Chapter 5

Z-Transforms.

4.1 Definitions
Unilateral Z-transform.

T
Let {f (n)} = f (0), f (1), f (2), · · · be any sequence defined for n = 0, 1, 2, · · · .Then the

IS
Z-transform of the given sequence { f(n) }is defined as
∞
X M
f (n)z −n
SR
Z[f (n)] =
n=0

where z is arbitrary complex variable.

h,
es

The right hand side of the above equation is a function of z and hence it is denoted by
Z[f (n)] = F (z).
ur

Bilateral Z-transform
.S
r.E

Let {f (n)}be a sequence defined for n = 0, ±1, ±2, · · · .Then the Z-transform the given
sequence { f(n) }is defined as
∞
D

X
Z[f (n)] = f (n)z −n
n=−∞

This transform is called as Bilateral Z-transform or two-sided Z-transform.

Note
1. Throughout this chapter we consider only the unilateral Z-transform.

Basic Formulae:
1. (1 − z)−1 = 1 + z + z 2 + z 3 + ...
2. (1 − z)−2 = 1 + 2z + 3z 2 + 4z 3 + ...
z2 z3 z4
3. − log(1 − z) = z + + + + ...
2 3 4
z z2 z3 z4
4. ez = 1 + + + + + ...
1! 2! 3! 4!

1
2

4.2 Z-transform of some well-known sequences

Property 1. Linear Property
‘

Z[af (n) + bg(n)] = aF (z) + bG(z)

Proof.
∞
X
Z[af (n) + bg(n)] = [af (n) + bg(n)]z −n
n=0
X∞ ∞
X
= af (n)z −n + bg(n)z −n
n=0 n=0
X∞ ∞
X
−n
= a f (n)z +b g(n)z −n
n=0 n=0

T
= aF (z) + bG(z)

IS
1. Z[1] =
z M
SR
z−1
Proof.
h,

∞ ∞  n
X
−n
X 1
Z[f (n)] = f (n)z = f (n)
es

n=0 n=0
z
∞  n  2  3
1 1 1 1
ur

X
Z[1] = =1+ + + + ···
z z z z
.S

n=0
 −1  −1
1 z−1 z
= 1− = =
r.E

z z z−1
D

z
2. Z[n] = , |z| > 1
(z − 1)2
Proof.
∞ ∞  n
X
−n
X 1
Z[f (n)] = f (n)z = f (n)
n=0 n=0
z
∞  n    2  3
X 1 1 1 1
Z[n] = n =0+1 +2 +3 + ···
n=0
z z z z
−2 −2
1 z2
 
1 1 1 z−1 z
= 1− = = 2 = , |z| > 1
z z z z z (z − 1) (z − 1)2
 3

z
3. Z[an ] = if |z| > |a|
z−a
Proof.
∞ ∞  n
X
−n
X 1
Z[f (n)] = f (n)z = f (n)
n=0 n=0
z
∞  n X ∞  
n 1 a n
X  a   a 2  a 3
n
Z(a ) = a = =1+ + + + ···
n=0
z n=0
z z z z
 −1
 a −1 z−a z
= 1− = = , |z| > |a|
z z z−a

z
4.Z[(−a)n ] = if |z| > |a|
z+a
Proof.
∞ ∞ n
X
−n
X 1
Z[f (n)] = f (n)z = f (n)

T
n=0 n=0
z

IS
∞  n X ∞  n
n
X n 1 −a  a   a 2  a 3
Z[(−a) ] = (−a) = =1− + − + ···
n=0
z n=0
z z M z z
SR
 −1
 a −1 z+a z
= 1+ = = , if |z| > |a|
z z z+a
h,

   
1 z
5. Z = log , |z| > 1
es

n z−1
ur

Proof.
∞ ∞
.S

 n
X
−n
X 1
Z[f (n)] = f (n)z = f (n)
z
r.E

n=0 n=0

1
D

Here the domain of the sequence is n = 1, 2, · · · . Hence we have

n
  ∞    n
1 X 1 1
Z =
n n=1
n z
   2  3
1 1 1 1 1 1
= + + + ···
1 z 2 z 3 z
   2  3
1 1 1
z z z
= + + + ···
1  2 3  
1 z−1
= − log 1 − = − log
z z
 −1  
z−1 z
= log = log
z z−1
4
   
1 z
6. Z = z log , |z| > 1
n+1 z−1
Proof.
∞ ∞  n
X
−n
X 1
Z[f (n)] = f (n)z = f (n)
n=0 n=0
z
  ∞     n
1 X 1 1
Z =
n+1 n=0
n+1 z
   2  3
1 1 1 1 1 1
= 1+ + + ···
2 z 3 z 4 z
Multiplying and divide by z
"   2  3  4 #
1 1 1 1 1 1 1
= z + + + ···
z 2 z 3 z 4 z
   
1 z−1
= −z log 1 − = −z log

T
z z

IS
 −1  
z−1 z
= z log = z log , |z| > 1
z
M z−1
SR
   
1 1 z
7. Z = log , |z| > 1
n−1 z z−1
h,

Proof.
es

∞ ∞  n
X
−n
X 1
Z[f (n)] = f (n)z = f (n)
ur

n=0 n=0
z
.S

  ∞     n  2  3  4
1 X 1 1 1 1 1 1 1 1
Z = = + + + ···
n−1 n − 1 z 1 z 2 z 3 z
r.E

n=2
"   2  3 #
1 1 1 1 1 1
= + + + ···
D

z z 2 z 3 z
   
1 1 z−1
= − log 1 − = − log
z z z
 −1  
1 z−1 1 z
= log = log , |z| > 1
z z z z−1

 
1 1
8. Z = ez
n!
Proof.
∞ ∞  n
X
−n
X 1
Z[f (n)] = f (n)z = f (n)
n=0 n=0
z
  ∞  n    2  3  4
1 X 1 1 1 1 1 1 1 1 1 1
Z = =1+ + + + + ···
n! n=0
n! z 1! z 2! z 3! z 4! z
 5
   2  3  4
1 1 1 1
z z z z 1
= 1+ + + + + · · · = ez
1! 2! 3! 4!

