TOPICS:

1. Z-Transforms –Elementary properties.

2. Inverse Z-Transform (using partial fractions
and residues)
3. Convolution theorem.
4 .Formation of difference equations
5.Solutions of difference equations using
Z-Transform
 Introduction
•The Z-Transform plays an important role in the
communication engineering.
•In communication engineering there are two
basic types of signals are encountered.
• They are continuous time signal and discrete
time signals.
•The continuous time signals are defined by the
independent variable time and are denoted by a
function f(t)..
• On the other hand, discrete time signals are
defined only at discrete set of values of the
independent variable and are denoted by a
sequence {x(n)}.

• For the continuous time signal, Laplace

transform and Fourier transform play important
role.

• Z- Transform plays an important role in discrete

time signal analysis.
Definition 1
Let {x(n)} be a sequence defined for n = 0, ±1, ±2, ±3, ⋅⋅⋅.
Then the two sided Z-transform of the sequence x(n) is
defined as ∞
Z{x(n)} = X(z) = ∑
n = −∞
x(n)z −n

where z is a complex variable in general.

Definition 2
If {x(n)} is a casual sequence, i.e., x(n) = 0 for n < 0, then
the Z-transform reduces to one-sided Z-transform and is
defined as ∞
Z{x(n)} = X(z) = ∑ x(n)z
−n

n =0
∞

Note: The infinite series ∑ x(n)z will be convergent only for

−n

n =0

certain values of z depending on the sequence x(n).

Definition 3
The inverse Z-transform of Z{x(n)} = X(z) is defined
as
Z-1{x(z)} = {x(n)}
Definition 4
The unit sample sequence δ(n) is defined as the
sequence with values
δ(n) = 1 for n = 0
0 for n ≠ 0

Definition 5
That unit step sequence u(n) has values
u(n) = 1 for n ≥ 0
0 for n < 0
Definition 6
If f(t) is a function defined for discrete values of t
where t = nT, n = 0, 1, 2, … T being the sampling
period, then Z-transform of f(t) is defined as
∞ ∞
Z[f(t)] = ∑ f(t)z
n =0
−n
= ∑ f(nT)z
n =0
−n

Now we follow the notations

(i) Z[f(t) = F(z) and
(ii) Z{x(n)} = X(z)
(iii) We shall mostly deal with one sided Z-
transform which will be here after referred to as Z-
transform.
Theorem 1 The Z-transform is linear
i.e., (i) Z[af(t)+bg(t)] = aZ[f(t)] + bZ[g(t)]
(ii) Z[a{x(n)} + b{y(n)}] = aZ{x(n)} + bZ{y(n)}
Proof
∞
(i) Z[af(t) + bg(t)] = ∑ [af(t) + bg(t)]z −n

n =0

∞ ∞
= a ∑ f(nT)z −n
+ b∑ g(nT)z −n
n =0 n =0

= aZ[f(t)] + bZ[g(t)]

= aF(z) + bG(z)
∞

(ii) Z[a{x(n)} + b{y(n)}] = ∑ [ax(n) + by(n)]z − n

n =0

∞ ∞
= a ∑ x(n)z −n
+ b∑ y(n)z−n
n =0 n =0

= aX(z) + bY(z)
= aZ{x(n)}+ bZ{y(n)}
Theorem 2 Frequency shifting
(i) Z[a f(t)] = F az 
n

z
(ii) Z[a n x(n)] = X 
a

Proof
∞

(i) Z[a f(t)] = ∑ a f(nT)z

n

n =0
n −n

∞ −n
z
= ∑ f(nT)  
n =0 a
z
= F 
a
∞
(ii) Z[a x(n)] = ∑ a n x(n)z −n
n

n =0
∞ −n
z
= ∑ x(n)  
n =0 a
z
= X 
a
Theorem 3
(i) Z[nf(t)] = −z d
dz Z[f(t)] = −z dzd F(z)

(ii) Z[nx(n)] = −z Z[x(n)] = −z X(z)

d
dz
d
dz

Proof
∞

(i) Z[f(t)] = ∑ f(nT)z

n =0
−n

Differentiating w.r.t ‘z’

∞
d
Z[f(t)] = ∑ f(nT) − nz −n −1
dz n =0

d ∞
z −n
F(z) = −∑ nf(nT)
dz n =0 z
∞
dF(z)
z = −∑ nf(nT)z −n
dz n =0

= − Z[nf(t)]
d
∴ Z[nf(t)] = −z F(z)
dz
 ∞

(ii) X(z) = Z{x(n)} = ∑ x(n)z , differentiating w.r.t ‘z’

n =0
−n

∞
d
X(z) = ∑ x(n)(− n) ⋅ z −n −1
dz n =0

1 ∞
= − ∑ nx(n)z −n
z n =0

1
= − Z{nx(n)
z

d
∴ Z{nx(n)} = −z X(z)
dz
Theorem 4
(i) Z[f(t + T)] = z[F(z) − f(0)]
 f(1 ⋅ T) f(z ⋅ T) f[(k − 1)T] 
(ii) Z[f(t + kT)] = z k F(z) − f(0 ⋅ T) − − −
 z z 2
z k −1 
∞

Proof (i) Z[f(t + T) = ∑ f(nT + T)z −n

n =0
∞
= ∑ f[(n + 1)T]z −n (Put n + 1 = m)
n =0
∞
= ∑ f(mT)z −(m−1)
m =1
∞
= ∑ f(mT)z − m ⋅ z
m =1

