maths ZT.pdf

Unit - II: Z - TRANSFORMS ...
Transformation OR Transform: A transform is a mathematical devices which
transforms one function to the another function. Most of the science and engineering
problems are ending with the mathematical equations like equations, ordinary
differential equations, partial differential equations and difference equations.
In the present topic, we discuss the basic definition of Z - transform, properties of Z -
transforms, Z - transforms of the standard functions, problems on Z - transforms,
initial value and final value theorem of the Z - transforms and problems, inverse Z -
transforms, definition of the inverse Z - transforms, formulas of the inverse Z
-transforms of standard functions and problems. Application of Z -transforms to
solve difference equations.
Dr. Vijaya kumar, MSRIT, Bengaluru - 560 054
Engineering Mathematics - IV - IM41 May 5, 2022 1 / 99

Continued Unit - II on Z - Transforms ...
Definition of the Z - transforms: The Z - transforms of the discrete function un,
defined for all n ≥ 0 and un = 0, for n < 0, is denoted by ZT [un] OR Z[un] OR U(z)
OR Ū(z) and is defined as,
ZT [un] = Z[un] =
∞
X
n=0
unz−n
= U(z) = Ū(z)
Where Z OR ZT is the Z - transforms operator and z is the Z - transforms parameter.
Definition of the Inverse Z - Transforms: The function U(z) is said to be
inverse Z - transforms of the function un. The inverse Z - transforms of the function
U(z) is denoted by Z−1
[U(z)] and is given by
un = Z−1
[U(z)]
where, Z−1
is the inverse Z - transforms operator.
Properties of the Z - transforms:
1. Linearity Property: If c1, c2, ..., cp all are constants then
Z[c1u1(n) + c2u2(n) + ... + cpup(n)] = c1Z[u1(n)] + c2Z[f2(x)] + ... + cpZ[up(x)]

Continued Unit - II on Properties of the Z - Transforms ...
Poof: We have, by definition of Z - transforms
Z[c1u1(n) + c2u2(n) + ... + cpup(n)] =
∞
X
n=0
(c1u1(n) + c2u2(n) + ... + cpup(n))z−n
Z[c1u1(n)+c2u2(n)+...+cpup(n)] = c1
∞
X
n=0
u1(n)z−n
+c2
∞
X
n=0
u2(n)z−n
+...+cp
∞
X
n=0
up(n)z−
Z[c1u1(n) + c2u2(n) + ... + cpup(n)] = c1Z[u1(n)] + c2Z[u2(n)] + ... + cpZ[up(n)]
2. Damping Property OR Damping rule: If Z[un = U(z)], then prove that (i)
Z[an
un] = U(z
a
), (ii) Z[a−n
un] = U(az).
Poof: (i) We have, by definition of Z - transforms
Z[un] =
∞
X
n=0
unz−n
= U(z)
Therefore,
Z[an
un] =
∞
X
n=0
(an
un)z−n

Z[an
un] =
∞
X
n=0
un
z−n
a−n
Z[an
un] =
∞
X
n=0
un(
z
a
)−n
Comparing this with the definition here instead of z is z
a
Z[an
un] = U(
z
a
)
(ii) We have, by definition of Z - transforms
Z[un] =
∞
X
n=0
unz−n
= U(z)
Therefore,
Z[a−n
un] =
∞
X
n=0
(a−n
un)z−n
Z[a−n
un] =
∞
X
n=0
un(az)−n

Comparing this with the definition here instead of z is az
Z[a−n
un] = U(az)
3. Left Shifting Property: If Z[un = U(z)], then prove that (i)
Z[un+1] = z[U(z) − u0],
(ii) Z[un+2] = z2
[U(z) − u0 − u1
z
],
(iii)Z[un+3] = z3
[U(z) − u0 − u1
z
− u2
z2 ] and
(iv) Z[un+k] = zk
[U(z) −
Pk−1
r=0 urzr
].
Poof: (i) We have, by definition of Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
∴
Z[un+1] =
∞
X
n=0
un+1z−n
Multiply and dividing by z in the RHS
Z[un+1] = (z)(
1
z
)
∞
X
n=0
un+1z−n

Z[un+1] = z
∞
X
n=0
un+1
1
z
z−n
Z[un+1] = z
∞
X
n=0
un+1z−1
z−n
Z[un+1] = z
∞
X
n=0
un+1z−n−1
Z[un+1] = z
∞
X
n=0
un+1z−(n+1)
Expanding the summation in the RHS
Z[un+1] = z[u1z−1
+ u2z−2
+ u3z−3
+ ...]
Now, adding and subtracting the missing term u0 in the RHS
Z[un+1] = z[u0 + u1z−1
+ u2z−2
+ u3z−3
+ ... − u0]
Z[un+1] = z[(u0 + u1z−1
+ u2z−2
+ u3z−3
+ ...) − u0]

Z[un+1] = z[
∞
X
n=0
unz−n
− u0]
Z[un+1] = z[U(z) − u0]
(ii) We have, by definition of Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
∴
Z[un+2] =
∞
X
n=0
un+2z−n
Multiply and dividing by z2
in the RHS
Z[un+2] = (z2
)(
1
z2
)
∞
X
n=0
un+2z−n
Z[un+2] = z2
∞
X
n=0
un+2
1
z2
z−n

Z[un+2] = z2
∞
X
n=0
un+2z−2
z−n
Z[un+2] = z2
∞
X
n=0
un+2z−n−2
Z[un+2] = z2
∞
X
n=0
un+2z−(n+2)
Z[un+2] = z2
[u2z−2
+ u3z−3
+ u4z−4
+ ...]
Now, adding and subtracting the missing terms u0 and u1z−1
in the RHS
Z[un+2] = z2
[u0 + u1z−1
+ u2z−2
+ u3z−3
+ ... − u0 − u1z−1
]
Z[un+2] = z2
[(u0 + u1z−1
+ u2z−2
+ u3z−3
+ ...) − u0 − u1z−1
]
Z[un+2] = z2
[
∞
X
n=0
unz−n
− u0 −
u1
z
]
Z[un+2] = z2
[U(z) − u0 −
u1
z
]

(iii) We have, by definition of Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
∴
Z[un+3] =
∞
X
n=0
un+3z−n
Multiply and dividing by z3
in the RHS
Z[un+3] = (z3
)(
1
z3
)
∞
X
n=0
un+3z−n
Z[un+3] = z3
∞
X
n=0
un+3
1
z3
z−n
Z[un+3] = z3
∞
X
n=0
un+3z−3
z−n
Z[un+3] = z3
∞
X
n=0
un+3z−n−3

