A Typical Feedback System. Disturbance. Reference. Input Controller Input. Plant. Sensor. Noise
|
|
- Posy Lucy Rich
- 6 years ago
- Views:
Transcription
1 A Typical Feedback System Disturbance Reference Input Error Controller Input Plant Output Sensor Noise
2 Typical Control Objectives Uncontrolled system (plant) may not behave satisfactorily Design a control system that yields satisfactory behavior for the controlled system Typical properties desired of a controlled system Stability Input-output stability: Bounded inputs should give bounded outputs Internal stability: All internal variables remain bounded in the absence of inputs Tracking: (Output Input) 0 as t Regulation: Output 0 as t Disturbance/Noise Rejection: Satisfactory performance in the presence of plant disturbances and measurement noise Robustness: Satisfactory performance inspite of unmodelled dynamics and parameter uncertainty/change
3 Review of Continuous-Time Systems All signals are analog signals A linear, time invariant, single-input-single-output (SISO) system is typically described by a 0 y (n) + a 1 y (n 1) + + a (n) y = b 0 u (m) + b 1 u (m 1) + + b m u Solution = initial condition response + input response Input response = u impulse response (convolution) Transfer function = L(impulse response) L(y) = T.F. L(u) for zero initial conditions Transient response decided by poles and zero; poles decide stability Frequency response analysis: Harmonic Analysis, Bode, Nyquist
4 An Overview of Control Activities ANALYSIS: Relate system theoretic properties to system behaviour. Eg. Poles and stability Need analysis tools, eg. Routh-Hurwitz test CONTROLLER DESIGN: Translate specs to system properties and design a controller (control law) that assigns these properties to the controlled system Eg. Pole placement controller Need design tools, eg. pole placement technique IMPLEMENTATION: Sensors, actuators, filters, processors
5 A Digital Controller Digital Controller Input (From sensor) A/D Digital Processor D/A Output (To actuator) Discrete time signals (Sequences of numbers) Continuous time signals (Functions of time)
6 Discrete-Time Signals Sequence {u(k)} k=0 of real numbers A real-valued function k u(k) of integers Right-sided sequence (signal) u(0), u(1), u(2),... Two-sided sequence (signal)..., u( 2), u( 1), u(0), u(1), u(2),...
7 Operators on Discrete-Time Signals Identity operator 1 Shift or unit delay operator S (Su)(k) = u(k 1), k 1 = 0 k 1 Unit advance operator S 1 (S 1 u)(k) = u(k + 1), k 0 Difference operator u(k) = u(k) u(k 1) = (u Su)(k) = 1 S, S = 1
8 Some Basic Discrete-Time Signals Unit pulse/impulse signal δ(k) = 1, k = 0, = 0, k > 0, Unit step signal s(k) = 1, k 0 s = δ Harmonic signals Exponential signals u(k) = sin(kθ) u(k) = r k Harmonic signals with exponential amplitudes u(k) = r k sin(kθ) = Re (re jθ ) k
9 Linear Difference Equations y(k) + a 1 y(k 1) a n y(k n) = b 0 u(k) + b 1 u(k 1) b m u(k m) In terms of the shift operator y(k) + a 1 Sy(k) a n S n y(k) = b 0 u(k) + b 1 Su(k) b m S m u(k) D(S)y = N(S)u Auto-Regressive Moving Averages (ARMA) model Causal: Output independent of future input Strictly causal if output depends only the past inputs Shift invariant (time invariant) Shifted input Su produces shifted output Sy Linear (Superposition + Homogeneity) To solve need n initial conditions + input
10 An Example To numerically compute y(t) = t u(τ)dτ 0 ẏ(t) = u(t), y(0) = 0 At instants 0, T, 2T,..., kt,..., y(kt ) = y((k 1)T ) + kt u(τ)dτ (k 1)T Use forward rectangular rule to approximate the integral y(kt ) = y((k 1)T ) + T u((k 1)T ), y(0) = 0 e(t) y = T Su, y(0) = 0 e(t) u(t) t k-1 k t
11 Vector Spaces A vector space V is a set whose elements can be added in some manner whose elements can multiplied by scalars in some manner which contains a zero element For example: V = set of all functions of time V = set of all right-sided sequences
12 Linear Independence A linear combination is a finite sum of the form α 1 v α n v n Linear independence every linear combination involving atleast one nonzero scalar is nonzero {v 1, v 2,..., v n } V forms a basis for V if v s are linearly independent and every vector in V is a linear combination of v s If V has a basis of n elements for some n, then V is n-dimensional, else infinite-dimensional A linear operator is a linear function V V
13 Vector Space of Discrete Signals The set of all discrete signals is a vector space with (y 1 + y 2 )(k) = y 1 (k) + y 2 (k), (αy)(k) = αy(k), Zero element y 0 y 1,..., y n are linearly dependent iff α 1,..., α n such that α 1 y 1 (k) α n y n (k) = 0 k For λ 1 λ 2 nonzero real, {λ k 1}, {λ k 2} are linearly independent For λ complex, {Re λ k } and {Im λ k } are l. i. For λ 1 λ 2 complex, {Re λ k 1,2} and {Im λ k 1,2} are l. i. For λ 1 nonzero real and λ 2 complex, λ k 1, {Re λ k 2} and {Im λ k 2} are l. i. No finite basis possible Linear operators S, D(S),, D( )
14 Homogeneous Linear Difference Equations y(k) + a 1 y(k 1) a n y(k n) = 0 D(S)y = 0 y( 1), y( 2),..., y( n) Zero initial conditions imply solution is zero The set of all solutions is a vector space since y 0 is a solution D(S)(α 1 y 1 + α 2 y 2 ) = α 1 D(S)y 1 + α 2 D(S)y 2 Theorem: The vector space of solutions has dimension n (a n 0) To prove, need to find a basis consisting of n solutions
15 A Basis of Solutions Idea Solutions sets of initial conditions If w i form a basis for all initial conditions, the corresponding solutions form a basis for all solutions Choose n sets of initial conditions as follows n y w 1 y w y n w n Claim: y 1,..., y n form a basis for all solutions...
16 A Basis of Solutions (cont d) y 1,..., y n are linearly independent If α 1 y 1 (k) + + α n y n (k) = 0 k, then k = i α i = 0 Every solution is a linear combination of y 1,..., y n Let y be any solution and consider y(k) = y( 1)y 1 (k) + + y( i)y i (k) + + y( n)y n (k) y is a solution satisfying the same initial conditions as y Hence y = y is a linear combination of y 1,..., y n
17 Solution of Linear Difference Equations Choose a basis of initial conditions and use corresponding solutions as a basis of solutions Try a solution of the form y(k) = λ k (Sy)(k) = λ k 1 = λ 1 y(k) for k 1 (S 2 y)(k) = λ 2 y(k) for k 2 D(S)y(k) = D(λ 1 )y(k) for k n y(k) = λ k is a solution of D(S)y = 0 if λ satisfies D(λ 1 ) = 0, that is 1 + a 1 λ 1 + a 2 λ a n λ n = 0 = λ n + a 1 λ n 1 + a 2 λ n a n = 0 λ n D(λ 1 ) = characteristic polynomial
18 Another Basis of Solutions Characteristic polynomial/equation factor as C(λ) = (λ p 1 ) m 1 (λ p 2 ) m 2... (λ p l ) m l The following functions form a basis for the solutions of the LDE For p i real, {p k i }, {kpk i }, {k2 p k i },..., {kmi 1 p k i } For p i = re jθ and p i complex, {r k sin(kθ)}, {kr k sin(kθ)}, {k 2 r k sin(kθ)},..., {k mi 1 r k sin(kθ)} {r k cos(kθ)}, {kr k cos(kθ)}, {k 2 r k cos(kθ)},..., {k mi 1 r k cos(kθ)} Initial conditions determine the constants in the linear combination
19 Stability of Initial Condition Response Real characteristic root p {p k } decays iff p < 1 {k j p k } decays iff p < 1 bounded if p = 1 unbounded if p > 1 unbounded if p 1 Complex characteristic root p {Re p k }, {Im p k } decay iff p < 1 {Re k j p k }, {Im k j p k } decay iff p < 1 Theorem bounded if p = 1 unbounded if p > 1 unbounded if p 1 All solutions are bounded iff all characteristic roots lie in the closed unit disc {λ : λ 1} and all roots with unit magnitude are simple (unrepeated) All solutions decay iff all characteristic roots lie in the open unit disc {λ : λ < 1}
