1 b a C L Passive all‐pass II order sections: 0 0 0 0 1 ; ; 1 ; e = e = e = e = Q C Q L Q L Q C b b a a s s s T I + o ÷ o = ) ( | + o + | + o ÷ = s s s s H s T II 2 2 ) ( For passive T II (s), H=1 while for active realisations H has an arbitrary value. EXAMPLE 1: Synthesise a coherently loaded (R T =R 0 =1O) lattice structure having the following all‐ pass TF: Solution: The poles and zeros are: For а<2 they are complex‐conjugated and are located in a quadrant symmetry (zeroes are on the right‐half of the s‐ plane) ‐ typical of the all‐pass TF which is of non‐minimum‐ phase type. Table 7‐1 shows that the TF can be synthesised as a lattice or grounded RLC‐structure. Here it is realised as a lattice LC structure. 1 1 ) ( 2 2 + + + ÷ = as s as s s T H U 4 1 2 ; 4 1 2 2 2 , 1 2 0 2 , 1 a j a s a j a s ÷ ± ÷ = ÷ ± = . 1 1 1 1 ) ( ; 1 1 ) ( 1 ) ( 1 ) ( 2 2 as a s s as Y s Y as a s as s s T s T s Y H H H a b H U H U a + = + = = + = + = ÷ + = 4. ACTIVE ALL‐PASS SECTIONS OF FIRST‐ AND SECOND‐ORDER 4.1. ACTIVE ALL‐PASS SECTIONS OF FIRST‐ORDER ( ) 1 ; I = o = ÷ + ÷ = + o ÷ o = C G sC G sC G s s s T A well‐known active all‐pass first‐order section is depicted in fig. 9‐4. When we compare the TF realised by the section to a typical all‐pass first‐order TF T I (s) the following scaled values for the conductivity G and the capacitor С are obtained: G G G C Fig. 9‐4 ( ) 2 1 5 4 3 2 1 6 1 5 1 4 2 1 3 2 2 1 5 4 3 6 2 5 4 1 1 2 1 3 2 2 6 2 2 II 1 1 1 1 1 C C G G G C C G G G C G C C G s s C C G G G G G G G C G C C G s s G G s s s s H s T + + ( ¸ ( ¸ + ÷ + | | . | \ | + + + + ( ¸ ( ¸ + + ÷ | | . | \ | + ÷ | | . | \ | + = | + o + | + o ÷ = C and G p are arbitrarily chosen to ensure that G n and G 4 will be real positives. Then , when we derive: 1 0 = | = e . 2 2 ; 1 G ; ; ; ; ) 2 ( 2 2 2 2 6 4 2 3 5 2 2 1 p p p p n n n p p n p n p CG G C CG G C G H H G H G G G G C G H G G G G G CG C G o + + o ÷ + = ÷ = ÷ = = = = o + = 4.2. . ACTIVE ALL‐PASS SECTIONS OF SECOND‐ORDER Fig. 9‐5 C 1 R 2 R 1 C 2 R 3 R 4 R 6 R 5 All‐pass second‐order TFs are typical biquadratic TFs and are synthesised as active amplitude‐correction second‐order sections but the coefficient h is h = ‐1. The section in fig. 8‐7 is used, given again for convenience (fig. 9‐5). Its TF is as follows: 5. DELAY CIRCUITS Delay circuits are two‐port circuits providing delays of signals by a specified time without changing the signal waveform. Delay circuits have a linear Phase Response and a constant Magnitude Response for a specified band of frequencies, i.e. they are all‐pass circuits. The delay circuit TF is obtained by Paden approximation, which uses Bessel polynomials. For example, the delay circuits are part of the devices to protect Hi‐Fi stereo‐ equipment. Being directly connected to the speakers, the stereo amplifiers can be damaged in the event of power supply instability. The protective device provides the necessary delay of the signal thus protecting the stereo‐equipment. Typical "pop off“ when turning stereo‐equipment on or off also can be avoided by ensuring the signals are delayed. Time shift