Z[un+3] = z3
∞
X
n=0
un+3z−(n+3)
Z[un+3] = z3
[u3z−3
+ u4z−4
+ u5z−5
+ ...]
Now, adding and subtracting the missing terms u0, u1z−1
and u2z−2
in the RHS
Z[un+3] = z3
[u0 + u1z−1
+ u2z−2
+ u3z−3
+ ... − u0 − u1z−1
− u2z−2
]
Z[un+3] = z3
[(u0 + u1z−1
+ u2z−2
+ u3z−3
+ ...) − u0 − u1z−1
− u2z−2
]
Z[un+3] = z3
[
∞
X
n=0
unz−n
− u0 −
u1
z
−
u2
z2
]
Z[un+3] = z3
[U(z) − u0 −
u1
z
−
u2
z2
]
Similarly, we can prove (iv) Z[un+k] = zk
[U(z) −
Pk−1
r=0 urzr
].

4. Right Shifting Property: If Z[un = U(z)], then prove that
Z[un−k] = z−k
U(z), k > 0.
Property - 05: If Z[un = U(z)], then prove that Z[np
] = −z d
dz
[Z(np−1
)], where p is
any positive integer.
Poof: We have, by definition of Z - transforms
Z[np
] =
∞
X
n=0
np
z−n
Multiply and dividing by n and z in the RHS
Z[np
] = (z)
∞
X
n=0
(n)(
1
n
)np
(
1
z
)z−n
Z[np
] = z
∞
X
n=0
nn−1
np
z−1
z−n
Z[np
] = z
∞
X
n=0
nnp−1
z−n−1
(1)

Again, by definition of Z - transforms
Z[np−1
] =
∞
X
n=0
np−1
z−n
Differentiating w.r.t. z on both sides
d
dz
Z[np−1
] =
d
dz
[
∞
X
n=0
np−1
z−n
]
d
dz
[Z(np−1
)] =
∞
X
n=0
d
dz
[np−1
z−n
]
d
dz
[Z(np−1
)] =
∞
X
n=0
np−1
(−n)z−n−1
d
dz
[Z(np−1
)] = −
∞
X
n=0
nnp−1
z−n−1
∞
X
n=0
nnp−1
z−n−1
= −
d
dz
[Z(np−1
)] (2)

Substituting equation (2) in equation (1), we get
Z[np
] = z[−
d
dz
[Z(np−1
)]]
Z[np
] = −z
d
dz
[Z(np−1
)]
Hence, the property is proved.

Continued Unit - II on Z - Transforms of the Standard Functions ...
1. Find the Z - transforms of 1.
Solution: We know that, by definition of the Z - transforms
Z[un] =
∞
X
n=0
unz−n
= U(z)
but, here un = 1
Z[1] =
∞
X
n=0
(1)z−n
Z[1] =
∞
X
n=0
1
zn
Z[1] =
∞
X
n=0
1
z
!n
We have, by Binomial expansion
(1 − x)−1
= 1 + x + x2
+ x3
+ ... =
∞
X
n=0
xn

here, x = 1
z
in the above equation
Z[1] = 1 −
1
z
!−1
Z[1] =
z − 1
z
!−1
Z[1] =
z
z − 1
Replace 1 by −1 , we get
Z[−1] =
z
z + 1
2. Find the Z - transforms of an
.
Z[un] =
∞
X
n=0
unz−n
= U(z)
but, here un = an
Z[an
] =
∞
X
n=0
(an
)z−n
n
∞
X an

Z[an
] =
∞
X
n=0
(
a
z
)n
(1 − x)−1
= 1 + x + x2
+ x3
+ ... =
∞
X
n=0
xn
here, x = a
z
Z[an
] = 1 −
a
z
!−1
Z[an
] =
z − a
z
!−1
Z[an
] =
z
z − a
Replace a by −a , we get
Z[(−a)n
] =
z
z + a

Similarly, the Z - transforms
(a)
Z[enθ
] = Z[(eθ
)n
] =
z
z − eθ
(b)
Z[e−nθ
] = Z[(e−θ
)n
] =
z
z − e−θ
(c)
Z[einθ
] = Z[(eiθ
)n
] =
z
z − eiθ
(d)
Z[e−inθ
] = Z[(e−iθ
)n
] =
z
z − e−iθ
(e) Z - transforms of any constant k
Z[k] =
kz
z − 1

We know that,
I.
einθ
= cos(nθ) + i sin(nθ)
II.
e−inθ
= cos(nθ) − i sin(nθ)
III.
cos(nθ) =
einθ
+ e−inθ
2
IV.
einθ
+ e−inθ
= 2 cos(nθ)
V.
sin(nθ) =
einθ
− e−inθ
2i
VI.
einθ
− e−inθ
= 2i sin(nθ)
VII.
sinh(nθ) =
enθ
− e−nθ
2
VIII.
cosh(nθ) =
enθ
+ e−nθ
2

3. Obtain the Z - transforms of sinh(nθ) and cosh(nθ).
Solution: (i) To obtain the Z - transforms of sinh(nθ)
We know that,
Z[enθ
] =
z
z − eθ
and
Z[e−nθ
] =
z
z − e−θ
Also, we have
sinh(nθ) =
enθ
− e−nθ
2
Applying the Z - transforms on the both sides
Z[sinh(nθ)] = Z[
enθ
− e−nθ
2
]
Z[sinh(nθ)] =
1
2
"
Z[enθ
] − Z[e−nθ
]
#
Substituting Z[enθ
] = z
z−eθ and Z[e−nθ
] = z
z−e−θ

Z[sinh(nθ)] =
1
2
"
z
z − eθ
−
z
z − e−θ
#
Z[sinh(nθ)] =
1
2
"
z(z − e−θ
) − z(z − eθ
)
(z − eθ)(z − e−θ)
#
Z[sinh(nθ)] =
1
2
[
z2
− ze−θ
− z2
+ zeθ
z2 − ze−θ − zeθ + eθe−θ
]
Z[sinh(nθ)] =
1
2
"
z(eθ
− e−θ
)
z2 − z(eθ + e−θ) + e0
#
Z[sinh(nθ)] =
1
2
"
z(2 sinh(θ))
z2 − z(2 cosh(θ)) + 1
#
Z[sinh(nθ)] =
1
1
"
z(sinh(θ))
z2 − 2z cosh(θ) + 1
#
Z[sinh(nθ)] =
z sinh(θ)
z2 − 2z cosh(θ) + 1

