EE422G Review 2 Solutions
1. Given the initial condition problem
 x  Ax  Bu

 x(t 0 )  x0
d
f ( , t )
f ( , t )d  f ( , t )  t  
d )

dt a

t
a
t
(1) What is its solution? (2) Prove it. (Note:
t
Solution:
t
(1) x(t )  e A(t t0 ) x0   e A(t  ) Bu ( )d
t0
t
(2) Let z (t )  e A(t t0 ) x0   e A(t  ) Bu ( )d
t0
t
Then z(t ) 
t
dz (t ) d A(t t0 )
d
d
 e
x0  { e A(t  ) Bu ( )d} = Ae A(t t0 ) x0  { e A(t  ) Bu ( )d}
dt
dt
dt t0
dt t0
Denote f ( , t )  e A(t  ) Bu ( )
d
f ( , t )
f ( , t )d  f ( , t )  t  
d

dt t0

t
t0
t
Then
t
t
 e A(t t ) Bu (t )   { Ae A(t  ) Bu ( )}d
t0
t
 Bu (t )  A e A(t  ) Bu ( )d
t0
t
Hence, z(t )  Ae A(t t0 ) x0  Bu (t )  A e A(t  ) Bu ( )d
t0
t
 A{e A(t t0 ) x0   e A(t  ) Bu ( )d}  Bu (t )  Az(t )  Bu (t )
t0
t0
Also, z (t 0 )  e A(t0 t0 ) x0   e A(t  ) Bu ( )d  x0
t0
t
Hence, x(t )  e A(t t0 ) x0   e A(t  ) Bu ( )d is the solution of the given initial value problem
t0
2.System’s initial condition, state equation, and output equation are given by
 x (t )  Ax(t )  Bu (t ), x(0)  0

y (t )  Cx(t )  Du (t )

Find the Laplace transform of the output.
Solution:
Take Laplace Transform for the state equation x (t )  Ax(t )  Bu (t ) :
sX ( s )  x0  AX ( s )  BU ( s )
( sI  A) X ( s )  x0  BU ( s )
X ( s )  ( sI  A) 1 x0  ( sI  A) 1 BU ( s )
Take Laplace Transform for the output equation y (t )  Cx(t )  Du (t ) :
Y ( s)  CX ( s)  DU ( s)  C[( sI  A) 1 x0  ( sI  A) 1 BU ( s)]  DU ( s)
 C ( sI  A) 1 x0  C ( sI  A) 1 BU ( s)  DU ( s)
When x0  x(0)  0 , Y (s)  C (sI  A) 1 BU (s)  DU (s) .


3. What is L1 ( SI  A) 1 where A is a square matrix.
Solution: e At
4. Find e
 0

 6

1


 5 
t
using two different methods.
Solution One: Using Inverse Laplace Transform
1  s  1 
 s 0  0
( sI  A)  




0 s   6  5 6 s  5
 a11
a
 21
a12 
a22 
1
 a22  a12 
 a
a11 
  21
a11 a22  a12 a21
 s  5 1
  6 s
s 1 


( sI  A) 1  


6
s

5
s
(
s

5
)

6


 s  5 1
s5

  6 s

  ( s  2)( s  3)
 

6
( s  2)( s  3) 
 ( s  2)( s  3)
1
1

( s  2)( s  3) 

s

( s  2)( s  3) 
s5
3
2


 3e  2t  2e  3t
( s  2)(s  3) s  2 s  3
1
1
1


 e  2t  e  3t
( s  2)(s  3) s  2 s  3
6
1 
 1
 6 

 6[e  2t  e  3t ]

( s  2)(s  3)
 s  2 s  3
s
2
3


 2e  2t  3e  3t
( s  2)(s  3) s  2 s  3
Hence, e
 0

 6



 5 
1
t
 3e 2t  2e 3t
 L [( sI  A) ]  
 2t
3t
 6e  6e
1
e 2t  e 3t
1
 2e
 2t
 3e
3t