∞ 
= z ∑ f(mT)z −m 
 m=1 
∞ 
= z  ∑ f(mT)z −m − f(0)
 m =0 
Extending this result, we get = z[F(z) − f(0)]
(ii) Z[f(t + nT)] = Z{f[(n + k)T]}
∞
= ∑ f[(n + k)T]z −n (Putn + k = m)
n =0

∞
= ∑ f(mT)z
m=k
−(m − k)

∞
=z k
∑ f(mT)z
m =k
−m

∞ k −1

= z  ∑ f(mT)z − ∑ f(mT)z −m 
k −m

 m =0 m =0 
 f(T) f(2T) f[(k − 1)T] 
= z k F(z) − f(0) − − 2 − ... −
 z z z k −1 
Theorem 5
Shifting theorem
If Z[f(t)] = F(z) then Z[e − aT f(t)] = F[ze aT ]

∞
Proof Z[e −aT
f(t)] = ∑ e −anT f(nT)z n
n =0

∞
= ∑ f(nT)(zeaT ) −n
n =0

= F[zeaT ]

= [F(z)]z→zeaT
Theorem 6
Initial value theorem
If Z[f(t)] = F(z) then f(0) = lim F(z)
z →∞

Proof F(z) = Z[f(t)]

∞
= ∑ f(nT)z −n
n =0

f(1⋅ T) f(2 ⋅ T)
= f(0 ⋅ T) + + 2
+ ...
z z
f(T) f(2T)
= f(0) + + 2 + ...
z z

Taking limit as z → ∞
lim F(z) = f(0)
z→∞
Theorem 7
Final value theorem
If Z[f(t)] = F(z) then lim f(t) = lim (z − 1)F(z)
t →∞ z →1

Proof ∞
Z [ f (t + T ) − f (t )] = ∑[ f (nT + T ) − f (nT )]z −n
n =0

∞
Z [ f (t + T )] − Z [ f (t )] = ∑[ f (nT + T ) − f (nT )]z −n
n =0

∞
zF ( z ) − zf (0) − F ( z ) = ∑[ f (nT + T ) − f (nT )]z −n
n =0
Taking limit as z → 1
∞
lim (z − 1)F(z) − f(0) = lim ∑ [f(nT + T) − f(nT)]z −n
z→1 z →1
n =0
∞
= ∑ [f(nT + T) − f(nT)]
n =0

= lim [f(T) − f(0) + f(2T) − f(T) + ... + f[(n + 1)T] − f(nT)]

n →∞

= lim f[(n + 1)T] − f(0)

n →∞

= f(∞) − f(0)
lim (z − 1)F(z) − f(0) = f(∞) − f(0)
z→1

∴ f(∞ ) = lim f(t) = lim (z − 1)F(z)

t →∞ z →1
 Convolution of Sequences
The convolution of two sequences {x(n)} and {y(n)}
is defined as x(n) * y(n) = w(n)
∞
w(n) = ∑ x(k)y(n − k)
k = −∞
(if the sequences are non-casual)

n
= ∑ x(k)y(n − k) (if the sequences are casual)
k =0

The convolution of two functions f(t) and g(t) is

defined by
n
f(t) ∗ g(t) = ∑ f(kT)g[(n − k)T]
k =0
Theorem 11 Convolution theorem
(i) if Z{x(n)} = X(z) and
Z{y(n)} = Y(z), then
Z{x(n) ∗ y(n)} = X(z) ⋅ Y(z)
(ii) if Z{f(t)} = F(z) and
Z{g(t)} = G(z), then
Z{f(t) ∗ g(t)} = F(z) ⋅ G(z)

Proof (i) Let Z{x(n)} = X(z)

Z{y(n)}= Y(z)
∞ −n  
∞

X(z) Y(z) = ∑ x(n)z  ∑ y(n)z−n 
 n =0   n =0 
∞
 n

= ∑ ∑ x(n)y(n − k) z −n
n = 0  k =0 
∞
n 
= ∑ ∑ x(n − k)y(n) z −n
n = 0  k =0  (1)
By definition
∞
Z[x(n) ∗ y(n)] = ∑ [x(n) ∗ y(n)]z−n
n =0

∞
n 
= ∑ ∑ x(k)y(n − k) z −n (2)
n =0  k =0 

From equation (1) and (2)

Z[x(n) ∗ y(n)] = X(z) ⋅ Y(z)

= Z[x(n)]⋅ Z[y(n)]

Note: Z −1[X(z)Y(z)] = x(n) ∗ y(n)

= Z −1[X(z)] ∗ Z −1[Y(z)]
(ii) If F(z) and G(z) are one sided Z-transform of f(t)
and g(t)
∞ −m  
∞

F(z)G(z) =  ∑ f(mT)z  ∑ g(nT)z −n 
 m =0   n =0 
∞ ∞
= ∑∑[f(mT)g(nT)z −m z −n ]
n =0 m =0

∞
n 
= ∑ ∑ f(kT)g{(n − k)T}z −n
n =0  k =0 
∞
= ∑ [f(t) ∗ g(t)] z −n
n =0

= Z[f(t)∗ g(t)]
n
Q f(t) ∗ g(t) = ∑ f(kT)g{(n − k)T}
k =0

∴ Z[f(t) ∗ g(t)] = F(z) ⋅ G(z)

z-Transforms of some basic functions
Result 1
Z{δ(n)} = 1

Proof
δ(n) = 1 for n = 0
0 for n ≠ 0

∞
∴ Z{δ(n)} = ∑ δ(n)z −n
n =0

=1
Result 2
Where u(n) is an unit step
sequence

Z[1]

z
Note : Z{k} = kZ{1} = k if z > 1
z −1
Result 3
z
Z{a n } = Z{a n } = if z > a
z−a