 
1 h 1 i
9. Z =z e −1
z
(n + 1)!
Proof.
∞ ∞  n
X
−n
X 1
Z[f (n)] = f (n)z = f (n)
n=0 n=0
z
  ∞   n
1 X 1 1
Z =
(n + 1)! n=0
(n + 1)! z
   2  3
1 1 1 1 1 1 1
= + + + + ···
1! 2! z 3! z 4! z

T
"    2  3  4 #
1 1 1 1 1 1 1 1
= z + + + + ···

IS
1! z 2! z 3! z 4! z

=
"
z 1+
 
1 1
+
 2
1 1 M +
 3
1 1
+
 4
1 1
+ ··· − 1
#
SR
1! z 2! z 3! z 4! z
h 1 i
= z ez − 1
h,
es

10. Evaluate Z [ean ]

ur

z
w.k.t Z[an ] =
z−a
.S

z
Z [ean ] = Z [(ea )n ] =
r.E

z − ea
D

11. Evaluate Z a−n

 

z
w.k.t Z[an ] =
z−a  n 
 −n   −n   −1
n  1 z az
Z a = Z a =Z a =Z = 1

=
a z− a az − 1

12. Evaluate Z [(−1)n ]

z z
w.k.t Z[an ] = ⇒ Z [(−1)n ] =
z−a z+1

13. Evaluate Z [n + 3]
   
z z
Z [n + 3] = Z [n] + Z [3] = Z [n] + 3Z [1] = +3
(z − 1)2 (z − 1)
6

14. Find Z [sin nθ] , Z [cos nθ]

Proof.

let a = eiθ
an = einθ = cos nθ + i sin nθ
z
w.k.t Z[an ] =
z−a  
 inθ   iθ
n  z
Z e = Z e =
z − eiθ
   
z z
= =
z − (cos θ + i sin θ) (z − cos θ) − i sin θ
   
z (z − cos θ) + i sin θ
= ×
(z − cos θ) − i sin θ (z − cos θ) + i sin θ
z(z − cos θ) + iz sin θ
=

T
(z − cos θ)2 + sin2 θ

IS
z(z − cos θ) + iz sin θ z(z − cos θ) + iz sin θ
Z [(cos nθ + i sin nθ)] = 2 =
2 z 2 − 2z cos θ + 1
M
(z − cos θ) + sin θ
z(z − cos θ) z sin θ
SR
Z [cos nθ] + iZ [sin nθ] = 2 +i 2
z − 2z cos θ + 1 z − 2z cos θ + 1
equating real and imaginary parts
(z 2 − z cos θ)
h,

Z [cos nθ] = 2
z − 2z cos θ + 1
es

z sin θ
Z [sin nθ] = 2
z − 2z cos θ + 1
ur
.S

z 2 − z cosh θ
15.Z [cosh nθ] =
r.E

z 2 − 2z cosh θ + 1
Proof. Simply
D

(z 2 − z cos θ)
Z [cos nθ] =
z 2 − 2z cos θ + 1
(z 2 − z cosh θ)
∴ Z [cosh nθ] = 2
z − 2z cosh θ + 1

z sinh θ
16.Z [sinh nθ] =
z2 − 2z cosh θ + 1
Proof. Simply
z sin θ
Z [sin nθ] =
z2
− 2z cos θ + 1
z sinh θ
∴ Z [sinh nθ] = 2
z − 2z cosh θ + 1
 7

h nπ i z2
17. Prove that Z cos = 2
2 z +1
Solution. We know that

z 2 − z cos θ
Z [cos nθ] =
z 2 − 2z cos θ + 1

z 2 − z cos π2


h nπ i z 2 − z(0) z2
Z cos = 2 = =
z − 2z cos π2 + 1


2 z 2 − 2z(0) + 1 z2 + 1

h nπ i z
18. Prove that Z sin = 2
2 z +1
Solution. We know that
z sin θ
Z [sin nθ] = .
z2 − 2z cos θ + 1

T
Hence

IS
z sin π2


nπ i z(1) z
h
Z sin
2
= 2 π


z − 2z cos 2 + 1
= 2
z − 2z(0) + 1 M
= 2
z +1
SR
h nπ i
19. Find Z sin2
h,

2
es

Solution. We know that

1 − cos 2θ
ur

w.k.tsin2 θ =
2
.S

2 nπ 1 h nπ i
h i
Z sin = Z 1 − cos 2
2 2 2
r.E

1
= [Z[1] − Z [cos nπ]]
2
D

1
= [Z[1] − Z [(−1)n ]]
2 
1 z z
= −
2 z−1 z+1
 
z
=
z2 − 1
8

Property 2. First shifting (or) Frequency shifting Property (or)

Damping Rule
z 
If Z[f (n)] = F (z) then (i) Z[an f (n)] = Z[f (n)]Z→ z = F
a a
Proof.
∞ ∞  n
X
−n
X 1
(i) Z[f (n)] = f (n)z = f (n)
n=0 n=0
z
∞  n
X 1
Z[an f (n)] = n
a f (n)
n=0
z
X∞  a n X ∞  z −n
= f (n) = f (n)
n=0
z n=0
a
z 
= F
a

T
20. Find the Z-transform of rn cos nθ

IS
Solution. M
SR
z(z − cos θ)
w.k.t Z[cos nθ] =
z2 − 2z cos θ + 1
Z[rn cos nθ] = Z[cos nθ]z→ az
h,

 
z(z − cos θ)
es

=
z 2 − 2z cos θ + 1 z→ z
a
ur

z z 
− cos θ
Z[rn cos nθ] = r r
.S

z2 z
2
− 2 cos θ + 1
r.E

r r
z(z − r cos θ)
=
z 2 − 2rz cos θ + r2
D

20. Obtain the Z-transform of rn sin nθ

Solution.
z sin θ
w.k.t Z[sin nθ] =
z2 − 2z cos θ + 1
Z[rn sin nθ] = Z[sin nθ]z→ zr
 
z sin θ
=
z 2 − 2z cos θ + 1 z→ z
r
z
sin θ
Z[rn sin nθ] = r
z2 z
2
− 2 cos θ + 1
r r
rz sin θ
=
z 2 − 2rz cos θ + r2
 9

21. Obtain the Z-transform of nan

Solution.
z
w.k.t Z[n] =
(z − 1)2
 z 
a
= az
Z[nan ] = Z[n]z→ az =  2
z
a
−1 (z − a)2

h nπ i az
22.Show that Z an sin = 2
2 z + a2
Solution.
z sin θ
w.k.t Z [sin nθ] =
z2
− 2z cosθ + 1
π
h nπ i z sin z(1) z
2π 