Solution Two: Cayley – Hamilton Theorem Based Method
(1) Find eigenvalues
 1
| I  A |
  (  5)  6  2  5  6  0
6  5
1  2,
2  3,
(2) Find  0 (t ) and 1 (t )
1  2  e 2t   0 (t )  1 (t )(2)
(1)
2  3  e  3t   0 (t )  1 (t )(3)
(2)
(1)  (2)  e  2t  e  3t  1 (t )
  0 (t )  e  3t  31 (t )  e  3t  3e  2t  3e  3t
 3e  2t  2e  3t
(3)
e At   0 (t ) I   1 (t ) A
3e  2t  2e 3t
 
0
0


 2t
 3t 
 2t
0
3e  2e   6e  6e 3t

 3e 2t  2e 3t
e  2 t  e  3t 


 2t
 3t
 2e  2t  3e 3t 
 6e  6e
e  2 t  e  3t 

 5e 2t  5e 3t 
5A. Obtain a state model for the network
Solution:
(1) Assign state variables for voltage of each capacitor and current for each inductor current
x1  vc , x2  i L
v  vL
(2) Write KVL for each capacitor and KCL for each inductor: v s1  R1iC  x1 , s 2
 x2
R2
(3) Replace undesirable variables using allowed variable: iC  Cx1 , v L  Lx 2
Therefore,
v s1  R1iC  x1  R1Cx1  x1
 x1  x1 (1 / R1C )  v s1 (1 / R1C )


 v s 2  v L v s 2  Lx 2

 x2
 x 2  x 2 ( R2 / L)  v s 2 (1 / L)
 R
R2
2

Hence, the state equation is
0   x1  1 / R1C
0  v s1 
 x1   1 / R1C

 
 x    0



 R2 / L  x2   0
1 / L  v s 2 
 2 
(4) Obtain the output equation
v s1 
x 
v0  vc  v L  x1  Lx 2  x1  x2 R2  v s 2  v0  1 R2  1   [0  1] 
 x2 
v s 2 
5B. Example 7-7
6. Obtain a state model for the system defined by
Y ( s)
s 1
Y ( s)
s 1
(1)
(2)


U ( s ) ( s  3) 3
U ( s) s( s  3)
Solution for (1):
U ( s)
U (s)
Y ( s)
s 1
A
B
B
, Y ( s)  A

 
s
s3
U ( s) s( s  3) s s  3
U (s)
U ( s)
, X 2 ( s) 
Let X 1 ( s ) 
, Y (s)  AX 1 (s)  BX 2 (s)
s
s3
Then, sX 1 (s)  U (s), sX 2 (s)  3X 2 (s)  U (s)
Inverse Laplace Transform gives
x1  u(t )  0 x1 (t )  u(t )
and x 2  3x2  u(t ) or x 2  3x2  u(t )
Hence, the state equation is
 x1  0 0   x1  1
 x   0  3   x   1u
 2   
 2 
x 
The output equation is y  [ A B] 1 
 x2 
Solution for (2):
Y ( s)
s 1
C
B
A




3
3
2
U ( s ) ( s  3)
s3
( s  3)
( s  3)
U ( s)
U (s)
U (s)
Y ( s)  C
B
A
3
2
s3
( s  3)
( s  3)
U (s)
Let X 1 ( s ) 
,
s3
U ( s)
1 U (s)
1
X 2 (s) 

{
}
X 1 ( s) ,
2
s3 s3
s3
( s  3)
U (s)
1
U (s)
1

{
}
X 2 (s)
3
2
s  3 ( s  3)
s3
( s  3)
Y ( s)  CX 3 ( s)  BX 2 ( s)  AX 1 ( s)
Then,
sX 1 (s)  3 X 1 (s)  U (s) ,
sX 2 (s)  3X 2 (s)  X 1 (s) ,
sX 3 (s)  3 X 3 (s)  X 2 (s)
Inverse Laplace Transform gives
x1  3x1 (t )  u(t )
x 2  3x2 (t )  x1 (t )
x 3  3x3 (t )  x2 (t )
Hence, the state equation is
0   x1  1 
 x1   3 0
 x    1  3
0   x2   0u
 2 
 x 3   0
1  3   x3  0
X 3 ( s) 
 x1 
The output equation is y  [ A B C ] x 2 
 x3 
7. x(t) is a continuous time signal. Its Fourier transform is X(f). xs(t) is the sampled signal of x(t)
by using the sampling function p(t) with sampling period T. Denote