(ii) To obtain the Z - transforms of cosh(nθ)
We know that,
Z[enθ
] =
z
z − eθ
and
Z[e−nθ
] =
z
z − e−θ
Also, we have
cosh(nθ) =
enθ
+ e−nθ
2
Applying the Z - transforms on the both sides
Z[cosh(nθ)] = Z[
enθ
+ e−nθ
2
]
Z[cosh(nθ)] =
1
2
"
Z[enθ
] + Z[e−nθ
]
#
Substituting Z[enθ
] = z
z−eθ and Z[e−nθ
] = z
z−e−θ

Z[cosh(nθ)] =
1
2
"
z
z − eθ
+
z
z − e−θ
#
Z[cosh(nθ)] =
1
2
"
z(z − e−θ
) + z(z − eθ
)
(z − eθ)(z − e−θ)
#
Z[cosh(nθ)] =
1
2
"
z2
− ze−θ
+ z2
− zeθ
z2 − ze−θ − zeθ + eθe−θ
#
Z[cosh θ)] =
1
2
"
2z2
− z(eθ
+ e−θ
)
z2 − z(eθ + e−θ) + e0
#
Z[cosh(nθ)] =
1
2
"
2z2
− z(2 cosh(θ))
z2 − z(2 cosh(θ)) + 1
#
Z[cosh(nθ)] =
1
1
"
z2
− z(cosh(θ))
z2 − 2z cosh(θ) + 1
#
Z[cosh(nθ)] =
z2
− z cosh(θ)
z2 − 2z cosh(θ) + 1

4. Obtain the Z - transforms of sin(nθ) and cos(nθ).
Solution: We know that,
einθ
= cos(nθ) + i sin(nθ)
Applying Z - transforms on both sides
Z[einθ
] = Z[cos(nθ) + i sin(nθ)]
Z[cos(nθ)] + iZ[sin(nθ)] = Z[einθ
]
Substituting Z[einθ
] = z
z−eiθ in the RHS
Z[cos(nθ)] + iZ[sin(nθ)] =
z
z − eiθ
z
z − (cos(θ) + i sin(θ))
z
(z − cos(θ)) − i sin(θ)
Now, rationalize RHS

z((z − cos(θ)) + i sin(θ))
((z − cos(θ)) − i sin(θ))((z − cos(θ)) + i sin(θ))
(z2
− z cos(θ)) + iz sin(θ)
(z − cos(θ))2 − (i sin(θ))2
(z2
z2 − 2z cos(θ) + cos2(θ) − i2 sin2
(θ)
(z2
z2 − 2z cos(θ) + cos2(θ) + sin2
(θ)
(z2
z2 − 2z cos(θ) + 1
Separate the real and imaginary parts in the RHS
z2
− z cos(θ)
z2 − 2z cos(θ) + 1
+ i
z sin(θ)
z2 − 2z cos(θ) + 1
Comparing the real and imaginary parts on both sides, we get

Z[sin(nθ)] =
z sin(θ)
z2 − 2z cos(θ) + 1
and
Z[cos(nθ)] =
z2
− z cos(θ)
z2 − 2z cos(θ) + 1
5. Find the Z - transforms of n.
Solution: We know that, by property
Z[np
] = −z
d
dz
[Z(np−1
)]
Substitute p = 1
Z[n1
] = −z
d
dz
[Z(n1−1
)]
Z[n] = −z
d
dz
[Z(n0
)]
Z[n] = −z
d
dz
[Z(1)]
substituting Z[1] = z
z−1
in the RHS

Z[n] = −z
d
dz
[
z
z − 1
]
Differentiate using the quotient rule
Z[n] = −z[
(z − 1)(1) − (z)(1)
(z − 1)2
]
Z[n] = −z[
z − 1 − z
(z − 1)2
]
Z[n] = −z[
−1
(z − 1)2
]
Z[n] =
z
(z − 1)2
6. Find the Z - transforms of n2
.
Z[np
] = −z
d
dz
[Z(np−1
)]
Substitute p = 2
Z[n2
] = −z
d
dz
[Z(n2−1
)]

Z[n2
] = −z
d
dz
[Z(n1
)]
Z[n2
] = −z
d
dz
[Z(n)]
substituting Z[n] = z
(z−1)2 in the RHS
Z[n2
] = −z
d
dz
[
z
(z − 1)2
]
Z[n2
] = −z[
(z − 1)2
(1) − (z)2(z − 1)(1)
(z − 1)4
]
Z[n2
] = −z(z − 1)[
z − 1 − 2z
(z − 1)4
]
Z[n2
] = −z[
−z − 1
(z − 1)3
]
Z[n2
] =
z2
+ z
(z − 1)3

.
Z[np
] = −z
d
dz
[Z(np−1
)]
Substitute p = 3
Z[n3
] = −z
d
dz
[Z(n3−1
)]
Z[n3
] = −z
d
dz
[Z(n2
)]
substituting Z[n2
] = z2
+z
(z−1)3 in the RHS
Z[n3
] = −z
d
dz
[fracz2
+ z(z − 1)3
]
Z[n3
] = −z[
(z − 1)3
(2z + 1) − (z2
+ z)3(z − 1)2
(1)
(z − 1)6
]
Z[n3
] = −z(z − 1)2
[
(z − 1)(2z + 1) − 3(z2
+ z)
(z − 1)6
]

Z[n3
] = −z[
2z2
+ z − 2z − 1 − 3z2
− 3z
(z − 1)4
]
Z[n3
] = −z[
−z2
− 4z − 1
(z − 1)4
]
Z[n3
] =
z3
+ 4z2
+ z
(z − 1)4
.
Z[np
] = −z
d
dz
[Z(np−1
)]
Substitute p = 4
Z[n4
] = −z
d
dz
[Z(n4−1
)]
Z[n4
] = −z
d
dz
[Z(n3
)]

substituting Z[n3
] = z3
+4z2
+z
(z−1)4 in the RHS
Z[n4
] = −z
d
dz
[
z3
+ 4z2
+ z
(z − 1)4
]
Z[n4
] = −z[
(z − 1)4
(3z2
+ 8z + 1) − (z3
+ 4z2
+ z)4(z − 1)3
(1)
(z − 1)8
]
Z[n4
] = −z(z − 1)3
[
(z − 1)(3z2
+ 8z + 1) − 4(z3
+ 4z2
+ z)
(z − 1)8
]
Z[n4
] = −z[
(z − 1)(3z2
+ 8z + 1) − 4z3
− 16z2
− 4z
(z − 1)5
]
Z[n4
] = −z[
3z3
+ 8z2
+ z − 3z2
− 8z − 1 − 4z3
− 16z2
− 4z
(z − 1)5
]
Z[n4
] = −z[
−z3
− 11z2
− 11z − 1
(z − 1)5
]
Z[n4
] =
z4
+ 11z3
+ 11z2
+ z
(z − 1)5