T
Z sin = = 2 = 2
2 z 2 − 2z cos +1 z − 2z(0) + 1 z +1

IS
2
By Scaling Property
Z[an f (n)] = F
z 
M
SR
a

z
h
n nπ i
a az
Z a sin = 2
= 2
2 z
a2
+1 z + a2
h,
es

h
nnπ i z2
23. Show that Z a cos = 2
2 z + a2
ur

Solution. We know that

.S

z 2 − z cos θ
r.E

Z [cos nθ] =
z 2 − 2z cos θ + 1
D

π 
2
h nπ i z − z cos z 2 − z(0) z2
Z cos =  π 2 = 2 = 2
2 z 2 − 2z cos +1 z − 2z(0) + 1 z +1
2

 2
z
h nπ i a2 z2
Z an cos = z2
=
z 2 + a2


2 a2
+1
10

Theorem
d
Z[nk ] = −z Z[nk−1 ] where k being any positive integer.
dz
Proof. By the definition of Z-transform we have

∞
X
k
Z[n ] = nk z −n (1)
n=0
X∞
Z[nk−1 ] = nk−1 z −n (2)
n=0

Differentiating (2) with respect to z we get

∞
! ∞
d d X
k−1 −n
X d −n

Z[nk−1 ] = n z = nk−1 z
dz dz n=0 n=0
dz
∞
X ∞
X
−n−1 −1
nk z −n

T
k−1
= n (−n)z = −z

IS
n=0 n=0
1
= − Z[nk ]
z M
SR
d
=⇒ Z(nk ) = −z Z(nk−1 )
dz
h,

Using the above the theorem we have the following

es

z
24. Z[n] =
ur

(z − 1)2
.S

d
w.k.t Z[nk ] = −z Z[nk−1 ]
dz
r.E

 
d d z
Z[n] = −z Z(1) = −z
dz dz z − 1
D

   
(z − 1)(1) − z(1) −1 z
= −z = −z =
(z − 1)2 (z − 1)2 (z − 1)2

z2 + z
25. Z[n2 ] =
(z − 1)3

d
w.k.t Z[nk ] = −z Z[nk−1 ]
dz
d
Z[n2 ] = −z Z(n)
dz 
(z − 1)2 (1) − z 2(z − 1)
  
d z
= −z = −z
dz (z − 1)2 (z − 1)4
z2 + z
   
z − 1 − 2z −z − 1
= −z = −z =
(z − 1)3 (z − 1)3 (z − 1)3
 11

2z
26. Z[n(n − 1)] =
(z − 1)3

Z[n(n − 1)] = Z[n2 − n] = Z[n2 ] − Z[n]

z2 + z z [z 2 + z − z(z − 1)]
= − =
(z − 1)3 (z − 1)2 (z − 1)3
[z 2 + z − z 2 + z] 2z
= 3 =
(z − 1) (z − 1)3

z 3 + 4z 2 + z
27. Z[n3 ] =
(z − 1)4

d
w.k.t Z[nk ] = −z Z[nk−1 ]
dz
d
Z[n3 ] = −z Z(n2 )
dz 

T
z2 + z

d

IS
= −z
dz (z − 1)3

M
(z − 1)3 (2z + 1) − (z 2 + z)3(z − 1)2
 
= −z
(z − 1)6
SR
(z − 1)(2z + 1) − 3(z 2 + z)
 
= −z
(z − 1)4
h,

 2
2z − z − 1 − 3z 2 − 3z

= −z
es

(z − 1)4
 2
z 3 + 4z 2 + z

−z − 4z − 1
−z
ur

= =
(z − 1)4 (z − 1)4
.S

1
r.E

28. Find the Z transform of

n(n + 1)
Solution.
D

1 A B
= +
n(n + 1) n n+1
A(n + 1) + Bn = 1

n=0 =⇒ A = 1
n = −1 =⇒ B = −1

     
1 1 1
Z = Z −Z
n(n + 1) n n+1
   
z z
= log − z log
z−1 z−1
 
z
= (1 − z) log
z−1
12

1
29. Find the Z transform of (A.U. Nov/Dec 2006)
(n + 1)(n + 2)
Solution. Let
1 A B
= +
(n + 1)(n + 2) n+1 n+2
A(n + 2) + B(n + 1) = 1

n = −1 =⇒ A = 1
n = −2 =⇒ B = −1

     
1 1 1
Z = Z −Z
(n + 1)(n + 2) n+1 n+2
∞
X 1 1 ∞
X 1 1
= −
n=0
n + 1 z n n=0 n + 2 z n

T
 
11 1 1 1 11 1 1

IS
= 1+ + + ··· − + + + ···
2 z 3 z2 2 3 z 4 z2

= z

1 1 1
+ +
z 2 z2 3 z3
1 1
M 
+ ··· − z

2 1 1
+
1 1
+
2 z2 3 z3 4 z4
1 1
+ ···

SR
     
1 2 1 1
= −z log 1 − − z − log 1 − −
z z z
h,

   
1 1
= −z log 1 − + z 2 log 1 − +z
es

z z
 
z−1
z + (z 2 − z) log
ur

=
z
.S
r.E

Property 4 Second Shifting Property (or) Translation theorem

D

Theorem . If Z [f (n)] = F (z) then

(i)Z[f (n + k)] = z k F (z) − f (0) − f (1)z −1 − f (2)z −2 · · · − f (k − 1)z −(k−1)

 

(ii)Z[f (n − k)] = z −k F (z)

 13

4.3 Inverse Z-Transforms

If Z[f (n)] = F (z) is the Z-transform of a given sequence {f (n)}, then the inverse Z-transform
is defined as

Z −1 [F (z)] = f (n)

Inverse Z-Transform by Partial Fraction

This is the standard method of finding the inverse transform. Let F (z) be the function for
which inverse transform to be find out. For this first construct F (z)
z
then decompose F (z)
z
into
partial fractions, multiply this with z then find the inverse transform.

Examples
 
−1 10z
Example 1. Find Z
(z − 1)(z − 2)

T
IS
Solution.
 