1 T / 2
Cn  
p(t )e  jn2 f s t dt where fs =1 / T. Prove X s( f )   Cn X ( f  nf s )
T T / 2
n  
8. Give the mapping of the following regions / points in s-plane to z-plane:
(1) left hand s-plane; (2) right hand s-plane; (3) jw axis; (4) s=0

 1 n  0
1 n0
9. Find the z-transform of  (n) 
, u ( n)  
, and
 0 otherwise
0 n  0
x(nT )  e nT  (e T ) n (n  0,  0)
10. Memorize z-transform pairs No. 1 to No. 5 in Table 8-1.
11. Given Z (e nT ) 
1
1  e T z 1
. Prove Z (nT ) 
Tz 1
Te T z 1
nT
Z
(
nTe
)

and
(1  z 1 ) 2
(1  e T z 1 ) 2
12. Determine the z-transform for the following sequences of samples
n
1
(A) x(nT )    u (n)
 5
n
 1
(B) x(nT )     u (n)
 5
n
 3
(C) x(nT )  u (n)    u (n  4)
 4
(D) x(nT )  2u (n)  2u (n  8)
n
 2
(E) x(nT )    u (n  4)
 3
 2
(F) x(nT )   
 3
n4
u (n  4)
1
1

1
1  0.2 z 1
1  z 1
5
1
1
Solution for (B): X ( z ) 

1
1  0.2 z 1
1  ( ) z 1
5
Solution for (A): X ( z ) 
n
 3
Solution for (C): Let x1 (nT )    u (n  4)
 4
4
 3  3
Then x1 (nT )     
 4  4
 3
Z{ 
 4
n4
 2
Z { 
 3
n4
n4
u ( n  4)
1
z 4
 3
u (n  4)}  z Z{  u (n)}  z 4

3
1  0.75 z 1
 4
1  z 1
4
4
n4
0.75 4 z 4
 3
 3
Hence, Z ( x1 (nT ))    Z {  u (n  4)} 
1  0.75 z 1
 4
 4
Therefore,
1
0.75 4 z 4
X ( z) 

1  z 1 1  0.75 z 1
2
2 z 8
2(1  z 8 )
Solution for (D): X ( z ) 


1  z 1 1  z 1
1  z 1
4
n4
 2  2
Solution for (E): x(nT )      u (n  4)
 3  3
n
4
n
1
 2
u (n  4)}  z Z {  u (n)}  z  4
2
 3
1  z 1
3
4
4
 2
 2
Hence, X (z )    Z { 
 3
 3
1
Solution for (F): X (z )  z 4
2
1  z 1
3
13. Find the inverse z-transform for
n4
u (n  4)} 
(2 / 3) 4 z 4
1  (2 / 3) z 1
(a) X ( z ) 
2
(1  z )(1  0.2 z 1 )
1
(b) X ( z ) 
2
(1  z )(1  z 1 )
1
1  0.3z 1
1  0.3z 1
(d)
X
(
z
)

(1  0.2 z 1 )(1  0.4 z 1 )
(1  0.2 z 1 )( 2  0.4 z 1 )
1
1
(e) X ( z ) 
(f) X ( z ) 
1
1 2
(1  z )(1  0.5 z )
(1  0.81z 1 ) 2
(c) X ( z ) 
(g) X ( z ) 
1
1  0.81z  2
(h) X ( z ) 
z2  z
( z  0.5)( z  0.25)
Note: The following are two z-transform pairs in Table 8-1
Tz 1
nT
(1  z 1 ) 2
Te  aT z 1
(1  e  aT z 1 ) 2
nTe  anT
You may modify them as
z 1
(1  z 1 ) 2
n
n( k )
kz 1
(1  kz 1 ) 2
n
(or
n( k )
n 1
z 1
)
(1  kz 1 ) 2
z 1
).
Now you can very easily tell Z (
(1  0.5 z 1 ) 2
A
B

