Unit5 : Z−TRANSFORM

Definition: The Z−Transform of a sequence (un ), denoted by Z(un ), is defined as

∞
X un
Z(un ) =
n=0
zn

provided the series on the right side converges.

Z−Transform of a geometric sequence:

z
Z(an ) =
z−a

Proof:

∞
X an
Z(an ) =
n=0
zn
∞  n
X a
=
n=0
z
∞
!
1 a X 1
= provided <1 ∵ rn = 1 + r + r2 + · · · = provided |r| < 1
1 − (a/z) z n=0
1−r
z
=
z−a

Corollary:

z
Z(1) =
z−1

Problem: Find
1
 
Z
n!
.
Solution:
∞
1 (1/n!)
  X
Z =
n! n=0
zn

∞
X (1/z)n
=
n=0
n!

∞
!
X xn x x2
= e1/z ∵ ex = =1+ + + ···
n=0
n! 1! 2!

1
Linearity

Z(aun + bvn ) = aZ(un ) + bZ(vn )

Proof:
∞
X aun + bvn
Z(aun + bvn ) =
n=0
zn

∞ ∞
X un X vn
= a n
+ b
n=0
z n=0
zn

= aZ(un ) + bZ(vn )

Corollary:
Putting a = 1 and b = 1,

Z(un + vn ) = Z(un ) + Z(vn )

Putting a = 1 and b = −1,

Z(un − vn ) = Z(un ) − Z(vn )

Putting b = 0,

Z(aun ) = aZ(un )

Is it true that Z(un vn ) = Z(un )Z(vn )? Answer: Resounding “NO”. A counter example will be given later.

Problem: Find Z(2 cos nπ + 3.2n − 4.5n ).

Solution:

Z(2 cos nπ + 3.2n − 4.5n ) = 2Z(cos nπ) + 3Z(2n ) − 4Z(5n ) (by linearity)

= 2Z((−1)n ) + 3Z(2n ) − 4Z(5n )

     
z z z
= 2 +3 −4
z+1 z−2 z−5

Damping rule:

2
 Z(an un ) = [Z(un )]z→z/a

Proof:
∞
X an un
Z(an un ) =
n=0
zn

∞
X un
=
n=0
(z/a)n

∞
" #
X un
=
n=0
zn
z→z/a

= [Z(un )]z→z/a

Problem: Find the Z−Transform of the sequence

h πi
4n sin (2n + 1) .
2
Solution:

 h π i z
Z sin (2n + 1) = Z((−1)n ) =
2 z+1
∴ By damping rule,
 h π i
Z 4n sin (2n + 1) = Z(4n (−1)n )
2

= [Z((−1)n )]z→z/4

 
z
=
z+1 z→z/4

z
=
z+4
Problem: Find Z(2n 3n ).
Solution: By damping rule,

Z(2n 3n ) = [Z(3n )]z→z/2

 
z
=
z−3 z→z/2

z
=
z−6

3
 Aliter:

Z(2n 3n ) = Z([2 × 3]n )

= Z(6n )

z
=
z−6

Multiplication by n

d
Z(nun ) = −z Z(un )
dz

Proof: By definition,
∞
X un
Z(un ) = .
n=0
zn
Differentiating with respect to z,
∞
d X (−n)un
Z(un ) =
dz n=0
z n+1
∞
d 1 X nun
=⇒ Z(un ) =
dz −z n=0 z n
d 1
=⇒ Z(un ) = Z(nun )
dz −z

Rewriting this,
d
Z(nun ) = −z Z(un )
dz
Corollary:

d
Z(np ) = −z Z(np−1 )
dz

Problem: Find Z(n).

Solution:
We have
d
Z(np ) = −z Z(np−1 )
dz
Putting p = 1,

4
 d
Z(n) = −z Z(1)
dz  
d z
= −z
dz z − 1

(z − 1) − z
 
= −z
(z − 1)2

z
=
(z − 1)2

Problem: Find Z(n2 ).

Solution:
We have
d
Z(np ) = −z Z(np−1 )
dz
Putting p = 2,

d
Z(n2 ) = −z Z(n)
dz  
d z
= −z
dz (z − 1)2

(z − 1)2 − 2z(z − 1)
 
= −z
(z − 1)4

(z − 1) − 2z)
 
= −z(z − 1)
(z − 1)4

z(z + 1)
=
(z − 1)3

Problem: Find Z(cos nθ) and Z(sin nθ) .