9. Find the Z - transforms of 1
n
, for n ≥ 1.
Z
"
1
n
#
=
∞
X
n=1
1
n
!
z−n
Z
"
1
n
#
=
1
1
!
z−1
+
1
2
!
z−2
+
1
3
!
z−3
+ ...
Z
"
1
n
#
=
1
z
+
1
2z2
+
1
3z3
+ ...
Z
"
1
n
#
=
(1
z
)
1
+
(1
z
)2
2
+
(1
z
)3
3
+ ...
Z[
1
n
] = −[−
1
z
!
1
−
1
z
!2
2
−
1
z
!3
3
− ...]
We know that, by logarithmic series
log(1 − x) = −x −
x2
−
x3
− ...

Here, x = 1
z
Z
"
1
n
#
= −
"
log 1 −
1
z
!#
Z
"
1
n
#
= − log
z − 1
z
!
Z
"
1
n
#
= log
z − 1
z
!−1
Z
"
1
n
#
= log
z
z − 1
!
10. Find the Z - transforms of 1
n+1
, for n ≥ 0.
Z
"
1
n + 1
#
=
∞
X
n=0
1
n + 1
!
z−n
Z
"
1
n + 1
#
=
1
1
!
z0
+
1
2
!
z−1
+
1
3
!
z−2
+
1
4
!
z−3
+ ...

Multiply and dividing by −z in the RHS
Z
"
1
n + 1
#
= (−z)[−
(1
z
)
1
−
(1
z
)2
2
−
(1
z
)3
3
− ...]
We know that, by logarithmic series
log(1 − x) = −x −
x2
2
−
x3
3
− ...
Z
"
1
n + 1
#
= (−z)[log(1 −
1
z
)]
Z
"
1
n + 1
#
= −z log
z − 1
z
!
Z
"
1
n + 1
#
= z log
z − 1
z
!−1
Z
"
1
n + 1
#
= z log
z
z − 1
!

11. Define unit step sequence for discrete and hence find its Z - transforms.
Solution:
Definition of the unit step sequence for discrete:
The unit step sequence for discrete is denoted by u(n) or H(n) and is defined as
u(n) = H(n) =
(
1, for n ≥ 0
0, for n < 0
To find the Z - transforms of the unit step sequence for discrete:
We have, by definition of Z - transforms
Z[H(n)] =
∞
X
n=0
(H(n))z−n
Z[H(n)] = H(0)z0
+ H(1)z−1
+ H(2)z−2
+ H(3)z−3
+ ...
Z[H(n)] = 1 +
1
z
!
+
1
z
!2
+
1
z
!3
+ ...

(1 − x)−1
= 1 + x + x2
+ x3
+ ...
Z[H(n)] = 1 −
1
z
!−1
Z[H(n)] =
z − 1
z
!−1
Z[H(n)] =
z
z − 1
12. Define unit impulse sequence for discrete and hence find its Z - transforms.
Solution:
Definition of the unit impulse sequence for discrete:
The unit impulse sequence for discrete is denoted by δ(n) and is defined as
δ(n) =
(
1, for n = 0
0, for n 6= 0

To find the Z - transforms of the unit impulse sequence for discrete:
We have, by definition of Z - transforms
Z[δ(n)] =
∞
X
n=0
(δ(n))z−n
Z[δ(n)] = δ(0)z0
+ δ(1)z−1
+ δ(2)z−2
+ δ(3)z−3
+ ...
Z[δ(n)] = 1 + (0) + (0) + (0) + ...
Z[δ(n)] = 1

Continued Unit - II on Formulas of the Z - Transforms of the Standard
Functions ...
1.
Z[1] =
z
z − 1
2.
Z[−1] =
z
z + 1
3.
Z[an
] =
z
z − a
4.
Z[(−a)n
] =
z
z + a
5.
Z[sinh(nθ)] =
z sinh(θ)
z2 − 2z cosh(θ) + 1
6.
Z[cosh(nθ)] =
z2
− z cosh(θ)
z2 − 2z cosh(θ) + 1

Functions ...
7.
Z[sin(nθ)] =
z sin(θ)
z2 − 2z cos(θ) + 1
8.
Z[cos(nθ)] =
z2
− z cos(θ)
z2 − 2z cos(θ) + 1
9.
Z[n] =
z
(z − 1)2
10.
Z[n2
] =
z2
+ z
(z − 1)3
11.
Z[n3
] =
z3
+ 4z2
+ z
(z − 1)4

Functions ...
12.
Z[n4
] =
z4
+ 11z3
+ 11z2
+ z
(z − 1)5
13.
Z[
1
n
] = log(
z
z − 1
)
14.
Z[
1
n + 1
] = z log(
z
z − 1
)
15.
Z[H(n)] =
z
z − 1
16.
Z[δ(n)] = 1
17.
Z[k] =
kz
z − 1

Continued Unit - II Problems on Z - Transforms ...
Problem - 01 : Find the Z - transforms of the 3n + sin(5n) − cosh(4n) + 5.
Solution: By data 3n + sin(5n) − cosh(4n) + 5
Applying the Z - transforms for the problem
Z[3n + sin(5n) − cosh(4n) + 5] = 3Z[n] + Z[sin(5n)] − Z[cosh(4n)] + Z[5]
Z[3n+sin(5n)−cosh(4n)+5] = 3(
z
(z − 1)2
)+(
z sin(5)
z2 − 2z cos(5) + 1
)−(
z2
− z cosh(4)
z2 − 2z cosh(4) + 1
)+
Z[3n+sin(5n)−cosh(4n)+5] =
3z
(z − 1)2
+
z sin(5)
z2 − 2z cos(5) + 1
−
z2
− z cosh(4)
z2 − 2z cosh(4) + 1
+
5z
z − 1
Problem - 02 : Find the Z - transforms of the cos(3n + 2).
Solution: By data cos(3n + 2)
cos(3n + 2) = cos(3n) cos(2) − sin(3n) sin(2)
Applying the Z - transforms on both sides
Z[cos(3n + 2)] = Z[cos(3n) cos(2) − sin(3n) sin(2)]
Z[cos(3n + 2)] = Z[cos(3n) cos(2)] − Z[sin(3n) sin(2)]
Z[cos(3n + 2)] = (cos(2))Z[cos(3n)] − (sin(2))Z[sin(3n)]