Proof
∞
Z{a } = ∑ a n z −n
n

n =0
∞ n
a
= ∑ 
n =0  z 

1 a
= if <1
1 − (a/z) z
z
= ,
z−a
z
Z{(a ) n } =
z−a
Result 4
z
Z{a n } = Z{a n u(n)} = if z > a
z−a

Proof ∞
Z{a u(n)} = ∑ a n z −n
n

n =0

∞ n
a
= ∑ 
n =0  z 

1 a
= if <1
1 − (a/z) z
z
= , 1< z
z +1

z
Z{(−1) n } = if z > 1
z +1
Result 5
z
Z{n} =
(z − 1) 2

Proof
d
Z{n} = z{n.1} = − z z{1}
dz
d
(by Theorem Z{nx(n)} = − z z{x(n)})
dz
d  z 
= −z  
dz  z − 1 
 (z − 1)1 − z ⋅ 1
= −z  
 (z − 1)
2

z
Z{n} =
(z − 1) 2
Result 6
az
Z{na n } =
(z − a) 2

Proof Z{na n } = −z
d
z{a n }
dz

d
by Theorem Z{nx(n)} = −z z{x(n)}
dz

d  z  z
= −z   ∴ Z{a n } =
dz  z − a  z−a

 (z − a)1 − z ⋅ 1
= −z  
 (z − a)
2


az
Z{na n } =
(z − a) 2
Result 7
z(z + 1)
Z{n2 } =
(z − 1)3

Proof
d
Z{n 2 } = z{n.n} = − z z{n}
dz
d
by Theorem Z{nx(n)} = −z z{x(n)}
dz
d  z 
= −z  
dz  (z − 1)2 
 (z − 1) 21 − z ⋅ 2(Z - 1) 
= −z  
 (z − 1) 4

Z(Z + 1)
Z{n 2 } =
(z − 1)3
Result 8 2z
Z{n(n – 1)} =
( z − 1) 3

Proof : Z{n(n – 1)} = Z ( n

2
− n ) = z(n 2 ) − z(n )
z ( z + 1) z
= −
( z − 1)3 ( z − 1) 2
z 2 + z − z ( z − 1)
=
( z − 1)3
2z
=
( z − 1)3
Result 10
Find the Z – transform of the sequences
fn = (n + 1)(n + 2) and gn = n(n – 1)
Proof z{f(n)} = z{(n + 1)(n + 2)}
= z{n2 + 3n + 2}
= z{n2} + 3z{n} + z(2)
z(z + 1) 3z 2z
= + +
(z − 1) 3 (z − 1) 2 z − 1

z 2 + z + 3z(z − 1) + 2z(z − 1) 2
=
(z − 1) 3
z 2 + z + 3z 2 − 3z + 2z 3 − 4z 2 + 2z
=
(z − 1) 3
2z 3
=
(z − 1) 3
Z{g(n)} = Z{n(n – 1)}
= Z{n2 – n}
= Z{n2} – Z{n}
z(z + 1) z
= −
(z − 1) 3 (z − 1) 2
z 2 + z − z(z − 1)
=
(z − 1) 3
2z
=
(z − 1) 3
Result 11: Find the Z – transform ( i ) { a n cos n θ } and ( ii ) {a n sin n θ }
( i ) We know that
Proof
z
Z {a n } =
z−a
put a = r e i θ , we get
z z
Z {( r e i θ ) n } = ( byresult 3 : z { a n
} = )
z − r eiθ z−a
z
Z {r n e i n θ } = iθ
( puta = re iθ )
z −re
z
Z { r n (cos n θ + i sin n θ )} =
z − r (cos θ + i sin θ )
z
Z { r n cos n θ + i r n sin n θ } =
( z − r cos θ ) − i r sin θ
z [( z − r cos θ ) + i r sin θ ]
=
[( z − r cos θ ) − i r sin θ ][( z − r cos θ ) + i r sin θ ]
z ( z − r cos θ ) + i z r sin θ
=
( z − r cos θ ) 2 + r 2 sin 2 θ
z ( z − r cos θ ) + i z r sin θ
=
z 2 − 2 zr cos θ + r 2
Equating R . P and I . P , we get
z ( z − r cos θ ) z r sin θ
Z {r n cos n θ } = and Z { r n
sin n θ } =
z 2 − 2 zr cos θ + r 2 z 2 − 2 zr cos θ + r 2
Result 12
Find Z(1/n)
Proof  1  ∞ 1 −n
Z  = ∑ z
 n  n =1 n
1 1 1
= + 2 + 3 +L x2 x3
− log(1 − x) = x + + + ..........
z 2z 3z 2 3
 1
= −log1 − 
 z
 z −1
= −log 
 z 
 z 
= log 
 z −1 
Result 13
Result 14

Result 15
Result 16

Result 17
 Inverse z-transforms
The inverse z-transform of X(z) defined as
Z-1[X(z)] = x(n)
When X(z) = Z[x(n)]. X(z) can be expanded in a
series of ascending powers of z-1, by binomial
exponential, logarithmic theorem, the coefficient
of z-n in the expansion gives Z-1[X(z)].
Z-1[X(z)] can be found out by any one of the
following methods.
Methods to find inverse Z-transform:
Z-1[X(z)] can be found out by any
one of the following methods.