Solution:

5
 Z(cos nθ + i sin nθ) = Z(einθ ) (∵ eix = cos x + i sin x)

= Z([eiθ ]n )

z
=
z − eiθ

z
=
z − (cos θ + i sin θ)

z
=
(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

=
(z − cos θ)2 + sin2 θ

z(z − cos θ) + iz sin θ

=
(z 2 − 2z cos θ + cos2 θ) + sin2 θ

z(z − cos θ) + iz sin θ

=
z 2 − 2z cos θ + 1

Therefore, we have

z(z − cos θ) z sin θ

Z(cos nθ) + iZ(sin nθ) = +i 2
z 2 − 2z cos θ + 1 z − 2z cos θ + 1

Equating the real and imaginary parts, we have

z(z − cos θ)
Z(cos nθ) =
z2 − 2z cos θ + 1

z sin θ
Z(sin nθ) =
z 2 − 2z cos θ + 1

Corollary: Letting θ = π/2,

h  nπ i z
Z sin = 2
2 z +1

6
 h  nπ i z2
Z cos = 2
2 z +1
Problem: Find Z(an n).
Solution:
By damping rule,

Z(an n) = [Z(n)]z→z/a

 
z
=
(z − 1)2 z→z/a

z
= a

2
z
a −1

az
=
(z − a)2

Shifting to the Left and to the Right:

Notations:
Let k be a positive integer.

(un ) : u0 , u1 , u2 , u3 , · · ·

(un+k ) : uk , uk+1 , uk+2 , uk+3 , · · · (Shifting to the left)

(un−k ) : 0, 0, 0, · · · 0, u , u , u , u , · · · (Shifting to the right)

| {z } 0 1 2 3
k terms

• (un+k ) is obtained by shifting the sequence (un ) to the left.

• (un−k ) is obtained by shifting the sequence (un ) to the right.

Result:

h u1 u2 uk−1 i
Z(un+k ) = z k U (z) − u0 − − 2 − · · · − k−1
z z z
where U (z) = Z(un ).

Corollary:

7
 Z(un+1 ) = z [U (z) − u0 ]

h u1 i
Z(un+2 ) = z 2 U (z) − u0 −
z

h u1 u2 i
Z(un+3 ) = z 3 U (z) − u0 − − 2
z z

Problem: Let
1
u0 = 0 and un = for n ≥ 1.
n
Find Z(un ) and hence Z(un+1 ).
Solution:

∞
X un
Z(un ) = .
n=0
zn

∞
X 1 1
= .
n=1
n zn

 2  3
1 1 1 1 1
= + + + ···
z 2 z 3 z

1 1 x2 x3
   
= − log 1 − provided <1 ∵ − log(1 − x) = x + + + ··· .
z z 2 3

We have

Z(un+1 ) = z [U (z) − u0 ]

= zU (z)

1
 
= −z log 1 −
z

Result:

Z(un−k ) = z −k U (z)

where U (z) = Z(un ).

8
Initial Value Theorem
If U (z) := Z(un ), then

u0 = lim U (z)
z→∞

Additional initial values can be found as follows:

u1 = lim z [U (z) − u0 ]
z→∞

h u1 i
u2 = lim z 2 U (z) − u0 −
z→∞ z

h u1 u2 i
u3 = lim z 3 U (z) − u0 − − 2
z→∞ z z

Problem: If the Z−Transform of a sequence is given by

2z 2 + 5z + 14
U (z) = ,
(z − 1)4

evaluate u2 and u3 .
Solution: By initial value theorem, we have

u0 = lim U (z)
z→∞

2z 2 + 5z + 14
= lim
z→∞ (z − 1)4

z 2 (2 + z5 + z142 )
= lim
z→∞ z 4 (1 − z1 )4

1 (2 + z5 + z142 )
= lim
z→∞ z 2 (1 − z1 )4

= 0.

9
 u1 = lim z [U (z) − u0 ]
z→∞

2z 2 + 5z + 14
 
= lim z
z→∞ (z − 1)4

z 3 (2 + z5 + z142 )
= lim
z→∞ z 4 (1 − z1 )4

1 (2 + z5 + z142 )
= lim
z→∞ z (1 − z1 )4

= 0.

h u1 i
u2 = lim z 2 U (z) − u0 −
z→∞ z

2z 2 + 5z + 14
 
= lim z 2
z→∞ (z − 1)4

z 4 (2 + z5 + z142 )
= lim
z→∞ z 4 (1 − z1 )4

(2 + z5 + z142 )
= lim
z→∞ (1 − z1 )4

= 2.

