Transform-domain response (CT)

Systems and Control Theory Lecture Notes

Laura Giarr´ e

Lesson 7: Response of LTI systems in the transform domain

 Laplace Transform

 Transform-domain response (CT)

 Transfer function

 Zeta Transform

 Transform-domain response (DT)

 Discrete Transfer function

Response of LTI systems - summary

 The solution of initial value problems (the response) for LTI systems can be written in the form:

 DT:

y (k) = CA ^k x (0) + C

k−1 

i=0

(A ^k−i−1 Bu (i)) + Du(k)

 CT:

y (t) = C exp(At)x(0) + C

 _t

o

exp (A(t − τ))Bu(τ)dτ + Du(t).

Response of LTI systems - summary

 The convolution integral (CT) and sum (DT) equation are hard to interpret, and do not offer much insight.

 In order to gain a better understanding, we will study the response to elementary inputs of a form that is particularly easy to analyze:



the output has the same form as the input.



very rich and descriptive: most signals/sequences can be written as linear combinations of such inputs.

 Then, using the superposition principle, we will recover the

response to general inputs, written as linear combinations of

the easy inputs.

The continuous-time case: elementary inputs

 Let us choose as elementary input u (t) = u ₀ e ^st , s ∈ C.

 If s is real, then u is a simple exponential.

 If s = jω is imaginary, then the elementary input must always be accompanied by the conjugate, i.e.,

u (t) + u ^∗ (t) = u 0 e ^jωt + u 0 e ^−jωt = 2u 0 cos (ωt)

 if s is imaginary, then u (t) = e ^st must be understood as a half of a sinusoidal signal with exponentially-changing amplitude

 if s = σ + jω:

u (t) + u ^∗ (t) =u ₀ (e ^σt e ^jωt + u ₀ e ^?t e ^−jωt )

= u ₀ (e ^σt (e ^jωt + e ^−jωt )) = 2u ₀ e ^σt cos (ωt)

Output response to elementary inputs 1

 Recall that

y (t) = Ce ^(At) x (0) + C

 _t

o

e ^{A(t−τ )} Bu (τ)dτ + Du(t).

 Plug in u (t) = u ₀ e ^st

y (t) =Ce ^At x (0) + C

 _t

o

e ^{A(t−τ )} Bu ₀ e ^sτ d τ + Du 0 e ^st

=Ce ^At x (0) + C(

 _t

o

e ^(sI−A)τ d τ)e ^At Bu ₀ + Du ₀ e ^st

 If (sI − A) is invertible (i.e., s is not an eigenvalue of A), then

y (t) = Ce ^At x (0) + C(sI − A) ⁻¹ e ^(sI−A)t − Ie ^At Bu ₀ + Du ₀ e ^st .

Output response to elementary inputs 2

 Rearranging:

y (t) = Ce ^At x (0) − C(sI − A) ⁻¹ e ^At Bu ₀

  

Transient response

+ [C(sI − A) ⁻¹ B + D]u ₀ e ^st

  

Steady-state response

 If the system is asymptotically stable, e ^At → 0, and the transient response will converge to zero.

 The steady state response can be written as:

y _ss = G(s)e ^st

Output response to elementary inputs 2

 G (s) ∈ C ^p×m

 G (s) = C(sI − A) ⁻¹ B + D is a complex matrix.

 The function G : s → G(s) is also known as the transfer

function: it describes how the system transforms an input into

the output.

Laplace Transform

 The (one-sided) Laplace transform F : C → C of a sequence f : R → R is defined as

L[f (t)] = F (s) =

 _+∞

0

f (t)e ^−st dt

for all s such that the series converges (region of convergence).

 Given the above definition, and recalling Laplace Transform properties, the steady state response is:

Y _ss (s) = G(s)U(s).

 G (s) is also the Laplace transform of the impulse response.