(i)Method-I Using Convolution theorem

(ii)Method-II UsingCauchys residue theorem
(iii)Method-III Using Partial Fractions method
Model I : Using convolution theorem
1. Using convolution theorem, find the inverse
1
Z – transform of ( z − 1)( z − 2)
Solution: Z  1  = Z  1  ∗ Z  1 
−1 −1 −1

 ( z − 1)( z − 2)   z − 1  z − 2
= 1 ∗ 2 n −1
2n 1 n
=1 ∗ = (2 ∗ 1)
2 2
1 n r n−r
= ∑ 2 .(1)
2 r =0
1
= [1 + 2 + 2 2 + 2 3 + .......... + 2 n ]
2
1 2 n +1 − 1
=
2 2 −1
2 n +1 − 1
=
2
2. Using convolution theorem, find the inverse
2
Z – transform of ( z +z a) 2

Solution:  z
−1
2
  z z 
−1
Z  2
= Z  .
 ( z + a)   z + a z + a 
 z  −1  z 
= Z −1  ∗ Z
 z + a   z + a 

= (− a ) n ∗ (− a ) n
n
= ∑ (− a ) r (− a ) n − r
r =0
n
= ∑ (−a) n
r =0

= (n + 1)(−a) n
3. Using convolution theorem, find the inverse
2
Z – transform of ( z + az)(z + b)
Solution: Z  z
−1 2
=Z
 z
.
z  −1
 ( z + a )( z + b)   z + a z + b 
 
 z  −1  z 
= Z −1  ∗ Z
 z + a   z + b 

= (−a) n ∗ (− b) n
n
= ∑ (−a) r (− b) n − r
r =0
n
= (− b) n
∑ (−a) (−b)
r =0
r −r

 −a  n r

= (− b) ∑  n

r =0  − b 
 r
n
a
= (− b) ∑  
n

r =0  b 

  a   a 
2
 a 
n

= (−b) 1 +   +   + ......... +   
n

  b   b   b  

  a  n +1 
1 −   
= (−b) n    
b
 a 
 1 −   
 b 

 (b n +1 − a n +1 )/b n +1 
= (−1) b 
n n

 (b − a)/b 

 b n +1 − a n +1 
= (−1) n

 b − a 
4. Using convolution theorem, find the inverse
2
12 z
Z – transform of (3z − 1)(4 z + 1)
Solution:  12 z   12 z
−1
2



 z−1

 2
−1
2
Z   =Z  =Z  
 (3 z − 1)(4 z + 1)   3 z − 1 4 z + 1    z − 1   z + 1  
   
3 

4     
3

4  
 z  −1  z 
= Z −1  ∗ Z
 z − 1/3   z + 1/4 

= (1/3) n ∗ (−1/4) n
= (−1/4) n ∗ (1/3) n
n
= ∑ (−1/4) r (1/3) n − r
r =0

 −1
n n r
1
= 
 3
∑   (3)
r =0  4 
r

 −3
n n r
1
= 
 3
∑
r =0

 4


 1
n
  − 3   − 3 2  −3 
n

=  1 +  +  + ......... +   
3   4   4   4  

  − 3  n +1 
n 1 −   
1   4  
= 
 3  1−  − 3 
   
  4  

 1  4   − 3   − 3 
n n

=   1 −    
   
3 7 4   4 

1
n
 4 3  − 3 n 
=   +   
 3  7 7  4  
5. Using convolution theorem, find the inverse
2
z
Z – transform of ( z − 4)( z − 3)
Solution: Z  z  = Z  z . z 
−1
2
−1

 (z − 4)(z − 3)   z − 4 z − 3
 z  −1  z 
= Z −1  ∗ Z
 z − 4   z − 3 

= (4) n ∗ (3) n
n
= ∑ (4) r (3) n − r
r =0
n
=3 n
∑ (4) (3)
r =0
r −r

r
4
n
= 3n ∑  
r =0  3 
   4   4 2 4 
n

= 3 1 +   +   + ......... +   
n

  3   3   3  

 (4/3)n +1 − 1
=3 n

 (4/3 ) − 1 
 (4 n +1 − 3n +1 )/3n +1 
=3 n

 1/3 

= 4 n +1 − 3n +1
6. Using convolution theorem, find the inverse
3
 z 
Z – transform of  z − 4 
Solution:  z  
−1
3
 z    z 
−1
2
−1
Z    = Z    ∗ Z   − − − − − (1)
 z − 4    z − 4    z − 4 

 z  2 
−1 −1  z  −1  z 
Z    = Z   ∗ Z  
 z − 4    z − 4    z − 4 
= 4n ∗ 4n
n n
= ∑ (4) (4) r n −r
= ∑ (4) n
r =0 r =0

= (n + 1)(4) n
Equation (1) becomes
 z 3 
−1
  = (n + 1) 4 ∗ 4
n n
Z 
 z − 4  
n
= ∑ ( r + 1)(4) r ( 4) n − r
r =0
n
= ∑ (r + 1)(4) n
r =0

= 4 n [1 + 2 + 3 + ......... + (n + 1)]

(n + 1)(n + 2)
= 4n
2
6. Using convolution theorem, find the inverse
3
Z – transform of ( z − 2)z ( z − 3)
2

Solution:

−1 z3  −1  z2 z 
Z   = Z .
 (z − 2) 2 z − 3 
 (z − 2) 2
(z − 3)   
 z2  −1 −1  z 
=Z  2
∗ Z  z − 3 
 ( z − 2) 
= (n + 1)(2) n ∗ (3) n
n
= ∑ (r + 1)(2) r (3) n − r
r =0
n
=3 n
∑ (r + 1)(2) (3)
r =0
r −r

r
2
n
= 3n ∑ (r + 1) 
r =0 3
 2 2
2
2
3
2 
n

= 3 1 + 2   + 3   + 4   + ......... + (n + 1)   
n

 3 3 3  3  

   2  n+1 n +1
2 
1 −   (n + 1)  
= 3n    2 − 3 
3
  2 2 
 1 −  1 − 
 3  3 
  2 n  2   2   2 
n

