8/3/2019 ENEE 660 HW Sol #7

1/8

1

Electrical and Computer Engineering Department

University of Maryland College Park

ENEE 660

System Theory

Fall 2008

Professor John S. Baras

Solutions to Homework Set #7

Problem 1

(a) 11T element: zeroes = {-1}, poles = {-3, -3}

12T element: zeroes = none, poles = none

21T element: zeroes = none, poles = none

22

T element: zeroes = {-2}, poles = {- 4, - 4}

31T element: zeroes = {-2}, poles = {-5, -5}

32T element: zeroes = {-1}, poles = {-5, -5}

(b) No to both questions, because of zero-pole cancellations.To find the zeroes and poles of the system we construct the McMillan canonical

form of T. We have2 2 2

2 2

2 2

2 2 2 2

2 2 2

0 1 2

1 2 1

( ) ( 3) ( 4) ( 5)

( 1)( 4) ( 5) 0

( ) 0 ( 2)( 3) ( 5) ,( 2)( 3) ( 4) ( 1)( 3) ( 4)

( ) 1, ( ) 1, ( ) ( 3) ( 4) ( 5)

10

( )

( ) 0 1

0 0

( ) ( ) 1, (

d s s s s

s s s

N s s s ss s s s s s

D s D s D s s s s

d s

T s

So s s

= + + +

+ + +

= + + + + + + + + +

= = = + + +

= =2) ( ), ( ) 1s d s s

= =

And therefore: system zeroes = none, system poles = {-3, -3, -4, -4, -5, -5}

(c) We have directly1

2

2

2

( 3) 0 0 1 0

( ) 0 ( 4) 0 0 2

0 0 ( 5) 2 1

s s

T s s s

s s s

+ + = + + + + +

8/3/2019 ENEE 660 HW Sol #7

2/8

2

And we easily check that the rank of2

2

2

( 3) 0 0 1 0

0 ( 4) 0 0 2

0 0 ( 5) 2 1

s s

s s

s s s

+ +

+ + + + +

is 3 for all s.

Thus the two polynomial matrices in the above MFD are left coprime.

Problem 2

(a) Yes. The model given is controllable since [ ] 0 11 0

b Ab

=

has rank 2. So the

poles of the closed loop system can be placed at any arbitrary location by state

feedback, exactly the type of control suggested. By placing the poles so that theirreal part is in the left half plane we can make the closed loop system

exponentially stable. Indeed for [ ]1 2K k k= the closed loop system becomes

[ ]

[ ]

1 2

0 1 0 0

( ) ( ) ( )1 0 1 1

( ) 0 1 ( )

x t k k x t r t

y t x t

= + +

=

The closed loop poles are the roots of the characteristic polynomial of the closedloop A matrix, which is:

2

2 1

1 2

0 1( ) det( ) ( 1)

1

= = + +

CLAs sI s k s k

k k

Clearly we can arbitrarily assign the closed loop poles to achieve any degree ofexponential stability we wish. Indeed for k1 = -26 and k2 = -10, both poles of the

closed loop are located at s = -5.

(b) No to both questions. The closed loop system for the given control becomes:[ ]

[ ]

0 1 0 0 0 1 0( ) 0 1 ( ) ( ) ( ) ( )

1 0 1 1 1 1

( ) 0 1 ( )

x t L x t r t x t r t L

y t x t

= + + = +

=

And the closed loop transfer function is

[ ]1

2

0 1 0 ( ) 0 1

1 1 1c

sT s sI

L s Ls

= =

Since the product of the closed loop poles is -1 for all values of L, one of the

poles has to be in the right half plane. So the closed system cannot be made stableor exponentially stable with such a feedback scheme.

(c) Yes to both questions. We will use an observer followed by linear feedback onthe state estimate controller. Indeed the given system is observable and

controllable. Controllability was shown in (a) above. For observability, observe

8/3/2019 ENEE 660 HW Sol #7

3/8

3

that the rank of0 1

1 0

c

cA

=

is 2. So we can construct a dynamical system to

estimate the state. The full dimension observer has the form

[ ]

1

2

0 1 0 ( ) ( ) ( ( ) ( ))

1 0 1 ( ) 0 1 ( )

hx x t u t y t y t

h y t x t

= + +

=

The resulting state estimate error equation is

1

2

0 1( ) ( )

1

he t e t

h

=

And by selecting h1 = 26 and h2 = 10, to place both eigenvalues at -5, we getexponential stability (exponential decay ) of the error equation. Then the observer

becomes

0 25 0 26 ( ) ( ) ( ) ( )

1 10 1 10

x t x t u t y t

= + +

To achieve closed loop exponential stability we use linear constant feedback of

the state estimate( ) ( ) ( )u t Kx t r t = +

with r(t) the scalar reference input. We select [ ]1 2K k k= to place both

eigenvalues of1 2

0 1

1 A BK

k k

+ = +

at -1 , which leads to [ ]2 2K= .

