Laplace Transform Analysis of Signals and Systems

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1 Laplace Transform Analysis of Signals and Systems

2 Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J. Roberts - All Rights Reserved 2

3 Transfer Functions A circuit can be described by a system of differential equations L d dt R t L d d 1i 1 i1t i2t ()+ ( ()) ( ()) vg t dt dt = () t d i t i t 1 2() ( 1 ) i2 d vc 0 R2i2 t 0 dt C λ λ ( ) () + ( ) + ( )+ ()= 0 5/10/04 M. J. Roberts - All Rights Reserved 3

4 Transfer Functions Using the Laplace transform, a circuit can be described by a system of algebraic equations ()+ () ()= () R I s sl I s sl I s V s sl I s sl I 1 s sc I s R I s () ()+ ()+ ()= 0 5/10/04 M. J. Roberts - All Rights Reserved 4 g

5 Transfer Functions A circuit can even be described by a block diagram. 5/10/04 M. J. Roberts - All Rights Reserved 5

6 Transfer Functions A mechanical system can be described by a system of differential equations () d 1 ( ) s1[ 1() 2() ]= 1 1 ( ) Ks1[ x1() t x2() t ] Ks2x2()= t m2x ( 2 t) f t K x t K x t x t m x t or a system of algebraic equations. () d 1() s1[ 1() 2() ]= 1 1() s1[ 1() 2() ] s2 2()= 2 2() 2 F s K sx s K X s X s m s X s 2 K X s X s K X s m s X s 5/10/04 M. J. Roberts - All Rights Reserved 6

7 Transfer Functions The mechanical system can also be described by a block diagram. Time Domain Frequency Domain 5/10/04 M. J. Roberts - All Rights Reserved 7

8 System Stability System stability is very important A continuous-time LTI system is stable if its impulse response is absolutely integrable This translates into the frequency domain as the requirement that all the poles of the system transfer function must lie in the open left half-plane of the s plane (pp ) Open left half-plane means not including the ω axis 5/10/04 M. J. Roberts - All Rights Reserved 8

9 System Interconnections Cascade Parallel 5/10/04 M. J. Roberts - All Rights Reserved 9

10 E(s) System Interconnections Error signal Feedback () H 1 s Forward path transfer function or the plant () H 2 s Feedback path transfer function or the sensor. ()= () () T s H1 s H2 s Loop transfer function H E()= s X() s H2() s Y() s Y()= s H1() s E() s Y s H s ()= s () X() s = 1() 1+ H1() s H2() s T()= s H1() s H2() s H1() s H()= s 1+ T() s 5/10/04 M. J. Roberts - All Rights Reserved 10

11 Analysis of Feedback Systems Beneficial Effects H ()= K s + KH () s If K is large enough that KH 2 ()>> s 1 then H() s. This H s means that the overall system is the approximate inverse of the system in the feedback path. This kind of system can be useful for reversing the effects of another system. 2 () 5/10/04 M. J. Roberts - All Rights Reserved 11

12 Analysis of Feedback Systems A very important example of feedback systems is an electronic amplifier based on an operational amplifier Let the operational amplifier gain be H 1 ()= s () Vo s A Ve () s = 1 0 s p 5/10/04 M. J. Roberts - All Rights Reserved 12

13 Analysis of Feedback Systems The amplifier can be modeled as a feedback system with this block diagram. The overall gain can be written as V V () s A Z f s () s = () s 1 + A0 Zi s 1 p 0 0 i ()+ s Zf() s p 5/10/04 M. J. Roberts - All Rights Reserved 13

14 Analysis of Feedback Systems If the operational amplifier low-frequency gain,, is very large (which it usually is) then the overall amplifier gain reduces at low-frequencies to V Z 0() s f s V() s () Z s i i () A 0 the gain formula based on an ideal operational amplifier. 5/10/04 M. J. Roberts - All Rights Reserved 14