20 Convolution Convolution of two right sided sequences u and g is the sequence k (u g)(k) = u(l)g(k l) l=0 = u(0)g(k) + u(1)g(k 1) + + u(k 1)g(1) + u(k)g(0) u g = g u (α 1 u 1 + α 2 u 2 ) g = α 1 (u 1 g) + α 2 (u 2 g) For a fixed g, u g is a linear operator on u S(u g) = u Sg = g Su D(S)(u g) = u D(S)g = g D(S)u (u g) = u g = g u u δ = u (u δ)(k) = u(0)δ(k) u(k 1)δ(1) + u(k)δ(0) = u(k)
21 Pulse Response and Input Response Pulse response g = zero initial condition response to a unit pulse D(S)g = N(S)δ, 0 = g( 1) = g( 2) = Fact: The response y of a linear time invariant system to an arbitrary input u under zero initial conditions is given by y = u g Proof: To show D(S)y = N(S)u D(S)y = D(S)(u g) = u D(S)g = u N(S)δ = N(S)(u δ) = N(S)u Step response = s g
22 Bounded-Input-Bounded-Output (BIBO) Stability A system is BIBO stable if the output to every bounded input is bounded A sequence y is said to be bounded if there exists M such that y(k) < M, k For a bounded sequence y, define Fact: y = sup y(k) = least upper bound of {y(k)} k>0 A system is BIBO stable if and only if there exists N such that for every nonzero input u, the corresponding output y satisfies Theorem: y u < N A system is BIBO stable iff the input response g is absolutely summable, that is, g(k) < k=0
23 BIBO Stability and Pulse Response Suppose the pulse response is absolutely summable k y(k) = (u g)(k) u(l) g(k l) l=0 k u g(k l) u l=0 l=0 Suppose the pulse response is not absolutely summable u k (l) = sign g(k l) l k, = 0, l > k g(l) < u k = 1 y k y k (k) = (u k g)(k) = y k k g(k) l=0
24 Output Zeroing Inputs D(S)y = N(S)u y(k) + a 1 y(k 1) + + a n y(k n) = b i u(k i) + + b m u(k m) An output zeroing input produces no response under under zero initial conditions, that is, satisfies u g = 0 Must satisfy the difference equation N(S)u = 0, that is, b i u(k i) + b m u(k m) = 0 Set of all null inputs is a vector space A basis can be found from the characteristic zeros, solutions of b i λ m i + b i+1 λ m i b m = 0 Zero z i of multiplicity m i contributes {zi k }, {kzi k },, {k mi 1 zi k } Dimension of this vector space is m i
25 Impulse Response and Initial Condition Responses Let d be the impulse response of the system D(S)y = u, that is, D(S)d = δ d is an initial condition response of the system D(S)d = N(S)u since (D(S)d)(k) = 0, k > 0 d is a linear combination of the initial condition responses corresponding to the characteristic roots d = α 1 y 1 + α 2 y α n y n Impulse response g of D(S)y = N(S)u is g = N(S)d g = α 1 N(S)y 1 + α 2 N(S)y α n N(S)y n Characteristic root p i affects g iff it is not a characteristic zero g(k) = α 1 N(S)p k 1 + α 2 N(S)kp k 1 + α 3 k 2 p k 1 + α 4 N(S)p k 2 g decays iff roots p 1 are also a zeros of equal or greater multiplicity
26 Roots, Zeros and BIBO Stability p k, k=0 kp k < k=0 p < 1 p k, kp k decay Since g involves p k, kp k, g is absolutely summable iff g decays System is BIBO stable iff pulse response decays Theorem: System is BIBO stable iff every characteristic root with p 1 is also a characteristic zero of equal or greater multiplicity
27 Z Transform The Z transform of a sequence is a function of the complex variable z Given a sequence y, its right sided Z transform is Z(y) : Y (z) = y(0) + y(1) z = y(k)z k k=0 + y(2) z 2 Z(y) is the Laurent expansion of the complex function Y + y(3) z 3 + Z(y) agrees with Y only in the region of convergence of the Laurent series Recall that if x < 1, then 1 + x + x 2 + = 1 1 x If z 1 < 1, then 1 + z 1 + z 2 + z 3 + = Z transform is linear: Z(α 1 y 1 + α 2 y 2 ) = α 1 Z(y 1 ) + α 2 Z(y 2 ) 1 1 z = z 1 z 1
28 Z Transforms of Some Common Sequences Unit pulse δ Z(δ) = 1 Unit step s Exponential sequence {p k } Harmonic signal {sin(kθ)} S(z) = 1 + z 1 + z 2 + = 1 + pz 1 + p 2 z 2 + = 1 1 z = z 1 z 1, z > pz = z 1 z p, pz 1 < 1 z sin θ z 2 2z cos θ + 1, z > 1 Exponentially modulated harmonic signals {r k sin(kθ)} rz sin θ z 2 2rz cos θ + r2, z > r
29 Properties of the Z Transform Delay Z(Su) = u( 1) + z 1 U(z) Z(S 2 u) = Su( 1) + z 1 Z(Su) = u( 2) + z 1 u( 1) + z 2 U(z) Z(S n u) = u( n) + z 1 u( n + 1) + + z n 1 u( 1) + z n U(z) Z(D(S)u) = D(z 1 )U(z) Advance Z(S 1 u) = zu(z) zu(0), Z(S 2 u) = z 2 U(z) z 2 u(0) zu(1) Z(S n u) = z n U(z) z n u(0) z n 1 u(1) zu(n 1) Difference Convolution Z( u) = Z(u) Z(Su) = z 1 U(z) u( 1) z Z(u g) = G(z)U(z)
30 Properties of the Z Transform (Contd.) Scaling in the complex plane Z({r k u(k)}) = U(z/r) Complex differentiation Initial value Final value theorem Z({ku(k)}) = z du dz (z) lim k provided the limit on the left exists u(0) = lim z U(z) u(k) = lim(z 1)U(z) z 1
31 Transfer Functions The transfer function G of the system D(S)y = N(S)u is the Z transform of its pulse response g G(z) = Z(g) If y is the input response (zero i.c.) to the input u, then y = (g u) Y (z) = G(z)U(z) G(z) = Y (z) U(z) transfer function = Z(output) Z(input) zero initial conditions To calculate the transfer function of D(S)y = N(S)u, take the Z transform on both sides Z(D(S)y) = Z(N(S)u) G(z) = Y (z) U(z) = N(z 1 ) D(z 1 )
32 Transfer Functions of Common Operators Unit Delay: y = Su Y (z) = Z(Su) = u( 1) + z 1 U(z), Y (z) U(z) = z 1 zero i.c. Unit advance: y = S 1 u Non causal. y(k) = u(k + 1) Difference operator: y(k) = u(k) u(k 1) Causal but not strictly causal Pulse Response = Z 1 (z 1 ) = {0, 1, 0, 0,...} Y (z) = zu(z) u(0), Y (z)/u(z) = z D(λ) = 1, N(λ) = 1 λ, T.f. = N(z 1 ) D(z 1 ) = 1 z 1 Impulse Response = {1, 1, 0, 0,...}
33 Inverse Z Transform Laurent expansion Perform long division for rational Y (z) Partial fraction expansion followed by look-up table Convolution property Y (z) = Y 1 (z)y 2 (z) = y = y 1 y 2 Solve numerically by forming a linear difference equation Y (z) = N(z 1 ) D(z 1 ) = y = pulse response of D(S)y = N(S)u
34 Partial Fractions Y (z) = N(z 1 ) D(z 1 ) = N(z 1 ) (1 p 1 z 1 )(1 p 2 z 1 ) 2 Unrepeated factor 1 pz 1 contributes to the expansion A 1 pz 1 Repeated factor (1 pz 1 ) m contributes A m (1 pz 1 ) + A m 1 m (1 pz 1 ) + + A 1 m 1 (1 pz 1 ) Unrepeated quadratic factor 1 2rz 1 cos θ + z 2 contributes Az 1 + B 1 2rz 1 cos θ + z 2 Repeated quadratic (1 2rz 1 cos θ + z 2 ) m factor contributes A m z 1 + B m (1 2rz 1 cos θ + z 2 ) m + + A 1 z 1 + B 1 1 2rz 1 cos θ + z 2 Expand in terms of z 1 (not z) in the usual fashion Inverse transform each term in the expansion using tables
35 The s z Correspondence y(t) = e σt : Y (s) has a pole at s = σ y(kt ) = e σkt = (e σt ) k = r k : Y (z) has a pole at z = r = e σt y(t) = e σt sin ωt: Y (s) has a pole at s = σ ± ıω y(kt ) = e σkt sin ωkt = (e σt ) k sin k(ωt ) = r k sin kθ: Y (z) has poles at z = re ±ıθ = e σt e ±ıωt = e (σ±ıω)t Suggests the correspondence z = e st for mapping poles of a s-domain signal to the poles of its sampled sequence in z-domain Where should z poles lie to get good transient behaviour (ζ, ω n )? Locate s poles using s domain experience for desired ζ, ω n Map s poles to z poles using z = e st
36 Jury s Test for Stability a 0 z n + a 1 z n a n 1 z + a n = 0, a 0 > 0 is said to be Hurwitz if all roots lie in the OLHP, Schur if all roots lie in the OUD z n z n 1 z n 2 z 2 z z 0 a 0 a 1 a 2 a n 2 a n 1 a n a n a n 1 a n 2 a 2 a 1 a 0 b 0 b 1 b 2 b n 2 b n 1 b n 1 b n 2 b n 3 b 1 b 0 c 0 c 1 c 2 c n 2 c n 2 c n 3 c n 4 c 0 b k = 1 a 0 a 0 a n a n k a k..., k = 0, 1,..., n 1, c k = 1 b 0 b 0 b n 1 b n 1 k b k, k = 0, 1,..., n 2 Stable if a 0 > 0, b 0 > 0, c 0 > 0,...
