LTI system response Daniele Carnevale Dipartimento di Ing. Civile - - PowerPoint PPT Presentation

Laplace and Zeta transforms State space Forced response

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

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Laplace and Zeta transforms State space Forced response Differential equations Difference equations

Laplace and Zeta transforms

Recalling Laplace and Zeta transforms L{f(t)}(s) +∞

0−

f(t)e−st dt Z{f(k)}(z) +∞ f(k)z−k e−sT z (sampling T) L

• th

h! eat

(s) =

1 (s−a)h+1

Z k h

• ak−h
• (z) =

z (z−a)h+1

L{f(t − T)}(s) = F(s)e−sT Z{f(k − n)}(z) = F(z)z−n L{ ˙ f(t)}(s) = sF(s) − F(0−) Z{f(k + n)}(z) = zn F(z)−n−1

h=0 f(h)z−h

F(s) = c0 + ν

i=1

ni

j=1 ci,j (s−pi)j , F (z) z

= c0 + ν

i=1

ni

j=1 ci,j (z−pi)j ,

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Laplace and Zeta transforms State space Forced response Differential equations Difference equations

Laplace transforms and differential equations

The Laplace transform can be used to obtain the solution of a linear differential equation. Example: As first transform ¨ y(t) + 3 ˙ y(t) + 2y(t) = ˙ u(t) + 2u(t) by Laplace transform into s2y(s) − sy(0−) − ˙ y(0−) + 3sy(s) − 3y(0−) + 2y(s) = su(s) − u(0−) + 2u(s) y(s) = sy(0−) + 3y(0−) + ˙ y(0−) s2 + 3s + 2

• free response

+ −u(0−) s2 + 3s + 2 + s + 2 s2 + 3s + 2 u(s)

• forced response

(1) ⇓ (u(t) = tδ−1(t) → u(s) = 1/s2, u(0−) = 0) y(s) = c1,1 s + 2 + c2,1 s + 1 + c3,1 s + c3,2 s2 (2) y(t) =    c1,1e−2t + c2,1e−t

• Plant modes

+ c3,1 + c3,2t

• Input modes

+ ?

• ther modes?

    δ−1(t) (3)

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Laplace and Zeta transforms State space Forced response Differential equations Difference equations

Zeta transforms and difference equations

The Zeta transform can be used to obtain the solution of a linear difference equation. Example: As first transform y(k + 2) + 3y(k + 1) + 2y(k) = u(k + 1) + 2u(k) by Zeta transform into z2y(z) − z2y(0) − zy(1) + 3zy(z) − 3zy(0) + 2y(z) = zu(z) − zu(0) + 2u(z) y(z) = z2y(0) + 3zy(0) + zy(1) z2 + 3z + 2

• free response

+ −zu(0) z2 + 3z + 2 + z + 2 z2 + 3z + 2 u(z)

• forced response

(4) ⇓ (u(k) = kδ−1(k) =→ u(z) = z/(z − 1)2, u(0) = 0) y(z) z = c1,1 z + 2 + c2,1 z + 1 + c3,1 z − 1 + c3,2 (z − 1)2 + c4,1 z (5) y(k) =    c1,1(−2)k + c2,1(−1)k

• Plant modes

+ c3,1 + c3,2k

• Input modes

+c4,1δ0(k − 1) ?

• ther modes?

    δ−1(t) (6)

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time System modes

Laplace transform for linear continuous time systems

Consider the system in state space form obtained via Laplace transform as

• ˙

x(t) = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t),

L(·)(s)

• →

sx(s) − x0 = Ax(s) + Bu(s), y(s) = Cx(s) + Du(s), (7) that through sx(s) − Ax(s) = x0 + Bu(s), (sI − A)x(s) = x0 + Bu(s), (sI − A)−1x(s) = (sI − A)−1(x0 + Bu(s)), [(sI − A) is invertible ] x(s) = (sI − A)−1x0 + (sI − A)−1Bu(s), (8) allows to write x(s) = Φ(s)x0 + H(s)u(s), y(s) = Ψ(s)x0 + W(s)u(s), ,        Φ(s) = (sI − A)−1, H(s) = (sI − A)−1B, Ψ(s) = C(sI − A)−1, W(s) = C(sI − A)−1B + D. (9)