Transform- Domain solution (CT) 1

 Starting from the state-space model, applying the Laplace transform, recalling that if g (t) = ^{df (t)} _dt , then

G (s) = sF (s) − f (0 ₋ ), we get

sX (s) − x(0−) =AX(s) + BU(s) Y (s) =CX(s) + DU(s)

 This is solved to yield

X (s) =(sI − A) ⁻¹ x (0 ₋ ) + (sI − A) ⁻¹ BU (s)

Y (s) =C(sI − A) ⁻¹ x (0 − ) + (C(sI − A)   ⁻¹ B + D)  transfer function

U (s)

Transform- Domain solution (CT) 2

 Note that (from properties of the State Transition matrix and Laplace transform properties)

L[Φ(t)] = L[e ^At ] = (sI − A) ⁻¹ = Φ(s)

X (s) =Φ(s)x(0−) + Φ(s)BU(s)

Y (s) =CΦ(s)x(0−) + (CΦ(s)B + D)    transfer function

U (s)

The discrete-time case: elementary inputs

 Let us choose as elementary input u (k) = u ₀ z ^k , z ∈ C.

 If z is real, then u is a simple geometric sequence.

 Recalling

y (k) = CA ^k x (0) + C

k−1 

i=0

(A ^k−i−1 Bu (i)) + Du(k)

 Plug in u [k] = u ₀ z ^k , and substitute l = k − i − 1:

y (k) =CA ^k x (0) + C

k−1 

i=0

(A ^l Bu ₀ z ^k−l−1 + Du 0 z ^l

=CA ^k x (0) + Cz ^k−1 (

k−1 

(Az − 1) ⁱ )Bu + Du z ^k

Matrix geometric series

 Recall the formula for the sum of a geometric series:

k−1 

i=0

m ⁱ = 1 − m ^k 1 − m For a matrix:  _k−1

i=0 M ⁱ = I + M + M ² + ...M ^k−1

 k−1 i=0

M ⁱ (I − M) = (I + M + M ² + ...M ^k−1 )(I − M) = I − M ^k i.e.,  _k−1

i=0 M ⁱ = (I − M ^k )(I − M) ⁻¹

Discrete Transfer Function

 Using the result in the previous slide, we get

y (k) =CA ^k x (0) + Cz ^k−1 (I − A ^k z ^−k )(I − Az ⁻¹ ) ^?1 Bu ₀ + Du ₀ z ^k

=CA ^k x (0) + C(z ^k I − A ^k )(zI − A) ^?1 Bu ₀ + Du 0 z ^k

Rearranging:

y (k) = CA ^k x (0) − (zI − A) ⁻¹ Bu ₀

  

Transient response

+ C(zI − A) ⁻¹ B + Du 0 z ^k

  

Steady-state Response

 If the system is asymptotically stable, the transient response

will converge to zero.

Discrete Transfer Function

 G (z) ∈ C

 G (z) = C(zI − A) ⁻¹ B + D is a complex number.

 The function G : z → G(z) is also known as the (discrete)

transfer function: it describes how the system transforms an

input into the output.

Z-Transform

 The (one-sided) z-transform F : C → C of a sequence f : N _ → R is defined as

Z[f ] = F (z) =

 +∞

k=0

f [k]z ^?k

for all z such that the series converges (region of convergence).

 Given the above dfinition, and the previous discussion,

Y _ss (z) = G(z)U(z)

 G (z) is the z-transform of the impulse response, i.e., the

response to the sequence u = (1, 0, 0, ...).

Transform Domain Solution (DT)

 Starting from the state-space model, applying the Zeta transform, recalling that if g (k) =

f (k − 1) k ≥ 1

0 k = 0 , then

G (z) = z ⁻¹ F (z), and that if g(k) = f (k + 1), then G (z) = z[F (z) − f (0)] we get

zX (z) − zx(0) =AX(z) + BU(z) Y (z) =CX(z) + DU(z)

 This is solved to yield

X (z) =z(zI − A) ⁻¹ x (0) + (zI − A) ⁻¹ BU (z)

Y (z) =Cz(zI − A) ⁻¹ x (0) + (C(zI − A)   ⁻¹ B + D)  transfer function

U (z)

Transform Domain Solution (DT)

 Note that (from properties of the State Transition matrix and Laplace transform properties)

Φ(z)Z[Φ(k)] = Z[A ^k ] = z(zI − A) ⁻¹

X (z) =zΦ(z)x(0) + Φ(z)BU(z)

Y (z) =zΦ(z)x(0) + (CΦ(z)B + D)    transfer function

U (z)

Thanks

DIEF- laura.giarre@unimore.it