1 −     (n + 1)    
= 3n      −  3   3 
3 3
 1 1 
 9 3 
 
 
  2 
n
 2  
 
  2 
n
 2 
= 3 91 −     − 3(n + 1)   
n

   3   3    3   3 
 2
n
2 
n

= 3 9 − 6   − 2 (n + 1)  
n

 3  3  

  2 n 
= 3 9 −   (6 + 2n + 2)
n

  3  
  2 n 
= 3 9 −   (2n + 8)
n

  3  
= 9.3n − 2 n (2n + 8)
7. Using convolution theorem, find the inverse
2
Z – transform of ( z − 1)z ( z − 2) 2

Solution: Z  z
−1 

2
= Z
 z
 .
z 

−1

 (z − 1) (z − 2)   (z − 1) z − 2
2 2

 z  −1  z 
= Z −1  2
∗ Z  z − 2 
 ( z − 1) 
= n (1) n ∗ (2) n = n ∗ 2 n
n
= ∑ r (2) n − r
r =0
n
=2 n
∑
r =0
r (2) − r
r
1 n
= 2 ∑ r 
n

r =0 2
  1   1 
2
 1 
3
 1 
n

= 2 0 + 1  + 2   + 3   + ......... + n   
n

 2 2 2  2  

n −1
1 1 
2
1 1
= 2   1 + 2   + 3   + ......... + n   
n

 2   2 2  2  
   1 n 1 
n

n 1 −   n  
2  2 2 
= −
2   1 2 −
1 
 1 −  1
2 
  2  
  1 n 1 
n

1 −   n  
2n   2 
−   
2
=
2  1 1 
 4 2 
 

2n    1  n    1  n 
= 41 −    − 2n   
2    2     2  

2n  1
n
1 
n

= 4 − 4   − 2 n   
2  2  2  
= 2.2 n − 2 − n
Model II : Using Cauchy’s residue theorem
By using the theory of complex variables, it can be
shown that the inverse Z-transform is given by
1 n −1
x(n) = ∫
2πi c
X(z).z dz

Where c is the closed contour which contains all

the isolated singularities of X(z) and containing the
origin of the Z-plane in the region of convergence.
By Cauchy’s Residue theorem.
x(n) = Sum of the residue of X(z) zn-1 at the isolated
singularities.
Where
1. Residue for simple pole z = a is
lim [(z − a)X(z).zn −1 ]
z →a

2. Residue or order r at the pole Z = a is

 1 d r −1 n −1 
lim  (z − a) r
X(z)z 
Z→a (r − 1)! dz r −1
 
  z ( z 2 − z + 2) 
−1
1. Find Z  using residue method.
2
 ( z + 1)( z − 1) 

Solution: Let Z { f ( z)} = f (n) = sum of the residues of

−1

 z ( z − z + 2)
2

 at its poles.
n −1
 . z
 ( z + 1 2
)( z − 1) 
(i.e.) f(n) = sum of the residues of  z( z (+z1)(−zz−+1)2)  at its
n 2

poles.
Poles of f(z).z n −1 are

(z + 1)(z − 1) 2 = 0

⇒ z = −1, 1

z = –1 is the simple pole

and z = 1 is the pole of order 2.
 z n ( z 2 − z + 2)
Re s ( z = −1) = lim ( z + 1)
z →−1 ( z + 1)( z − 1) 2
z n ( z 2 − z + 2)
= lim
z →−1 ( z − 1) 2
(−1) n (1 + 1 + 2)
=
4
= (−1)n
1  d  2 z ( z − z + 2) 
n 2

Re s ( z = 1) = lim  ( z − 1) 
1! z →1  d z  ( z + 1)( z − 1) 2 

 d  z n ( z 2 − z + 2) 
= lim   
z →1 d z
  ( z + 1) 
 ( z + 1){z n (2 z − 1) + ( z 2 − z + 2).n z n −1} − z n ( z 2 − z + 2)(1) 
= lim  
z →1
 ( z + 1) 2 
 (2){1 + (2).n} − (2)(1) 
= 
 (2) 2 
2 + 4n − 2
=
4
=n
∴ f (n) = Re s ( z = −1) + Re s ( z = 1)
= (−1) n + n
 z ( z + 1)
2. Find the inverse Z – transform of by
( z − 1) 3
residue method.
Solution: Let Z { f ( z )} = f (n) = sum of the residues of
−1

 z ( z + 1)
n −1 
 . z  at its poles.
 ( 3
z − 1) 
(i.e.) f(n) = sum of the residues of  z( z(−z 1+)1)  at its
n

3
 
poles.
Poles of f ( z ).z n −1 are

( z − 1) 3 = 0
⇒ z =1

z = 1 is the pole of order 3.