15 Analysis of Feedback Systems 7 If A = 10 and p= 100 then H ( j100)= j If A = 10 and p= 100 then H ( j100)= j The change in overall system gain is about 0.001% for a change in open-loop gain of a factor of 10. The half-power bandwidth of the operational amplifier itself is 15.9 Hz (100/2π). The half-power bandwidth of the overall amplifier is approximately 14.5 MHz, an increase in bandwidth of a factor of approximately 910,000. 5/10/04 M. J. Roberts - All Rights Reserved 15

16 Analysis of Feedback Systems Feedback can stabilize an unstable system. Let a forward-path transfer function be 1 H 1()= s, p > s p This system is unstable because it has a pole in the right halfplane. If we then connect feedback with a transfer function, K, a constant, the overall system gain becomes H s 1 s p K ()= + and, if K > p, the overall system is now stable. 0 5/10/04 M. J. Roberts - All Rights Reserved 16

17 Analysis of Feedback Systems Feedback can make an unstable system stable but it can also make a stable system unstable. Even though all the poles of the forward and feedback systems may be in the open left half-plane, the poles of the overall feedback system can be in the right half-plane. A familiar example of this kind of instability caused by feedback is a public address system. If the amplifier gain is set too high the system will go unstable and oscillate, usually with a very annoying high-pitched tone. 5/10/04 M. J. Roberts - All Rights Reserved 17

18 Analysis of Feedback Systems As the amplifier gain is increased, any sound entering the microphone makes a stronger sound from the speaker until, at some gain level, the returned sound from the speaker is a large as the originating sound into the microphone. At that point the system goes unstable (pp ). Public Address System 5/10/04 M. J. Roberts - All Rights Reserved 18

19 Analysis of Feedback Systems Stable Oscillation Using Feedback Prototype Feedback System Feedback System Without Excitation 5/10/04 M. J. Roberts - All Rights Reserved 19

20 Analysis of Feedback Systems Stable Oscillation Using Feedback Can the response be non-zero when the excitation is zero? Yes, if the overall system gain is infinite. If the system transfer function has a pole pair on the ωaxis, then the transfer function is infinite at the frequency of that pole pair and there can be a response without an excitation. In practical terms the trick is to be sure the poles stay on the ω axis. If the poles move into the left half-plane the response attenuates with time. If the poles move into the right half-plane the response grows with time (until the system goes nonlinear). 5/10/04 M. J. Roberts - All Rights Reserved 20

21 Analysis of Feedback Systems Stable Oscillation Using Feedback A real example of a system that oscillates stably is a laser. In a laser the forward path is an optical amplifier. The feedback action is provided by putting mirrors at each end of the optical amplifier. 5/10/04 M. J. Roberts - All Rights Reserved 21

22 Analysis of Feedback Systems Stable Oscillation Using Feedback Laser action begins when a photon is spontaneously emitted from the pumped medium in a direction normal to the mirrors. 5/10/04 M. J. Roberts - All Rights Reserved 22

23 Analysis of Feedback Systems Stable Oscillation Using Feedback If the round-trip gain of the combination of pumped laser medium and mirrors is unity, sustained oscillation of light will occur. For that to occur the wavelength of the light must fit into the distance between mirrors an integer number of times. 5/10/04 M. J. Roberts - All Rights Reserved 23

24 Analysis of Feedback Systems Stable Oscillation Using Feedback A laser can be modeled by a block diagram in which the K s represent the gain of the pumped medium or the reflection or transmission coefficient at a mirror, L is the distance between mirrors and c is the speed of light. 5/10/04 M. J. Roberts - All Rights Reserved 24

25 Analysis of Feedback Systems The Routh-Hurwitz Stability Test The Routh-Hurwitz Stability Test is a method for determining the stability of a system if its transfer function is expressed as a ratio of polynomials in s. Let the numerator be N(s) and let the denominator be D D 1 D()= s a s + a s + L+ a s+ a D D /10/04 M. J. Roberts - All Rights Reserved 25