Z[cos(3n + 2)] = (cos(2))(
z2
− z cos(3)
z2 − 2z cos(3) + 1
) − (sin(2))(
z sin(3)
z2 − 2z cos(3) + 1
)
Z[cos(3n + 2)] =
z2
cos(2) − z cos(3) cos(2) − z sin(3) sin(2)
z2 − 2z cos(3) + 1
Z[cos(3n + 2)] =
z2
cos(2) − z(cos(3) cos(2) + sin(3) sin(2))
z2 − 2z cos(3) + 1
Z[cos(3n + 2)] =
z2
cos(2) − z(cos(3 − 2))
z2 − 2z cos(3) + 1
Z[cos(3n + 2)] =
z2
cos(2) − z cos(1)
z2 − 2z cos(3) + 1
Problem - 03 : Find the Z - transforms of the sin(π
4
− nπ
2
).
Solution: Given sin(π
4
− nπ
2
)
sin(
π
4
−
nπ
2
) = sin(
π
4
) cos(
nπ
2
) − cos(
π
4
) sin(
nπ
2
)
But, sin(π
4
) = 1
√
2
and cos(π
4
) = 1
√
2

sin(
π
4
−
nπ
2
) = (
1
√
2
) cos(
nπ
2
) − (
1
√
2
) sin(
nπ
2
)
Z[sin(
π
4
−
nπ
2
)] = (
1
√
2
)Z[cos(
nπ
2
) − sin(
nπ
2
)]
Z[sin(
π
4
−
nπ
2
)] =
1
√
2
[(
z2
− z cos(π
2
)
z2 − 2z cos(π
2
) + 1
) − (
z sin(π
2
)
z2 − 2z cos(π
2
) + 1
)]
sin(π
2
) = 1 and cos(π
2
) = 0
Z[sin(
π
4
−
nπ
2
)] =
1
√
2
[(
z2
− z(0)
z2 − 2z(0) + 1
) − (
z(1)
z2 − 2z(0) + 1
)]
Z[sin(
π
4
−
nπ
2
)] =
z2
− z
√
2(z2 + 1)
Problem - 04 : Find the Z - transforms of the 4n2
+ cos(nπ
4
) + 10a3
.
Solution: Given 4n2
+ cos(nπ
4
) + 10a3
Applying the z - transforms for the given
problem
Z[4n2
+ cos(
nπ
4
) + 10a3
] = 4Z[n2
] + Z[cos(
nπ
4
)] + Z[10a3
]

Z[4n2
+ cos(
nπ
4
) + 10a3
] = 4(
z2
+ z
(z − 1)3
) + (
z2
− z cos(π
4
)
z2 − 2z cos(π
4
) + 1
) + (
10a3
z
z − 1
)
but, cos(π
4
= 1
√
2
Z[4n2
+ cos(
nπ
4
) + 10a3
] =
4(z2
+ z)
(z − 1)3
+
z2
− z( 1
√
2
)
z2 − 2z( 1
√
2
) + 1
+
10a3
z
z − 1
Z[4n2
+ cos(
nπ
4
) + 10a3
] =
4(z2
+ z)
(z − 1)3
+
√
2z2
− z
√
2z2 − 2z +
√
2
+
10a3
z
z − 1
Problem - 05 : Find the Z - transforms of the (3n + 2)2
.
Solution: By data (3n + 2)2
(3n + 2)2
= (3n)2
+ (2)2
+ 2(3n)(2)
(3n + 2)2
= 9n2
+ 4 + 12n
Applying the z - transforms on both sides
Z[(3n + 2)2
] = Z[9n2
+ 4 + 12n]
Z[(3n + 2)2
] = 9Z[n2
] + 4Z[1] + 12Z[n]

Z[(3n + 2)2
] = 9(
z2
+ z
(z − 1)3
) + 4(
z
z − 1
) + 12(
z
(z − 1)2
)
Z[(3n + 2)2
] =
9(z2
+ z)
(z − 1)3
+
4z
z − 1
+
12z
(z − 1)2
Problem - 06 : Find the Z - transforms of the (2n − 1)3
.
Solution: By data (2n − 1)3
(2n − 1)3
= (2n)3
− 3(2n)2
(1) + 3(2n)(1)2
− (1)3
(2n − 1)3
= 8n3
− 12n2
+ 6n − 1
Applying the z - transforms on both sides
Z[(2n − 1)3
] = Z[8n3
− 12n2
+ 6n − 1]
Z[(2n − 1)3
] = 8Z[n3
] − 12Z[n2
] + 6Z[n] − Z[1]
Z[(2n − 1)3
] = 8(
z3
+ 4z2
+ z
(z − 1)4
) − 12(
z2
+ z
(z − 1)3
) + 6(
z
(z − 1)2
) − (
z
z − 1
)
Z[(2n − 1)3
] =
8(z3
+ 4z2
+ z)
(z − 1)4
−
12(z2
+ z)
(z − 1)3
+
6z
(z − 1)2
−
z
z − 1

Problem - 07 : Find the Z - transforms of the an
n.
Solution: By data an
n
We know that, by damping property
Z[an
un] = U(
z
a
)
Here, un = n and U(z) = Z[n] = z
(z−1)2 , Replace z by z
a
Z[an
n] = [
z
(z − 1)2
]z→ z
a
Z[an
n] = [
(z
a
)
(z
a
− 1)2
]
Z[an
n] =
az
(z − a)2
Problem - 08 : Find the Z - transforms of the a−n
n2
.
Solution: By data a−n
n2
Z[a−n
un] = U(az)
Here, un = n2
and U(z) = Z[n2
] = z2
+z
(z−1)3 , Replace z by az

Z[a−n
n2
] = [
z2
+ z
(z − 1)3
]z→az
Z[a−n
n2
] = [
(az)2
+ az
(az − 1)3
]
Z[a−n
n2
] =
a2
z2
+ az
(az − 1)3
n2
.
n2
Z[an
un] = U(
z
a
)
Here, un = n2
and U(z) = Z[n2
] = z2
+z
(z−1)3 , Replace z by z
a
Z[an
n2
] = [
z2
+ z
(z − 1)3
]z→ z
a