 1  d2  3 z ( z + 1) 
n

Re s ( z = 1) = lim  2 ( z − 1) 
2 ! z →1  d z  ( z − 1) 3 
1  d2 
= lim  2 {z n ( z + 1)}
2 ! z →1  d z 
1  d 
= lim  {z n (1) + ( z + 1) n z n −1}
2 ! z →1  d z 
1
= lim [n z n −1 + n( z + 1).(n − 1) z n − 2 + n z n −1 (1)]
2 ! z →1
1
= [n + 2n (n − 1) + n ]
2
1
= [ n + 2n 2 − 2 n + n]
2
= n2
∴ f (n) = Re s ( z = 1)
= n2
Model III : Using Partial Fractions Method
When X(z) is a rational function in which the
denominator is factorisable, X(z) is resolved into
partial fractions and then Z-1[X(z)] is derived as the
sum of the inverse Z-transforms of the partial
fractions.
 z 
1. Find Z−1
 ( z − 1)( z − 2) 
 
z A B
Solution: = +
( z − 1)( z − 2) z − 1 z − 2

z = A( z − 2) + B( z − 1)
Put z = 1, we get 1 = A(−1) + 0
⇒ A = −1
Put z = 2, we get 2 = 0 + B(1)
⇒ B=2

z −1 2
= +
( z − 1)( z − 2) z − 1 z − 2
 z  −1  1  −1  1 
∴ Z −1   = − Z  z − 1 + 2 Z  z − 2 
 ( z − 1)( z − 2) 
= −(1) n −1 + 2(2) n −1
2n
= −1 + 2
2
= −1 + 2 n
 
−1 z3 
2. Find Z  using partial fraction method.

 ( z − 1) ( z − 2) 
2

z3
Solution: Let f ( z ) =
( z − 1) 2 ( z − 2)
f ( z) z2 A B C
= = + +
z ( z − 1) 2 ( z − 2) z − 1 ( z − 1) 2 z − 2
z 2 = A( z − 1)( z − 2) + B( z − 2) + C ( z − 1) 2
Put z = 1, we get 1 = 0 + B (−1) + 0
⇒ B = −1
Put z = 2, we get 4 = 0 + 0 + C (1)
⇒ C=4
Coeff . of z 2 , 1= A+C
1= A+ 4
⇒ A = −3
  −1 z2 
3. Find Z  
 ( z + 2)( z + 4) 
2 by the method of partial
fractions.
z2
Solution: Let f ( z ) =
( z + 2)( z 2 + 4)
f ( z) z A Bz +C
= = +
z ( z + 2)( z 2 + 4) z + 2 z 2 + 4

z = A( z 2 + 4) + ( B z + C )( z + 2)
Put z = −2, we get − 2 = A(4 + 4) + 0
− 2 = 8A
1
⇒ A=−
4
Coeff . of z 2 , 0 = A+ B
1
0=− +B
4
1
⇒B =
4
 Coeff . of z , 1 = 2B + C
2
1= +C
4
1 1
⇒C =1− =
2 2
f ( z) − 1 / 4 1 / 4 z + 1 / 2
= +
z z+2 z2 + 4
1 z 1 z2 1 z
f ( z) = − + +
4 z + 2 4 z2 + 4 2 z2 + 4

−1 1 −1  z  1 −1  z 2  1 −1  2 z 
∴ Z { f ( z )} = − Z   + Z  2  + Z  2
4  z + 2 4  z + 4 4  z + 4 
1 1 nπ 1 n nπ
= − (−2) n + 2 n cos + 2 sin
4 4 2 4 2
 z 3 + 3z
4. Find the inverse Z-transform of ( z − 1) 2 ( z 2 + 1)

z 3 + 3z
Solution: Let f ( z ) =
( z − 1) 2 ( z 2 + 1)
f ( z) z2 + 3 A B Cz+D
= = + +
z ( z − 1) 2 ( z 2 + 1) z − 1 ( z − 1) 2 z 2 +1
z 2 + 3 = A( z − 1)( z 2 + 1) + B( z 2 + 1) + (C z + D)( z − 1) 2
Put z = 1, we get 4 = 0 + B(2) + 0
⇒ B=2
Coeff . of z 3 , 0 = A + C − − − − − (1)
Coeff . of z 2 , 1 = − A + B − 2C + D
1 = − A + 2 − 2C + D
A + 2C − D = 1 − − − − − (2)
 (2) ⇒ A + 2C = 1 - - - - - (3)
(3)-(1) ⇒ C = 1
(1) ⇒ A = -1
f ( z) −1 2 z+0
= + +
z z − 1 ( z − 1) 2 z 2 + 1
−z 2z z2
f ( z) = + +
z − 1 ( z − 1) 2 z 2 + 1
 z  −1  z  −1  z 
2
−1 −1
∴ Z { f ( z )} = − Z  + 2 Z  ( z − 1) 2  + Z  z 2 + 1
 z − 1    
nπ
= −1 + 2n + cos
2
 z n + 2 − 20 z n
Res(z = 4) = lim (z − 4)
z→4 (z − 2) 3 (z − 4)
z n + 2 − 20 z n
= lim
z →4 ( z − 2) 3
4 n + 2 − 20 4 n
=
(2) 3
4n
= (16 − 20)
8
4n
=−
2
∴ f (n) = Re s ( z = 2) + Re s ( z = 4)
2n 4n
= (2n + 1) −
2

2 2
 Applications of z-transform in Solving
Finite Difference Equations
Z-transform can be applied in solving difference
equation.
Using the relations
(i) Z [ x(n − m)] = z − m X ( z )
(ii) Z [ y n+1 ] = z[Y ( z ) − y0 ]
 y 
(iii) Z [ y n+ 2 ] = z 2 Y ( z ) − y0 − 1 
 z
 y y 
(iv) Z [ y n +3 ] = z 3 Y ( z ) − y 0 − 1 − 22 
 z z 