26 Analysis of Feedback Systems The Routh-Hurwitz Stability Test The first step is to construct the Routh array. D a a a L a D D 2 D 4 0 D 1 a a a L 0 D 1 D 3 D 5 D 2 b b b L 0 D 2 D 4 D 6 D 3 c c c L 0 D 3 D 5 D 7 M M M M M M 2 d d e f D even D a a a L a D D 2 D 4 1 D 1 a a a L a D 1 D 3 D 5 0 D 2 b b b L 0 D 2 D 4 D 6 D 3 c c c L 0 D 3 D 5 D 7 M M M M M M 2 d d e f D odd 5/10/04 M. J. Roberts - All Rights Reserved 26

27 Analysis of Feedback Systems The Routh-Hurwitz Stability Test The first two rows contain the coefficients of the denominator polynomial. The entries in the following row are found by the formulas, b D 2 = a a D a a a D 2 D 1 D 3 D 1 b D 4 = a a D a a a D 4 D 1 D 5 The entries on succeeding rows are computed by the same process based on previous row entries. If there are any zeros or sign changes in the a D column, the system is unstable. The number of sign changes in the column is the number of poles in the right half-plane (pp ). D /10/04 M. J. Roberts - All Rights Reserved 27

28 Analysis of Feedback Systems Root Locus Common Type of Feedback System System Transfer Function H()= s 1 KH1() s + KH () s H () s 1 2 ()= () () Loop Transfer Function T s KH s H s 1 2 5/10/04 M. J. Roberts - All Rights Reserved 28

29 Analysis of Feedback Systems Root Locus Poles of H(s) Zeros of 1 + T(s) T is of the form T P s K s Q s ()= () () () () P Poles of H(s) Zeros of 1 + K s Q s Poles of H(s) Q()+ s KP()= s 0 or Q() s + P ()= s 0 K 5/10/04 M. J. Roberts - All Rights Reserved 29

30 Analysis of Feedback Systems Root Locus K can range from zero to infinity. For K approaching zero, using Q s KP s ()+ ()= 0 the poles of H are the same as the zeros of Q()= s 0 which are the poles of T. For K approaching infinity, using Q() s + P ()= s 0 K the poles of H are the same as the zeros of P()= s 0 which are the zeros of T. So the poles of H start on the poles of T and terminate on the zeros of T, some of which may be at infinity. The curves traced by these pole locations as K is varied are called the root locus. 5/10/04 M. J. Roberts - All Rights Reserved 30

31 Analysis of Feedback Systems Root Locus K Let H 1 ()= s and let H. s+ 1 s 2 2 ()= s 1 Then ()= T s ( )( + ) K s+ 1( s 2) Root Locus ( ) + No matter how large K gets this system is stable because the poles always lie in the left half-plane (although for large K the system may be very underdamped). 5/10/04 M. J. Roberts - All Rights Reserved 31

32 Analysis of Feedback Systems Root Locus Let K H 1 ()= s s+ 1 s 2 s 3 ()= and let H 2 s 1. ( )( + )( + ) Root Locus At some finite value of K the system becomes unstable because two poles move into the right half-plane. 5/10/04 M. J. Roberts - All Rights Reserved 32

33 Analysis of Feedback Systems Root Locus Four Rules for Drawing a Root Locus 1. Each root-locus branch begins on a pole of T and terminates on a zero of T. 2. Any portion of the real axis for which the sum of the number of real poles and/or real zeros lying to its right on the real axis is odd, is a part of the root locus. 3. The root locus is symmetrical about the real axis... 5/10/04 M. J. Roberts - All Rights Reserved 33