Z[an
n2
] = [
(z
a
)2
+ (z
a
)
(z
a
− 1)3
]
Z[an
n2
] =
az2
+ a2
z
(z − a)3
ebn
.
ebn
Z[a−n
un] = U(az)
Here, un = ebn
and U(z) = Z[ebn
] = z
z−eb , Replace z by az
Z[a−n
ebn
] = [
z
z − eb
]z→az
Z[a−n
ebn
] = [
az
az − eb
]
Z[a−n
ebn
] =
az
az − eb

cos(nθ).
cos(nθ)
Z[an
un] = U(
z
a
)
Here, un = cos(nθ) and U(z) = Z[cos(nθ)] = z2
−z cos(θ)
z2−2z cos(θ)+1
, Replace z by z
a
Z[an
cos(nθ)] = [
z2
− z cos(θ)
z2 − 2z cos(θ) + 1
]z→ z
a
Z[an
cos(nθ)] =
(z
a
)2
− (z
a
) cos(θ)
(z
a
)2 − 2(z
a
) cos(θ) + 1
Z[an
cos(nθ)] =
z2
− az cos(θ)
z2 − 2az cos(θ) + a2
sin(nθ).
sin(nθ)
Z[a−n
un] = U(az)
Here, un = sin(nθ) and U(z) = Z[sin(nθ)] = z sin(θ)
z2−2z cos(θ)+1
, Replace z by az

Z[a−n
sin(nθ)] = [
z sin(θ)
z2 − 2z cos(θ) + 1
]z→az
Z[a−n
sin(nθ)] =
az sin(θ)
(az)2 − 2az cos(θ) + 1
Z[a−n
sin(nθ)] =
az sin(θ)
a2z2 − 2az cos(θ) + 1
sinh(nθ).
sinh(nθ)
Z[an
un] = U(
z
a
)
Here, un = sinh(nθ) and U(z) = Z[sinh(nθ)] = z sinh(θ)
z2−2z cosh(θ)+1
, Replace z by z
a
Z[an
sinh(nθ)] = [
z sinh(θ)
z2 − 2z cosh(θ) + 1
]z→ z
a
Z[an
sinh(nθ)] =
(z
a
) sinh(θ)
(z
a
)2 − 2(z
a
) cosh(θ) + 1

Z[an
sinh(nθ)] =
az sinh(θ)
z2 − 2az cosh(θ) + a2
cosh(nθ).
cosh(nθ)
Z[a−n
un] = U(az)
Here, un = cosh(nθ) and U(z) = Z[cosh(nθ)] = z2
−z cosh(θ)
z2−2z cosh(θ)+1
, Replace z by az
Z[a−n
cosh(nθ)] = [
z2
− z cosh(θ)
z2 − 2z cosh(θ) + 1
]z→az
Z[a−n
cosh(nθ)] =
(az)2
− (az) cosh(θ)
(az)2 − 2(az) cosh(θ) + 1
Z[a−n
cosh(nθ)] =
a2
z2
− az cosh(θ)
a2z2 − 2az cos(θ) + 1

Problem - 15 : Find the Z - transforms of the 1
n(n+1)
.
Solution: By data 1
n(n+1)
, adding and subtracting n in the numerator
1
n(n + 1)
=
n + 1 − n
n(n + 1)
1
n(n + 1)
=
(n + 1) − n
n(n + 1)
1
n(n + 1)
=
(n + 1)
n(n + 1)
−
n
n(n + 1)
1
n(n + 1)
=
1
n
−
1
n + 1
Z[
1
n(n + 1)
] = Z[
1
n
] − Z[
1
n + 1
]
We know that
Z[
1
n
] = log(
z
z − 1
)
and
Z[
1
n + 1
] = z log(
z
z − 1
)

Using these results in the above equation
Z[
1
n(n + 1)
] = (log(
z
z − 1
)) − (z log(
z
z − 1
))
Z[
1
n(n + 1)
] = (1 − z) log(
z
z − 1
)
Problem - 16 : Find the Z - transforms of the n
n+1
.
Solution: By data n
n+1
, adding and subtracting 1 in the numerator
n
n + 1
=
n + 1 − 1
n + 1
n
n + 1
=
(n + 1) − 1
n + 1
n
n + 1
=
(n + 1)
n + 1
−
1
n + 1
n
n + 1
= 1 −
1
n + 1
Z[
n
n + 1
] = Z[1] − Z[
1
n + 1
]

We know that
Z[1] =
z
z − 1
and
Z[
1
n + 1
] = z log(
z
z − 1
)
Using these results in the above equation
Z[
n
n + 1
] =
z
z − 1
− z log(
z
z − 1
)
Problem - 17 : Show that ZT [ 1
n!
] = e
1
z , and hence evaluate (i) ZT [ 1
(n+1)!
],
(ii) ZT [ 1
(n+2)!
].
Solution: We have, by definition of Z - transform
ZT [
1
n!
] =
∞
X
n=0
(
1
n!
)z−n
ZT [
1
n!
] = [(
1
0!
)z0
+ (
1
1!
)z−1
+ (
1
2!
)z−2
+ (
1
3!
)z−3
+ ...]

ZT [
1
n!
] = 1 +
(1
z
)
1!
+
(1
z
)2
2!
+
(1
z
)3
3!
+ ...
We know that, by exponential series
ex
= 1 +
x
1!
+
x2
2!
+
x3
3!
+ ...
Here x = 1
z
ZT [
1
n!
] = e
1
z
(i) To evaluate ZT [ 1
(n+1)!
]
Taking un = 1
n!
, un+1 = 1
(n+1)!
, u0 = 1
0!
= 1, and U(z) = ZT [un] = ZT [ 1
n!
] = e
1
z
We have, by shifting property
ZT [un+1] = z[U(z) − u0]
Now, substituting all results
ZT [
1
(n + 1)!
] = z[e
1
z − 1]

(ii) To evaluate ZT [ 1
(n+2)!
]
Taking un = 1
n!
, un+2 = 1
(n+2)!
, u0 = 1
0!
= 1, u1 = 1
1!
= 1, and
U(z) = ZT [un] = ZT [ 1
n!
] = e
1
z
ZT [un+2] = z2
[U(z) − u0 −
u1
z
]
Now, substituting all results
ZT [
1
(n + 2)!
] = z2
[e
1
z − 1 −
1
z
]

