Control Systemslanari/ControlSystems/CS... · Partial fraction expansion (Heaviside) for rational...

Control Systems

System responseL. Lanari

prelim

inary

versio

n

Monday, November 3, 2014

Lanari: CS - System response 2

Outline

What we are going to see:

• how to compute in the s-domain the forced response (zero-state response) using the transfer function

• how to inverse transform the resulting responses so to obtain the time response

• analyze the relationship between eigenvalues and poles in deeper details



Zero State Response (forced response)

see table of common Laplace transforms

(rational functions)

(rational function).(rational function) = rational function

YZS(s) = W (s)U(s)

yZS(t)

YZS(s)

Basic idea: write the function, we want to find the inverse Laplace transform of, as a linear combination of “easy to transform” terms and then use the linearity property of the inverse transformation

Partial fraction expansion (Heaviside)for rational functions

For rational functions: represent a complicated fraction as the sum of simpler fractions for which we know the inverse Laplace transform (partial fraction expansion or decomposition)



Partial fraction expansion

Let

be a strictly proper rational function with coprime N(s) and D(s) and with distinct roots of D(s)

F (s) =N(s)

D(s)

D(s) = an

n�

i=1

(s− pi) ⇒ F (s) =N(s)

an�n

i=1(s− pi)

F (s) =n�

i=1

Ris− pi

Ri = [(s− pi)F (s)]��s=pi

i.e.

then F(s) can be expanded as

with the residues Ri computed as

(distinct roots case)



Partial fraction expansionexample: given the transfer function W(s) of a system S, find its impulse response w(t)

W (s) =s+ 2

s(s+ 1)(s+ 10)




Let

be a strictly proper rational function with coprime N(s) and D(s). Let D(s) have m roots each with multiplicity ni that is

F (s) =N(s)

D(s)

m�

i=1

ni = n

then F(s) can be expanded as

with the residues Rij computed as

(general case)

F (s) =m�

i=1

ni�

j=1

Rij(s− pi)j

Rij =

�1

(ni − j)!dni−j

dsni−j{(s− pi)niF (s)}

�

s=pi



Partial fraction expansionexample: find the zero-state output response (or output forced response) of the system characterized by the transfer function W(s) to the input u(t) = t (remember the function u(t) is assumed to be zero for t < 0)

W (s) =s− 1

(s+ 1)(s+ 10)

U(s) = L[u(t)] = 1s2

⇒ Y (s) = W (s)U(s) = s− 1(s+ 1)(s+ 10)

1

s2

Y (s) =R11s

+R12s2

+R2s+ 1

+R3

s+ 10

y(t) = L−1(Y (s)) =�R11 +R12t+R2e

−t +R3e−10t� δ−1(t)

R11 =

�1

(2− 1)!d2−1

ds2−1�s2Y (s)

��

s=0

=

�d

ds

�s− 1

(s+ 1)(s+ 10)

��

s=0

=21

100

R12 =

�1

(2− 2)!d2−2

ds2−2�s2Y (s)

��

s=0

=

�s− 1

(s+ 1)(s+ 10)

�

s=0

= − 110

R2 = [(s+ 1)Y (s)]s=−1 =

�s− 1

s2(s+ 10)

�

s=−1= −2

9

R3 = [(s+ 10)Y (s)]s=−10 =

�s− 1

s2(s+ 1)

�

s=−10=

11

900




F (s) =N(s)

D(s)=

N(s)

skD�(s)

F (s) =R11s

+R12s2

+ · · ·+ R1ksk

+�� Rij

(s− pi)j

R1k =

�1

(k − k)!dk−k

dsk−k�skF (s)

�� s=0

=�skF (s)

� ��s=0

=N(0)

D�(0)

special case: F(s) has k poles in s = 0

then

with

(this result will be useful for the steady-state response to an order k input )

leadingcoefficient

(generic rational function)

tk

k!




p = α+ jβ

p∗ = α− jβ

R

(s− p)k +R∗

(s− p∗)k =As+B

((s− α)2 + β2)k

for complex poles R = a+ jb

R∗

residue�




R

s− p +R∗

s− p∗ =R(s− p∗) +R∗(s− p)

(s− α)2 + β2 =s(R+R∗)− (Rp∗ +R∗p)