34 Analysis of Feedback Systems Root Locus.. 4. If the number of finite poles of T exceeds the number of finite zeros of T by an integer, m, then m branches of the root locus terminate on zeros of T which lie at infinity. Each of these branches approaches a straight-line asymptote and the angles of these asymptotes are at the angles, kπ, k = 135,,,... m with respect to the positive real axis. These asymptotes intersect on the real axis at the location, σ = 1 ( finite poles finite zeros) m 5/10/04 M. J. Roberts - All Rights Reserved 34

35 Analysis of Feedback Systems Root Locus Examples 5/10/04 M. J. Roberts - All Rights Reserved 35

36 Analysis of Feedback Systems Gain and Phase Margin Real systems are usually designed with a margin of error to allow for small parameter variations and still be stable. That margin can be viewed as a gain margin or a phase margin. System instability occurs if, for any real ω, T( jω)= 1 a number with a magnitude of one and a phase of -π radians. 5/10/04 M. J. Roberts - All Rights Reserved 36

37 Analysis of Feedback Systems Gain and Phase Margin So to be guaranteed stable, a system must have a T whose magnitude, as a function of frequency, is less than one when the phase hits -π or, seen another way, T must have a phase, as a function of frequency, more positive than - π for all T greater than one. The difference between the a magnitude of T of 0 db and the magnitude of T when the phase hits - π is the gain margin. The difference between the phase of T when the magnitude hits 0 db and a phase of - π is the phase margin. 5/10/04 M. J. Roberts - All Rights Reserved 37

38 Analysis of Feedback Systems Gain and Phase Margin 5/10/04 M. J. Roberts - All Rights Reserved 38

39 Analysis of Feedback Systems Steady-State Tracking Errors in Unity-Gain Feedback Systems A very common type of feedback system is the unity-gain feedback connection. The aim of this type of system is to make the response track the excitation. When the error signal is zero, the excitation and response are equal. 5/10/04 M. J. Roberts - All Rights Reserved 39

40 Analysis of Feedback Systems Steady-State Tracking Errors in Unity-Gain Feedback Systems The Laplace transform of the error signal is E()= s () X s + H () s 1 1 The steady-state value of this signal is (using the finalvalue theorem) lim e lim E lim X () s ()= t s ()= s s t s 0 s 0 1 H s u( ) + () If the excitation is the unit step, A t, then the steadystate error is A lim e t lim t s 0 1 H s ()= + () 1 1 5/10/04 M. J. Roberts - All Rights Reserved 40

41 Analysis of Feedback Systems Steady-State Tracking Errors in Unity-Gain Feedback Systems If the forward transfer function is in the common form, N N 1 2 bn s + bn 1s + Lb2s + bs 1 + b0 H 1 ()= s D D 1 2 as D + ad 1s + Las 2 + as 1 + a0 then 1 a0 lim e()= t lim N N 1 2 = t s 0 bn s + bn 1s + Lb2s + bs 1 + b0 a0 + b0 1+ D D 1 2 as D + ad 1s + Las 2 + as 1 + a0 If a 0 = 0 and b 0 0 the steady-state error is zero and the forward transfer function can be written as N N 1 2 bn s + bn 1s + b2s + bs 1 + b0 H 1 ()= s 1 2 sas ( D D D + ad 1s + as 2 + a) L 1 which has a pole at s = 0. 5/10/04 M. J. Roberts - All Rights Reserved 41 L

42 Analysis of Feedback Systems Steady-State Tracking Errors in Unity-Gain Feedback Systems If the forward transfer function of a unity-gain feedback system has a pole at zero and the system is stable, the steady-state error with step excitation is zero. This type of system is called a type 1 system (one pole at s = 0 in the forward transfer function). If there are no poles at s = 0, it is called a type 0 system and the steady-state error with step excitation is non-zero. 5/10/04 M. J. Roberts - All Rights Reserved 42