Then substituting into the system and observer controller equation we get theclosed-loop state space model

[ ]

0 1 0 0 0

( ) ( ) ( ) ( )1 0 2 2 1

0 25 26 0 ( ) ( ) ( ) ( )

1 12 10 1

( ) 0 1 ( )

4 dim mod

0 1 0 0

( ) 1 0 2 2

0 26 0 25( )

0 10 1 12

x t x t x t r t

x t x t y t r t

y t x t

or as a ensional state el

x t

x t

= + +

= + +

=

=

[ ]

0

( ) 1( )

( ) 0

1

( )( ) 0 1 0 0

( )

x tr t

x t

x ty t

x t

+

=

Which has two eigenvalues at 2, and two eigenvalues at 5, by construction.Thus exponential stability, which cannot be accomplished by static output

feedback can be obtained by dynamic output feedback.

8/3/2019 ENEE 660 HW Sol #7

4/8

4

We can furthermore compute the closed loop transfer function

[ ]

1

3 2 2

4 3 2 2 2 2

1 0 0 0

1 2 2 1 ( ) 0 1 0 0

0 26 25 0

0 10 1 12 1

10 25 ( 5)

12 46 60 25 ( 1) ( 5) ( 1)

c

s

sT s

s

s

s s s s s s

s s s s s s s

=

+ + + +

= = =+ + + + + + +

Problem 3

We will use the general dynamic compensation scheme for multivariable systemsdescribed in Figure 6.3.4 , pp. 259-260 of the Lecture Notes.

The given transfer function has the right MFD

122

1

2

10

1 0 0

( ) ( ) ( )1 1 1 1 1 1

1

r r

s

s ssT s N s D s s

s s s

+ +

= = = +

And since the rank of

2

1 0

1 1

0

1 1

s

s

s

+

+

is 2 for all s, this is a right coprime MFD. Thus clearly ( ) 3T = .

Furthermore Dr is column regular. Following the design from the Notes we willchoose L, K(s),H(s) and Q(s) as described in page 260 so as to make the closed

loop transfer function of figure 6.3.4 equal to1

, , ( ) ( ) ( ) , ( ) ( ) ( )c r r F r r F

T s N s D s L where F s D s D s= = .

We will design the compensator so that

2

10

2 2 ( )1

01

c

s

s sT s

s

+ + +

= +

which follows from the desired closed loop poles (this indeed has the desired

poles), the capabilities of the design scheme and the fact that Nr (s) will notchange. That means that, by takingL to be non-singular

2

1

,

2 2 0( )

1 1r F

s sL D s

s s

+ += + +

.

By the properties of the scheme the matrix of highest coefficients column-wise

fromDr andDr,F are the same. Let ( ( ))c A s denote the matrix of highest power

coefficients column-wise of the polynomial matrixA(s).Then we have from the above

8/3/2019 ENEE 660 HW Sol #7

5/8

5

1 1 1

, ,

1 1

( ( )) ( ( )) ( ( ))

1 0 1 0 1 0,

0 1 0 1 0 1

c r F c r F c r L D s L D s L D s

L so L L

= =

= = =

Thus

2 2

,

2 2

1 0 2 2 0 2 2 0( )0 1 1 1 1 1

2 2 00 2 2 0( )

2 21 1 1 1

r F

s s s sD ss s s s

ss s sand F s

ss s s

+ + + + = = + +

+ + = = + +

The only step remaining is to solve the Diofantine equation (6.3.31)1 1( ) ( ) ( ) ( ) ( ) ( ), . . ,r rK s D s H s N s Q s F s s t Q K Q H

+ = are stable and proper.

SinceNr,Dr are coprime, we know that such a solution exists and is given via the

block Sylvester matrix.

We select Q(s) to have poles at 1 2s j=

as

5

( ) 1 2

s

Q s s

=

+ , which results

in2

2

5 2 2 0 2 3 10 10( ) ( )

1 2 2 2 6 6 2 4

s s s sQ s F s

s s s s s

+ + = = + + + + +

And then by solving the Diofantine equation we get

3 10 5 0( ) , ( )

2 2 2 4 1 0

s H s K s

s s

= = + +

And this completes the design.

Problem 4

(a) We will develop first a right coprime MFD for the given T . The lcm of thedenominators of the first column is s. The lcm of the denominators of the second

column is s+1. So we will take as "denominator" polynomial matrix

0( )

0 1r

sD s

s

= +

.

Then the "numerator" polynomial matrix is

1 1

2( )

1 02

r

s

s sN s

=

.

Then we have1( ) ( ) ( )r rT s N s D s

= .

This right MFD is right coprime because the rank of the matrix

8/3/2019 ENEE 660 HW Sol #7

6/8

6

1 1

2

10

2

00 1

s

s s

ss

+

is 2 for all complex s.