10
 h u1 u2 i
u3 = lim z 3 U (z) − u0 − − 2
z→∞ z z

2z 2 + 5z + 14 2
 
= lim z 3
− 2
z→∞ (z − 1)4 z

z 2 (2z 2 + 5z + 14) − 2(z − 1)4

 
= lim z 3
z→∞ (z − 1)4 z 2

z 2 (2z 2 + 5z + 14) − 2(z 4 − 4z 3 + 6z 2 − 4z + 1)

 
= lim z 3
[∵ (a − b)4 = a4 − 4a3 b + 6a2 b2 − 4ab3 + b4 ]
z→∞ (z − 1)4 z 2

13z 3 + 2z 2 + 8z − 2
 
= lim z 3
z→∞ z 2 (z − 1)4

13 + z2 + z82 − 2
 
= lim z 6 z3
z→∞ z 6 (1 − z1 )4

13 + z2 + z82 − 2
 
= lim z3
z→∞ (1 − z1 )4

= 13.

Problem: Given that un = 2n + n + 2, verify initial value theorem for Z−Transform.

Solution: Initial value theorem for Z−Transform states that

u0 = lim U (z)
z→∞

LHS = u0 = 3.
Further, we have

U (z) = Z(un )

= Z(2n + n + 2)

= Z(2n ) + Z(n) + Z(2)

= Z(2n ) + Z(n) + 2Z(1)

z z 2z
= + +
z − 2 (z − 1) 2 (z − 1)

11
 RHS = lim U (z)
z→∞

z z 2z
= lim + +
z→∞ z − 2 (z − 1)2 (z − 1)

z z 2z
= lim 2 + 2 1 2 +
z→∞ z(1 − z ) z (1 − z ) z(1 − z1 )

1 1 1 2
= lim + +
z→∞ (1 − z2 ) z (1 − z1 )2 (1 − z1 )

= 3.

Final Value Theorem

If U (z) := Z(un ), then

lim un = lim (z − 1)U (z)

n→∞ z→1

Problem: Verify final value theorem for the sequence un = 1 + (1/2)n .

Solution: Final value theorem states that

lim un = lim (z − 1)U (z)

n→∞ z→1

LHS = lim un
n→∞

 n
1
= lim 1 +
n→∞ 2

= 1.

12
Further, we have

U (z) = Z(un )

 n 
1

= Z 1+
2

 n 
1
= Z(1) + Z
2

z z
= +
z − 1 z − (1/2)

z 2z
= +
z − 1 2z − 1)

RHS = lim (z − 1)U (z)

z→1

2z
 
z
= lim (z − 1) +
z→1 z − 1 2z − 1

2z
 
z
= lim (z − 1) +
z→1 z − 1 2z − 1

4z 2 − 3z
 
= lim (z − 1)
z→1 (z − 1)(2z − 1)

4z 2 − 3z
= lim
z→1 2z − 1

= 1.

CONVOLUTION
Problem: Furnish an example to show that

Z(un vn ) 6= Z(un )Z(vn )

in general.
Solution: Let un = 1 and vn = 1. Then, we have

z z
Z(un ) = Z(1) = & Z(vn ) = Z(1) = .
z−1 z−1

Therefore, we have
z z z2
Z(un )Z(vn ) = =
z−1z−1 (z − 1)2

13
On the other hand, we have
z
Z(un vn ) = Z(1) = .
z−1
Hence, it can be concluded that
Z(un vn ) 6= Z(un )Z(vn )

Definition: The convolution of two sequences (un ) and (vn ), denoted by (un ∗ vn ), is defined as
n
X
un ∗ vn = uk vn−k = u0 vn + u1 vn−1 + u3 vn−2 + · · · + un v0 .
k=0

Some Properties:

• Convolution is commutative, that is, un ∗ vn = vn ∗ un .

• Convolution is associative, that is, un ∗ (vn ∗ wn ) = (un ∗ vn ) ∗ wn .

• un ∗ 1 6= un in general.