43 Analysis of Feedback Systems Steady-State Tracking Errors in Unity-Gain Feedback Systems The steady-state error with ramp excitation is Infinite for a stable type 0 system Finite and non-zero for a stable type 1 system Zero for a stable type 2 system (2 poles at s = 0 in the forward transfer function) 5/10/04 M. J. Roberts - All Rights Reserved 43

44 Block Diagram Reduction It is possible, by a series of operations, to reduce a complicated block diagram down to a single block. Moving a Pick-Off Point 5/10/04 M. J. Roberts - All Rights Reserved 44

45 Block Diagram Reduction Moving a Summer 5/10/04 M. J. Roberts - All Rights Reserved 45

46 Block Diagram Reduction Combining Two Summers 5/10/04 M. J. Roberts - All Rights Reserved 46

47 Block Diagram Reduction Move Pick-Off Point 5/10/04 M. J. Roberts - All Rights Reserved 47

48 Block Diagram Reduction Move Summer 5/10/04 M. J. Roberts - All Rights Reserved 48

49 Block Diagram Reduction Combine Summers 5/10/04 M. J. Roberts - All Rights Reserved 49

50 Block Diagram Reduction Combine Parallel Blocks 5/10/04 M. J. Roberts - All Rights Reserved 50

51 Block Diagram Reduction Combine Cascaded Blocks 5/10/04 M. J. Roberts - All Rights Reserved 51

52 Block Diagram Reduction Reduce Feedback Loop 5/10/04 M. J. Roberts - All Rights Reserved 52

53 Block Diagram Reduction Combine Cascaded Blocks 5/10/04 M. J. Roberts - All Rights Reserved 53

54 Definitions: Mason s Theorem Number of Paths from Input to Output - N p Number of Feedback Loops - N L Transfer Function of ith Path from Input to Output - P i () s Loop transfer Function of ith Feedback Loop - T i () s N L i i j i j k i= 1 ith loopand ith, jth, kth j th loop not loops not sharing a signal sharing a signal ()= + ()+ () ()+ () () ()+ s 1 T s T s T s T s T s T s L 5/10/04 M. J. Roberts - All Rights Reserved 54

55 Mason s Theorem The overall system transfer function is H()= s N p i= 1 P () s () s i () s i () where i () s is the same as s except that all feedback loops which share a signal with the ith path, P i () s, are excluded. 5/10/04 M. J. Roberts - All Rights Reserved 55

56 Mason s Theorem 10 P 1 ()= s s ()= + 1 P 2 s s 3 ()= + s H()= s N p i= 1 N p = 2 N L = ss+ 8 = 1+ 3 s s+ Pi() s i() s = () s ( 8) s s ss+ 8 ( ) ()= ()= 1 s 2 s 1 s+ 8 11s 30 2 s+ 3 s 8s 3 5/10/04 M. J. Roberts - All Rights Reserved 56 = ( )( + ) ( ) + + ( )

57 System Responses to Standard Signals N s Let H s be proper in s. Then the Laplace transform D s ()= () () of the unit step response is Y()= s H ()= s 1 N() s N s sd() s = 1() D s () + If the system is stable, the inverse Laplace transform of is called the transient response and the steady-state ( ) H0 response is. s Unit Step Response K s K = H0 ( ) () () N s 1 D s 5/10/04 M. J. Roberts - All Rights Reserved 57

58 System Responses to Standard Signals N s Let H s be proper in s. If the Laplace transform of the D s ()= () () excitation is some general excitation, X(s), then the Laplace transform of the response is N s Y D X N ()= s () s s D () s ()= () () s N D x x () s N s N x s () s = 1() D() s + 1() Dx() s same poles as system same poles as excitation 5/10/04 M. J. Roberts - All Rights Reserved 58

59 System Responses to Standard Signals Unit Step Response A Let H()= s. Then the unit step response is y s 1 p ()= ( ) () pt t A 1 e u t 5/10/04 M. J. Roberts - All Rights Reserved 59

