Control Systems (ECE411) Lectures 11 & 12 11... · 2016. 10. 17. · State-Space Representation...

State-Space Representation Analysis of Control Systems

Control Systems (ECE411)Lectures 11 & 12

M.R. Azimi, Professor

Department of Electrical and Computer EngineeringColorado State University

Fall 2016

M.R. Azimi Control Systems


State-Space Representation: Solution

Remark: For a diagonal (parallel) state space formulation where

A =

λ1 0. . .

0 λn

,

the state transition matrix can easily be formed using

φ(t) =

eλ1t 0. . .

0 eλnt

This is because φ(t) = L −1{(sI −A)−1} =

1

s−λ1 0. . .

0 1s−λn

This result can directly be used without any proofs for diagonal state spacerepresentations. An example is given next.




Example for diagonal case: Given an LTI system:

x(t) =

[−1 00 −2

]x(t) +

[11

]u(t) x(0) =

[10

]y(t) =

[3 1

]u(t) = us(t)

Find (a) φ(t), (b) h(t), (c) yh(t), and (d) yp(t)

Part a: From the Remark above φ(t) =

[e−t 00 e−2t

]Part b: For h(t), we use

h(t) = Cφ(t)B =[3 1

] [e−t 00 e−2t

] [11

]h(t) = (3e−t + e−2t), ∀t ≥ 0

Part c: For yh(t), we use

yh(t) = cφ(t)x(0) =[3 1

] [e−t 00 e−2t

] [11

]yh(t) = 3e−t, ∀t ≥ 0 M.R. Azimi Control Systems



Part d: For yp(t) for a unit step input, we use

yp(t) = L −1{H(s)U(s)}

H(s) = C(sI −A)−1B =[3 1

] [ 1s+1 0

0 1s+2

] [11

]=[

3s+1

1s+2

] [11

]= 3

s+1 + 1s+2 = 4s+7

(s+1)(s+2)

Yp(s) = H(s)U(s) = 4s+7s(s+1)(s+2)

yp(t) = 72 − 3e−t − 1

2e−2t ∀t ≥ 0



Characteristic Equation and Eigenvalues of Matrix A

Recall, for an LTI system described by nth order differential equation,

n∑i=0

aidiy(t)

dti=

m∑j=0

bjdjx(t)

dtj, a0 = 1, m < n

transfer function is

H(s) =bms

m + bm−1sm−1 + · · ·+ b0

sn + an−1sn−1 + · · ·+ a0=B(s)

ρ(s)

Characteristic Equation: ρ(s) = sn + an−1 + · · ·+ a0 = 0That is roots of ρ(s) are poles of H(s).

On the other hand, for the same LTI represented by state-space equations weknow that (d = 0 assume strictly proper case),

H(s) = C(sI −A)−1B = C[Adj(sI−A)]Bdet(sI−A)

which implies that Characteristic equation: ρ(s) = det(sI −A) = |sI −A|.But roots of |sI −A| = 0 are the eigenvalues of matrix A.Thus, eigenvalues of A are poles of H(s).



Characteristic Equation and Eigenvalues of Matrix A-Cont.

Remark:

For PVCF, the coefficients of ρ(s) are elements of last row of A with negativesign:

A =

0 1 0. . .

. . .

0 0 1−a0 −a1 · · · −an

,

ρ(s) = sn + an−1sn−1 + · · ·+ a0 = 0

Additionally, if d = 0 i.e. strictly proper case the transfer function can easily beformed using

H(s) = C

1s...

sn−1

1

ρ(s)=

C︷︸︸︷[b0 · · · bn−1]

1s...

sn−1

1

ρ(s)=bn−1s

n−1 + · · ·+ b0ρ(s)

i.e without using the direct approach H(s) = C(sI −A)−1B.M.R. Azimi Control Systems


Characteristic Equation and Eigenvalues of Matrix A-Cont.

Let λi be an eigenvalue of matrix A (i.e. roots of |λI −A| = 0), then thecorresponding eigenvectors are obtained using λipi = Ap

i, where p

iis the

eigenvector associated with λi.

λipi = Ai

That is, matrix P = [p1, . . . , p

n] formed of all these eigenvectors diagonalizes

matrix A,

PΛ = AP =⇒ Λ = P−1AP

where Λ is a diagonal matrix containing all the eigenvalues of A,

Λ =

λ1 0. . .

0 λn

Example: Find eigenvalues and eigenvectors of matrix A =

[3 11 3

].



For eigenvalues, we use

|λI −A| =∣∣∣∣λ− 3 −1−1 λ− 3

∣∣∣∣ = 0

λ2 − 6λ+ 8 = 0⇒ λ1 = 2, λ2 = 4

For eigenvector p1, we use

λ1p1 = Ap1

Let p1

=

[α1

α2

], =⇒ 2

[α1

α2

]=

[3 11 3

] [α1

α2

]=⇒

{2α1 = 3α1 + α2

2α2 = α1 + 3α2

=⇒ −α1 = α2

Pick: α1 = 1, α2 = −1 =⇒ p1

=

[1−1

]Similarly for p

2using λ2p2 = Ap

2gives =⇒ p

2=

[11

]Thus, the diagonalizing matrix is P =

[1 1−1 1

], and Λ =

[2 00 4

]It is simple to verify that P−1AP = Λ.



State-Space Representation

Similarity Transformations

For a given transfer function, H(s), state-space representation is not unique,and there are infinite possible representations that are related to each other vianon-singular similarity transformations.