Continued Unit - II on Initial Value of the Z - Transforms ...
Initial Value Theorem of the Z - transforms:
Statement: If Z[un = U(z)], then prove that u0 = lim
z→
U(z). Hence prove that
(i) u1 = lim
z→
z[U(z) − u0],
(ii) u2 = lim
z→
z2
[U(z) − u0 − u1
z
],
(iii) u3 = lim
z→
z3
[U(z) − u0 − u1
z
− u2
z2 ].
Proof: We know that, by definition of the Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
U(z) = u0z0
+ u1z−1
+ u2z−2
+ ...
U(z) = u0(1) +
u1
z
+
u2
z2
+ ...
Applying the limit z → ∞ on both sides
lim
z→∞
U(z) = lim
z→∞
[u0 +
u1
z
+
u2
z2
+ ...]
lim
z→∞
U(z) = [u0 + (0) + (0) + ...]

u0 = lim
z→∞
U(z)
Hence initial value theorem is proved (i) To prove this u1 = lim
z→
z[U(z) − u0],
We know that, by definition of the Z - transforms
Z[un] = U(z) =
∞
X
n=0
unz−n
U(z) = u0z0
+ u1z−1
+ u2z−2
+ u3z−3
+ ...
U(z) − u0 =
u1
z
+
u2
z2
+
u3
z3
+ ...
Multiplying throughout by z
z[U(z) − u0] = u1 +
u2
z
+
u3
z2
+ ...
lim
z→∞
z[U(z) − u0] = lim
z→∞
[u1 +
u2
z
+
u3
z2
+ ...]
lim
z→∞
z[U(z) − u0] = [u1 + (0) + (0) + ...]
u1 = lim
z→∞
z[U(z) − u0]

(ii) To prove this result
u2 = lim
z→
z2
[U(z) − u0 − u1
z
],
Z[un] = U(z) =
∞
X
n=0
unz−n
U(z) = u0z0
+ u1z−1
+ u2z−2
+ u3z−3
+ u4z−4
+ ...
U(z) − u0 −
u1
z
=
u2
z2
+
u3
z3
+
u4
z4
+ ...
Multiplying throughout by z2
z2
[U(z) − u0 −
u1
z
] = u2 +
u3
z
+
u4
z2
+ ...
lim
z→∞
z2
[U(z) − u0 −
u1
z
] = lim
z→∞
[u2 +
u3
z
+
u4
z2
+ ...]
lim
z→∞
z2
[U(z) − u0 −
u1
z
] = [u2 + (0) + (0) + ...]

u2 = lim
z→∞
z[U(z) − u0 −
u1
z
]
(iii) To prove this result
u3 = lim
z→
z3
[U(z) − u0 − u1
z
− u2
z2 ],
Z[un] = U(z) =
∞
X
n=0
unz−n
U(z) = u0z0
+ u1z−1
+ u2z−2
+ u3z−3
+ u4z−4
+ ...
U(z) − u0 −
u1
z
−
u2
z2
=
u3
z3
+
u4
z4
+ ...
Multiplying throughout by z3
z3
[U(z) − u0 −
u1
z
−
u2
z2
] = u3 +
u4
z
+
u5
z2
+ ...
lim
z→∞
z3
[U(z) − u0 −
u1
z
−
u2
z2
] = lim
z→∞
[u3 +
u4
z
+
u5
z2
+ ...]

Continued Unit - II on Initial Value and Final Value Theorem of the Z -
Transforms ...
lim
z→∞
z3
[U(z) − u0 −
u1
z
−
u2
z2
] = [u3 + (0) + (0) + ...]
u3 = lim
z→∞
z3
[U(z) − u0 −
u1
z
−
u2
z2
]
Final Value Theorem of the Z - Transforms:
Statement: If Z[un] = U(z), then prove that lim
n→∞
un = lim
z→1
(z − 1)U(z).

Continued Unit - II Problems on Initial Value of the Z - Transforms ...
Problem - 01: Find the values of the u0, u1, u2 and u3, given that
U(z) = 5z2
+3z+12
(z−1)4 .
Solution: By data U(z) = 5z2
+3z+12
(z−1)4
(i) To find the value of u0
We know that, by initial value of Z - transforms
u0 = lim
z→∞
U(z)
Substituting U(z) in the RHS
u0 = lim
z→∞
[
5z2
+ 3z + 12
(z − 1)4
] = lim
z→∞
z2
[
5 + 3
z
+ 12
z2
z4(1 − 1
z
)4
] = lim
z→∞
[
5 + 3
z
+ 12
z2
z2(1 − 1
z
)4
]
u0 = lim
z→∞
1
z2
[
5 + 3
z
+ 12
z2
(1 − 1
z
)4
] = (0)[
5 + (0) + (0)
(1 − 0)4
] = (0)(5)
u0 = 0
(ii) To find the value of u1
u1 = lim
z→∞
z[U(z) − u0]

Substituting U(z) and u0 in the RHS
u1 = lim
z→∞
z[
5z2
+ 3z + 12
(z − 1)4
− 0] = lim
z→∞
z[
5z2
+ 3z + 12
(z − 1)4
] = lim
z→∞
z3
[
5 + 3
z
+ 12
z2
z4(1 − 1
z
)4
]
u1 = lim
z→∞
[
5 + 3
z
+ 12
z2
z(1 − 1
z
)4
] = lim
z→∞
1
z
[
5 + 3
z
+ 12
z2
(1 − 1
z
)4
] = (0)[
5 + (0) + (0)
(1 − 0)4
] = (0)(5)
u1 = 0
(iii) To find the value of u2
u2 = lim
z→∞
z2
[U(z) − u0 −
u1
z
]
Substituting U(z), u0 and u1 in the RHS
u2 = lim
z→∞
z2
[
5z2
+ 3z + 12
(z − 1)4
− 0 − 0] = lim
z→∞
z2
[
5z2
+ 3z + 12
(z − 1)4
] = lim
z→∞
z4
[
5 + 3
z
+ 12
z2
z4(1 − 1
z
)4
]
u2 = lim
z→∞
[
5 + 3
z
+ 12
z2
(1 − 1
z
)4
] = [
5 + (0) + (0)
(1 − 0)4
] =
5
1
u1 = 5