Convolution Theorem
Z −1 [U (z)V (z)] = Z −1 [U (z)] ∗ Z −1 [V (z)]

Equivalently,

Z(un ∗ vn ) = Z(un )Z(vn )

Problem: Use convolution theorem to evaluate

z2
 
Z −1
(z − a)(z − b)

Solution:

14
 z2
    
z z
Z −1
= Z −1
(z − a)(z − b) z−a z−b

   
z z
= Z −1 ∗ Z −1
z−a z−b

= an ∗ bn

= un ∗ vn where un = an and vn = bn

n
X
= uk vn−k
k=0

n
X
= ak bn−k
k=0

n
X
= bn ak b−k
k=0

n  k
X a
= bn
b
k=0

  a   a 2  a n 
= b 1+ n
+ + ···
b b b

Case1: a 6= b
In this case,

z2 (a/b)n+1 − 1 rn+1 − 1 1 − rn+1

     
Z −1 = bn ∵ 1 + r + r2 + · · · + rn = =
(z − a)(z − b) (a/b) − 1 r−1 1−r

(an+1 − bn+1 )/bn+1

 
= b n
(a − b)/b

an+1 − bn+1
=
a−b

Case2: a = b
In this case,
z2
 
Z −1
= (n + 1)an .
(z − a)2

Problem: Use convolution theorem to evaluate

z2
 
−1
Z
(2z − 1)(4z + 1)

15
Solution:

z2
    
z z
Z −1
= Z −1
(2z − 1)(4z + 1) 2z − 1 4z + 1

1 −1
  
z z
= Z
8 z − 1/2 z + 1/4

1
    
z z
= Z −1
∗Z −1
8 z − 1/2 z + 1/4

 n  n 
1 1 1
= ∗ −
8 2 4

 n n
1 1 1

= [un ∗ vn ] where un = and vn = −
8 2 4

n
X
= uk vn−k
k=0

n  k  n−k
1X 1 1
= −
8 2 4
k=0

n  k 
n X −k
1 1 1 1

= − −
8 4 2 4
k=0

n  k 
n X k
1 1 1 1

= − / −
8 4 2 4
k=0

n 
n X k
1 1 1

= − × (−4)
8 4 2
k=0

n
n X
1 1

= − (−2)k
8 4
k=0

n
1 1

= 1 + (−2) + (−2)2 + (−2)3 · · · + (−2)n
 
−
8 4

n 
1 1 (−2)n+1 − 1
 
= −
8 4 (−2) − 1

n
1 1

= − − [(−2)n+1 − 1]
24 4

Problem:
1 n
Let un = and vn = cos nπ. Use convolution theorem to evaluate the Z−Transform of (un ∗ vn ).


2

Solution:

16
  n 
1
Z(un ) = Z
2

z
=
z − (1/2)

2z
=
2z − 1

Z(vn ) = Z[cos nπ]

= Z[(−1)n ]

z
=
z − (−1)

z
=
z+1

By convolution theorem,

Z(un ∗ vn ) = Z(un )Z(vn )

2z z
=
2z − 1 z + 1

2z 2
=
(2z − 1)(z + 1)

17
INVERSE Z−TRANSFORM
Defition: If U (z) = Z(un ), the inverse Z−Transform of U (z), denoted by Z −1 [U (z)], is defined as

Z −1 [U (z)] = un .

Example:
   
z z
Z −1 = 2n ∵ Z (2n ) = .
z−2 z−2

   
z z
Z −1
= n ∵ Z(n) = .
(z − 1)2 (z − 1)2

Inverse Z−Transorm by Partial Fractions Method

Problem: Find the inverse Z−Transform of

2z 2 + 3z
.
(z + 2)(z − 4)

Solution: Let us assume that

2z 2 + 3z
U (z) = (The right side rational function is improper.)
(z + 2)(z − 4)

U (z) 2z + 3
∴ = (The right side rational function is proper)
z (z + 2)(z − 4)

Resolving the right side rational function into partial fractions,

2z + 3 A B
= +
(z + 2)(z − 4) z+2 z−4

A(z − 4) + B(z + 2)
=
(z + 2)(z − 4)

Therefore, we obtain that

A(z − 4) + B(z + 2) = 2z + 3.