37 Stability through s z Transformation z = 1 + st/2 1 st/2, s = 2 T (z 1) (z + 1) OLHP open unit disc imaginary axis unit circle Given a polynomial p(z), p(z) = p ( ) 1 + st/2 1 st/2 = n(s) d(s) zeros of p(z) zeros of n(s) p is Schur iff n is Hurwitz Apply Routh-Hurwitz test to n(s)
38 BIBO Stability of Transfer Functions A system given by a transfer function G(z) is BIBO stable if and only if the impulse response g is absolutely summable if and only if the impulse response g decays if and only if all poles (after cancellation) of G(z) lie in the interior of the unit disc, the open unit disc (OUD) {z : z < 1} We call a transfer function stable if all its poles lie in the OUD
39 Step Response Y (z) = G(z)(1 z 1 ) 1 Bounded if (after pole-zero cancellation) all poles of G(z) lie in {z : z 1} no repeated poles on the unit circle no pole at z = 1 Approaches a limit if all poles of G(z) lie in {z : z < 1} Decays to zero if Steady state value = lim k y(k) = lim z 1 (z 1)Y (z) = G(1) all poles of G(z) lie in {z : z < 1} and z = 1 is a zero, that is, G(1) = 0 For asymptotic tracking of a step input, need stability + G(1) = 1 For asymptotic rejection of a step disturbance, need stability + G(1) = 0
40 Harmonic Response of Stable Transfer Functions u(k) = sin(kωt ) U(z) = z sin ωt z 2 2z cos ωt + 1 = z 1 sin ωt (1 e jωt z 1 )(1 e jωt z 1 ) Y (z) = G(z)U(z) = a 1 1 e jωt z 1 + a 2 1 e jωt z 1 + b 1 1 p 1 z 1 + a 1 = G(z)U(z)(1 e jωt z 1 ) z=e jωt = 1 2j G(ejωT ) = 1 2j rejφ a 2 = ā 1 = 1 2j re jφ a 1 Y ss (z) = 1 e jωt z + a e jωt z 1 y ss (k) = a 1 (e jωt ) k + a 2 (e jωt ) k = Im re jφ (e jωt ) k y ss (k) = r sin(kωt + φ), r = G(e jωt ), φ = G(e jωt ) Amplification at ω is G(e jωt ), phase difference is G(e jωt ) Frequency response is periodic in frequency
41 Digital Analog Conversion Same response at ω and ω + ω s A/D Computer D/A A/D r(t) T {r(kt)} Sampler D/A ZOH Zero Order Hold ZOH
42 Analysis of Sample-and-Hold Operation r(t) T {r(kt)} ZOH r(t) Let s(t) = unit step function, s(t kt ) = unit step function delayed by kt r(t) = r(0)[s(t) s(t T )] + r(t )[s(t T ) s(t 2T )] + = r(kt )[s(t kt ) s(t kt T )] k=0 L(s(t)) = s 1, L(s(t kt )) = s 1 e skt [ ] [ ] 1 e R(s) = r(kt )(e st ) k st s k=0 }{{}}{{} R G (s) ZOH (s) Z(R(s)) def = Z(sampled sequence of r(t)) R (s) = Z(R(s)) z=e st
43 Sampler as an Impulse Modulator Let δ(t) = unit impulse in continuous time L(δ(t)) = 1, L(δ(t kt )) = e skt ( ) R (s) = L r(kt )δ(t kt ) = L(r (t)) Define δ T (t) = k=0 δ(t kt ) k=0 δ T is an impulse train r (t) = r(t)δ T (t) r is a modulated impulse train δ T (t) r * (t) Ideal sampler = impulse modulator
44 An Ideal ZOH G ZOH (s) = 1 s e st s = L[s(t) s(t T )] = L[ impulse response of ZOH] ZOH NOTE: No transfer function possible for a ZOH
45 Frequency Domain Analysis of an Impulse Train δ T (t) is a periodic function = expand in a Fourier series δ T (t) = c n e j2πnt/t n= c n = 1 T T/2 δ T (t)e j2πnt/t dt T/2 Fourier transform of δ T (t) = 1 T δ T = n= 1 T ejnω st 3ω s 2ω s ω s 0 ω s 2 ω s 3 ω s
46 Frequency Domain Analysis of a Modulated Impulse Train R (s) = L(r (t)) = L(r(t)δ T (t)) = 1 r(t)δ T (t)e st dt T = 1 T? = 1 T = 1 T 0 0 r(t) n= n= e jnωst e st dt n= 0 r(t)e (s jω sn)t dt R(s jω s n) Fourier transform R (jω) is periodic in ω with period ω s R (jω) obtained by superimposing scaled copies of R(jω) shifted by multiples of ω s
47 Aliasing R(jω) R *(jω) ω 2ω s ω s ω s 2ω s ω Contributions at ω due to R(jω), R(jω ± njω s ) Frequencies ω ± nω s aliases of ω, show up at ω after sampling
48 An Example of Aliasing y 1 (t) = sin t, y 2 (t) = sin 7t, ω s = 6, T = π/3 y 1 (kt ) = y 2 (kt ) = sin kt
49 Anti-Aliasing R(jω) R(jω) ω m R *(jω) ω ω m R *(jω) ω 2ω s ω s ω s 2ω s ω 2ω s ω s ω s 2ω s ω 2ω m < ω s 2ω m > ω s Nyquist s/shannon s Sampling Theorem: A signal can be recovered from its samples if the sampling frequency is more than twice the highest frequency in the signal To minimise the effect of aliasing, sampling is preceded by a low-pass antialias filter Eliminates frequencies above the Nyquist frequency
50 Signal Reconstruction from Samples Possible if signal is band limited and ω m < ω s /2 To recover R(jω) from R (jω), need a filter Lsuch that R(jω) = L(jω)R (jω) R (jω) = 1 T R(jω) + 1 R(jω njω s ) T n=,n 0 }{{} frequencies>ω s /2 L(jω) = T, ω [ ω s /2, ω s /2], L(jω) = 0 = 0, elsewhere everywhere L(j ω) T ω s ω s ω
51 Impulse Response of an Ideal Low-Pass Filter Inverse Fourier transform of L(jω) l(t) = 1 2π = 1 2π ωs /2 ω s /2 = sin(ω st/2) ω s t/2 L(jω)e jωt dω T e jωt dω = sinc(ω s t/2) sinc(π t/t) t/t Note: L is a noncausal filter
52 Reconstruction Using a Low-Pass Filter r(t) = (l r )(t) = = r(τ)δ T (τ)l(t τ)dτ r(kt )sinc(ω s (t kt )/2) k= RHS is the unique band limited signal that has ω m < ω s /2 Same samples as r Reconstruction is noncausal present value depends on future samples Cannot be implemented online, can be used for offline reconstruction
53 Antialias Filtering R(jω) Antialias ω Sample R *(jω) 2ω s ω s ω s 2ω s ω
54 Frequency Domain Analysis of Zero-Order Hold G ZOH (jω) = 1 e jωt jω = T e jωt/2sin(ωt/2) (ωt/2) = T e jωt/2 sinc(ωt/2) sincx = sin x x, x 0, = 1, x = 0 Magnitude : G ZOH (jω) = T sinc(ωt/2) Phase : G ZOH (jω) = ωt 2 + π at every sign change of sinc
55 Frequency Response of G ZOH 1 sinc(ω T/2) sinc(ω T/2) Phase/π ω T/2π
56 Harmonic Response of Sample and Hold R(jω) ω ω R *(jω) 2ω s ω s ω s 2ω s ω Input sinusoid of frequency ω < ω s /2 R(jω) First harmonic of output has magnitude : sinc(ωt/2) phase : ωt/2 First harmonic Imposter First harmonic is sinc(ωt/2) sin ω(t T/2)
57 Harmonic Response of Sample and Hold: An Example r(t)
58 Harmonic Response of Sample and Hold: An Example r(t) r(t)
59 Harmonic Response of Sample and Hold: An Example r(t) First harmonic r(t)
60 Harmonic Response of Sample and Hold: An Example r(t) r(t) First harmonic
61 Higher-Order Hold Functions Interpolation using Taylor series r(t) = r(nt ) + ṙ(nt )(t nt ) r(nt )(t nt )2 +, nt t < (n + 1)T Zero-order hold: Truncate at first term r(t) = r(nt ), nt t (n + 1)T First-order hold: Truncate at first-order term r(t) = r(nt ) + ṙ(nt )(t nt ), nt t (n + 1)T To find ṙ(nt ), extrapolate to (n 1)T t (n + 1)T, put t = (n 1)T ṙ(nt ) = r(nt ) r((n 1)T ) T
62 First-Order Hold Pulse Response (n 1)T nt Pulse Response = s(t) + t T s(t) 2 s(t T ) 2 (t T ) s(t T ) T + s(t 2T ) + 1 (t 2T ) s(t 2T ). T G FOH (s) = 1 s 1 s 2e st + 1 s e 2sT + 1 T s 2(1 2e st + e 2sT ) ( ) (1 + T s) 1 e st 2 = T s
63 Analysis of a Sampler and First-Order Hold r(t) T {r(kt)} FOH r(t) R(s) = R (s)g FOH (s) 1.5 G FOH (jω) Magnitude G ZOH (jω) ω T/2π Phase G FOH (jω) G ZOH (jω) ω T/2π
64 Analysis of a Sample, Process and Hold e(t) e(kt) H(z) u(kt) ZOH u(t) U(s) = U (s)g ZOH (s) = U(z) z=e st G ZOH (s) = [H(z)E(z)] z=e st G ZOH (s) = H (s)e (s) }{{} G ZOH (s) convolved impulse trains
65 ZOH Equivalent u(kt) ZOH u(t) G(s) y(t) y(kt) Transfer Function possible Let u(kt ) = δ(kt ), unit pulse sequence ū(t) = s(t) s(t τ) y(t) = w(t) w(t τ), w = unit step response of.g(s) y(kt ) = w(kt ) w((k 1)T ) Y (z) = (1 z 1 )W (z) W (z) = Z[L 1 (s 1 G(s))] = Z(s 1 G(s)). Transfer Function = (1 z 1 )Z(s 1 G(s)) }{{} ZOH equivalent Y (s) = Y (z) z=e st = (1 e st )Z(s 1 G(s)) z=e st U (s) Y (s) = G(s)Ū(s) = G(s)G ZOH(s)U (s)
66 A Glossary of Notation Given U(s) G ZOH (s) = 1 e st s Z(U(s)) = Z transform of sampled u(t) U (s) = Z(U(s)) z=e st Given U(z) U (s) = U(z) z=e st Given G(s) G h0 (z) def = ZOH equivalent of G(s) = (1 z 1 )Z(s 1 G(s))
67 An Example e(t) u(kt) u(t) y(t) y(kt) H(z) ZOH G(s) No transfer function possible between y(t) and e(t) Can find Y (s) Transfer function possible between y(kt ) and e(kt )