(s− α)2 + β2

=s(R+R∗)− α(R+R∗)− jβ(R∗ −R)

(s− α)2 + β2

R+R∗ = 2a, R∗ −R = −2jb

R

s− p +R∗

s− p∗ =2as− 2(aα+ bβ)(s− α)2 + β2 =

As+B

(s− α)2 + β2

A = 2a B = −2(aα+ bβ)

for k = 1

with As+B = A(s− α) + β(Aα+B)/β

R

s− p +R∗

s− p∗ =A(s− α)

(s− α)2 + β2 +β(Aα+B)/β

(s− α)2 + β2

eαt [A cosβt+ (Aα+B)/β sinβt]in t




H(s) =1

(s2 + 1)(s− 2)2 H(s) =R11

s− 2 +R12

(s− 2)2 +R2

s− j +R3

s+ j

H(s) =R11

s− 2 +R12

(s− 2)2 +As+B

s2 + 1, with A =

4

25, B = − 6

50

h(t) =�R11e

2t +R12te2t +A cos t+B sin t

�δ−1(t)

R11 =

�d

ds

�1

(s2 + 1)

��

s=2

= − 425

R12 =

�1

(s2 + 1)

�

s=2

=1

5

R2 = [(s− j)H(s)]s=j =�

1

(s+ j)(s− 2)2

�

s=j

=1

8 + 6j=

2

25− j 3

50

R3 = [(s+ j)H(s)]s=−j =

�1

(s− j)(s− 2)2

�

s=−j=

1

8 + 6j=

2

25+ j

3

50= R∗2



W (s) =N(s)

D(s)

Poles & eigenvalues

W (s) =1

det(sI −A)N�(s)

from def?

if det(sI −A) and N �(s) coprime {poles} = {eigenvalues}

characteristicpolynomial

if det(sI −A) and N �(s) not coprime {poles} subset of {eigenvalues}

Input/Outputrepresentation

Input/State/Outputrepresentation

vs

here visible

Transferfunction

Statespace

�

we need to understand when & why this happens(so to understand when we can consider the transfer function equivalent to a state space representation)



we have a “hidden mode” associated to the eigenvalue ¸j(see structural properties)

Poles & eigenvalues (distinct eigenvalues of A case)

n = state space dimension = dimension of A = number of eigenvaluesnp = number of poles in W(s)

W (s) =N(s)

D(s)=

N(s)�npi=1(s− pi)

=

np�

i=1

Ris− pi

if the eigenvalue ¸j does not appear as a poleand/or

vTj B = 0

Cuj = 0

W (s) = L[w(t)] = L[CeAtB] = L[C

n�

j=1

eλjtujvTj

B] =n�

j=1

Cuj vTj B

s− λj

done the same analysis in t for n = 2

NB distinct eigenvalues is different from diagonalizable




eAtB = H(t)

CeAt = Ψ(t)

vTj B = 0

Cuj = 0

implies the corresponding mode will not appear in the state impulsive response

the corresponding mode (or eigenvalue) is said to be uncontrollable

implies the corresponding mode will not appear in the output transition matrix

the corresponding mode (or eigenvalue) is said to be unobservable

If for an eigenvalue ¸j we have that




Theorem

Every pole is an eigenvalue.

An eigenvalue ¸i becomes a pole if and only if it is both controllable and observable

or equivalently the following two PBH rank tests are both verified

vTi B �= 0 Cui �= 0

rank�A− λiI B

�= n rank

�A− λiI

C

�= n

Popov-Belevitch-Hautuscontrollability test

Popov-Belevitch-Hautusobservability test

and

(the PBH test could be tested for a generic ¸ but matrix A - ¸I loses rank only for ¸=¸i )




Where does the PBH test comes from?