Letting z = −2, we obtain that

1
6A = −6 =⇒ A = .
6
Letting z = 4, we obtain that

11
6B = 11 =⇒ B = .
6

18
Therefore, we have

U (z) 1 1 11 1
= +
z 6z+2 6 z−4

1 z 11 z
U (z) = +
6z+2 6 z−4

Applying inverse Z−Transform,

1 −1 11
   
z z
un = Z + Z −1
6 z+2 6 z−4

1 11
= (−2)n + 4n
6 6

Problem: Find the inverse Z−Transform of

z(z 2 − z + 2)
.
(z + 1)(z − 1)2

Solution: Let us assume that

z(z 2 − z + 2)
U (z) = (The right side rational function is improper.)
(z + 1)(z − 1)2

U (z) (z 2 − z + 2
∴ = (The right side rational function is proper)
z (z + 1)(z − 1)2

Resolving the right side rational function into partial fractions,

z2 − z + 2 A B C
= + +
(z + 1)(z − 1)2 z + 1 z − 1 (z − 1)2

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

=
(z + 1)(z − 1)2

Therefore, we obtain that

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

Letting z = 1, we obtain that

2C = 2 =⇒ C = 1.

Letting z = −1, we obtain that

19
 4A = 4 =⇒ A = 1.

Letting z = 0, we obtain that

A − B + C = 2 =⇒ B = 0.

Therefore, we have

U (z) 1 1
= +
z z + 1 (z − 1)2

z z
U (z) = +
z + 1 (z − 1)2

Applying inverse Z−Transform,

   
z z
un = Z −1 + Z −1
z+1 (z − 1)2

= (−1)n + n

Problem: Find the inverse Z−Transform of

3z 2 − 18z + 26
.
(z − 2)(z − 3)(z − 4)

Solution: Let us assume that

3z 2 − 18z + 26)
U (z) =
(z − 2)(z − 3)(z − 4)
Resolving the right side rational function into partial fractions,

3z 2 − 18z + 26 A B C
= + +
(z − 2)(z − 3)(z − 4) z−2 z−3 z−4

A(z − 3)(z − 4) + B(z − 2)(z − 4) + C(z − 2)(z − 3)

=
(z − 2)(z − 3)(z − 4)

Therefore, we obtain that

20
 A(z − 3)(z − 4) + B(z − 2)(z − 4) + C(z − 2)(z − 3) = 3z 2 − 18z + 26.

Letting z = 2, we obtain that

2A = 2 =⇒ A = 1.

Letting z = 3, we obtain that

−B = −1 =⇒ B = 1.

Letting z = 4, we obtain that

2C = 2 =⇒ C = 1.

Therefore, we have

1 1 1
U (z) = + + .
z−2 z−3 z−4
Applying inverse Z−Transform,

1 1 1
     
un = Z −1 + Z −1 + Z −1
z−2 z−3 z−4

1
   
= 2 n−1
+3 n−1
+4 n−1
∵Z −1
=an−1
(n ≥ 1)
z−a
Problem: Find the inverse Z−Transform of
4z 2 − 2z
.
z3 − 5z 2 + 8z − 4
Solution: The given rational function is

4z 2 − 2z 2z(2z − 1)
= .
z3 − 5z 2 + 8z − 4 (z − 1)(z − 2)2
Let us assume that
2z(2z − 1)
U (z) =
(z − 1)(z − 2)2

U (z) 2(2z − 1)
=
z (z − 1)(z − 2)2

By partial fractions,
2z − 1 A B C
= + +
(z − 1)(z − 2)2 z − 1 z − 2 (z − 2)2

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

=
(z − 1)(z − 2)2

21
Therefore, we have
A(z − 2)2 + B(z − 1)(z − 2) + C(z − 1) = 2z − 1.

Putting z = 2, we obtain that C = 3.

Putting z = 1, we obtain that A = 1.
Putting z = 0, we obtain that 4A + 2B − C = −1 and hence B = −1.
Therefore, we have

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

3z
 
z z
∴ U (z) = 2 − +
z − 1 z − 2 (z − 2)2

Applying inverse Z−Transform,

3 2z
      
z z
un = 2 Z −1 − Z −1 + Z −1
z−1 z−2 2 (z − 2)2

   
az
= 2−2 n+1
+ 3n.2 n
∵Z −1
= na n
(z − a)2

Inverse Z−Transorm by Residue Method

(Note that Residue Method is also known as inversion integral method)
Let U (z) denote the Z−Transform of a sequence (un ). Let

f (z) = z n−1 U (z).