Z −1 10z
(z − 1)(z − 2) M
= Z −1 [F (z)]
SR
10z
F (z) =
(z − 1)(z − 2)
h,

F (z) 10
=
z (z − 1)(z − 2)
es

10 A B
= +
ur

(z − 1)(z − 2) z−1 z−2

=⇒ A(z − 2) + B(z − 1) = 10
.S

Put z = 1 A = −10 ,
r.E

z=2 B = 10
F (z) −10 10
D

= +
z z−1 z−2
−10z 10z
F (z) = +
z − 1 z − 2   
−1 −1 z −1 z
Z [F (z)] = −10Z + 10Z
z−1 z−2
−1 n n n n
Z [F (z)] = 10(−1 + 2 ) = 10(2 − 1 )

z
Example 2. Find the inverse transform of
z2 − 7z + 12
Solution.
 
−1 z
Z 2
= Z −1 [F (z)]
z − 7z + 12
z z
F (z) = =
z2 − 7z + 12 (z − 4)(z − 3)
14

F (z) 1
=
z (z − 4)(z − 3)

1 A B
= +
(z − 4)(z − 3) z−4 z−3
A(z − 3) + B(z − 4) = 1
Put z = 3 B = −1 ,
z=4 A=1
F (z) 1 1
= −
z z−4 z−3
z z
F (z) = −
z − 4 z − 3  
−1 −1 z −1 z
Z [F (z)] = Z −Z
z−4 z−3
−1 n n
Z [F (z)] = 4 − 3

T
IS
z(z 2 − z + 2)
 
Example 3. Find Z −1
(z + 1)(z − 1)2 M
SR
Solution.
z(z 2 − z + 2)
 
h,

−1
Z = Z −1 [F (z)]
(z + 1)(z − 1)2
es

z(z 2 − z + 2)
F (z) =
(z + 1)(z − 1)2
ur

F (z) z2 − z + 2
=
.S

z (z + 1)(z − 1)2
z2 − z + 2
r.E

A B C
2
= + +
(z + 1)(z − 1) z + 1 (z − 1) (z − 1)2
D

=⇒ A(z − 1)2 + B(z + 1)(z − 1) + C(z + 1) = z 2 − z + 2

Substituting z = 1 2C = 2 =⇒ C = 1 ,
z = −1 4A = 4 =⇒ A = 1
z=0 A − B + C = 2,
1 − B + 1 = 2 =⇒ B = 0
F (z) 1 1
= +0+
z z+1 (z − 1)2
z z
F (z) = +
z + 1 (z − 1)2
   
z z
Z −1 [F (z)] = Z −1
−Z −1
z+1 (z − 1)2
Z −1 [F (z)] = (−1)n + n
 15

z2 − z
 
−1
Example 4. Find Z
(z + 1)(z 2 + 1)
Solution.
z2 − z
 
−1
Z = Z −1 [F (z)]
(z + 1)(z 2 + 1)
z2 − z
F (z) =
(z + 1)(z 2 + 1)
F (z) z−1
=
z (z + 1)(z 2 + 1)
z−1 A Bz + C
= +
(z + 1)(z 2 + 1) z+1 z2 + 1

A(z 2 + 1) + (Bz + C)(z + 1) = z − 1

Substituting z = −1 2A = −2 =⇒ A = −1 ,

T
z=0 A + C = −1

IS
−1 + C = −1 =⇒ C = 0
z=1
M
2A + 2B + 2C = 0
SR
−2 + 2B + 0 = 0 =⇒ B = 1
F (z) 1 z
= − + 2
z z+1 z +1
h,

z z2
F (z) = − + 2
es

z + 1 z + 1  2 
z z
Z −1 [F (z)] = −Z −1
+Z −1
ur

z+1 2
z +1
nπ
.S

Z −1 [F (z)] = −(−1)n + cos

2
r.E

z(z + 1)
Example 5. Find the inverse z-transform of .
D

(z − 1)3
Solution.
z(z + 1)
F (z) =
(z − 1)3
F (z) (z + 1) (z − 1 + 2) (z − 1) 2
= = = 3 +
z (z − 1)3 (z − 1)3 (z − 1) (z − 1)3
F (z) 1 2
= 2 +
z (z − 1) (z − 1)3
z 2z
F (z) = 2 +
(z − 1) (z − 1)3
   
z 2z
Z −1 [F (z)] = Z −1
+Z −1
(z − 1)3 (z − 1)3
Z −1 [F (z)] = n + n(n − 1) = n2
16

4.4 Difference equations

A difference equation is a relation between the differences of an unknown function at one or
more general values of the argument.

Formation of Difference Equations

Example 1. Form yn = a(2)n + b(−2)n , derive a difference equation by eliminating the con-
stants.

Solution. Given yn = a(2)n + b(−2)n ,

Solving Difference equations using Z-Transform

Points to Remember

T
Z[y(n)] = F (z)

IS
Z [y(n + 1)] = zF (z) − zy(0)
Z [y(n + 2)] = z 2 F (z) − z 2 y(0) − zy(1)
Z [y(n + 3)] = M
z 3 F (z) − z 3 y(0) − z 2 y(1) − zy(2)
SR

Examples
h,
es

Example 1. Solve un+2 − 5un+1 + 6un = 4n with u0 = 0, u1 = 1.

ur

Solution. The given difference equation is

.S

un+2 − 5un+1 + 6un = 4n , u0 = 0, u1 = 1.

r.E

Consider the difference equation as

D

y(n + 2) − 5y(n + 1) + 6y(n) = 4n , y(0) = 0, y(1) = 1

Z[y(n)] = F (z)
Z [y(n + 1)] = zF (z) − zy(0)
Z [y(n + 2)] = z 2 F (z) − z 2 y(0) − zy(1)

Taking Z-transform on both side of the equation we get

Z [y(n + 2) − 5 y(n + 1) + 6y(n)] = Z [4n ]

z
z 2 F (z) − y(0) − y(1)z −1 − 5z [F (z) − y(0)] + 6F (z) =
 
z−4
z
(z 2 − 5z + 6)F (z) − z =
z−4
z z 2 − 4z + z z 2 − 3z
(z − 2)(z − 3)F (z) = +z = =
z−4 z−4 z−4
 17

z(z − 3) z
F (z) = =
(z − 2)(z − 3)(z − 4) (z − 2)(z − 4)
F (z) 1 A B
= = +
z (z − 2)(z − 4) (z − 2) (z − 4)

⇒ A(z − 4) + B(z − 2) = 1

1
Put z = 2 ⇒ −2A = 1 ⇒ A=−
2
1
Put z = 4, ⇒ 2B = 1 ⇒ B=
2

T
F (z) 1 A 1 B
= −

IS
+
z 2 (z − 2) 2 (z − 4)

M
   
1 z 1 z
F (z) = − +
2 z−2 2 z−4
SR
   
−1 1 −1 z 1 −1 z
Z [F (z)] = − Z + Z
2 z−2 2 z−4
h,

1 1
y(n) = − (2)n + (4)n
2 2
es
ur

Example 2. Solve yn+2 + 6yn+1 + 9yn = 2n with y0 = y1 = 0.