60 System Responses to Standard Signals Let Unit Step Response ()= Η s s 2 2 Aω 0 + 2ζω s+ ω (pp ) 5/10/04 M. J. Roberts - All Rights Reserved 60

61 System Responses to Standard Signals Let ()= Η s s 2 2 Aω 0 + 2ζω s+ ω Unit Step Response /10/04 M. J. Roberts - All Rights Reserved 61

62 System Responses to Standard Signals ()= () () N s Let H s be proper in s. If the excitation is a suddenly- D s applied, unit-amplitude cosine, the response is N s Y()= s () D s s which can be reduced and inverse Laplace transformed into (pp ) y()= t N D L 1 1 () s () s If the system is stable, the steady-state response is a sinusoid of same frequency as the excitation but, generally, a different magnitude and phase. s 2 () + 2 ω 0 ( ) () H jω cos ω t H jω u t + ( 0) 0 + ( 0) 5/10/04 M. J. Roberts - All Rights Reserved 62

63 Pole-Zero Diagrams and Frequency Response If the transfer function of a system is H(s), the frequency response is H(jω). The most common type of transfer function is of the form, ( H s A s z s z s z 1) ( 2) L( N ) ()= s p s p L s p Therefore H(jω) is ( )( ) ( ) 1 2 ( )( ) ( N ) ( )( ) ( ) H j A j ω z j z j z 1 ω 2 L ω ( ω)= jω p jω p L jω p 1 2 D D 5/10/04 M. J. Roberts - All Rights Reserved 63

64 Let H s H( jω)= Pole-Zero Diagrams and ()= + 3 Frequency Response s s 3 jω 3 jω + 3 The numerator, jω, and the denominator, jω + 3, can be conceived as vectors in the s plane. ( ) = H jω 3 jω jω + 3 ( ) H( jω)= { 3+ jω jω + 3 = 0 5/10/04 M. J. Roberts - All Rights Reserved 64

65 Pole-Zero Diagrams and Frequency Response lim H( jω) = lim ω 0 ω 0 3 jω jω + 3 = 0 lim H( jω) = lim + + ω 0 ω 0 3 jω jω + 3 = 0 lim H( j ω) = ω lim ω 3 jω jω + 3 = 3 lim H( j ω) = ω + lim ω + 3 jω jω + 3 = 3 5/10/04 M. J. Roberts - All Rights Reserved 65

66 Pole-Zero Diagrams and Frequency Response π π lim H( jω)= 0 = ω lim π H ( jω )= 0 = + ω 0 2 π 2 lim ω π π H( j ω)= 2 2 = 0 lim H ( j π π ω )= = 0 ω /10/04 M. J. Roberts - All Rights Reserved 66

67 Butterworth Filters The squared magnitude of the transfer function of an nth order, unity-gain, lowpass Butterworth filter with a corner frequency of 1 radian/s is 2 ( ) = 1 + ω H jω 1 2n This is called a normalized Butterworth filter because its gain is normalized to one and its corner frequency is normalized to 1 radian/s. 5/10/04 M. J. Roberts - All Rights Reserved 67

68 Butterworth Filters A Butterworth filter transfer function has no finite zeros and the poles all lie on a semicircle in the left-half plane whose radius is the corner frequency in radians/s and the angle between the pole locations is always π/n radians. 5/10/04 M. J. Roberts - All Rights Reserved 68

69 Butterworth Filters Frequency Transformations A normalized lowpass Butterworth filter can be transformed into an unnormalized highpass, bandpass or bandstop Butterworth filter through the following transformations (pp ). Lowpass to Highpass s ω c s Lowpass to Bandpass s 2 s + ωω L s ω ω H H ( ) L Lowpass to Bandstop s ( ) s ωh ω 2 s + ωω L L H 5/10/04 M. J. Roberts - All Rights Reserved 69