Then, we have

Z −1 [U (z)] = Sum of the residues of f (z)

Remark:
Let f (z) be a given function.

• If z = a is a pole of order 1 (Simple pole), then

Resz=a f (z) = lim (z − a)f (z).

z→a

• If z = a is a pole of order 2, then

1
 
d
Resz=a f (z) = lim (z − a) f (z) .
2
1! z→a dz

22
• If z = a is a pole of order 3, then

1
 2 
d
Resz=a f (z) = lim (z − a) f (z) .
3
2! z→a dz 2

• In general, if z = a is a pole of order m, then

1
 m−1 
d
Resz=a f (z) = lim (z − a) f (z) .
m
(m − 1)! z→a dz m−1

Problem: Use Residue method to find the inverse Z−Transform of

3z
.
(z − 1)(z − 2)
Solution:

3z
Let U (z) = .
(z − 1)(z − 2)

3z n
Let f (z) = U (z)z n−1 = .
(z − 1)(z − 2)

Then, f (z) has simple poles (poles of order 1) at z = 1 and z = 2.

Resz=1 f (z) = lim (z − 1)f (z)

z→1

3z n
= lim (z − 1)
z→1 (z − 1)(z − 2)

3z n
= lim
z→1 (z − 2)

= −3.

Resz=2 f (z) = lim (z − 2)f (z)

z→2

3z n
= lim (z − 2)
z→2 (z − 1)(z − 2)

3z n
= lim
z→2 (z − 1)

= 3.2n .

23
By residue method,

Z −1 [U (z)] = Sum of the residues of f (z)

= −3 + 3.2n .

Problem: Use Residue method to find the inverse Z−Transform of

2z
.
(z − 1)(z 2 + 1)
Solution:

2z
Let U (z) = .
(z − 1)(z 2 + 1)

2z n
Let f (z) = U (z)z n−1 = .
(z − 1)(z 2 + 1)

Then, f (z) has simple poles (poles of order 1) at Z = 1, i, −i.

Resz=1 f (z) = lim (z − 1)f (z)

z→1

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

2z n
= lim
z→1 (z 2 + 1)

= 1.

24
Resz=i f (z) = lim (z − i)f (z)
z→i

2z n
 
= lim (z − i)
z→i (z − 1)(z 2 + 1)

2z n
 
= lim (z − i)
z→i (z − 1)(z − i)(z + i)

2z n
= lim
z→i (z − 1)(z + i)

2in
=
(i − 1)2i

in
=
(i − 1)i

in
=
(−1 − i)

in
= −
(1 + i)

25
 Resz=−i f (z) = lim (z + i)f (z)
z→−i

2z n
 
= lim (z + i)
z→−i (z − 1)(z 2 + 1)

2z n
 
= lim (z + i)
z→−i (z − 1)(z − i)(z + i)

2z n
= lim
z→−i (z − 1)(z − i)

2(−i)n
=
(−i − 1)(−2i)

(−i)n
=
(i + 1)i

(−i)n
=
(−1 + i)

(−i)n
= −
(1 − i)

By residue method,

Z −1 [U (z)] = Sum of the residues of f (z)

in (−i)n
= 1− − .
(1 + i) (1 − i)

Problem: Use Residue method to find the inverse Z−Transform of

z(z 2 − z + 2)
.
(z + 1)(z − 1)2
Solution:

z(z 2 − z + 2)
Let U (z) = .
(z + 1)(z − 1)2

z n (z 2 − z + 2)
Let f (z) = U (z)z n−1 = .
(z + 1)(z − 1)2

f (z) has a simple pole (pole of order 1) at z = −1 and a double pole (pole of order 2) at z = 1.

26
 Resz=−1 f (z) = lim (z + 1)f (z)
z→−1

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

z n (z 2 − z + 2)
= lim
z→−1 (z − 1)2

= (−1)n .

 
d 
Resz=1 f (z) = lim (z − 1)2 f (z)
z→1 dz

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

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

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

(z + 1) (n + 2)z n+1 − (n + 1)z n + 2nz n−1 − [z n+2 − z n+1 + 2z n ]

 
= lim
z→1 (z + 1)2

= n.

By residue method,

Z −1 [U (z)] = Sum of the residues of f (z)

= (−1)n + n.