.S

Solution. The given difference equation is

r.E

yn+2 + 6yn+1 + 9yn = 2n , y0 = y1 = 0

D

Consider the difference equation as

y(n + 2) + 6y(n + 1) + 9y(n) = 2n , y(0) = 0, y(1) = 1

Z[y(n)] = F (z)
Z [y(n + 1)] = zF (z) − zy(0)
Z [y(n + 2)] = z 2 F (z) − z 2 y(0) − zy(1)

Taking Z-transform on both side of the equation we get

Z[y(n + 2) + 6y(n + 1) + 9y(n)] = Z[2n ]

z
z 2 [F (z) − y(0) − y(1)z −1 ] + 6z[F (z) − u0 ] + 9F (z) =
z−2
2 z
(z + 6z + 9)F (z) =
z−2
18

z
F (z) = Z [y(n)] =
(z − 2)(z 2
+ 6z + 9)
F (z) 1 1
= 2
=
z (z − 2)(z + 6z + 9) (z − 2)(z + 3)2
F (z) A B C
= + +
z z − 2 z + 3 (z + 3)2

⇒ A(z + 3)2 + B(z − 2)(z + 3) + C(z − 2) = 1

1
Put z = −3 ⇒ −5C = 1 ⇒ C=−
5
1
Put z = 2, ⇒ 25A = 1 ⇒ A=
25

Equating the coefficient of z 2 on both sides we get

T
1

IS
A + B = 0 ⇒ B = −A ⇒ B = −
25
M
SR
     
F (z) 1 1 1 1 1 1
= − −
z 25 z − 2 25 z + 3 5 (z + 3)2
     
1 z 1 z 1 z
h,

F (z) = − −
25 z − 2 25 z + 3 5 (z + 3)2
es

     
−1 1 −1 z 1 −1 z 1 z
Z [F (z)] = Z − Z −
25 z−2 25 z+3 5 (z + 3)2
ur

(2)n (−3)n 1 1
 
(−3)z
.S

−1
= − − Z
25 25 5 (−3) (z + 3)2
r.E

 
az
∵Z −1
= n(a)n
(z − a)2
D

1 1 1
= (2)n − (−3)n + n(−3)n
25 25 15

Example 3. Solve yn+2 + 4yn+1 + 3yn = 3n with y0 = 0, y1 = 1.

Solution. The given difference equation is

yn+2 + 4yn+1 + 3yn = 3n , y0 = 0, y1 = 1

Consider the difference equation as

y(n + 2) + 4y(n + 1) + 3y(n) = 3n , y(0) = 0, y(1) = 1

Z[y(n)] = F (z)
Z [y(n + 1)] = zF (z) − zy(0)
 19

Z [y(n + 2)] = z 2 F (z) − z 2 y(0) − zy(1)

Taking Z-transform on both side of the equation we get

Z[y(n + 2) + 4y(n + 1) + 3y(n)] = Z[3n ]
z
z 2 [F (z) − y(0) − y(1)z −1 ] + 4z[F (z) − y(0)] + 3F (z) =
z−3
2 z
(z + 4z + 3)F (z) − z =
z−3
z
(z + 1)(z + 3)F (z) = +z
z−3
z + z(z − 3) z 2 − 2z
= =
z−3 z−3

z (z − 2)
F (z) =
(z − 3)(z + 1)(z + 3)

T
F (z) (z − 2) A B C
= = + +

IS
z (z − 3)(z + 1)(z + 3) (z − 3) (z + 1) (z + 3)

M
⇒ A(z + 3)(z − 3) + B(z + 1)(z − 3) + C(z + 1)(z + 3) = z − 2
SR
3
Put z = −1 ⇒ −8A = −3 ⇒ A=
8
h,

5
es

Put z = −3, ⇒ 12B = −5 ⇒ B=−

12
ur

1
Put z = 3, ⇒ 24C = 1 ⇒ C=
.S

24
r.E

     
F (z) 3 1 5 1 1 1
∴ = − +
z 8 z+1 12 z + 3 24 z − 3
D

     
−1 3 −1 z 5 −1 z 1 −1 z
Z [F (z)] = Z − Z + Z
8 z+1 12 z+3 24 z−3
     
−1 1 −1 z 1 −1 z 1 z
Z [F (z)] = Z − Z −
25 z−2 25 z+3 5 (z + 3)2
3 5 1
Z −1 [F (z)] = (−1)n − (−3)n + (3)n
8 12 24

Example 4. Solve y(k + 2) − 4y(k + 1) + 4y(k) = 0 with y(0) = 1 and y(1) = 0.

Solution. The given difference equation is

y(k + 2) − 4y(k + 1) + 4y(k) = 0, y(0) = 1, y(1) = 0
Solution. Consider the difference equation as
y(n + 2) − 4y(n + 1) + 4y(n) = 0, y(0) = 1, y(1) = 0
20

Z[y(n)] = F (z)
Z [y(n + 1)] = zF (z) − zy(0)
Z [y(n + 2)] = z 2 F (z) − z 2 y(0) − zy(1)

Taking Z-transform on both side of the equation we get

Z[y(n + 2) − 4y(n + 1) + 4y(n)] = 0

z [F (z) − y(0) − y(1)z −1 ] − 4z[F (z) − y(0)] + 4F (z)
2
= 0
z 2 [F (z) − 1 − 0] − 4z[F (z) − 1] + 4F (z) = 0
(z 2 − 4z + 4)F (z) − z 2 + 4z = 0
(z − 2)2 F (z) = z 2 − 4z

z (z − 4)
F (z) =
(z − 2)2

T
F (z) (z − 4) z−2−2

IS
= 2 =
z (z − 2) (z − 2)2
F (z)
=
(z − 2)
2 −
2 M2 =
1
−
2
SR
z (z − 2) (z − 2) z − 2 (z − 2)2
   
z 2z
F (z) = −
z−2 (z − 2)2
h,

   
z 2z
Z −1 [F (z)] = Z −1
−Z −1
es

z−2 (z − 2)2
n n n
y(n) = (2) − n(2) = (2) (1 − n)
ur
.S

Example 5. Solve the difference equation y(n + 3) − 3y(n + 1) + 2y(n) = 0 given that
r.E

y(0) = 4, y(1) = 0 and y(2) = 8, by the method of Z - transform.