68 An Example e(t) u(kt) u(t) y(t) y(kt) H(z) ZOH G(s) No transfer function possible between y(t) and e(t) Can find Y (s) Transfer function possible between y(kt ) and e(kt ) Y (s) = G(s)U(s) = G(s)G ZOH (s)u (s) = G(s)G ZOH (s)h (s)e (s)
69 An Example e(t) u(kt) u(t) y(t) y(kt) H(z) ZOH G(s) No transfer function possible between y(t) and e(t) Can find Y (s) Transfer function possible between y(kt ) and e(kt ) Y (s) = G(s)U(s) = G(s)G ZOH (s)u (s) = G(s)G ZOH (s)h (s)e (s) Y (z) E(z) = H(z)G h 0 (z) = H(z)(1 z 1 )Z(s 1 G(s))
70 Block Diagram Manipulation For Sampled Data System: An Example r(t) e(t) e(kt) C(z) u(kt) ZOH u(t) G(s) y(t) r(t) r(kt) u(kt) u(t) C(z) ZOH G(s) y(t) y(kt) u(kt) ZOH u(t) G(s) y(t) r(t) r(kt) C(z) G h0(z) y(kt) Y (z) E(z) = C(z)G h 0 (z) 1 + C(z)G h0 (z) Y (s) = G(s)G ZOH (s)u (s) = G(s)G ZOH (s)c (s)(r (s) Y (s))
71 Another Example r(t) e1(t) e2(kt) G1(s) H(z) e2(t) u(kt) u(t) y(t) ZOH G(s) r(t) y(t) r1(t) G1(s) H(z) ZOH G(s) G1(s) r1(kt) y1(kt) Y 1 (z) R 1 (z) = H(z)(GG 1) h0 (z) 1 + H(z)(GG 1 ) h0 (z), Y 1 (s) = ( Y1 (z) R 1 (z)) Y (s) = G(s)G ZOH (s)h (s)[r1(s) Y1 (s)] Discrete-Time Equivalents of Continuous-Time Controllers Design controller in continuous time R 1(s) = (GR 1 ) (s) G (s)r (s) Numerically implement a discrete-time equivalent z=e st R 1(s) Example H(s) = 1 s + a = ẏ + ay = u
72 = y(t) = t = y(kt ) = y(kt T ) + 0 [u(τ) ay(τ)]dτ kt (k 1)T [u(τ) ay(τ)]dτ Each numerical approximation for the integral gives a discrete-time equivalent
73 Backward Rectangular Rule kt (k 1)T u(τ)dτ u(kt )T y(kt ) = y(kt T ) + kt (k 1)T [u(τ) ay(τ)]dτ y(kt ) = y(kt T ) + T u(kt ) at y(kt ) Y (z) U(z) = T 1 z 1 + at = 1 ( ) 1 z 1 + a H B (z) = H(s) s=(1 z 1 )/T s 1 z 1, z 1 T 1 T s T
74 Stability Regions Under Backward Rule z 1 2 = T s 2 1 T s Re s < 0 = z 1 2 < 1 2 Im Image of OLHP 0 1 Re
75 Forward Rectangular Rule kt (k 1)T u(τ)dτ u(kt T )T y(kt ) = y(kt T ) + kt (k 1)T [u(τ) ay(τ)]dτ y(kt ) = y(kt T ) + T u(kt T ) at y(kt T ) Y (z) U(z) = T z 1 + at = 1 ( z 1 ) + a H F (z) = H(s) s=(z 1)/T s z 1 T, z 1 + T s T
76 Stability Regions Under Forward Rule z = 1 + T s Re s < 0 = Re z < 1 Re Im 0 1 Image of OLHP
77 Trapezoidal Rule kt (k 1)T u(τ)dτ T [u(kt T ) + u(kt )] 2 y(kt ) = y(kt T ) + kt (k 1)T [u(τ) ay(τ)]dτ y(kt ) = y(kt T ) + T 2 [u(kt T ) + u(kt )] at 2 [y(kt T ) + y(kt )] Y (z) U(z) = 1 ( ) 2 1 z 1 T + a 1+z 1 H T (z) = H(s) s= 2 T 1 z 1 1+z 1 : Tustin s Rule s 2 T 1 z T s/2, z 1 + z 1 1 T s/2
78 Stability Regions Under Tustin s Rule z = 1 + T s/2 1 T s/2 Re s < 0 = z < 1 ORHP Re Im 0 1 OLHP
79 Discrete-Time Equivalent by Impulse Invariance Find Ĥ(z) such that pulse response of Ĥ(z) is the sampled sequence of the impulse response of H(s) H(s) ^ H(z) SAME Ĥ(z) = Z(H(s))
80 Discrete-Time Equivalence by Step Invariance Find H(s) Ĥ(z) such that step response of Ĥ(z) is the sampled sequence of the step response of H(s) H(z) ^ ( ) H(s) Ĥ(z) = (1 z 1 )Z = H h0 (z) s SAME
81 Equivalence at a Frequency When will the steady state response of Ĥ(z) to {cos kωt } equal the sampled sequence of the steady state response of H(s) to cos ωt? H(s) H(z) ^ SAME If and only if H(jω) = Ĥ(ejωT )
82 Tustin s Rule and Equivalence at a Frequency H(s) = a s + a, H T(z) = 2 T a z 1 z+1 + a H(ja) = j, H T(e jat 1 ) = 1 + j 2 at tan at 2 The discrete equivalent does not match the original at the corner frequency Tustin s rule causes frequency distortion Distortion is reduced if at/2 << 1
83 Tustin s Rule with Pre-warping Pre-warp the continuous system such that on applying Tustin s rule, matching is obtained at the selected frequency Substitute Recover Tustin s rule if b = 2/T s = b z 1 z + 1 Same as applying Tustin s rule to the pre-warped transfer function H pre warped (s) = H(bT s/2) Choose b to get matching at the desired frequency
84 Pole-Zero Mapping Equivalent Map all poles of H(s) according to z = e st Map all finite zeros of H(s) by z = e st Map zeros at to zeros at 1 1 s + a 1 1 e at z 1 (s + a) 1 e at z 1 1 s 1 + z 1 To get a strictly causal system, map one s 1 factor to z 1 Choose gain factor to get matching at a specified frequency Usually ω = 0, that is, matching at DC H(jω) = H zp (e jωt )
85 Root Locus r(t) K* H(z) ZOH G(s) y(t) K* H(z) Gh0(s) y(kt) Root locus = locus of roots of 1 + KH(z)G h0 (z) = 0 as K varies fro 0 to Plotted in the same way as for continuous-time systems
86 Mapping Theorem Based on Mapping Theorem z traces a simple closed curve C clockwise in the complex plane The no. clockwise of encirclements of the origin by F (z) equals no. of zeros of F enclosed by C no of poles of F enclosed by C Application to closed-loop stability analysis Choose F (z) = 1 + G(z)H(z) = closed-loop characteristic polynomial Choose C to enclose all possible unstable poles
87 Nyquist Contour Choose C to enclose the exterior of the open unit disc Im ω=ω s /2 OUD Re All encirclements are contributed by portion along the unit circle
88 Nyquist Criterion + G(z) H(z) Nyquist Criterion: Loop transfer function L(z) = G(z)H(z) Z = N + P P = no. of unstable open-loop poles (unstable poles of L(z)) Z = no. of closed-loop unstable poles (unstable roots of 1 + L(z) = 0) N = no. of clockwise encirclements of 1 by L(e jωt ), ω [0, ω s ]
89 Gain and Phase Margins 1 1/GM PM
90 Frequency Response Analysis with W-Transform Frequency response in terms of Z-transform is Periodic in ω Difficult to draw by hand (s-domain rules do not apply) Use W-transform to map OUD into OLHP using w = 2 T (z 1) (z + 1), z = 1 + wt/2 1 wt/2 Ĝ(w) = G(z) z= 1+wT/2 1 wt/2 Bode plots of Ĝ(w) can be drawn using s-domain rules Nyquist criterion can be applied to Ĝ(w) as in s-domain Controller Ĥ designed for Ĝ can be transformed back and applied to G Ĝ and G yield the same gain and phase margins
91 Closed-Loop Asymptotic Tracking of Reference Inputs r(k) e(k) y(k) + H(z) G(z) Asymptotic Tracking: Want E(z) R(z) = G(z)H(z) lim e(k) = 0 k lim k e(k) exists if and only if all poles of E(z) lie in the OUD except possibly for one pole at z = 1 lim k e(k), if it exists, equals lim z 1 (z 1)E(z)
92 Tracking of Step Inputs r(k) = 1, R(z) = E(z) = z z 1 z 1 (z 1) [1 + G(z)H(z)] For lim k e(k) to exist, all closed-loop poles must lie in the OUD lim e(k) = lim k z 1 z 1 + G(z)H(z) = lim z 1 G(z)H(z) For lim k e(k) = 0, the (open) loop transfer function must have a pole at z = 1 No. of poles of G(z)H(z) at z = 1 is the type of the open-loop system Define position error constant For perfect tracking, need K p = K p = lim z 1 G(z)H(z) For perfectly tracking step inputs, need closed-loop stability + type 1 open-loop system
93 Tracking of Ramp Inputs E(z) = r(k) = kt, R(z) = T z (z 1) 2 T z (z 1) 2 1 [1 + G(z)H(z)] = T z (z 1)[z 1 + (z 1)G(z)H(z)] For lim k e(k) to exist, all closed-loop poles must lie in the OUD, and open-loop system must be of type 1 lim e(k) = lim k z 1 T z (z 1)[1 + G(z)H(z)] = T lim z 1 (z 1)G(z)H(z) For lim k e(k) = 0, the (open) loop transfer function must have at least two poles at z = 1 Define velocity error constant For perfect tracking, need K v = K v = lim z 1 (z 1)G(z)H(z)/T For perfectly tracking ramp inputs, need closed-loop stability + type 2 open-loop system
94 Tracking of Sinusoidal Inputs r(k) = A sin(kωt ), R(z) has poles at e ±jωt For e(k) to converge to a steady state behavior, closed-loop must be (BIBO) stable Steady-state error amplitude = G(e jωt )H(e jωt ) For lim k e(k) = 0, G(z)H(z) must have at least one pole at z = e jωt For perfectly tracking {A sin kωt }, need closed-loop stability + open-loop poles at z = e ±jωt