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time System modes

State and output response in Laplace domain

x(s) = xl(s) + xf(s), y(s) = yl(s) + yf(s), ,        xl(s) = (sI − A)−1x0 = Φ(s)x0, xf(s) = (sI − A)−1Bu(s) = H(s)u(s), yl(s) = C(sI − A)−1x0 = Ψ(s)x0, yf(s) = (C(sI − A)−1B + D)u(s) = W(s)u(s), (10) where xl(s) is the free state response in s, xf(s) is the forced state response in s, yl(s) is the free output response in s and yf(s) is the forced output-response in s. The rational function W(s) is called the system transfer function and is defined as y(s) u(s) = W(s) C(sI − A)−1B + D when y(i)(0−) = 0 ⇐ ⇒ x0 = 0. (11) Since the Lagrange solution for continuous-time LTI systems is given by (prove it) x(t) = eA(t−t0)x0 + t

t0

eA(t−τ)Bu(τ) dτ (12) and x(t) = L−1 (sI − A)−1x0 + (sI − A)−1Bu(s)

• (t),

(13) then...

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time System modes

Continuous time state response: free evolution

... with t0 = 0 it holds true that L−1{(sI − A)−1}(t) = L−1{Φ(s)}(t) = eAt, (14) L−1{(sI − A)−1Bu(s)}(t) = L−1{H(s)u(s)}(t) = t eA(t−τ)Bu(τ) dτ, (15) in accordance also with the Laplace convolution theorem. Then, by mean of the residual technique iterated for each component of the state x and the matrix elements, it is possible to write the Laplace state transform of the free state response as xl(s) = (sI − A)−1x0 =

ν

• i=1

mi

• h=1

1 (s − λi)h Ri,h x0, (16) where Ri,h is the generalized residual matrix of (sI − A)−1 and λi ∈ σ{A} are the eigenvalues of A. Then by inverse Laplace transform it is possible to obtain xl(t) = δ−1(t)

ν

• i=1

mi

• h=1

th−1 (h − 1)! eλit

• system modes

Ri,h x0. (17)

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time System modes

Discrete time state response: free evolution

Consider the discrete time system and the Z-transform as x(k + 1) = Ax(k) + Bu(k), y(k) = Cx(k) + Du(k),

Z(·)(s)

• →

zx(z) − zx0 = Ax(z) + Bu(z), y(z) = Cx(z) + Du(z), (18) that through zx(z) − Ax(z) = zx0 + Bu(z), (zI − A)x(z) = zx0 + Bu(z), (zI − A)−1x(z) = (zI − A)−1(zx0 + Bu(z)), [(zI − A) is invertible ] x(z) = z(zI − A)−1x0 + (zI − A)−1Bu(z), (19) allows to write x(z) = Φ(z)x0 + H(z)u(z), y(z) = Ψ(z)x0 + W(z)u(z), ,        Φ(z) = z(zI − A)−1, H(z) = (zI − A)−1B, Ψ(z) = zC(zI − A)−1, W(s) = C(zI − A)−1B + D. (20)

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time System modes

State and output response in the Z domain

x(z) = xl(z) + xf(z), y(z) = yl(z) + yf(z), ,        xl(z) = z(zI − A)−1x0 = Φ(z)x0, xf(z) = (zI − A)−1Bu(z) = H(z)u(z), yl(z) = zC(zI − A)−1x0 = Ψ(z)x0, yf(z) = (C(zI − A)−1B + D)u(z) = W(z)u(z), (21) where xl(z) is the free state response in z, xf(z) is the forced state response in z, yl(z) is the free output response in z and yf(z) is the forced output-response in z. The rational function W(z) is called the system transfer function and is defined as y(z) u(z) = W(z) C(zI − A)−1B + D when y(i) = 0 ⇐ ⇒ x0 = 0. (22) Since the Lagrange solution for discrete time LTI systems is given by (prove it) x(k) = Akx0 +

k−1

• h=0

Ak−h−1Bu(h), (23) and x(k) = Z−1 z(zI − A)−1x0 + (zI − A)−1Bu(z)

• (k),

(24) then...