D

Solution. The given difference equation is

y(n + 3) − 3y(n + 1) + 2y(n) = 0, y(0) = 4, y(1) = 0, y(2) = 8

Z[y(n)] = F (z)
Z [y(n + 1)] = zF (z) − zy(0)
Z [y(n + 2)] = z 2 F (z) − z 2 y(0) − zy(1)
Z [y(n + 3)] = z 3 F (z) − z 3 y(0) − z 2 y(1) − zy(2)

Taking Z-transform on both side of the equation we get

Z[y(n + 3)] − 3Z[y(n + 1)] + 2Z[y(n)] = 0

3
F (z) − y(0) − y(1)z − y(2)z −2 − 3z[F (z) − y(0)] + 2F (z) = 0
−1
 
z
using y(0) = 4, y(1) = 0, y(2) = 8
z F (z) − 4 − 8z −2 − 3z[F (z) − 4] + 2F (z) = 0
3
 
 21

(z 3 − 3z + 2)F (z) − 4z 3 − 8z + 12z = 0

(z 3 − 3z + 2)F (z) = 4z 3 − 4z

4z (z 2 − 1)
F (z) = Z [y(n)] =
(z 3 − 3z + 2)
F (z) 4(z 2 − 1) 4(z + 1)(z − 1)
= = 2 =
z (z − 1) (z + 2) (z − 1)2 (z + 2)
F (z) 4(z + 1) A B
= = +
z (z − 1)(z + 2) (z − 1) (z + 2)
⇒ 4(z + 1) = A(z + 2) + B(z − 1)

8
Put z = 1 ⇒ 3A = 8 ⇒ A=
3
4

T
Put z = −2, ⇒ −3B = −4 ⇒ B=
3

IS
F (z)
=
8

1

+
4

1 M

SR
z 3 z−1 3 z+2
   
8 z 4 z
F (z) = +
3 z−1 3 z+2
h,

   
−1 8 −1 z 4 −1 z
Z [F (z)] = Z + Z
es

3 z−1 3 z+2
8 n 4
(1) + (−2)n
ur

y(n) =
3 3
.S

Example 6.Solve the difference equation y(n) + 3y(n − 1) − 4y(n − 2) = 0, n > 2 given that
y(0) = 3, y(1) = −2, by the method of Z - transform.
r.E

Solution. The given difference equation is

D

y(n) + 3y(n − 1) − 4y(n − 2) = 0

Replace n by n + 2, we get

y(n + 2) + 3y(n + 1) − 4y(n) = 0

Z[y(n)] = F (z)
Z [y(n + 1)] = zF (z) − zy(0)
Z [y(n + 2)] = z 2 F (z) − z 2 y(0) − zy(1)

Taking Z-transform on both side of the equation we get

Z[y(n + 2) + 3y(n + 1) − 4y(n)] = 0

z [F (z) − y(0) − y(1)z −1 ] + 3z[F (z) − y(0)] − 4F (z) = 0
2
22

z 2 [F (z) − 3 + 2z −1 ] + 3z[F (z) − 3] − 4F (z) = 0

(z 2 + 3z − 4)F (z) − 3z 2 + 2z − 9z = 0
(z 2 + 3z − 4)F (z) = 3z 2 + 7z

z (3z + 7)
F (z) =
z 2 + 3z − 4
F (z) (3z + 7) A B
= = +
z (z + 4)(z − 1) (z + 4) (z − 1)

⇒ 3z + 7 = A(z − 1) + B(z + 4)

Put z = 1 ⇒ 5B = 10 ⇒ B=2
Put z = −4, ⇒ −5A = −5 ⇒ A=1

T
   
F (z) 1 1

IS
= 1 +2
z z+4 z−1

F (z) = 1

z

+2

zM 
SR
z+4 z−1
   
−1 −1 z −1 z
Z [F (z)] = Z + 2Z
z+4 z−1
h,

n n
y(n) = (−4) + 2(1)
es
ur

Inverse Z-Transform by Residue Theory

.S

Z −1 [F (z)] = sum of residues of f (z) at its poles lies inside C.

r.E

where f (z) = z n−1 F (z) and C is the contour which include all the poles of f (z) .
D

Methods to find Residues

Case (i) : If z = a is a simple pole, then

R(a) = lim (z − a) z n−1 F (z)

z→a

Case (ii) : If z = a is a pole of order m simple pole, then

1 dm−1 
(z − a)m z n−1 F (z)

R(a) = lim m−1
z→a (m − 1)! dz

Examples
10z
Example 1. Find the inverse z-transform of by residue method.
(z − 1)(z − 2)
 23

10z
Solution. Let F (z) =
(z − 1)(z − 2)
−1
Z [F (z)] = sum of the residues , z = 1, 2 are the simple poles (or) pole of order 1

R(1) = lim(z − 1)z n−1 F (z)

z→1
 
n−1 10z
= lim (z − 1)z
z→1 (z − 1)(z − 2)
10z n
 
= lim (z − 1)
z→1 (z − 1)(z − 2)
n
 
10z
= lim
z→1 (z − 2)

R(1) = −10(1n )

R(2) = lim(z − 2)z n−1 F (z)

z→2

T
 
10z

IS
n−1
= lim (z − 2)z
z→2 (z − 1)(z − 2)

= lim (z − 2)

M10z n
(z − 1)(z − 2)

SR
z→1
n
 
10z
R(1) = lim = 10(2)n
z→2 (z − 1)
h,

=⇒ Z −1 [F (z)] = R(1) + R(2) = 10(2n − 1n )

es
ur

z(z + 1)
.S

Example 2. Find the inverse z-transform of by residue method.

(z − 1)3
r.E

z(z + 1)
Solution. Let F (z) =
(z − 1)3
D

Z −1 [F (z)] = sum of the residues , z = 1 is a pole of order 3.

