721700-11-17E AID:256255 | 02/08/2020

Z-transform is used to convert the discrete time signal into a complex frequency domain.
The z-transform for a discrete time signal x(n) is:

X  z   x  n z
n 
n

(a)

Consider the discrete time signal as:

x  n   u  n 
x  n
The bilateral z-transform of is:

X  z   x  n z
n 
n


  u  n  z
n 
n

0
 z
n 
n


  zn
n 0

X  z   1 z  z2 
X  z
Since is a sum of geometric progression therefore;
1
X  z   z 1
1 z
1
X  z 
Hence, the z-transform of discrete-time signal is 1  z and region of convergence
z 1
is .

(b)

Consider the discrete time signal as:

x  n   a nu   n 
x  n
The bilateral z-transform of is:

X  z   x  n z
n 
n


  a u  n  z
n 
n n

0
 a
n 
n n
z

   a 1 z 
n

n0

X  z   1   a 1 z    a 1 z   
2

X  z
Since is a sum of geometric progression therefore;
1
X  z   z  a 1
1  a 1 z
1 1
X  z   z 
z a
1
a
1
X  z 
z
1
Hence, the z-transform of discrete-time signal is a and region of
1
z 
a
convergence is .

(c)

Consider the discrete time signal as:

x  n    0.5  u   n    0.3 u  n 
n n

x  n
The bilateral z-transform of is:

X  z    0.5 u  n    0.3 u  n   z
n n n

n 


  0.5 u  n  z   0.3 u  n  z  n
n n n

n 
 
X  z    0.5 u  n  z  n    0.3 u  n  z n
n n

n  n  …… (1)

X  z
To calculate the , the above equation is expressed as:
X  z   X1  z   X 2  z 
X1  z  X2  z
Solve and separately as:
 
X1  z     0.5 u  n  z
n n

n 
0

  0.5
n
 zn
n 

   0.51 z 
n

n 0

X 1  z   1   0.51 z    0.51 z   
2

X1  z 
Since is a sum of geometric progression therefore;
1
X1  z    z   0.5 
1
1
1  0.5 z
1
X1  z    z 2
1 2z

Similarly,

X2  z    0.3 u  n  z
n n

n 

   0.3 z  n
n

n0

   0.3z 1 
n

n0

X 2  z   1   0.3 z 1    0.3 z 1   
2

X  z
Since is a sum of geometric progression therefore;
1
X2  z   z  0.3
1  0.3z 1
z
X2  z   z  0.3
z  0.3
Therefore the z-transform of discrete time signal is:
z 1 1
X  z   , 0.3  z 
z  0.3 1  2 z 2
2 z  2 z  0.3
2
1
X  z  2 , 0.3  z 
2 z  1.6 z  0.3 2
2 z 2  2 z  0.3
X  z 
Hence, the z-transform of discrete-time signal is 2 z 2  1.6 z  0.3 and region of
1
0.3  z 
convergence is 2 .

(d)

Consider the discrete time signal as:

 2n 
x  n    1.5 cos   u  n 
n

 8 
Discrete time signal can be expressed as:
j 2 n j 2 n

e 8
e 8
x  n    1.5   u  n 
n

2
j 2 n
1 2
 1.5  u  n  
n
e 8
j 2
, z 
2  3
1 e z 8
3
j 2 n
n  1 2
  1.5  e 8
u  n   j 2
, z 
2 8 3
1 e z
3

x  n
The bilateral z-transform of is:
 
2 n   
 1.5  cos 
1 1 1
 
n
 u  n    j 2 
 8  2  2  j 28 2 8 
1 e z 1 e z
 3 3 
0.3143  z  1.414  2
X  z  2 , z 
z  0.9427 z  0.444 3
0.3143  z  1.414 
X  z 
Hence, the z-transform of discrete-time signal is z 2  0.9427 z  0.444 and
2
z 
region of convergence is 3 .