Laplace Transform Analysis of Signals and Systems


 Blaise Hunt
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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 continuoustime 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 halfplane of the s plane (pp ) Open left halfplane 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 lowfrequency gain,, is very large (which it usually is) then the overall amplifier gain reduces at lowfrequencies 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 openloop gain of a factor of 10. The halfpower bandwidth of the operational amplifier itself is 15.9 Hz (100/2π). The halfpower 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 forwardpath 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 halfplane, the poles of the overall feedback system can be in the right halfplane. 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 highpitched 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 nonzero 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 halfplane the response attenuates with time. If the poles move into the right halfplane 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 roundtrip 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 RouthHurwitz Stability Test The RouthHurwitz 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 RouthHurwitz 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 RouthHurwitz 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 halfplane (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 halfplane (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 halfplane. 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 rootlocus 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 straightline 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 SteadyState Tracking Errors in UnityGain Feedback Systems A very common type of feedback system is the unitygain 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 SteadyState Tracking Errors in UnityGain Feedback Systems The Laplace transform of the error signal is E()= s () X s + H () s 1 1 The steadystate 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 SteadyState Tracking Errors in UnityGain 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 steadystate 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 SteadyState Tracking Errors in UnityGain Feedback Systems If the forward transfer function of a unitygain feedback system has a pole at zero and the system is stable, the steadystate 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 steadystate error with step excitation is nonzero. 5/10/04 M. J. Roberts  All Rights Reserved 42
43 Analysis of Feedback Systems SteadyState Tracking Errors in UnityGain Feedback Systems The steadystate error with ramp excitation is Infinite for a stable type 0 system Finite and nonzero 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 PickOff 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 PickOff 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 steadystate ( ) 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, unitamplitude 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 steadystate 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 PoleZero 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)= PoleZero 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 PoleZero 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 PoleZero 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, unitygain, 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 lefthalf 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 canonicalform 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 secondorder 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 StateSpace 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 StateSpace Analysis State space is an Ndimensional 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 StateSpace Analysis There are several advantages to statespace 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 StateSpace Analysis To illustrate statespace 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 StateSpace 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 StateSpace 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 StateSpace 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 StateSpace 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 StateSpace 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 StateSpace Analysis The timedomain 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 StateSpace 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 StateSpace 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 StateSpace 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 StateSpace Analysis In a system in which the initial conditions are zero (the zeroinput 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 StateSpace 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 StateSpace 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
Software Engineering 3DX3. Slides 8: Root Locus Techniques
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