1 d2 n−1 2
 
3z (z + z)
R(1) = lim (z − 1)
z→1 2! dz 2 (z − 1)3
1 d2  n+1
+ zn

= lim 2
z
z→1 2! dz
1 d 
(n + 1)nz n + nz n−1

= lim
z→1 2! dz
1 
(n + 1)nz n−1 + n(n − 1)z n−2

= lim
z→1 2!
1
= [(n + 1)n + n(n − 1)]
2!
1 2
= (n + n + n2 − n) = n2
2!
24

4.5 Convolution Theorem

Convolution. Let {f (n)} and {g(n)} be any two sequences. Then the convolution of these
sequence is defined as another sequence given by

n
X
f (n) ∗ g(n) = f (r)g(n − r)
r=0

Theorem. Let Z [f (n)] = F (z) and Z [g(n)] = G(z). Then

Z [f (n) ∗ g(n)] = F (z) G(z)

Z −1 [F (z)G(z)] = Z −1 [F (z)] ∗ Z −1 [G(z)]

Proof. From the definition of Z-transform we have

T
F (z) G(z) = Z[f (n)] Z[g(n)]

IS
"∞ #" ∞ #
X X
= f (n)z −n g(n)z −n
n=0 n=0 M
SR
∞
X
= [f (0)g(n) + f (1)g(n − 1) + f (2)g(n − 2) + · · · + f (n)g(0)] z −n
n=0
h,

∞
" n #
X X
= f (r)g(n − r) z −n
es

n=0 r=0
n
ur

X
w.k.t f (n) ∗ g(n) = f (r)g(n − r)
.S

r=0
∞
X
[f (n) ∗ g(n)] z −n = Z [f (n) ∗ g(n)]
r.E

=
n=0
D

Formula
1 − an+1
1 + a + a2 + a3 + · · · + an =
1−a
 25

Inverse Z-Transform by Convolution

Examples
z2
 
−1
Example 1. Find Z .
(z − a)(z − b)
   
−1 z n −1 z
Solution. We rnow that Z = a and Z = bn
z−a z−b

z2
   
−1 −1 z z
Z = Z
(z − a)(z − b) (z − a) (z − b)

By Convolution Theorem
   
−1 z −1 z
= Z ∗Z
(z − a) (z − b)
= an ∗ b n
By Defn. Convolution

T
n

IS
X
f (n) ∗ g(n) = f (r)g(n − r)
r=0
n M n n  a r
SR
X X X
n n r n−r r n −r n
a ∗b = ab = ab b = b
r=0 r=0
b r=0
  a   a 2  a n 
= bn 1 + + + ··· +
h,

b b b
es

bn+1 − an+1
 a n+1 
  
ur

1−
= bn  b   n bn+1
a =b 
 
.S

b−a
 
1−
b b
r.E

 n+1 n+1

b − a
= bn+1 n
b b (b − a)
D

2 n+1
− an+1
 
−1 z b
Z =
(z − a)(z − b) b−a

z2
 
−1
Example 2. Using convolution theorem evaluate Z
(z − 1)(z − 3)
Solution.
Derive the solution of Example 1

z2 bn+1 − an+1
 
−1
Z =
(z − a)(z − b) b−a
Put a = 1, b = 3 we get

z2 3n+1 − 1n+1 3n+1 − 1

 
−1
Z = =
(z − 1)(z − 3) 3−1 2
26

8z 2
 
−1
Example 3. Using convolution theorem evaluate Z
(2z − 1)(4z − 1)
Solution.

 
8z 2 z2
 
−1
= Z −1 
 
Z    
(2z − 1)(4z − 1)  1 1 
z− z−
2 4
Derive the solution of Example 1

z2 bn+1 − an+1
 
−1
Z =
(z − a)(z − b) b−a

T
1 1
Put a = , b = we get

IS
2 4
 n+1  n+1  n+1  n+1

z2

1
4
−
1
2
M 1
4
−
1
2
SR
−1
 
Z      =     =  
 1 1  1 1 1
z− z− − −
2 4 4 2 4
h,

"   2(n+1) #
n+1
 n+1  n+1
1 1 1 1
− −
es

2 4 2 2
=  2 =  2
1 1
ur

2 2
.S

"  #
n+1−2  2n+2−2
1 1
= −
r.E

2 2
"   2n #
n−1
8z 2
 
1 1
D

−1
Z = −
(2z − 1)(4z − 1) 2 2

8z 2
 
−1
Example 4. Using convolution theorem evaluate Z
(2z − 1)(4z + 1)
Solution.
8z 2 z2
   
−1 −1
Z = Z
(2z − 1)(4z + 1) (z − 12 )(z + 14 )
 n "  n+1 #
2 1 1
= 1− −
3 2 2

Derive the solution of Example 1

z2 bn+1 − an+1
 
−1
Z =
(z − a)(z − b) b−a
 27

1 1
Put a = , b = − , we get
2 4
 n+1  n+1  n+1  n+1
1 1 1 1
 
2 − − − −
z 4 2 4 2
Z −1 
 
     =     =  
 1 1  1 1 3
z− z+ − − −
2 4 4 2 4
"  #
n+1  n+1    n    n 
4 1 1 4 1 1 1 1
= − − = − − −
3 2 4 3 2 2 4 4
n n n n
8z 2
            
−1 2 1 1 1 2 1 2 1
Z = + − = + −
(2z − 1)(4z + 1) 3 2 2 4 3 2 3 4

 
 1 
Example 7. Using convolution theorem and evaluate Z −1 
 −1
 
−1 
z z

T
1− 1+
2 4

IS
Solution.
M
SR
   
1 1
Z −1   = Z −1  
   

z −1 z −1 
      
h,

 1 1 
1− 1+ 1− 1+
2 4 2z 4z
es

 
2
ur

 
−1
 1  −1 8z
= Z  2z − 1   4z + 1   = Z
 
(2z − 1) (4z + 1)
.S

2z 4z
r.E

FromExample 8. we get the Solution.