…………….
where Y(z) = Z[yn]
 Applications of z-transform in Solving
Finite Difference Equations
1. Solve un+2 + 6un+1 + 9un = 2n with u0 = u1 = 0
using Z-transform.
Solution: Given un+2 + 6un+1 + 9un = 2n
Taking Z – transform on both sides, we get
Z [u n + 2 ] + 6 Z [u n +1 ] + 9 Z [u n ] = Z (2 n )

z
{z 2u ( z ) − z 2u (0) − z u (1)} + 6 {z u ( z ) − z u (0)} + 9 u ( z ) =
z−2

z
{z 2u ( z ) − 0 − 0} + 6 {z u ( z ) − 0} + 9 u ( z ) =
z−2
 z
( z 2 + 6 z + 9)u ( z ) =
z−2
z
( z + 3) 2 u ( z ) =
z−2
z
u ( z) =
( z − 2)( z + 3) 2
u ( z) 1
=
z ( z − 2)( z + 3) 2
1 A B C
= + +
( z − 2)( z + 3) 2 z − 2 z + 3 ( z + 3) 2
1 = A ( z + 3) 2 + B ( z − 2)( z + 3) + C ( z − 2)
Put z = 2, we get 1 = A(5) 2 + 0 + 0
1
⇒ A=
25
Put z = −3, we get 1 = 0 + 0 + C (−5)
1
⇒ C=−
5
Coeff . of z 2 , 0 = A+ B
1
0= +B
25
1
⇒ B=−
25
u ( z ) 1 / 25 1 / 25 1/ 5
= − −
z z − 2 z + 3 ( z + 3) 2
1 z 1 z 1 z
u ( z) = − −
25 z − 2 25 z + 3 5 ( z + 3) 2
1 −1  z  1 −1  z  1 −1  − 3 z 
∴ u n = Z −1{u ( z )} = Z   − Z   + Z  2
25  z − 2  25  z + 3  15  ( z + 3) 
1 n 1 1
(i.e.) un = .2 − (−3) n + .n (−3) n
25 25 15
2. Solve un+2 – 2un+1 + un = 2n with u0 = 2, u1 = 1
using Z - transform.
Solution: Given un+2 – 2un+1 + un = 2n
Taking Z – transform on both sides, we get
Z [u n + 2 ] − 2 Z [u n +1 ] + Z [u n ] = Z (2 n )
z
{z 2u ( z ) − z 2u (0) − z u (1)} − 2 {z u ( z ) − z u (0)} + u ( z ) =
z−2
z
{z 2u ( z ) − 2 z 2 − z} − 2{z u ( z ) − 2 z} + u ( z ) =
z−2
z
( z 2 − 2 z + 1)u ( z ) = + 2 z 2 − 3z
z−2
z + z (2 z − 3)( z − 2)
( z − 1) 2 u ( z ) =
z−2
z [1 + 2 z 2 − 7 z + 6]
u ( z) =
( z − 2)( z − 1) 2
 u ( z) 2 z 2 − 7 z + 7
=
z ( z − 2)( z − 1) 2
2z 2 − 7z + 7 A B C
= + +
( z − 2)( z − 1) 2 z − 2 z − 1 ( z − 1) 2
2 z 2 − 7 z + 7 = A ( z − 1) 2 + B ( z − 2)( z − 1) + C ( z − 2)
Put z = 2, we get 8 − 14 + 7 = A(1) 2 + 0 + 0
⇒ A =1
Put z = 1, we get 2 − 7 + 7 = 0 + 0 + C (−1)
⇒ C = −2
2
Coeff . of z , 2 = A+ B
2 =1+ B
⇒ B =1
u ( z) 1 1 2
= + −
z z − 2 z − 1 ( z − 1) 2
z z 2z
u ( z) = + −
z − 2 z − 1 ( z − 1) 2
 z  −1  z  −1  z 
∴ u n = Z −1{u ( z )} = Z −1  + Z − 2 Z  ( z − 1) 2 
 z − 2   z − 1
 
(i.e.) un = 2 n + 1 − 2n
3. Solve yn+2 + 4yn+1 + 3yn = 3n with y0 = 0, y1 = 1
using Z - transform.
Solution: Given yn+2 + 4yn+1 + 3yn = 3n
Taking Z – transform on both sides, we get
Z [ yn + 2 ] + 4 Z [ yn +1 ] + 3 Z [ yn ] = Z (3n )
z
{z 2 y ( z ) − z 2 y (0) − z y (1)} + 4 {z y ( z ) − z y (0)} + 3 y ( z ) =
z −3
z
{z 2 y ( z ) − 0 − z} + 4{z y ( z ) − 0} + 3 y ( z ) =
z −3
z
( z 2 + 4 z + 3) y ( z ) = +z
z −3
z + z ( z − 3)
( z + 1)( z + 3) y ( z ) =
z −3
z [1 + z − 3]
y( z) =
( z + 1)( z + 3)( z − 3)
 z 2 − 2z
y( z) =
( z + 1)( z + 3)( z − 3)
z 2 − 2z A B C
= + +
( z + 1)( z + 3)( z − 3) z + 1 z + 3 z − 3
z 2 − 2 z = A ( z + 3)( z − 3) + B ( z + 1)( z − 3) + C ( z + 1)( z + 3)
Put z = 3, we get 9 − 6 = 0 + 0 + C (4)(6)
3 1
⇒ C= =
24 8
Put z = −1, we get 1 + 2 = A(2)(−4) + 0 + 0
3
⇒ A=−
8
Coeff . of z 2 , 1= A+ B +C
3 1
1= − + B +
8 8
3 1
⇒ B =1+ −
8 8
8 + 3 − 1 10 5
⇒ B= = =
8 8 4
 − 3 / 8 5 / 4 1/ 8
y( z) = + +
z +1 z + 3 z − 3
3  1  5 −1  1  1 −1  1 
∴ yn = Z −1{ y ( z )} = − Z −1   + Z   + Z 
8  z + 1 4  z + 3 8  z − 3 
3 5 1
yn = − (−1) n −1 + (−3) n −1 + (3) n −1
8 4 8
3 (−1) n 5 (−3) n 1 (3) n
yn = − + +
8 (−1) 4 (−3) 8 3
3 5 3n
(i.e.) yn = (−1) − (−3) +
n n