Solution for (a): X ( z ) 
, x(nT )  Au(n)  B(0.2) n u(n)
1
1 z
1  0.2 z 1
A
B

Solution for (b): X ( z ) 
, x(nT )  Au(n)  B(1) n u(n)
1
1
1 z
1 z
A
B

Solution for (c): X ( z ) 
, x(nT )  A(0.2) n u(n)  B(0.4) n u(n)
1
1
1  0.2 z
1  0. 4 z
1  0.3z 1
0.5  0.15 z 1
Solution for (d): X ( z ) 

(1  0.2 z 1 )( 2  0.4 z 1 ) (1  0.2 z 1 )(1  0.2 z 1 )
A
B
X ( z) 

,
1
1  0.2 z
1  0.2 z 1
x(nT )  A(0.2) n u(n)  B(0.2) n u(n)
Solution for (e):
1
X ( z) 
A
B
Cz 1
A
B
(C / 0.5T )(T 0.5 z 1 )





1  z 1 1  0.5 z 1 (1  0.5 z 1 ) 2 1  z 1 1  0.5 z 1
(1  0.5 z 1 ) 2
x(nT )  Au(n)  B(0.5) n u (n)  (C / 0.5T )nT 0.5 n u (n)
 Au(n)  B(0.5) n u (n)  (C / 0.5)n0.5 n u (n)
Solution for (f): X ( z ) 
1
(1  0.81z 1 ) 2

A
Bz 1

1  0.81z 1 (1  0.81z 1 ) 2

A
( B / 0.81)(0.81z 1 )

1  0.81z 1
(1  0.81z 1 ) 2
x(nT )  A(0.81) n u(n)  ( B / 0.81)n(0.81) n u(n)  A(0.81) n u(n)  Bn (0.81) n1 u(n)
Solution for (g):
1
1
1/ 2
1/ 2
X ( z) 



2
1
1
1
1  0.81z
(1  j 0.9 z )(1  j 0.9 z ) 1  j 0.9 z
1  j 0.9 z 1
x(nT )  0.5[( j 0.9) n  ( j 0.9) n ]u(n)  0.5( j 0.9) n {1  (1) n }u(n)
Hence,
(1) n / 2 0.9 n u (n) n : even
x(nT )  
n : odd
0
Solution for (h):
z2  z
( z 2  z ) z 2
X ( z) 

( z  0.5)( z  0.25) ( z  0.5) z 1 ( z  0.25) z 1

1  z 1
A
B


1
1
1
(1  0.5 z )(1  0.25 z ) 1  0.5 z
1  0.25 z 1
x(nT )  A(0.5) n u(n)  B(0.25) n u(n)
n
n
1
1
14. Find y (nT )  x(nT ) * h(nT ) where x(nT )    u (n  3) and h(nT )    u (n  5)
 4
 3
Solution: Y ( z )  X ( z ) H ( z )  Z ( x(nT )) Z (h(nT ))
n
3
1
1 1
X ( z )  Z {  u (n  3)}  Z {   
 4
 4  4
3
1
1
   Z { 
 4
 4
n 3
n
5
3
5
n 5
u (n  3)}
z 3
1
u (n  3)}   
1
 4  1  (1 / 4) z
1
1 1
H ( z )  Z {  u (n  5)}  Z {   
 3
 3  3
1
1
   Z { 
 3
 3
n 3
n 5
u (n  5)}
z 5
1
u (n  5)}   
1
 3  1  (1 / 3) z
5
z 3
z 5
1
1
Y ( z)   


1
1
 4  1  (1 / 4) z  3  1  (1 / 3) z
3
5
3
5
3
5
1
 1 1
     z 8
1
(1  (1 / 4) z )(1  (1 / 3) z 1 )
 4  3
A
B
 1 1
     z 8 {