95 Internal Model Principle Requirements for closed-loop tracking Input to be tracked Requirements for tracking Input Input poles Open-loop poles Closed-loop poles Step z = 1 z = 1 OUD Ramp z = 1, 1 z = 1, 1 OUD Sinusoidal z = e ±jωt z = e ±jωt OUD Internal Model Principle: A closed-loop system will track an input perfectly asymptotically if and only if The closed-loop system is stable and The open-loop system contains a model of the input
96 Continuous-Time LTI State-Space Systems ẋ = Ax + Bu, y = Cx + Du, state dynamics measurement equation x = vector of state/internal variables y = vector of output measurements u = vector of inputs Linear: A, B, C, D independent of x, y Time Invariant: A, B, C, D independent of time To find output, need Initial state vector and input
97 Matrix Exponential Given a square matrix A, define e At def = I + At + 1 2! A2 t ! A3 t 3 + Well defined (series converges sufficiently nicely) Satisfies d e A0 dt eat e A(t+τ) = I = Ae At = e At A = e At e Aτ (e At ) 1 = e At Note: (e At ) ij e A ijt
98 Examples ÿ + ω 2 ny = 1 m u d ẏ = 0 1 ẏ + 1 k u dt y ωn 2 0 y 0 }{{}}{{} A B e At = cos ω 1 nt ω n sin ω n t ω n sin ω n t cos ω n t ÿ = u A = 0 1, e At = t 0 1
99 Solution of the State Equation State response Output response Impulse response Transfer matrix y(t) = x(t) = t e At x(0) + e }{{} A(t τ) Bu(τ)dτ 0 Natural Response }{{} Forced Response e At = state transition matrix t Ce At x(0) + Ce }{{} A(t τ) Bu(τ)dτ + Du(t) 0 Natural Response }{{} g(t) = Ce At B + Dδ(t) C(sI A) 1 B + D Forced Response
100 ZOH Equivalent of a State Space System u(kt) ZOH u(t) State Space System y(t) y(kt) x(k + 1) = e AT x(k) + kt [ T = e AT x(k) + u(t) = u(k), t [kt, kt + T ) kt +T 0 e A(kT +T τ) Bu(k)dτ ] e A(T σ) Bdσ u(k), σ = τ kt Discrete-time state space model of the ZOH equivalent x(k + 1) = Φx(k) + Γu(k) y(k) = Hx(k) + Ju(k) Φ = e AT, Γ = [ T 0 ] e A(T σ) Bdσ, H = C, J = D
101 State Space Realizations G(z) = b 0 + b 1 z 1 + b n z n 1 + a 1 z a n z n Let Ŷ (z) = U(z)[1 + a 1z a n z n ] 1, so that Y (z) = [b 0 + b 1 z 1 + b n z n ]Ŷ (z) ŷ(k) + a 1 ŷ(k 1) + + a n ŷ(k n) = u(k) Choose x(k) = ŷ(k n) ŷ(k n + 1). ŷ(k 1), x(k+1) = ŷ(k n + 1) ŷ(k n + 2). ŷ(k) = x 2 (k) x 3 (k). a n x 1 (k) a 1 x n (k) + u(k) y(k) = b 0 ŷ(k) + b 1 ŷ(k 1) + + b n ŷ(k n) = b 0 ŷ(k) + b 1 x n (k) + + b n x 1 (k) = (b n b 0 a n )x 1 (k) + (b n 1 b 0 a n 1 )x 2 (k) + + (b 1 b 0 a 1 )x n (k) + b 0 u(k)
102 State Space Realizations (cont d) First companion form x(k + 1) = y(k) = x(k) +. u(k) } a n a n 1 a n 2 {{ a 1 } 1 }{{} [ A ] B (b n b 0 a n ) (b 1 b 0 a 1 ) x(k) + [b 0 ] u(k) }{{}}{{} D C
103 Transfer Matrix from State Space Model State dynamics equations in Laplace domain Poles are eigenvalues of A zx(z) zx(0) = AX(z) + BU(z) X(z) = z(zi A) 1 x(0) }{{} Natural response Y (z) = CX(z) + DU(z) + (zi A) 1 BU(z) }{{} Forced response = Cz(zI A) 1 x(0) + [C(zI A) 1 B + D] U(s) }{{} Transfer matrix Transfer matrix = C(zI A) 1 B + D
104 Transformations of State Space Models State transformation ˆx = Sx Input-output relation is unchanged x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) + Du(k) ˆx(k + 1) = SAS 1 }{{} Â y(k) = CS 1 }{{} Ĉ ˆx(k) + }{{} SB u(k) B ˆx(k) + }{{} D u(k) D C(zI A) 1 B + D = Ĉ(zI Â) 1 B + D Two state-space models are equivalent if they yield the same transfer matrix Every input-output system has several equivalent state space representations/realizations
105 State Evolution in Discrete-Time System x(k + 1) = Ax(k) + Bu(k) x(1) = Ax(0) + Bu(0) x(2) = A 2 x(0) + ABu(0) + Bu(1) x(3) = A 3 x(0) + A 2 Bu(0) + ABu(1) + Bu(2). x(k) = y(k) = A k x(0) + }{{} natural response CA k x(0) + }{{} natural response k 1 l=0 A k l 1 Bu(l) } {{ } forced response k 1 l=0 CA k l 1 Bu(l) + Du(k) } {{ } forced response
106 Impulse Response Response to impulse input u(k) = u 0 δ(k) y(k) = CA k 1 Bu 0 + Du 0 δ(k) Impulse response sequence = {Du 0, CBu 0, CABu 0,...} Impulse response matrix General response H(k) = D, k = 0 = CA k 1 B, k > 0 y(k) = CA k x(0) + (H u)(k) Compare with forced response in z-domain Y (z) = [C(zI A) 1 B + D]U(z) = Z(H) = C(zI A) 1 B + D = D + z 1 CB + z 2 CAB +
107 Reachable Sets Which states can be reached from a given initial condition by using all possible inputs? Reachable set from x 0 at step k R(k, x 0 ) = {states reachable from x 0 in k steps} { } k 1 = A k x 0 + A k l 1 Bu(l) : u(0), u(1),..., u(k 1) arbitrary l=0 System is controllable if every state can be reached from every other state in a finite (but possibly large) number of steps System is controllable iff R(k, x 0 ) = R n for sufficiently large k
108 Facts on Reachable Sets Fact 1 : R(k, x 0 ) = R(k, 0) + A k x 0 If a R(n, 0), then Fact 2 : R(n, 0) = Range C, C = [B AB A 2 B A n 1 B] a = Bu(n 1) + ABu(n 2) + + A n 1 Bu(0) = C If a Range C, then u(n 1). u(0) a = Cb = Bb 1 + ABb A n 1 Bb n R(n, 0) Range C
109 Facts on Reachable Sets (cont d) Fact 3 : R(k, 0) = R(n, 0), k n Clearly R(n, 0) R(k, 0) For k > n, an element of R(k, 0) is of the form Bu(k 1) + + A n 1 Bu(k n) + A n Bu(k n + 1) + + A k 1 Bu(0) By Cayley-Hamilton theorem, powers of A higher than n 1 can be written as combinations of powers of A upto n 1 Elements of R(k, 0) are contained in R(n, 0) System is controllable iff rank C = n
110 Unobservable Sets Can we guess the initial state by observing only the output? y 1 (k) = CA k x 1 + (H u)(k) y 2 (k) = CA k x 2 + (H u)(k) Can distinguish x 1 from x 2 iff CA k x 1 CA k x 2 for some k Unobservable set from x 0 at step k U(k, x 0 ) = {states that yield the same output as x 0 upto step k 1} = {x : CA i x = CA i x 0, i = 0, 1,, k 1} System is observable if every state can be distinguished from every other state in a finite (but possibly large) number of steps System is observable iff U(k, x 0 ) = {x 0 } for sufficiently large k
111 Facts on Unobservable Sets Fact 1 : U(k, x 0 ) = U(k, 0) + x 0 Fact 2 : U(k, 0) = U(n, 0), k n Fact 3 : U(n, 0) = kernel O, O = C CA. CA n 1 System is observable iff rank O = n
112 Hautus Test for Controllability Eigenvalue λ C of A is controllable if rank [λi A B] = n Fact: System is controllable iff every eigenvalue of A is controllable If λ is not controllable, then there exists x C n such that x A = λx, x B = 0 = x A i B = 0 = x C = 0 = rank C < n Controllability remains invariant under state transformation Uncontrollable eigenvalues are unaffected by control If λ is an uncontrollable eigenvalue, and the feedback u = Kx is used, then λ also appears as a closed-loop eigenvalue
113 Hautus Test for Observability Eigenvalue λ C of A is unobservable if rank λi A C = n Fact: System is observable iff every eigenvalue of A is observable If λ is not observable, then there exists x C n such that Ax = λx, Cx = 0 = CA i x = 0 = Ox = 0 = rank O < n Observability remains unchanged under state transformations Unobservable eigenvalues cannot be detected through the output
114 A Two-Dimensional Example A = C = λ 1 0, B = 0 λ 2 [ ] c 1 c 2 b 1 b 2 u b b 1 1 (z λ c 1 ) (z ) λ 2 1 c 2 + y By the Hautus test, need b 1 0 b 2 for controllability and c 1 0 c 2 for observability
115 Kalman Decomposition Theorem Fact: Every state space system can be transformed into A co A 12 A 13 x co (k) x(k + 1) = 0 A co A 23 x co (k) A c x c (k) [ ] y(k) = 0 C co C c x(k) + Du(k) B co B co 0 u(k) G(s) = C(zI A) 1 B + D = C co (zi A co ) 1 B co + D The controllable and observable part yields a smaller realization Eigenvalues that are either unobservable or uncontrollable are not poles
116 Minimal Realizations A minimal realization is one having the least no. of states Desirable for implementation A minimal realization has as many states as the number of poles A realization is minimal iff it is controllable and observable All minimal realizations are equivalent
