West Virginia University
Collage of Engineering and Mineral Resources
Lane Department of Computer Science and Electrical Engineering
HW # 7
Solution
Signals and Systems 1, EE 327 - Fall 2002
Student Name:
ID#:
Nov. 15, 2002
Q#1
Grade:
Find the transfer functions of the following discrete time systems:
a)
b)
y (k )  0.5 y (k  1)  2 x(k ).
y (k )  2 y (k  1)  y (k  2)  2 x(k )  x(k  1)  2 x(k  2).
Solution:
a)
y (k )  0.5 y (k  1)  2 x(k ).
Y ( z )(1  0.5 z 1 )  2 X ( z )  H ( z ) 
2
2z

1  0.5 z 1 z  0.5
Solution:
b)
y (k )  2 y (k  1)  y (n  2)  2 x(n)  x(n  1)  2 x(k  2).
Y ( z )(1  2 z 1  z  2 )  X ( z )( 2  z 1  2 z  2 )
2  z 1  2 z  2 2 z 2  z  2
H ( z) 
 2
1
2
1  2z  z
z  2z  1
Q#2
------------------------------------------------------------------------------------------------A) Plot the poles and zeros of the following transfer functions and B) determine
the stability and C) Determine the unit step response for the following transfer
functions.
( z  0.5)
( z 2  1)
a) H ( z ) 
,
b) H ( z )  2
( z  0.75)
( z  0.25)
z ( z  1)
( z  0.5)( z  0.5)
c) H ( z )  2
,
d ) H ( z) 
( z  0.5 z  0.5)
( z 2  z  0.75)
Solution:
We may rewrite them as follows:
1
a)
H ( z) 
( z  0.5)
,
( z  0.75)
pole at
 0.75,
zero at
0.5
( z 2  1)
( z  j )( z  j )
b) H ( z )  2

( z  0.25) ( z  0.5)( z  0.5)
Poles at  j , zeros at  0.5.
Plots:
z ( z  1)
z ( z  1)

( z 2  0.5 z  0.5) ( z  1)( z  0.5)
Poles at  0.5 and 1, zeros at 0 and 1.
c)
H ( z) 
d)
H ( z) 
( z  0.5)( z  0.5)
( z  0.5)( z  0.5)

( z 2  z  0.75)
( z  0.5  07 j )( z  0.5  07 j )
Poles at 0.86126o ,
zeros at  0.5.
Plots:
C) For step response use MATLAB. It gives the plot of step response.
>>step([1, -0.5],[1, 0.75]) % for part (a).
>>step([1, 0, 1],[1, 0, -0.25]) % for part (b).
>>step([1, -1, 0],[1, -0.5, -0.5]) % for part (c).
>>step([1, -0.25],[1, 1, 0.75]) % for part (d).
2
Q#3
Perform the graphical convolution for x[n]*h[n] where:
x[n]  u[n]  u[n  4], and
h[n]  0.5n u[n]
Solution:
1-Find the z transform of the following signals:
Q#4
x[n]  u[n]  u[n  4], and
h[n]  0.5n u[n]
Solution:
a)
x[n]  u[n]  u[n  4],
3
X ( z )   z  n  1  z 1  z  2  z  3
n0
and
b) h[n]  0.5n u[n]

X ( z )   0 .5 z
n
n
n0
use

  (0.5 z 1 ) n
n0
geometric series
 an 
n0

1
1 a
if
convergence :
a  1  X ( z) 
1
1  0.5 z 1
3
2- Find inverse Z transform of Y(z)=X(z).H(z) of part 1 (i. e. y(n)=inv{Y[z]}
and plot y(n). (hint: your plot should be same as plot of convolution in Q#3)
Solution:
Y(z)=X(z).H(z) is as same as y(k)=x(k)*h(k), therefore one way is using
formula of convolution as follows:
x[n] * h[n] 
k 
 x[n]h[n  k ]
k  

k 
 (u[k ]  u[k  4])0.5
nk
u[n  k ]
k  
n
if
0  n  4    0.5
k 0
4
if
n  4    0.5
k 0
nk
nk
1  2n 1
 0.5
 (0.5n  2)
1 2
n
1  25
 0.5
 0.5n  0.5n  5
1 2
n
Therefore for k=0, 1, 2, 3, … we get 1, 1.5, 1.75, … the same plot as problem
in Q#3. The other way is inverse (i. e. y(n)=inv{Y[z]}.
3- Find inverse Z-transform of the following signals:
( z  1)( z  0.8)
a) H ( z ) 
,
( z  0.5)( z  0.2)
( z 2  1)( z  0.8)
b) H ( z ) 
( z  0.5) 2 ( z  0.2)
Solution:
( z  1)( z  0.8)
,
( z  0.5)( z  0.2)
C3
X ( z ) C1
C2



z
z z  0.5 z  0.2
X ( z)
X ( z)
C1 
z z 0  8,
C2 
( z  0.5) z 0.5  1.857,
z
z
X ( z)
C3 
( z  0.2) z 0.2  5.143,
z
1.857 z 5.143z
X ( z)  8 

z  0.5 z  0.2
x[n]  8 [n]  1.857(0.5) n u[n]  5.143(0.2) n u[n]
a)
H ( z) 
4
( z 2  1)( z  0.8)
b) H ( z ) 
( z  0.5) 2 ( z  0.2)
C3
X ( z ) C1
C2
C4




z
z z  0.2 z  0.5 ( z  0.5) 2
X ( z)
X ( z)
C1 
z z 0  16,
C2 
( z  0.2) z 0.2  5.88,
z
z
X ( z)
C4 
( z  0.5) z 0.5  2.79,
z
d  ( z 2  1)( z  0.8) 

C3  
 11.12
dz  z ( z  0.2)  z 0.5
X ( z )  16 
5.88 z 11.12 z
2.79 z


z  0.2 z  0.5 ( z  0.5) 2
x[n]  16 [n]  5.88(0.2) n u[n]  11.12(0.5) n u[n]  2.79n(0.5) n u[n].
4- Solve the following difference equation using classical method and Ztransform method. y[k]+3y[k-1]+2y[k-2]=2x[k]-x[k-1]
where:
y[-1]=0;, y[-2]=1 and x[k]=u[k]
Solution:
See HW # 8 Solution later.
Test # 3 November 18 from chapters 6, 7 and 8.
5