}
1
1  (1 / 4) z
1  (1 / 3) z 1
 4  3
n
n
A
B
1
1
Z 1{

}  A  u (n)  B  u (n)
1
1
1  (1 / 4) z
1  (1 / 3) z
 4
 3
3
5
n 8
3
5
 1 1  1
 1 1 1
y (nT )  A      u (n  8)  B     
 4  3  4
 4  3  3
15. Determine H(z) and h(nT) for following systems
n 8
u (n  8)
(a) y (nT )  y (nT  T )  x(nT )
(b) y (nT )  2 y (nT  T )  y (nT  2T )  x(nT )  3x(nT  3T )
Solution for (a):
Z ( y (nT )  y (nT  T ))  Z ( x(nT ))
Y ( z )  z 1Y ( z )  X ( z )
1
H ( z)  Y ( z) / X ( z) 
, h(nT )  Z 1 ( H ( z ))  u(n)
1
1 z
Solution for (b):
Z ( y (nT )  2 y (nT  T )  y (nT  2T ))  Z ( x(nT )  3x(nT  3T ))
Y ( z )  2 z 1Y ( z )  z 2Y ( z )  X ( z )  3z 3 X ( z )
1  3z 3
1
z 1
2
H ( z)  Y ( z) / X ( z) 

 3z
1  2 z 1  z 2 (1  z 1 ) 2
(1  z 1 ) 2
1
A
Bz 1
1
z 1




Because
(1  z 1 ) 2 1  z 1 (1  z 1 ) 2 1  z 1 (1  z 1 ) 2
1
1
z 1
1
]Z [

]  u (n)  nu (n)
Hence Z [
(1  z 1 ) 2
1  z 1 (1  z 1 ) 2
1
z 1
z 1
1
2
}  nu (n) . Thus Z {z
}  (n  2)u (n  2)
Also, Z {
(1  z 1 ) 2
(1  z 1 ) 2
Therefore,
h(nT )  u (n)  nu (n)  3(n  2)u (n  2)
1
16. Determine the range of K which ensures the following systems stable:
(a) y (nT )  Ky(nT  T )  K 2 y (nT  2T )  x(nT )
(b) y (nT )  2 Ky(nT  T )  K 2 y (nT  2T )  x(nT )
Solution for (a):
Z ( y(nT )  Ky(nT  T )  K 2 y(nT  2T ))  Z ( x(nT ))
Y ( z )  Kz 1Y ( z )  K 2 z 2Y ( z )  X ( z )
1
H ( z) 
1
1  Kz  K 2 z  2
Find the poles for H (z ) :
1  Kz 1  K 2 z 2  0 , z 2  Kz  K 2  0
3
3
z 2  Kz  K 2  z 2  Kz  ( K / 2) 2  K 2  ( z  K / 2) 2  K 2  0
4
4
3
( z  K / 2) 2   K 2
4
z  K / 2  j 3K / 2
z  ( K / 2) 2  ( 3K / 2) 2  K 2
To be stable, z  1 . Hence, the range of K for stability is  1  K  1 .
Solution for (b):
Z ( y(nT )  2Ky(nT  T )  K 2 y(nT  2T ))  Z ( x(nT ))
Y ( z )  2Kz 1Y ( z )  K 2 z 2Y ( z )  X ( z )
1
H ( z) 
1
1  2 Kz  K 2 z  2
Find the poles for H (z ) :
1  2 Kz 1  K 2 z 2  0 ,
(z  K )2  0
zK
z  K2
To be stable, z  1 . Hence, the range of K for stability is  1  K  1 .
17. Find the frequency response for the system: y (nT )  x(nT )  x(nT  2T )
Solution: Z ( y (nT ))  Z ( x(nT )  x(nT  2T ))
Y ( z )  X ( z )  z 2 X ( z )
H ( z )  Y ( z ) / X ( z )  1  z 2
Frequency response
H (e j 2r )  1  e  j 4r  (e j 2r  e  j 2r )e  j 2r  2 cos( 2r )e  j 2r
Amplitude response A(r )  2 cos(2r )
 0.5  r  0.5
cos( 2r )  0
 2r
Phase response  (r )  
  2r cos( 2r )  0