D

z2
 
−1
Example 9. Find Z .
(z − a)2
 
−1 z
Solution. We know that Z = an
z−a

z2
   
−1 −1 z z
Z = Z
(z − a)2 (z − a) (z − a)
   
−1 z −1 z
= Z ∗Z
(z − a) (z − a)
n
X Xn
= an ∗ an = ar an−r = ar an a−r
r=0 r=0
= an + an + · · · + an (n + 1) times
= (n + 1)an
28

z2
 
−1
Example 10. Find Z .
(z + a)2
 
−1 z
Solution. We know that Z = (−a)n
z+a

z2
   
−1 −1 z z
Z = Z
(z + a)2 (z + a) (z + a)
   
−1 z −1 z
= Z ∗Z
(z + a) (z + a)
Xn
= (−a)n ∗ (−a)n = (−a)r (−a)n−r
r=0
= (−a) + (−a) + · · · + (−a)n
n n
(n + 1) times
= (n + 1)(−a)n

z2
 
−1
Example 11. Find Z .

T
(z − 1)3

IS
 
−1 z
Solution. We rnow that Z = 1n
z−1
M
SR
z2
   
−1 −1 z z
Z = Z
(z − 1)3 (z − 1) (z − 1)2
   
z z
h,

−1 −1
= Z ∗Z
(z − 1) (z − 1)2
es

= 1n ∗ n = n ∗ 1n
n
ur

X
= r1n−r = 1 + 2 + · · · + n
.S

r=0
n(n + 1)
=
r.E

2
D

z3 z3
   
−1 −1
Example 12. Find Z and hence deduce Z .
(z − a)3 (z − 1)3
z2
   
−1 z n −1
Solution. We know that Z = a ,Z = (n + 1)an
z−a (z − a)2

z3 z2
   
−1 −1 z
Z = Z
(z − a)3 (z − a)2 (z − a)
z2
   
−1 −1 z
= Z ∗Z
(z − a)2 (z − a)
n n
= (n + 1)a ∗ a
Xn X n n
X
r n−r r n −r n
= a (r + 1)a = a (r + 1)a a = a (r + 1)
r=0 r=0 r=0
n
= a [1 + 2 + · · · + (n + 1)]
3
 
z (n + 1)(n + 2)
Z −1 = an
(z − a)3 2
 29

z3
 
−1 (n + 1)(n + 2)
Z = 1n
(z − 1)3 2

z3
 
−1
Example 13. Find Z using Convolution Theorem.
(z − 2)2 (z − 3)
z2
   
−1 z n −1
Solution. We know that Z = a ,Z = (n + 1) an
z−a (z − a)2
z3 z2
   
−1 −1 z
Z = Z .
(z − 2)2 (z − 3) (z − 2)2 (z − 3)
z2
   
−1 −1 z
= Z ∗Z
(z − 2)2 (z − 3)
n n
= (n + 1) 2 ∗ 3
n n n  r
X
r n−r
X
r n −r n
X 2
= (r + 1) 2 .3 = (r + 1) 2 .3 .3 = 3 (r + 1)
r=0 r=0 r=0
3

T
"    2  3  n #
n 2 2 2 2
= 3 1+2 +3 +4 + · · · + (n + 1)

IS
3 3 3 3

Let x =
 
2 M
SR
3

S = 1 + 2x + 3x2 + 4x3 + · · · + (n + 1)xn

 
h,

xS = x + 2x2 + 3x3 + 4x4 + · · · + nxn + (n + 1)xn+1

 

S − xS = 1 + x + x2 + x3 + · · · + xn − (n + 1)xn+1
 
es

1 − xn+1
 
(1 − x) S = − (n + 1)xn+1
ur

1−x
.S

(1 − xn+1 ) (n + 1)xn+1
S = −
(1 − x)2 (1 − x)
r.E

 
n+1
1 − 23


n+1
(n + 1) 23
=
2 −
D

1 − 23


1 − 23
"  n+1 #  n+1
2 2
= 9 1− − 3(n + 1)
3 3
 n+1  n+1
2 2
= 9−9 − 3(n + 1)
3 3
 n+1
2
= 9 − (9 + 3n + 3)
3
 n+1
2
= 9 − (3n + 12)
3
 n+1
2
S = 9 − 3 (n + 4)
3

z3
 
−1
Z = 3n .S
(z − 2)2 (z − 3)
30
"  n+1 #
2
= 3n 9 − 3 (n + 4)
3
 n+1
n+2 n+1 2
= 3 −3 (n + 4)
3
n+2 n+1
= 3 − (n + 4) 2

4.6 Initial and Final value Theorem

Initial value Theorem
If Z[f (n)] = F (z) then f (0) = lim F (z).
z→∞

Proof. By the definition of Z transform we have

F (z) = Z[f (n)]

T
∞

IS
X
= f (n)z −n
n=0
M
= f (0) + f (1)z −1 + f (2)z −2 + · · · +
SR
f (1) f (2)
= f (0) + + 2 + ···
z z
h,

As z → ∞ we get
es

 
f (1) f (2)
lim F (z) = lim f (0) + + 2 + ···
z→∞ z→∞ z z
ur

= f (0)
.S

Final value Theorem

r.E

If Z[f (n)] = F (z) then lim f (n) = lim(z − 1)F (z).

D

n→∞ z→1

Proof. By the definition we have

∞
X ∞
X
−n
Z[f (n)] = f (n)z , Z[f (n + 1)] = f (n + 1)z −n
n=0 n=0

∞
X
Z[f (n + 1) − f (n)] = [f (n + 1) − f (n)]z −n
n=0
∞
X
Z[f (n + 1)] − Z[f (n)] = [f (n + 1) − f (n)]z −n
n=0
X∞
z[F (z) − f (0)] − F (z) = [f (n + 1) − f (n)]z −n
n=0
∞
X
(z − 1)F (z) − zf (0) = [f (n + 1) − f (n)]z −n
n=0
 31

As z → 1 we get
∞
X
lim (z − 1)F (z) − 1f (0) = lim [f (n + 1) − f (n)]z −n
z→1 z→1
n=0
∞
X
= [f (n + 1) − f (n)]
n=0
n
X
= lim [f (n + 1) − f (n)]
n→∞
k=0
= lim [[f (1) − f (0)] + [f (2) − f (1)] + · · · [f (n + 1) − f (n)]]
n→∞
= lim [f (n + 1) − f (0)]
n→∞
lim (z − 1)F (z) − f (0) = lim [f (n + 1)] − f (0)
z→1 n→∞
lim (z − 1)F (z) = lim f (n + 1)

T
z→1 n→∞

IS
Since lim f (n + 1) = lim f (n) we have
n→∞ n→∞
M
lim f (n) = lim(z − 1)F (z)
SR
n→∞ z→1
h,
es
ur
.S
r.E
D