117 Jordan Form Every matrix can be reduced to its Jordan form through a similarity transformation λ λ λ λ λ λ λ λ λ 4 characteristic polynomial = (z λ 1 ) 3 (z λ 2 ) 3 (z λ 3 ) 2 (z λ 4 ) λ 1 and λ 4 have 1 eigenvector each, λ 2 and λ 3 have 2 eigenvectors each λ 3 and λ 4 are semisimple, λ 4 is simple
118 Internal Stability Internal stability refers to the natural response of state (internal) variables A state space system is (internally) Lyapunov stable if every initial condition response is bounded Asymptotically stable if every initial condition response decays to zero Unstable if it is not Lyapunov stable Stability depends on the elements of J k x(k) = A k x(0) = T J k T 1 x(0), J = Jordan form
119 Powers of Jordan Blocks System is Lyapunov stable iff J = λ 0 = J k = λk 0 0 λ 0 λ k J = λ 1 = J k = λk kλ k 1 0 λ 0 λ k λ 1 0 λ k kλ k 1 k(k 1) 2 λ k 2 J = 0 λ 1 = J k = 0 λ k kλ k λ 0 0 λ k All eigenvalues CUD and All eigenvalues of unit magnitude are semisimple System is asymptotically stable iff all eigenvalues OUD
120 BIBO Stability and Internal Stability System is BIBO stable iff every input vector with bounded components gives an output vector with bounded components System is BIBO stable if and only if every pole OUD (internal) asymptotic stability = BIBO stability Converse does not hold in general Fact: A controllable, observable, BIBO stable system is asymptotically stable Fact: A system is BIBO stable iff every minimal realization is asymptotically stable
121 Positive-Definite Matrices P R n n, symmetric, is positive-definite (P > 0) if x T P x > 0 for every x C n, x 0 A symmetric positive-definite matrix has real eigenvalues that are positive Every symmetric positive-definite matrix gives rise to the quadratic function V P (x) = x T P x If P > 0, then the level sets of V P are hyper-ellipsoids, eg. P = x x 1 Level curves of V P
122 Lyapunov Function How does a given quadratic function change along the natural state response? x(k + 1) = Ax(k) V P (x(k)) = x T (k)p x(k) V P (x(k + 1)) = x T (k + 1)P x(k + 1) = x T (k)a T P Ax(k) V P (x(k + 1)) V P (x(k)) = x T (k)[a T P A P ]x(k) Idea: If P is positive definite and V P (x(k)) decreases with k, then x(k) 0 Such a V P is called a Lyapunov function Want P > 0 and A T P A P = Q, where Q > 0
123 Lyapunov Equation Fact: If there exist P > 0 and Q > 0 satisfying the Lyapunov equation below, then system is asymptotically stable A T P A P = Q Fact: System is asymptotically stable iff for every Q > 0, there exists a positive-definite solution P to the Lyapunov equation For an asymptotically stable system, the solution P is unique To prove stability or instability, pick Q > 0 (eg. definiteness of P Q = I), solve for P and check sign OR check the feasibility of the linear matrix inequalities (LMIs) A T P A + P > 0 P > 0 Can be done using efficient numerical algorithms
124 Full-State Feedback x(k + 1) = Ax(k) + Bu(k) Open loop system u(k) = Kx(k) + r(k) Full state feedback x(k + 1) = (A BK)x(k) + Br(k) Closed loop system Pole-placement problem Can we design a gain matrix K such that A BK has desired eigenvalues? Fact: Every uncontrollable open-loop eigenvalue is a closed-loop eigenvalue Assume Complete controllability Single input
125 Pole Placement using Companion Form Idea: Use state transformation ˆx = Sx such that  = SAS 1 = a n a n 1 a n 2 a 1, B = SB = }{{} controller canonical form Use feedback u = K ˆx = [ˆk n ˆkn 1 ˆk 1 ]ˆx  B K = a n ˆk n a n 1 ˆk n 1 a n 2 ˆk n 2 a 1 ˆk 1
126 Pole Placement using Companion Form (cont d) Characteristic polynomial of A and  z n + a 1 z n a n Characteristic polynomial of  B K z n + (a 1 + ˆk 1 )z n (a n + ˆk n ) Desired characteristic polynomial z n + α 1 z n α n Choose K = [α n a n α n 1 a n 1 α 1 a 1 ] Feedback in terms of original states Ackermann s Formula: u = K ˆx = }{{} KS x K K = e T nc 1 [A n + α 1 A n α n I] e T n = [0 0 1]
127 Proof of Ackermann s Formula Define s 1 = e T nc 1, and consider S = s 1 s 1 A. s 1 A n 1 Claim: SA = ÂS Claim: S is invertible Claim: SB = B S yields the transformation to the controller canonical form K = KS yields Ackermann s formula Multi-input case: redundant degrees of freedom in choosing K Can be used to assign eigenstructure
128 Estimation/Observation Online reconstruction of state vector from the output x(k + 1) = Ax(k) + Bu(k) } y(k) = Cx(k) {{ } Observable single output system output injection {}}{ L(y(k) ŷ(k)), x(k + 1) = A x(k) + Bu(k) + ŷ(k) = C ˆx(k) }{{} Luenberger observer u A, B x C A, B x^ C y y^ + O b s e r v L e r Error dynamics e(k + 1) = (A LC)e(k), e = x x
129 Observer Design by Pole Placement Can we choose observer gain matrix L such that A LC is Schur? Yes, if the system is observable (A, C) is observable iff (A T, C T ) is controllable There exists a gain matrix K such that A T C T K has desired eigenvalues K = e T n[c T C T A T C T A (n 1)T ] 1 [A nt + α 1 A (n 1)T + + α n I] Letting L = K T, A LC has desired eigenvalues L = [A n + α 1 A n α n I]O 1 e n
130 Dynamic Output-Feedback Compensation Idea: In a full-state feedback controller, use state estimate generated by an observer in place of the actual state r + u x A, B C y K + x^ A, B C y^ Dynamic output feedback controller L
131 Seperation Principle x(k + 1) = Ax(k) + Bu(k) x(k + 1) = A x(k) + Bu(k) + L(y(k) C x(k)) y(k) = Cx(k) u(k) = K ˆx(k) + r(k) x(k + 1) = A BK x(k) + B r(k) x(k + 1) LC A LC BK x(k) B }{{} Closed loop system Choose state vector as [x T e T ] T, e = x x x(k + 1) = A BK BK x(k) + B e(k + 1) 0 A LC e(k) 0 r(k) Seperation Principle: Output-feedback controller can be obtained by combining an independently designed Regulator that uses full state feedback with An observer
Modelling and Control of Dynamic Systems. Stability of Linear Systems. Sven Laur University of Tartu
Modelling and Control of Dynamic Systems Stability of Linear Systems Sven Laur University of Tartu Motivating Example Naive open-loop control r[k] Controller Ĉ[z] u[k] ε 1 [k] System Ĝ[z] y[k] ε 2 [k]
More information1. Z-transform: Initial value theorem for causal signal. = u(0) + u(1)z 1 + u(2)z 2 +
1. Z-transform: Initial value theorem for causal signal u(0) lim U(z) if the limit exists z U(z) u(k)z k u(k)z k k lim U(z) u(0) z k0 u(0) + u(1)z 1 + u(2)z 2 + CL 692 Digital Control, IIT Bombay 1 c Kannan
More informationProperties of Open-Loop Controllers
Properties of Open-Loop Controllers Sven Laur University of Tarty 1 Basics of Open-Loop Controller Design Two most common tasks in controller design is regulation and signal tracking. Regulating controllers
More informationAutomatique. A. Hably 1. Commande d un robot mobile. Automatique. A.Hably. Digital implementation
A. Hably 1 1 Gipsa-lab, Grenoble-INP ahmad.hably@grenoble-inp.fr Commande d un robot mobile (Gipsa-lab (DA)) ASI 1 / 25 Outline 1 2 (Gipsa-lab (DA)) ASI 2 / 25 of controllers Signals must be sampled and
More informationAnalysis of Discrete-Time Systems
TU Berlin Discrete-Time Control Systems 1 Analysis of Discrete-Time Systems Overview Stability Sensitivity and Robustness Controllability, Reachability, Observability, and Detectabiliy TU Berlin Discrete-Time
More informationSystems and Control Theory Lecture Notes. Laura Giarré
Systems and Control Theory Lecture Notes Laura Giarré L. Giarré 2017-2018 Lesson 7: Response of LTI systems in the transform domain Laplace Transform Transform-domain response (CT) Transfer function Zeta
More informationChapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steady-state error, and transient response for computer-controlled systems. Transfer functions,
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response
.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........