70 Standard Realizations of Systems There are multiple ways of drawing a system block diagram corresponding to a given transfer function of the form, N k bs k N N Y s k bn s bn s bs b H()= s () 1 N N N X() s = = 0 = L+ + 1 k s + an 1s + L+ a1s+ a0 as k = 0 k 1 0, a N = 1 5/10/04 M. J. Roberts - All Rights Reserved 70

71 Standard Realizations of Systems Canonical Form The transfer function can be conceived as the product of two transfer functions, and H H 2 1 ()= s () Y1 s 1 N N 1 X() s = s + a s + L+ a s+ a 1 N Y s N N 1 ()= s () bn s bn 1s bs b Y () s = L 1 0 5/10/04 M. J. Roberts - All Rights Reserved 71

72 Standard Realizations of Systems Canonical Form The system can then be realized in this form, 5/10/04 M. J. Roberts - All Rights Reserved 72

73 Standard Realizations of Systems Cascade Form The transfer function can be factored into the form, H s A s ()= z s p 1 1 s z s p 2 s zn L L s p s p s p s p 2 N N+ 1 N+ 2 D and each factor can be realized in a small canonical-form subsystem of either of the two forms, and these subsystems can then be cascade connected. 5/10/04 M. J. Roberts - All Rights Reserved 73

74 Standard Realizations of Systems Cascade Form A problem that arises in the cascade form is that some poles or zeros may be complex. In that case, a complex conjugate pair can be combined into one second-order subsystem of the form, 5/10/04 M. J. Roberts - All Rights Reserved 74

75 Standard Realizations of Systems Parallel Form The transfer function can be expanded in partial fractions of the form, K1 K2 KD H()= s + + L+ s p s p s p 1 Each of these terms describes a subsystem. When all the subsystems are connected in parallel the overall system is realized. 2 D 5/10/04 M. J. Roberts - All Rights Reserved 75

76 Standard Realizations of Systems Parallel Form 5/10/04 M. J. Roberts - All Rights Reserved 76

77 State-Space Analysis In larger systems it is important to keep the analysis methods systematic to avoid errors One popular method for doing this is through the use of state variables State variables are signals in a system which, together with the excitations, completely characterize the state of the system As the system changes dynamically the state variables change value and the system moves on a trajectory through state space 5/10/04 M. J. Roberts - All Rights Reserved 77

78 State-Space Analysis State space is an N-dimensional space where N is the order of the system The order of a system is the number of state variables needed to characterize it State variables are not unique, there can be multiple correct sets 5/10/04 M. J. Roberts - All Rights Reserved 78

79 State-Space Analysis There are several advantages to state-space analysis Reduction of the probability of analysis errors Complete description of the system signals Insight into system dynamics Can be formulated using matrix methods and the system state can be expressed in two matrix equations Combined with transform methods it is very powerful 5/10/04 M. J. Roberts - All Rights Reserved 79

80 State-Space Analysis To illustrate state-space methods, let the system be this circuit Let the state variables be the capacitor voltage and the inductor current and let the output signals be v t and i t. out () () R 5/10/04 M. J. Roberts - All Rights Reserved 80

81 State-Space Analysis Two differential equations in the two state variables, called the state equations, characterize the circuit, i ()= t v () t L 1 L C 1 v i v i C t G 1 C C ()= t () ()+ t () t C L C in Two more equations called the output equations define the responses in terms of the state variables, v out ()= t v () t i t Gv t C R ()= () C 5/10/04 M. J. Roberts - All Rights Reserved 81

82 State-Space Analysis The state equations can be written in matrix form as i v L() t () t C = 0 1 C 1 L G C i v () t () t C [ iin() t ] and the output equations can be written in matrix form as v i out R () t () t i = G v L C L C () t () t iin t + 0 () 0 [ ] 5/10/04 M. J. Roberts - All Rights Reserved 82

83 State-Space Analysis A block diagram for the system can be drawn directly from the state and output equations. 5/10/04 M. J. Roberts - All Rights Reserved 83

