finals02 - Sample Final Exam Covering Chapters...

Sample Final Exam Covering Chapters 10-17 (finals02)
1
Sample Final Exam (
finals02)
Covering Chapters 10-17 of Fundamentals of Signals & Systems
Problem 1 (30 marks)
Consider the system depicted below used for discrete-time processing of continuous-time signals. The
sampling period is 100 milliseconds (
0.1 T =
s).
The discrete-time filter
( )
d
H z
is given by the following causal difference equation initially at rest:
1 2 0 1 2
[ ] [ 1] [ 2] [ ] [ 1] [ 2] y n a y n a y n b x n b x n b x n + − + − = + − + −
(a) [8 marks] Find the controllable canonical state-space realization of the filter
( )
d
H z
, i.e., sketch the
block diagram and give the state-space equations.
Answer:
The state equation:
CT/DT
Τ
DT/CT
Τ
[ ]
d
x n
x t
c
( )
y t
c
( )
[ ]
d
y n
( )
d
H z
+
+
+
+
-
-
+
+
[ ] y n
z
− 1
[ ] u n
z − 1
1
a
2
a
0
b
1
b
2
b

Sample Final Exam Covering Chapters 10-17 (finals02)
2
N
1 1
2 2 1 2
[ 1] 0 1 [ ]
0
[ ]
[ 1] [ ] 1
B
A
x n x n
u n
x n a a x n
+
       
= +
       
+ − −
 
     
±²³²´
.
The output equation:
[ ]
N
2 2 0 1 1 0 0
[ ] [ ] [ ]
D
C
y n b a b b a b x n b u n = − − +
±²²²³²²²´
(b) [10 marks] The filter
( )
d
H z
is designed to approximate the unity-gain, second-order, continuous-
time, causal LTI filter
2
2
( ) , Re{ } 1
3 2
H s s
s s
= > −
+ +
. Find the values of the parameters
1 2 0 1 2
{ , , , , } a a b b b
of
( )
d
H z
using the "c2d" transformation. Specify the ROC of
( )
d
H z
.
Answer:
The state equation for
( ) H s
:
N
1 1
2 2
( ) ( ) 0 1 0
( )
( ) ( ) 2 3 1
B
A
x t x t
u t
x t x t
       
= +
       
− −
       
µ
µ
±²³²´
.
The output equation for
( ) H s
:
[ ]
( ) 2 0 ( )
C
y t x t =
±³´
Diagonalize to compute matrix exponential:
1
1
1 0 1 1
,
0 2 1 2
A V AV V −
− −
   
= = =
   
− −
   
1
1
2
2 2
2 2 2 2
1
2
1 1 2 1
0
1 2 1 1
0
1 1 0.9909 0.0861
2 2
1 2 0.1722 0.7326
2 2 2
3/ 2 1/ 2 2
1 0
T
A T AT
d
T
T T T T T T
T T T T T T
AT
d
e
A e Ve V
e
e e e e e e
e e e e e e
e
B A e I B
−
−
−
− − − − − −
− − − − − −
−
−  
   
= = =
     
−
   
 
−
    − −    
= = =
   
   
− −
− + − +
   
   
− −
 
 
= − =
 
 
 
2 2
2 2
2 2
2
2
0 1
1
2 2 2 1
1 1
3/ 2 1/ 2 0.0045
2 2
1 0 0.0861
2 1
T T T T
T T T T
T T T T
T T
T T
d
e e e
e e e e
e e e e
e e
e e
C C
− − − −
− − − −
− − − −
− −
− −
  − − −
 
   
− + − + −
 
 
 
− −
  − + + −
   
 
= = =
 
   
 
− + −
   
 
−
 
=
Transfer function:

Sample Final Exam Covering Chapters 10-17 (finals02)
3
[ ]
[ ]
1
1
( ) ( )
0.9909 0.0861 0.0045
2 0
0.1722 0.7326 0.0861
0.0045( 0.7326) (0.0861)(0.0861)
1
2 0
( 0.1722)(0.0045) (0.0861)( 0.9909) ( 0.9909)( 0.7326) 0.0148
0.009( 0.732
d n d d d
z C zI A B D
z
z
z
z z z
z
−
−
= − +
− −
   
=
   
−
   
− +
 
=
 
− + − − − +
 
−
=
H
2
1 2
1 2
6) 2(0.0861)(0.0861)
( 0.9909)( 0.7326) 0.0148
0.009 0.0082
1.7235 0.7407
0.009 0.0082
1 1.7235 0.7407
z z
z
z z
z z
z z
− −
− −
+
− − +
+
=
− +
+
=
− +
Hence,
0 1 2 1 2
0, 0.009, 0.0082, 1.7235, 0.7407 b b b a a = = = = − =
(c) [5 marks] Sketch the pole-zero plot of
( )
d
H z
. Compute the gain of