Consider:x(t) = Ax(t) +Bu(t)y(t) = Cx(t) + du(t)

Let T be a nonsingular matrix (i.e. T is invertible). Define:x(t) = Tz(t), or z(t) = T−1x(t), where z(t) is the new transformed statevector. Now, substituting for x(t) in the state equations,{

T z(t) = ATz(t) +Bu(t)

y(t) = CTz(t) + du(t)=⇒

{z(t) = T−1ATz(t) + T−1Bu(t)

y(t) = CTz(t) + du(t)

which is a new set of state equations with z(t) as the state vector andA = T−1AT C = CTB = T−1B d = d



State-Space Representation: Similarity Transformations

Then: {z(t) = Az(t) + Bu(t)

y(t) = Cz(t) + du(t)

Now, since one can find infinite nonsingular matrices T s, there will be infinite

state-space representation for the same system. However, they all have thesame transfer function.Invariance Theorem: H(s) of an LTI system is invariant under similaritytransformation.Proof: That is, we need to show that H(s) = H(s) where H(s) is the transferfunction of the above mapped state space system with z(t) as the state vector.But,

H(s) = C(sI − A)−1B + d

where A, B, C, and d were defined before. Using the expressions for these andmatrix inversion property (PQ)−1 = Q−1P−1, we get



State-Space Representation: Similarity Transformations

H(s) = CT (sI − T−1AT )−1T−1B + d

= CT ((sT −AT ))−1B + d

= C(sI −A)−1B + d = H(s)



Time-Domain Analysis

Time-domain analysis involves studying the behavior of the control system forsome useful reference signals such as unit step (due to instantaneous change att = 0) or unit ramp (due to linear increase with time). This entails studying (a)Steady-state behavior; and (b) Transient behavior of the system.

Assume you are the pilot of an F-22 fighter jet.

Desired Specs:1 Fast response / rise time2 Good relative stability (small overshoot)3 Accuracy (low steady-state error)4 Less sensitivity5 Noise immunity

These specs cannot be met at the same time (trade-off).M.R. Azimi Control Systems


Time-Domain Analysis-Cont.

Consider a closed-loop control system

The response c(t) to a unit step input generally,c(t) = ctr(t)︸︷︷︸

transient

+ css(t)︸︷︷︸steaty-state

Transient Response: Part of the response that goes to zero when t goes toinfinity, i.e. limt→∞ ctr(t)→ 0.Steady-State Response: Part of the response that remains well-behaved orconstant when t goes to infinity, i.e.limt→∞ c(t) = css(t)

where css(t) is constant or a well-behaved function of time. Here, we typicallydeal with the former case i.e. css a constant.



Steady-State Analysis

Steady-state analysis is used to determine the accuracy of the system infollowing the reference signal. Types of inputs typically used are: step, ramp.

Since c(t) and r(t) cannot be directly compared (when H(s) 6= 1) we caninstead compare f(t) and r(t) or study the steady-state behavior of the errorsignale(t) = r(t)− f(t)

Then using the Final Value Theorem (FVT),

ess = limt→∞

e(t) = lims→0

sE(s)

where E(s) = L {e(t)} .

Note: To use FVT, sE(s) should not have poles on the jω axis or in RHS ofs-plane.E(s) = R(s)− F (s) = R(s)− C(s)H(s) = R(s)−G(s)H(s)E(s)

=⇒ E(s) =R(s)

1 +G(s)H(s)



Steady-State Analysis-Cont.

Thus, we have

ess = lims→0

sR(s)

1 +G(s)H(s)

As can be seen the steady-state error ess depends of (a) the input signal R(s)and the type of loop transfer function G(s)H(s). More specifically, the LoopTransfer Function is said to be Type j if

G(s)H(s) =K(1 + α1s) · · · (1 + αMs)

sj(s+ βs) · · · (1 + βNs)

sj represents j consecutive integral operations in the loop.



Steady-State Error Analysis

(a) Steady-State Error Due to Step Input:

Let r(t) = Rus(t) =⇒ R(s) = Rs

ess = lims→0

sE(s) = lims→0

sR(s)

1 +G(s)H(s)= lims→0

R

1 +G(s)H(s)

=R

1 + lims→0G(s)H(s)=

R

1 +Kp

where Kp = lims→0G(s)H(s) is called Step Error Constant .

To achieve ess = 0, we need Kp →∞, which in turn requireslims→0G(s)H(s)→∞.

If the system is:Type 0: Kp = K =⇒ ess = R

1+K 6= 0Type 1 or Higher: Kp →∞ =⇒ ess → 0

Thus, for ess = 0 we need a Type 1 or higher system. If G(s)H(s) is not Type1, using an integral controller can make it a Type 1. See figure next page.



Steady-State Error Analysis-Cont.

This is called a Proportional Integrator (PI).

Note: Adding an integrator introduces a lag, so you need a differentiator tocompensate (PID controller).

(b) Steady-State Error Due to Ramp Input:

Let r(t) = Rtus(t) =⇒ R(s) = Rs2



Steady-State Error Analysis-Cont.

ess = lims→0

sE(s) = lims→0

R

s+ sG(s)H(s)=

R

Kv

Kv = lims→0

sG(s)H(s)

where Kv is called the ramp error constant.

To achieve ess = 0, we need Kv →∞, which in turn requires Type 2 or highersystem (j ≥ 2).

Type 0: Kv = 0 =⇒ ess →∞Type 1: Kv = constant =⇒ ess = R

Kv

Type 2 or Higher: Kv →∞ =⇒ ess → 0


Control Systems (ECE411) Lectures 11 & 12 11... · 2016. 10. 17. · State-Space Representation...

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