Furthermore we have for the above rD , that the column degrees are 1 2 1d d= = . Since the

matrix of maximum degree column-wise is1 0

0 1

, rD is column regular. Also the

degree of each column of rN is the degree of the corresponding column of rD .

(b) The dynamic compensation scheme follows the diagram below.( )r s ( )u s ( )s ( ) y s

( )z s

Here we have

1 1

1 1 1

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

r

r

r r

y s N s s

D s s u s

u s r s Q s K s u s Q s H s y s

r s Q s K s u s Q s H s N s D s u s

=

=

= + +

= + +

Therefore1 1 1 1

1 1 1 1 1

1

( ) [ ( ) ( ) ( ) ( ) ( ) ( )] ( )

( ) ( ) ( )[ ( ) ( ) ( ) ( ) ( ) ( )] ( )

( )[ ( ) ( ) ( ) ( ) ( ) ( )] ( ) ( )

r r

r r r r

r r r r

u s I Q s K s Q s H s N s D s r s

y s N s D s I Q s K s Q s H s N s D s r s

N s Q s D s K s D s H s N s Q s r s

=

=

=

Dr-1 Nr

K H

Q-1

8/3/2019 ENEE 660 HW Sol #7

7/8

7

Let , ( )r FD s be the "desired" denominator polynomial matrix, which is arbitrary, except

for the constraint that the polynomial matrix ,( ) ( ) ( )r r FF s D s D s = has degree of each

column degree of the corresponding column of ( )rD s . Given the properties of the

MFD we established in (a) above we can find polynomial matrices K(s) andH(s) such

that the Matrix Bezout Equation (MBE) is satisfied

( ) ( ) ( ) ( ) ( ) ( )r rH s N s K s D s Q s F s + = (MBE)

where, Q(s) is a (stable) polynomial matrix with zeroes of det(Q(s)) in the left half

complex plane { ; Re 0}s s < . We can actually find such K(s) and H(s) so that Q-

1(s)K(s) and

Q-1

(s)H(s) are stable proper transfer function matrices. The key property that guarantees

existence of such K(s) and H(s) is the right coprimeness of andr rN D .

But then the close loop transfer function from above is given by

1 1 1 1

, , ( ) ( ) ( ) ( ) ( ) ( ) ( )F r r F r r F T s N s D s Q s Q s N s D s

= = .

The solution, in terms of computing the coefficients of K(s) and H(s) in terms of the

coefficients of Q(s), F(s), is obtained via the generalized Sylvester matrix ofNr(s)and Dr(s), and the associated generalized resultant matrix.

(c) The present problem is simple and we can solve (*) directly without using thegeneralized resultants matrix. We have that the "desired" denominator polynomial

matrix is

and so, ,1 0 1 0

( ) , ( ) ( ) ( )0 1 0 0

r F r r F

sD s F s D s D s

s

+ = = = +

.

We next select Q(s) , to be stable and "fast"; i.e. the zeroes of det (Q(s)) more in the

left half plane than the zeroes of det(Dr, F(s)). Let us select5 0

( )0 5

sQ s

s

+ = +

.

Then (*) becomes

1 10 5 0 1 0 ( 5) 02

( ) ( )0 1 0 5 0 0 0 01

02

ss s ss s

K s H s s s

+ + + = = + +

Simple degree considerations suggest that we should have as solutions polynomial

matrices K(s) and H(s) of degree 0, i.e. constant. Therefore the equation we want to

solve becomes

8/3/2019 ENEE 660 HW Sol #7

8/8

8

11 12 11 12 13

21 22 21 22 23

11 12 13 11 12 1111 12 12

21 22 2221 22 23 21 22 21

1 1

0 ( 5) 02

0 1 0 010

2

1( 2 )(2

1( 2 )

2

s

K K s H H H s s s

K K s H H H

H H s H H H s H K s K s K

K s K s K H H s H H H s H

+ + = +

+ + + + + = + + + +

11 11 12 13 11 12 12 12 11

21 21 22 23 21 22 22 22 21

5) 0

0 0

1( 2 ) ( )

( 5) 02

1 0 0( 2 ) ( )

2

s

K H H s H H K H s K H s

K H H s H H K H s K H

+

+ + + + + +

= + + + + +

Equating coefficients of equal powers generates the following equations

11 11 12

13 11

12 12

12 11

21 21 22

23 21

22 22

22 21

2 1

15

2

0

0

2 0

10

2

0

0

K H H

H H

K H

K H

K H H

H H

K H

K H

+ + =

=

=

+ =

+ + =

=

=

+ =

Setting as many coefficients as possible to zero we obtain the solution

1 0 0 0 10,

0 0 0 0 0K H

= =

.

If we let the output of the dynamic compensator be denoted by ( )z s , we see that the

resulting controller is very simple and the diagram simplifies accordingly:

1 1 1

2

1 ( ) ( ( ) 10 ( ))5

( ) 0

z s u s y s s

z s

= ++

=

.