Observability (sketch):

rank

�A− λiI

C

�< n

means that the rectangular (n+1) x n matrix has not full column rank and therefore it has a non-zero nullspace, that is there exists a n vector ui such that

�A− λiI

C

�ui = 0

(A− λiI)ui = 0 Aui = λiui

Cui = 0 Cui = 0

� �

that is there exists an eigenvector which belongs to the nullspace of C (or the corresponding mode is unobservable)



example

A =

�−1 10 1

�B =

�10

�C =

�0 1

�

(sI −A)−1 =� 1

s+11

(s+1)(s−1)0 1s−1

�= M1

1

s+ 1+M2

1

s− 1

M1 =�(s+ 1)(sI −A)−1

�s=−1 =

�1 −1/20 0

�

M2 =�(s− 1)(sI −A)−1

�s=1

=

�0 1/20 1

�

fractional decompositionworks also for rational matrices

with

eAt = M1e−t +M2e

ta different way to compute the matrix exponential

both natural modes appear (as it should be) in the state transition matrix



(sI −A)−1B =�

1s+10

�

C(sI −A)−1 =�0 1s−1

�

u1 / (A− λ1I)u1 = 0�0 10 2

�u1 = 0 u1 =

�10

�

u2 / (A− λ2I)u2 = 0�−2 10 0

�u2 = 0 u2 =

�12

�vT1 =

�1 −1/2

�

vT2 =�0 1/2

�

Cu1 = 0 Cu2 �= 0vT1 B �= 0 vT2 B = 0

mode corresponding to ¸2 does not appear

mode corresponding to ¸1 does not appear

equivalently

eAtB

w(t) = CeAtB = 0

CeAt

forced response will always be zero independently from the input applied

(look at the 2 first order ODE)

example

W (s) = 0

no poles



example

equivalently with the PBH rank test

rank

�0 1 10 2 0

�= 2 = n

rank

�−2 1 10 0 0

�= 1 < n

rank

0 10 2

0 1

= 1 < n rank

−2 10 0

0 1

= 2 = n

controllability test

observability test

mode corresponding to ¸2 is uncontrollable

mode corresponding to ¸1 is controllable

mode corresponding to ¸1is unobservable

mode corresponding to ¸2is observable



Poles & eigenvalues (general case)

Theorem

Every pole is an eigenvalue.

An eigenvalue ¸i becomes a pole with the multiplicity ma (algebraic multiplicity) if and only if

both PBH rank tests are verified

rank�A− λiI B

�= n rank

�A− λiI

C

�= n

NB - If one of the two conditions is not verified then the eigenvalue ¸i will appear as a pole with multiplicity strictly less than the algebraic multiplicity, possibly even 0 (in this case we will have a hidden eigenvalue). In particular the eigenvalue will appear at most as a pole with multiplicity equal to its index (dimension of the largest Jordan block).

controllability observability



example

A0 =

λ1 1 00 λ1 10 0 λ1

(sI −A0)−1 =1

(s− λ1)3

(s− λ1)2 (s− λ1) 1

0 (s− λ1)2 (s− λ1)0 0 (s− λ1)2

B =

∀∀�= 0

C =��= 0 ∀ ∀

�

B1 =

001

C1 =�1 0 0

�F1(s) =

1

(s− λ1)3

B2 =

100

C2 = C1 =�1 0 0

�F2(s) =

(s− λ1)2

(s− λ1)3=

1

s− λ1

B3 = B2 =

100

C3 =�0 0 1

�F3(s) = 0

PBH rank testverified for

easily seen from A0 − λ1I =

0 1 00 0 10 0 0



A4 =

λ1 1 00 λ1 00 0 λ1

A4 − λ1I =

0 1 00 0 00 0 0

(sI−A4)−1 =1

(s− λ1)3

(s− λ1)2 (s− λ1) 0

0 (s− λ1)2 00 0 (s− λ1)2

= 1(s− λ1)2

(s− λ1) 1 0

0 (s− λ1) 00 0 (s− λ1)

B4 =

010

C4 =�1 0 0

�F4(s) =

1

(s− λ1)2

B5 =

100

C5 = C1 =�1 0 0

�

B6 = B1 =

001

C6 = C1 =�1 0 0

�F6(s) = 0

F5(s) =s− λ1

(s− λ1)2=

1

s− λ1

example

since

the PBH rank test will never be satisfied independently from B and C. At most the eigenvalue will appear as a pole with multiplicity = index of ¸1 = 2



example

A7 =

λ1 0 00 λ1 00 0 λ1

(sI −A7)−1 =1

s− λ1

1 0 00 1 00 0 1

etc ...

the PBH rank test will never be satisfied independently from B and C. At most the eigenvalue will appear as a pole with multiplicity = index of ¸1 = 1


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