More informationIdentification Methods for Structural Systems
Prof. Dr. Eleni Chatzi System Stability Fundamentals Overview System Stability Assume given a dynamic system with input u(t) and output x(t). The stability property of a dynamic system can be defined from
More information4F3 - Predictive Control
4F3 Predictive Control - Discrete-time systems p. 1/30 4F3 - Predictive Control Discrete-time State Space Control Theory For reference only Jan Maciejowski jmm@eng.cam.ac.uk 4F3 Predictive Control - Discrete-time
More informationEL 625 Lecture 11. Frequency-domain analysis of discrete-time systems
EL 625 Leture 11 1 EL 625 Leture 11 Frequeny-domain analysis of disrete-time systems Theorem: If S is a fixed linear disrete-time system, the zero-state response, S{z k } = H(z)z k (1) Proof: S{z k } =
More informationMethods for analysis and control of. Lecture 6: Introduction to digital control
Methods for analysis and of Lecture 6: to digital O. Sename 1 1 Gipsa-lab, CNRS-INPG, FRANCE Olivier.Sename@gipsa-lab.inpg.fr www.lag.ensieg.inpg.fr/sename 6th May 2009 Outline Some interesting books:
More informationAnalysis of Discrete-Time Systems
TU Berlin Discrete-Time Control Systems TU Berlin Discrete-Time Control Systems 2 Stability Definitions We define stability first with respect to changes in the initial conditions Analysis of Discrete-Time
More informationControl System Design
ELEC4410 Control System Design Lecture 19: Feedback from Estimated States and Discrete-Time Control Design Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science
More informationEEE582 Homework Problems
EEE582 Homework Problems HW. Write a state-space realization of the linearized model for the cruise control system around speeds v = 4 (Section.3, http://tsakalis.faculty.asu.edu/notes/models.pdf). Use
More informationControls Problems for Qualifying Exam - Spring 2014
Controls Problems for Qualifying Exam - Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function
More informationLecture: Sampling. Automatic Control 2. Sampling. Prof. Alberto Bemporad. University of Trento. Academic year
Automatic Control 2 Sampling Prof. Alberto Bemporad University of rento Academic year 2010-2011 Prof. Alberto Bemporad (University of rento) Automatic Control 2 Academic year 2010-2011 1 / 31 ime-discretization
More informationDiscrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture
Discrete Systems Mark Cannon Hilary Term 22 - Lecture 4 Step response and pole locations 4 - Review Definition of -transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k},
More informationSTABILITY ANALYSIS TECHNIQUES
ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to control-system design: 1 Stability, 2 Steady-state response, 3 Transient response
More informationControl Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control D-MAVT ETH Zürich
Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:
More informationIntroduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
Introduction Classical Control Robust Control u(t) y(t) G u(t) G + y(t) G : nominal model G = G + : plant uncertainty Uncertainty sources : Structured : parametric uncertainty, multimodel uncertainty Unstructured
More informationCDS Solutions to Final Exam
CDS 22 - Solutions to Final Exam Instructor: Danielle C Tarraf Fall 27 Problem (a) We will compute the H 2 norm of G using state-space methods (see Section 26 in DFT) We begin by finding a minimal state-space
More informationGATE EE Topic wise Questions SIGNALS & SYSTEMS
www.gatehelp.com GATE EE Topic wise Questions YEAR 010 ONE MARK Question. 1 For the system /( s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) s (C)
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More informationSAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015
FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequency-domain analysis and control design (15 pt) Given is a
More informationRaktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries
. AERO 632: of Advance Flight Control System. Preliminaries Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Preliminaries Signals & Systems Laplace
More informationEL 625 Lecture 10. Pole Placement and Observer Design. ẋ = Ax (1)
EL 625 Lecture 0 EL 625 Lecture 0 Pole Placement and Observer Design Pole Placement Consider the system ẋ Ax () The solution to this system is x(t) e At x(0) (2) If the eigenvalues of A all lie in the
More informationLTI system response. Daniele Carnevale. Dipartimento di Ing. Civile ed Ing. Informatica (DICII), University of Rome Tor Vergata
LTI system response Daniele Carnevale Dipartimento di Ing. Civile ed Ing. Informatica (DICII), University of Rome Tor Vergata Fondamenti di Automatica e Controlli Automatici A.A. 2014-2015 1 / 15 Laplace
More informationDepartment of Electronics and Instrumentation Engineering M. E- CONTROL AND INSTRUMENTATION ENGINEERING CL7101 CONTROL SYSTEM DESIGN Unit I- BASICS AND ROOT-LOCUS DESIGN PART-A (2 marks) 1. What are the
More informationEECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 5-8 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3
More informationDiscrete and continuous dynamic systems
Discrete and continuous dynamic systems Bounded input bounded output (BIBO) and asymptotic stability Continuous and discrete time linear time-invariant systems Katalin Hangos University of Pannonia Faculty
More informationLecture Note #6 (Chap.10)
System Modeling and Identification Lecture Note #6 (Chap.) CBE 7 Korea University Prof. Dae Ryook Yang Chap. Model Approximation Model approximation Simplification, approximation and order reduction of
More informationModule 07 Controllability and Controller Design of Dynamical LTI Systems
Module 07 Controllability and Controller Design of Dynamical LTI Systems Ahmad F. Taha EE 5143: Linear Systems and Control Email: ahmad.taha@utsa.edu Webpage: http://engineering.utsa.edu/ataha October
More informationLinear dynamical systems with inputs & outputs
EE263 Autumn 215 S. Boyd and S. Lall Linear dynamical systems with inputs & outputs inputs & outputs: interpretations transfer function impulse and step responses examples 1 Inputs & outputs recall continuous-time
More informationECE504: Lecture 9. D. Richard Brown III. Worcester Polytechnic Institute. 04-Nov-2008
ECE504: Lecture 9 D. Richard Brown III Worcester Polytechnic Institute 04-Nov-2008 Worcester Polytechnic Institute D. Richard Brown III 04-Nov-2008 1 / 38 Lecture 9 Major Topics ECE504: Lecture 9 We are
More informationControl Systems Lab - SC4070 Control techniques
Control Systems Lab - SC4070 Control techniques Dr. Manuel Mazo Jr. Delft Center for Systems and Control (TU Delft) m.mazo@tudelft.nl Tel.:015-2788131 TU Delft, February 16, 2015 (slides modified from
More informationModule 03 Linear Systems Theory: Necessary Background
Module 03 Linear Systems Theory: Necessary Background Ahmad F. Taha EE 5243: Introduction to Cyber-Physical Systems Email: ahmad.taha@utsa.edu Webpage: http://engineering.utsa.edu/ taha/index.html September
More informationModule 9: State Feedback Control Design Lecture Note 1
Module 9: State Feedback Control Design Lecture Note 1 The design techniques described in the preceding lectures are based on the transfer function of a system. In this lecture we would discuss the state
More informationControl Systems I. Lecture 5: Transfer Functions. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control D-MAVT ETH Zürich
Control Systems I Lecture 5: Transfer Functions Readings: Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich October 20, 2017 E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/2017
More informationControl Systems I. Lecture 4: Diagonalization, Modal Analysis, Intro to Feedback. Readings: Emilio Frazzoli
Control Systems I Lecture 4: Diagonalization, Modal Analysis, Intro to Feedback Readings: Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich October 13, 2017 E. Frazzoli (ETH)
More informationKalman Decomposition B 2. z = T 1 x, where C = ( C. z + u (7) T 1, and. where B = T, and
Kalman Decomposition Controllable / uncontrollable decomposition Suppose that the controllability matrix C R n n of a system has rank n 1
More informationEE Control Systems LECTURE 9
Updated: Sunday, February, 999 EE - Control Systems LECTURE 9 Copyright FL Lewis 998 All rights reserved STABILITY OF LINEAR SYSTEMS We discuss the stability of input/output systems and of state-space
More informationẋ n = f n (x 1,...,x n,u 1,...,u m ) (5) y 1 = g 1 (x 1,...,x n,u 1,...,u m ) (6) y p = g p (x 1,...,x n,u 1,...,u m ) (7)
EEE582 Topical Outline A.A. Rodriguez Fall 2007 GWC 352, 965-3712 The following represents a detailed topical outline of the course. It attempts to highlight most of the key concepts to be covered and
More informationEE451/551: Digital Control. Chapter 3: Modeling of Digital Control Systems
EE451/551: Digital Control Chapter 3: Modeling of Digital Control Systems Common Digital Control Configurations AsnotedinCh1 commondigitalcontrolconfigurations As noted in Ch 1, common digital control
More informationLinear System Theory
Linear System Theory Wonhee Kim Chapter 6: Controllability & Observability Chapter 7: Minimal Realizations May 2, 217 1 / 31 Recap State space equation Linear Algebra Solutions of LTI and LTV system Stability
More informationState will have dimension 5. One possible choice is given by y and its derivatives up to y (4)
A Exercise State will have dimension 5. One possible choice is given by y and its derivatives up to y (4 x T (t [ y(t y ( (t y (2 (t y (3 (t y (4 (t ] T With this choice we obtain A B C [ ] D 2 3 4 To
More informationECE 388 Automatic Control
Controllability and State Feedback Control Associate Prof. Dr. of Mechatronics Engineeering Çankaya University Compulsory Course in Electronic and Communication Engineering Credits (2/2/3) Course Webpage:
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a linear-time-invariant
More informationDigital Control Systems
Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist
More information1 Continuous-time Systems
Observability Completely controllable systems can be restructured by means of state feedback to have many desirable properties. But what if the state is not available for feedback? What if only the output
More informationECEN 420 LINEAR CONTROL SYSTEMS. Lecture 2 Laplace Transform I 1/52
1/52 ECEN 420 LINEAR CONTROL SYSTEMS Lecture 2 Laplace Transform I Linear Time Invariant Systems A general LTI system may be described by the linear constant coefficient differential equation: a n d n
More informationLinear State Feedback Controller Design
Assignment For EE5101 - Linear Systems Sem I AY2010/2011 Linear State Feedback Controller Design Phang Swee King A0033585A Email: king@nus.edu.sg NGS/ECE Dept. Faculty of Engineering National University
More informationControl Systems Design
ELEC4410 Control Systems Design Lecture 13: Stability Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 13: Stability p.1/20 Outline Input-Output
More informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control D-MAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
More informationDistributed Real-Time Control Systems
Distributed Real-Time Control Systems Chapter 9 Discrete PID Control 1 Computer Control 2 Approximation of Continuous Time Controllers Design Strategy: Design a continuous time controller C c (s) and then
More informationEEE 184: Introduction to feedback systems
EEE 84: Introduction to feedback systems Summary 6 8 8 x 7 7 6 Level() 6 5 4 4 5 5 time(s) 4 6 8 Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step)
More informationCBE507 LECTURE III Controller Design Using State-space Methods. Professor Dae Ryook Yang
CBE507 LECTURE III Controller Design Using State-space Methods Professor Dae Ryook Yang Fall 2013 Dept. of Chemical and Biological Engineering Korea University Korea University III -1 Overview States What
More informationWhy do we need a model?