84 State-Space Analysis The state and output equations can be written compactly as q ( t)= Aq()+ t Bx() t where ()= q t i v L C () t () t ()= y t y()= t Cq()+ t Dx() t 1 0 A = L B = G C C C vout() t R() t C = 0 1 i 0 G D = 0 0 [ in ] x()= t i () t 5/10/04 M. J. Roberts - All Rights Reserved 84

85 State-Space Analysis The solution of the state and output equations can be found using the Laplace transform. sq() s q( + 0 )= AQ()+ s BX() s or [ si A] Q()= s BX()+ s q( + 0 ) Multiplying both sides by si A ()= [ ] 1 () [ ] 1 [ ] 1 ()+ ( 0 + ) [ ] Q s si A BX s q The matrix, si A, is conventionally designated by the symbol, Φ s. Then ()= ()[ ()+ ( )]= () ()+ ()( ) + + Q s Φ s BX s q 0 Φ s BX s Φ s q zero state response zero input response 5/10/04 M. J. Roberts - All Rights Reserved 85

86 State-Space Analysis The time-domain solution is then + ()= () ()+ ()( ) q t φ t Bx t φ t q zero state response zero input response L where φ() t Φ s and φ t is called the state transition matrix. () () 5/10/04 M. J. Roberts - All Rights Reserved 86

87 State-Space Analysis To make the example concrete, let the excitation and initial conditions be and let i()= t Au() t q( + 0 )= 1 R=, C = 1, L = 1 3 ( ) = Φ()= s si A 1 1 s L 1 C s G + C 1 i v = ( + 0 ) L + C 0 ( ) 0 = 1 G 1 s + C L 1 s C 2 G s + C s 1 + LC 5/10/04 M. J. Roberts - All Rights Reserved 87

88 State-Space Analysis Solving for the states in the Laplace domain, ()= Q s slc s C s G C s 1 G L s LC C s LC 1 s + G C s 1 G + + s + LC C s 1 + LC Substituting in numerical component values, ()= Q s 1 ss ( + s+ ) s + 3s+ 1 1 s s + s s + 3s+ 1 5/10/04 M. J. Roberts - All Rights Reserved 88

89 State-Space Analysis Inverse Laplace transforming, the state variables are ()= q t t e e t e e t t u() t and the response is t e e y()= t q x t G + = e e t t u() t or ()= y t t e e t e e t t u() t 5/10/04 M. J. Roberts - All Rights Reserved 89

90 State-Space Analysis In a system in which the initial conditions are zero (the zero-input response is zero), the matrix transfer function can be found from the state and output equations. or The response is () ( )= ()+ () + sq s q 0 AQ s BX s 123 = 0 1 ()=[ ] ()= () () Q s si A BX s Φ s BX s ()= ()+ ()= () ()+ ()= () + Y s CQ s DX s CΦ s BX s DX s CΦ s B D X s and the matrix transfer function is ()= () + H s CΦ s B D [ ] () 5/10/04 M. J. Roberts - All Rights Reserved 90

91 State-Space Analysis Any set of state variables can be transformed into another valid set through a linear transformation. Let q 1 () t be the initial set and let q 2 t be the new set, related to by Then 1 and using q t T q t, where () q 1 () t q ()= t Tq () t 2 1 ( 2 t)= 1 ( t)= ( 1 1()+ t 1 () t )= 1 1()+ t 1 () t ()= () q Tq T A q B x TA q TB x ( t)= ()+ t ()= t ()+ t () t q TA T q TB x A q B x 2 1 A = TA T B = TB 2 1 5/10/04 M. J. Roberts - All Rights Reserved 91

92 State-Space Analysis By a similar argument, where C = ()= ()+ () y t C q t D x t C T Transformation to a new set of state variables does not change the eigenvalues of the system. D = D 2 1 5/10/04 M. J. Roberts - All Rights Reserved 92

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