8 12 24
4. Using Z-transform solve y(n)+3y(n-1)–4y(n-2)=0
n ≥ 2 given that y(0) = 3, y(1) = -2.
Solution: Changing n into n+2 in the given
equation, it becomes
y(n+2) + 3y(n+1) – 4y(n) = 0, n ≥ 0
Taking Z – transform on both sides, we get
Z [ y (n + 2)] + 3 Z [ y (n + 1)] − 4 Z [ y (n)] = Z (0)

{z 2 y ( z ) − z 2 y (0) − z y (1)} + 3{z y ( z ) − z y (0)} − 4 y ( z ) = 0

{z 2 y ( z ) − 3 z 2 + 2 z} + 3{z y ( z ) − 3 z} − 4 y ( z ) = 0

( z 2 + 3z − 4) y ( z ) = 3 z 2 + 7 z
( z + 4)( z − 1) y ( z ) = z (3z + 7)
 y( z) 3z + 7
=
z ( z − 1)( z + 4)
3z + 7 A B
= +
( z − 1)( z + 4) z − 1 z + 4
3 z + 7 = A( z + 4) + B( z − 1)
Put z = 1, we get 3 + 7 = A(5) + 0
10
⇒ A= =2
5
Put z = −4, we get − 12 + 7 = 0 + B(−5)
− 5 = −5 B
⇒ B =1
y( z) 2 1
= +
z z −1 z + 4
2z z
y( z) = +
z −1 z + 4
 z  −1  z 
∴ yn = Z −1{ y ( z )} = 2 Z −1  + Z
 z − 1  z + 4 

(i.e.) yn = 2 + (−4) n
5. Using Z-transform method solve yn+2 + yn = 2
given that y0 = y1 = 0.
Solution: Given yn+2 + yn = 2
Taking Z – transform on both sides, we get
Z [ y n + 2 ] + Z [ yn ] = Z (2)
2z
{z 2 y ( z ) − z 2 y (0) − z y (1)} + y ( z ) =
z −1
2z
{z 2 y ( z ) − 0 − 0} + y ( z ) =
z −1
2z
( z 2 + 1) y ( z ) =
z −1
2z
y( z) =
( z − 1)( z 2 + 1)
y( z) 2
=
z ( z − 1)( z 2 + 1)
 2 A Bz + C
= +
( z − 1)( z 2 + 1) z − 1 z 2 + 1
2 = A( z 2 + 1) + ( Bz + C )( z − 1)
Put z = 1, we get 2 = A(2) + 0
⇒ A =1
Coeff . of z 2 , 0 = A+ B
0 =1+ B
B = −1
Coeff . of z , 0 = −B + C
0 =1+ C
⇒ C = −1
y ( z) 1 − z −1
= + 2
z z −1 z +1
z z2 z
y( z) = − 2 − 2
z −1 z +1 z +1
 z  −1  z 
2
−1 −1 −1  z 
∴ yn = Z { y ( z )} = Z  − Z  z 2 + 1 − Z
 z − 1  
 z 2 + 1

 nπ   nπ 
(i.e.) yn = 1 − cos   − sin  
 2   2 
 6. Form the difference equation whose solution is
yn = (A + Bn)2n
Solution: Given
yn = (A + Bn)2n = A2n + Bn2n --------- (1)
yn+1 = [A + B(n+1)]2n+1 = 2[A + B(n+1)]2n
= 2A2n + 2B(n+1)2n --------(2)
yn+2 = [A + B(n+2)]2n+2 = 4[A + B(n+2)]2n
= 4A2n + 4B(n+2)2n -------(3)
Eliminating A and B from equations (1), (2) and (3),
we have
 yn 1 n
y n+1 2 2(n + 1) = 0
yn+2 4 4( n + 2)

yn [8(n + 2) − 8(n + 1)] − yn+1[4(n + 2) − 4n] + yn+2 [2(n + 1) − 2n] = 0

yn (16 − 8) − yn+1 (8) + yn+ 2 (2) = 0

8 yn − 8 yn+1 + 2 yn+ 2 = 0

(i.e.) yn+ 2 − 4 yn+1 + 4 yn = 0

7. Derive the difference equation from
yn = (A + Bn)(-3)n
Solution: Given
yn = (A + Bn)(-3)n = A(-3)n + Bn(-3)n --------- (1)
yn+1 = [A + B(n+1)](-3)n+1
= -3[A + B(n+1)](-3)n
= -3A(-3)n - 3B(n+1)(-3)n --------(2)
yn+2 = [A + B(n+2)](-3)n+2
= 9[A + B(n+2)](-3)n
= 9A(-3)n + 9B(n+2)(-3)n -------(3)
Eliminating A and B from equations (1), (2) and (3),
we have
 yn 1 n
yn+1 − 3 − 3(n + 1) = 0
y n+ 2 9 9(n + 2)

yn [−27(n + 2) + 27(n + 1)] − yn+1[9(n + 2) − 9n] + yn+2 [−3(n + 1) + 3n] = 0

yn (−54 + 27) − yn+1 (18) + yn+2 (−3) = 0

− 27 yn − 18 yn+1 − 3 yn+2 = 0

(i.e.) yn+2 + 6 yn+1 + 9 yn = 0