Problem: Use Residue method to find the inverse Z−Transform of

z2
(a 6= b).
(z − a)(z − b)
Solution:

z2
Let U (z) = .
(z − a)(z − b)

z n+1
Let f (z) = U (z)z n−1 = .
(z − a)(z − b)

27
f (z) has simple poles (poles of order 1) at z = a and z = b.

Resz=a f (z) = lim (z − a)f (z)

z→a

z n+1
 
= lim (z − a)
z→a (z − a)(z − b)

z n+1
= lim
z→a (z − b)

an+1
=
(a − b)

Similarly,
bn+1
Resz=b f (z) = .
b−a
By residue method,

Z −1 [U (z)] = Sum of the residues of f (z)

an+1 bn+1
= +
(a − b) (b − a)

an+1 bn+1
= −
(a − b) (a − b)

an+1 − bn+1
=
a−b

Problem: Use Residue method to find the inverse Z−Transform of

z2
.
(z − a)2
Solution:

z2
Let U (z) = .
(z − a)2

z n+1
Let f (z) = U (z)z n−1 = .
(z − a)2

Then , f (z) has a double pole (pole of order 2) at z = a.

28
  
d 
Resz=a f (z) = lim (z − a)2 f (z)
z→a dz

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

 
d n+1
= lim {z }
z→a dz

= lim [(n + 1)z n ]

z→a

= (n + 1)an .

By residue method,

Z −1 [U (z)] = Sum of the residues of f (z)

= (n + 1)an .

Problem: Use Residue method to find the inverse Z−Transform of

z
.
(z − a)
Solution:

z
Let U (z) = .
(z − a)

zn
Let f (z) = U (z)z n−1 = .
(z − a)

f (z) has a simple pole (pole of order 1)at z = a.

Resz=a f (z) = lim (z − a)f (z)

z→a

zn
 
= lim (z − a)
z→a z−a

= lim z n
z→a

= an

29
By residue method,

Z −1 [U (z)] = Sum of the residues of f (z)

= an .

Problem: Use Residue method to find the inverse Z−Transform of

1
.
(z − a)
Solution:

1
Let U (z) = .
(z − a)

z n−1
Let f (z) = U (z)z n−1 = .
(z − a)
f (z) has a simple pole (pole of order 1)at z = a.

Resz=a f (z) = lim (z − a)f (z)

z→a

z n−1
 
= lim (z − a)
z→a (z − a)

= lim z n−1
z→a

= an−1

By residue method,

Z −1 [U (z)] = Sum of the residues of f (z)

= an−1 .

APPLICATION TO DIFFERENCE EQUATIONS

(Climax of the unit)
Problem: Using the Z−Transform, solve the difference equation

un+2 + 4un+1 + 3un = 3n with u0 = 0, u1 = 1.

Solution: Applying Z−Transform,

Z(un+2 ) + 4Z(un+1 ) + 3Z(un ) = Z(3n )

h u1 i z
z 2 U (z) − u0 − + 4z [U (z) − u0 ] + 3U (z) =
z z−3

30
Using the given initial conditions,

1
 
z
z U (z) −
2
+ 4zU (z) + 3U (z) =
z z−3

z
(z 2 + 4z + 3)U (z) − z =
z−3

z
(z 2 + 4z + 3)U (z) = z+
z−3

1
 
(z 2 + 4z + 3)U (z) = z 1+
z−3

z−2
 
z
U (z) =
z + 4z + 3
2 z−3

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

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

31
 Resolving the right side rational function into partial fractions,

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

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

Letting z = −1, A = 83 .
Letting z = −3, B = −5
12 .

Letting z = 3, C = 1
24 .

Therefore, we have
U (z) 3 1 5 1 1 1
= − + .
z 8 z + 1 12 z + 3 24 z − 3
3 z 5 z 1 z
U (z) = − + .
8 z + 1 12 z + 3 24 z − 3
Applying inverse Z−Transform, we obtain that

3 5 1
un = (−1)n − (−3)n + 3n .
8 12 24

Problem: Using the Z−Transform, solve the difference equation

un+2 − 5un+1 + 6un = 4 with u0 = 0 and u1 = 0.

Solution: Applying Z−Transform,

Z(un+2 ) − 5Z(un+1 ) + 6Z(un ) = Z(4)

h u1 i
z 2 U (z) − u0 − − 5z [U (z) − u0 ] + 6U (z) = 4Z(1)
z

z
(z 2 − 5z + 6)U (z) = 4
z−1

z
U (z) = 4
(z 2 − 5z + 6)(z − 1)

U (z) 1
= 4
z (z − 1)(z − 2)(z − 3)

Resolving the right side rational function into partial fractions,

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

Therefore, we have
A(z − 2)(z − 3) + B(z − 1)(z − 3) + C(z − 1)(z − 2) = 1.