Why do we need a model? For simulation (study the system output for a given input) Ex. Thermal study of a space shuttle when it enters the atmosphere (étude thermique concernant la rentrée dans l atmosphère
More informationControl Systems. Laplace domain analysis
Control Systems Laplace domain analysis L. Lanari outline introduce the Laplace unilateral transform define its properties show its advantages in turning ODEs to algebraic equations define an Input/Output
More informationIdentification Methods for Structural Systems. Prof. Dr. Eleni Chatzi System Stability - 26 March, 2014
Prof. Dr. Eleni Chatzi System Stability - 26 March, 24 Fundamentals Overview System Stability Assume given a dynamic system with input u(t) and output x(t). The stability property of a dynamic system can
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design Method
.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response- Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to
More informationProblem Set 5 Solutions 1
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.245: MULTIVARIABLE CONTROL SYSTEMS by A. Megretski Problem Set 5 Solutions The problem set deals with Hankel
More informationSTABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable
ECE4510/5510: Feedback Control Systems. 5 1 STABILITY ANALYSIS 5.1: Bounded-input bounded-output (BIBO) stability Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated
More informationControl Design. Lecture 9: State Feedback and Observers. Two Classes of Control Problems. State Feedback: Problem Formulation
Lecture 9: State Feedback and s [IFAC PB Ch 9] State Feedback s Disturbance Estimation & Integral Action Control Design Many factors to consider, for example: Attenuation of load disturbances Reduction
More informationDIGITAL CONTROLLER DESIGN
ECE4540/5540: Digital Control Systems 5 DIGITAL CONTROLLER DESIGN 5.: Direct digital design: Steady-state accuracy We have spent quite a bit of time discussing digital hybrid system analysis, and some
More informationIt is common to think and write in time domain. creating the mathematical description of the. Continuous systems- using Laplace or s-
It is common to think and write in time domain quantities, but this is not the best thing to do in creating the mathematical description of the system we are dealing with. Continuous systems- using Laplace
More information6.241 Dynamic Systems and Control
6.241 Dynamic Systems and Control Lecture 12: I/O Stability Readings: DDV, Chapters 15, 16 Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology March 14, 2011 E. Frazzoli
More informationControl Systems. Frequency domain analysis. L. Lanari
Control Systems m i l e r p r a in r e v y n is o Frequency domain analysis L. Lanari outline introduce the Laplace unilateral transform define its properties show its advantages in turning ODEs to algebraic
More informationChapter 6: The Laplace Transform. Chih-Wei Liu
Chapter 6: The Laplace Transform Chih-Wei Liu Outline Introduction The Laplace Transform The Unilateral Laplace Transform Properties of the Unilateral Laplace Transform Inversion of the Unilateral Laplace
More informationLecture 3. Chapter 4: Elements of Linear System Theory. Eugenio Schuster. Mechanical Engineering and Mechanics Lehigh University.
Lecture 3 Chapter 4: Eugenio Schuster schuster@lehigh.edu Mechanical Engineering and Mechanics Lehigh University Lecture 3 p. 1/77 3.1 System Descriptions [4.1] Let f(u) be a liner operator, u 1 and u
More informationIntroduction to Modern Control MT 2016
CDT Autonomous and Intelligent Machines & Systems Introduction to Modern Control MT 2016 Alessandro Abate Lecture 2 First-order ordinary differential equations (ODE) Solution of a linear ODE Hints to nonlinear
More informationELEC2400 Signals & Systems
ELEC2400 Signals & Systems Chapter 7. Z-Transforms Brett Ninnes brett@newcastle.edu.au. School of Electrical Engineering and Computer Science The University of Newcastle Slides by Juan I. Yu (jiyue@ee.newcastle.edu.au
More informationCONTROL OF DIGITAL SYSTEMS
AUTOMATIC CONTROL AND SYSTEM THEORY CONTROL OF DIGITAL SYSTEMS Gianluca Palli Dipartimento di Ingegneria dell Energia Elettrica e dell Informazione (DEI) Università di Bologna Email: gianluca.palli@unibo.it
More informationLecture 9 Infinite Impulse Response Filters
Lecture 9 Infinite Impulse Response Filters Outline 9 Infinite Impulse Response Filters 9 First-Order Low-Pass Filter 93 IIR Filter Design 5 93 CT Butterworth filter design 5 93 Bilinear transform 7 9
More informationIntro. Computer Control Systems: F8
Intro. Computer Control Systems: F8 Properties of state-space descriptions and feedback Dave Zachariah Dept. Information Technology, Div. Systems and Control 1 / 22 dave.zachariah@it.uu.se F7: Quiz! 2
More informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review
Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the s-plane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
More informationSignals and Systems. Problem Set: The z-transform and DT Fourier Transform
Signals and Systems Problem Set: The z-transform and DT Fourier Transform Updated: October 9, 7 Problem Set Problem - Transfer functions in MATLAB A discrete-time, causal LTI system is described by the
More informationRichiami di Controlli Automatici
Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici
More information16.31 Fall 2005 Lecture Presentation Mon 31-Oct-05 ver 1.1
16.31 Fall 2005 Lecture Presentation Mon 31-Oct-05 ver 1.1 Charles P. Coleman October 31, 2005 1 / 40 : Controllability Tests Observability Tests LEARNING OUTCOMES: Perform controllability tests Perform
More informationChapter 3 Data Acquisition and Manipulation
1 Chapter 3 Data Acquisition and Manipulation In this chapter we introduce z transf orm, or the discrete Laplace Transform, to solve linear recursions. Section 3.1 z-transform Given a data stream x {x
More informationDiscrete-time Controllers
Schweizerische Gesellschaft für Automatik Association Suisse pour l Automatique Associazione Svizzera di Controllo Automatico Swiss Society for Automatic Control Advanced Control Discrete-time Controllers
More information7.2 Relationship between Z Transforms and Laplace Transforms
Chapter 7 Z Transforms 7.1 Introduction In continuous time, the linear systems we try to analyse and design have output responses y(t) that satisfy differential equations. In general, it is hard to solve
More informationChapter 5 Frequency Domain Analysis of Systems
Chapter 5 Frequency Domain Analysis of Systems CT, LTI Systems Consider the following CT LTI system: xt () ht () yt () Assumption: the impulse response h(t) is absolutely integrable, i.e., ht ( ) dt< (this
More informationME 234, Lyapunov and Riccati Problems. 1. This problem is to recall some facts and formulae you already know. e Aτ BB e A τ dτ
ME 234, Lyapunov and Riccati Problems. This problem is to recall some facts and formulae you already know. (a) Let A and B be matrices of appropriate dimension. Show that (A, B) is controllable if and
More informationEC Signals and Systems
UNIT I CLASSIFICATION OF SIGNALS AND SYSTEMS Continuous time signals (CT signals), discrete time signals (DT signals) Step, Ramp, Pulse, Impulse, Exponential 1. Define Unit Impulse Signal [M/J 1], [M/J
More information2.161 Signal Processing: Continuous and Discrete Fall 2008
MIT OpenCourseWare http://ocw.mit.edu 2.6 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS
More informationLecture 3 - Design of Digital Filters
Lecture 3 - Design of Digital Filters 3.1 Simple filters In the previous lecture we considered the polynomial fit as a case example of designing a smoothing filter. The approximation to an ideal LPF can
More informationLecture 1: Feedback Control Loop
Lecture : Feedback Control Loop Loop Transfer function The standard feedback control system structure is depicted in Figure. This represend(t) n(t) r(t) e(t) u(t) v(t) η(t) y(t) F (s) C(s) P (s) Figure
More informationIngegneria dell Automazione - Sistemi in Tempo Reale p.1/28
Ingegneria dell Automazione - Sistemi in Tempo Reale Selected topics on discrete-time and sampled-data systems Luigi Palopoli palopoli@sssup.it - Tel. 050/883444 Ingegneria dell Automazione - Sistemi in
More informationJoão P. Hespanha. January 16, 2009
LINEAR SYSTEMS THEORY João P. Hespanha January 16, 2009 Disclaimer: This is a draft and probably contains a few typos. Comments and information about typos are welcome. Please contact the author at hespanha@ece.ucsb.edu.
More informationNonlinear Control Systems
Nonlinear Control Systems António Pedro Aguiar pedro@isr.ist.utl.pt 7. Feedback Linearization IST-DEEC PhD Course http://users.isr.ist.utl.pt/%7epedro/ncs1/ 1 1 Feedback Linearization Given a nonlinear
More informationModule 08 Observability and State Estimator Design of Dynamical LTI Systems
Module 08 Observability and State Estimator Design of Dynamical LTI Systems Ahmad F. Taha EE 5143: Linear Systems and Control Email: ahmad.taha@utsa.edu Webpage: http://engineering.utsa.edu/ataha November
More informationDiscrete-Time David Johns and Ken Martin University of Toronto
Discrete-Time David Johns and Ken Martin University of Toronto (johns@eecg.toronto.edu) (martin@eecg.toronto.edu) University of Toronto 1 of 40 Overview of Some Signal Spectra x c () t st () x s () t xn
More informationProblem Set 3: Solution Due on Mon. 7 th Oct. in class. Fall 2013
EE 56: Digital Control Systems Problem Set 3: Solution Due on Mon 7 th Oct in class Fall 23 Problem For the causal LTI system described by the difference equation y k + 2 y k = x k, () (a) By first finding
More informationD(s) G(s) A control system design definition
R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure
More information6.241 Dynamic Systems and Control
6.241 Dynamic Systems and Control Lecture 24: H2 Synthesis Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology May 4, 2011 E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May
More informationSome solutions of the written exam of January 27th, 2014
TEORIA DEI SISTEMI Systems Theory) Prof. C. Manes, Prof. A. Germani Some solutions of the written exam of January 7th, 0 Problem. Consider a feedback control system with unit feedback gain, with the following
More informationTheory of Linear Systems Exercises. Luigi Palopoli and Daniele Fontanelli
Theory of Linear Systems Exercises Luigi Palopoli and Daniele Fontanelli Dipartimento di Ingegneria e Scienza dell Informazione Università di Trento Contents Chapter. Exercises on the Laplace Transform
More informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 2010-2011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 2010-2011
More informationH 2 Optimal State Feedback Control Synthesis. Raktim Bhattacharya Aerospace Engineering, Texas A&M University
H 2 Optimal State Feedback Control Synthesis Raktim Bhattacharya Aerospace Engineering, Texas A&M University Motivation Motivation w(t) u(t) G K y(t) z(t) w(t) are exogenous signals reference, process
More information