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time System modes

Discrete time state response: free evolution

...then Z−1{z(zI − A)−1}(k) = Z−1{Φ(z)}(k) = Ak, (25) Z−1{(zI − A)−1Bu(z)}(k) = Z−1{H(z)u(z)}(k) =

k−1

• h=0

Ak−h−1Bu(h), (26) in accordance also with the Zeta transform convolution theorem. Then, by mean of the residual technique iterated for each component of the state x and the matrix elements, it is possible to write the Laplace state transform of the free state response as xl(z) = z(zI − A)−1x0 =

ν

• i=1

mi

• h=1

z (z − λi)h Ri,h x0, (27) where Ri,h is the generalized residual matrix of (zI − A)−1 and λi ∈ σ{A} are the eigenvalues of A. Then by inverse Zeta transform it is possible to obtain xl(z) = δ−1(k)

ν

• i=1

mi

• h=1
• k

h − 1

• λk−h+1

i

• system modes

Ri,h x0. (28)

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time System modes

System (natural) modes

The continuous time system modes th−1 (h − 1)! eλit are converging to zero iff Re{λi} < 0 (for any h ∈ N), non converging/diverging iff Re{λi} = 0 and h = 1, diverging if Re{λi} > 0 or Re{λi} = 0 and h > 1. The discrete time system modes

• k

h − 1

• λk−h+1

i

are converging to zero iff |λi| < 1 (for any h ∈ N), non converging/diverging iff |λi| = 1 and h = 1, diverging if |λi| > 1 or |λi| = 1 and h > 1.

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time

Input modes and forced response

Consider an input as u(t) =

r

• i=1

ri

• h=1

βi th−1 (h − 1)! eαit, (29) whose Laplace transform is u(s) =

r

• i=1

ri

• h=1

βi 1 (s − αi)h , (30) where αi ∈ C and βi ∈ C. This yields that the forced state response xf(t) in the Laplace domain writes as xf(s) = (sI − A)−1Bu(s) = H(s)u(s) =

ν

• i=1

mi

• h=1

ξi,h (s − λi)h +

r

• i=1

ri

• h=1

χi,h (s − αi)h , (31) where (ξi,h, χi,h) ∈ C are generalized residuals. If αi = λj for all i and j, then mi = mi and rj = rj, whereas if some i-th mode of the input is the same of the j-th mode of system, then we can let mi = 0 and rj = rj + 1 (resonant effect).

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time

Input modes and forced response cont’d

The forced state response is obtained by inverse Laplace transform of (31) as xf(t) = L−1{(sI − A)−1Bu(s)}(t), =

ν

• i=1

mi

• h=1

ξi,h th−1 (h − 1)! eλi t +

r

• i=1

ri

• h=1

χi,h th−1 (h − 1)! eαi t (32) It is possible to define the permanent state response xp(t) as xp(t) =

r

• i=1

ri

• h=1

χi,h th−1 (h − 1)! eαi t, (33) under the assumption that all the system modes are converging (Re{λi} < 0) and the input are nor converging nor diverging. Then it is possible define the transient state response xt(t) as xt(t) = x(t) − xp(t) =

ν

• i=1

mi

• h=1

th−1 (h − 1)! eλi t(Ri,hx0 + ξi,h). (34)

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time

Input modes and forced response

Consider an input as u(k) =

r

• i=1

ri

• h=1

βi

• k

h − 1

• λk−h+1

i

, (35) whose Zeta transform is u(z) =

r

• i=1

ri

• h=1

βi z (z − αi)h , (36) where αi ∈ C and βi ∈ C. This yields that the forced state response xf(k) in the Zeta domain writes as xf(k) = (zI − A)−1Bu(s) = H(z)u(z) =

ν

• i=1

mi

• h=1

ξi,hz (z − λi)h +

r

• i=1

ri

• h=1

χi,hz (z − αi)h , (37) where (ξi,h, χi,h) ∈ C are generalized residuals. If αi = λj for all i and j, then mi = mi and rj = rj, whereas if some i-th mode of the input is the same of the j-th mode of system, then we can let mi = 0 and rj = rj + 1 (resonant effect).

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Laplace and Zeta transforms State space Forced response Continuous time Discrete time

Input modes and forced response cont’d

The forced state response is obtained by inverse Zeta transform of (37) as xf(k) = Z−1{(zI − A)−1Bu(z)}(k), =

ν

• i=1

mi

• h=1

ξi,h

• k

h − 1

• λk−h+1

i

+

r

• i=1

ri

• h=1

χi,h

• k

h − 1

• αk−h+1

i

. (38) It is possible to define the permanent state response xp(t) as xp(t) =

r

• i=1

ri

• h=1

χi,h

• k

h − 1

• αk−h+1

i

, (39) under the assumption that all the system modes are converging (Re{λi} < 0) and the input are nor converging nor diverging. Then it is possible define the transient state response xt(k) as xt(k) = x(k) − xp(k) =

ν

• i=1

mi

• h=1
• k

h − 1

• λk−h+1

i

(Ri,hx0 + ξi,h). (40)

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