Letting z = 1, A = 21 .

32
 Letting z = 2, B = −1.
Letting z = 3, C = 12 .
Therefore, we have

U (z) 1 1 1 1 1
 
=4 − +
z 2z−1 z−2 2z−3

U (z) 1 1 1
=2 −4 +2 .
z z−1 z−2 z−3
z z z
U (z) = 2 −4 +2 .
z−1 z−2 z−3
Applying inverse Z−Transform, we obtain that

un = 2 − 4.2n + 2.3n .

Problem: Using the Z−Transform, solve the difference equation

un+2 − 3un+1 + 2un = 0 with u0 = 0 and u1 = 1.

Solution: Applying Z−Transform,

Z(un+2 ) − 3Z(un+1 ) + 2Z(un ) = Z(0)

h u1 i
z 2 U (z) − u0 − − 3z [U (z) − u0 ] + 2U (z) = 0
z

1
 
z 2 U (z) − − 3zU (z) + 2U (z) = 0
z

(z 2 − 3z + 2)U (z) − z = 0

(z 2 − 3z + 2)U (z) = z

U (z) 1
=
z (z 2 − 3z + 2)

U (z) 1
=
z (z − 1)(z − 2)

Resolving the right side rational function into partial fractions,

1 A B
= + .
(z − 1)(z − 2) z−1 z−2

33
 ∴ A(z − 2) + B(z − 1) = 1.

Letting z = 1, A = −1.
Letting z = 2, B = 1.
Therefore, we have
U (z) 1 1
=− + .
z z−1 z−2
z z
U (z) = − + .
z−1 z−2
Applying inverse Z−Transform, we obtain that

un = −1 + 2n .

Problem: Using the Z−Transform, solve the difference equation

un+2 = un+1 + un with u0 = 0 and u1 = 1.

(This difference equation defines the Fibonacci sequence.)

Solution: The given difference equation is

un+2 − un+1 − un = 0.

Applying Z−Transform,

Z(un+2 ) − Z(un+1 ) − Z(un ) = Z(0)

h u1 i
z 2 U (z) − u0 − − z [U (z) − u0 ] − U (z) = 0
z

1
 
z U (z) −
2
− zU (z) − U (z) = 0
z

(z 2 − z − 1)U (z) − z = 0

z
U (z) =
z2 − z − 1

Let assume that

zn
f (z) := U (z)z n−1 = .
z2 −z−1
The poles of the function f (z) are given by z 2 − z − 1 = 0, that is, z = α, β where

√ √
1− 5 1+ 5
α= & β= .
2 2

34
 Resz=α f (z) = lim (z − α)f (z)
z→α

zn
 
= lim (z − α)
z→α z −z−1
2

zn
 
= lim (z − α)
z→α (z − α)(z − β)

zn
= lim
z→α (z − β)

αn
=
(α − β)

Similarly, it can be seen that

βn
Resz=α f (z) = .
(β − α)
Therefore, we have

un = Sum of the residues of f (z)

αn βn
= +
(α − β) (β − α)

αn βn
= −
(α − β) (α − β)

αn − β n
=
α−β

Formation of Difference Equations

Problem: Form a difference equation by eliminating the arbitrary constants from

yn = A2n + B3n

Solution:

35
 yn = A2n + B3n

yn+1 = A2n+1 + B3n+1 = 2[A2n ] + 3[B3n ]

yn+2 = A2n+2 + B3n+2 = 4[A2n ] + 9[B3n ]

Eliminating A and B, we obtain that

yn 1 1
yn+1 2 3 =0
yn+1 4 9
Therefore, we have
yn+2 − 5yn+1 + 6yn = 0.

Problem: Form a difference equation by eliminating the arbitrary constants from

yn = A + B3n

Solution:

yn = A + B3n

yn+1 = A + B3n+1 = A + 3[B3n ]

yn+2 = A + B3n+2 = A + 9[B3n ]

yn 1 1
yn+1 1 3 =0
yn+1 1 9
Therefore, we have
yn+2 − 4yn+1 + 3yn = 0.

36