(iv) To find the value of u3
u3 = lim
z→∞
z3
[U(z) − u0 −
u1
z
−
u2
z2
]
Substituting U(z), u0, u1 and u2 in the RHS
u3 = lim
z→∞
z3
[
5z2
+ 3z + 12
(z − 1)4
− 0 − 0 −
5
z2
] = lim
z→∞
z3
[
5z2
+ 3z + 12
(z − 1)4
−
5
z2
]
u3 = lim
z→∞
z3
[
5z4
+ 3z3
+ 12z2
− 5(z − 1)4
z2(z − 1)4
]
But (z − 1)4
= (z − 1)2
(z − 1)2
(z − 1)4
= (z2
− 2z + 1)(z2
− 2z + 1)
(z − 1)4
= (z4
− 2z3
+ z2
− 2z3
4z2
− 2z + z2
− 2z + 1)
(z − 1)4
= (z4
− 4z3
+ 6z2
− 4z + 1)
u3 = lim
z→∞
z3
[
5z4
+ 3z3
+ 12z2
− 5(z4
− 4z3
+ 6z2
− 4z + 1)
z2(z − 1)4
]

u3 = lim
z→∞
z3
[
5z4
+ 3z3
+ 12z2
− 5z4
+ 20z3
− 30z2
+ 20z − 5
z2(z − 1)4
]
u3 = lim
z→∞
z3
[
23z3
− 18z2
+ 20z − 5
z2(z − 1)4
] = lim
z→∞
z6
[
23 − 18
z
+ 20
z2 − 5
z3
z6(1 − 1
z
)4
]
u3 = lim
z→∞
[
23 − 18
z
+ 20
z2 − 5
z3
(1 − 1
z
)4
] = [
23 − (0) + (0) − (0)
(1 − 0)4
] =
23
1
u3 = 23
Problem - 02: Find the values of the u0, u1, u2 and u3, when U(z) = 2z2
+4z+12
(z−1)4 .
Solution: By data U(z) = 2z2
+4z+12
(z−1)4
(i) To find the value of u0
u0 = lim
z→∞
U(z)
Substituting U(z) in the RHS
u0 = lim
z→∞
[
2z2
+ 4z + 12
(z − 1)4
] = lim
z→∞
z2
[
2 + 4
z
+ 12
z2
z4(1 − 1
z
)4
] = lim
z→∞
[
2 + 4
z
+ 12
z2
z2(1 − 1
z
)4
]

u0 = lim
z→∞
1
z2
[
2 + 4
z
+ 12
z2
(1 − 1
z
)4
] = (0)[
2 + (0) + (0)
(1 − 0)4
] = (0)(2)
u0 = 0
(ii) To find the value of u1
u1 = lim
z→∞
z[U(z) − u0]
Substituting U(z) and u0 in the RHS
u1 = lim
z→∞
z[
2z2
+ 4z + 12
(z − 1)4
− 0] = lim
z→∞
z[
2z2
+ 4z + 12
(z − 1)4
] = lim
z→∞
z3
[
2 + 4
z
+ 12
z2
z4(1 − 1
z
)4
]
u1 = lim
z→∞
[
2 + 4
z
+ 12
z2
z(1 − 1
z
)4
] = lim
z→∞
1
z
[
2 + 4
z
+ 12
z2
(1 − 1
z
)4
] = (0)[
2 + (0) + (0)
(1 − 0)4
] = (0)(2)
u1 = 0
(iii) To find the value of u2
u2 = lim
z→∞
z2
[U(z) − u0 −
u1
z
]

Substituting U(z), u0 and u1 in the RHS
u2 = lim
z→∞
z2
[
2z2
+ 4z + 12
(z − 1)4
− 0 − 0] = lim
z→∞
z2
[
2z2
+ 4z + 12
(z − 1)4
] = lim
z→∞
z4
[
2 + 4
z
+ 12
z2
z4(1 − 1
z
)4
]
u2 = lim
z→∞
[
2 + 4
z
+ 12
z2
(1 − 1
z
)4
] = [
2 + (0) + (0)
(1 − 0)4
] =
2
1
u1 = 2
(iv) To find the value of u3
u3 = lim
z→∞
z3
[U(z) − u0 −
u1
z
−
u2
z2
]
Substituting U(z), u0, u1 and u2 in the RHS
u3 = lim
z→∞
z3
[
2z2
+ 4z + 12
(z − 1)4
− 0 − 0 −
2
z2
] = lim
z→∞
z3
[
2z2
+ 4z + 12
(z − 1)4
−
2
z2
]
u3 = lim
z→∞
z3
[
2z4
+ 4z3
+ 12z2
− 2(z − 1)4
z2(z − 1)4
]

But (z − 1)4
= (z − 1)2
(z − 1)2
(z − 1)4
= (z2
− 2z + 1)(z2
− 2z + 1)
(z − 1)4
= (z4
− 2z3
+ z2
− 2z3
4z2
− 2z + z2
− 2z + 1)
(z − 1)4
= (z4
− 4z3
+ 6z2
− 4z + 1)
u3 = lim
z→∞
z3
[
2z4
+ 4z3
+ 12z2
− 2(z4
− 4z3
+ 6z2
− 4z + 1)
z2(z − 1)4
]
u3 = lim
z→∞
z3
[
2z4
+ 4z3
+ 12z2
− 2z4
+ 8z3
− 12z2
+ 8z − 2
z2(z − 1)4
]
u3 = lim
z→∞
z3
[
12z3
+ 8z − 2
z2(z − 1)4
] = lim
z→∞
z6
[
12 + 8
z2 − 2
z3
z6(1 − 1
z
)4
]
u3 = lim
z→∞
[
12 + 8
z2 − 2
z3
(1 − 1
z
)4
] = [
12 − (0) + (0) − (0)
(1 − 0)4
] =
12
1
u3 = 12

Problem - 03: If ZT (un) = z
z−1
− z
z2+1
, then find ZT (un+1).
Solution: By data ZT (un) = U(z) = z
z−1
− z
z2+1
ZT (un+1) = z2
[U(z) − u0 −
u1
z
]
To find the value of u0, by initial value theorem
u0 = lim
z→∞
U(z)
u0 = lim
z→∞
[
z
z − 1
−
z
z2 + 1
] = lim
z→∞
[
z
z(1 − 1
z
)
−
z
z2(1 + 1
z2 )
]
u0 = lim
z→∞
[
1
(1 − 1
z
)
− (
1
z
)
1
(1 + 1
z2 )
] = [
1
(1 − 0)
− (0)
1
(1 + 0)
] = [
1
(1)
− (0)(
1
(1)
)]
u0 = 1
To find the value of u1, by initial value theorem
u1 = lim
z→∞
z[U(z) − u0]
u1 = lim
z→∞
z[
z
z − 1
−
z
z2 + 1
− 1] = lim
z→∞
z[
z(z2
+ 1) − z(z − 1) − (z − 1)(z2
+ 1)
(z − 1)(z2 + 1)
]

u1 = lim
z→∞
z[
z3
+ z − z2
+ z − z3
− z + z2
+ 1
(z − 1)(z2 + 1)
] = lim
z→∞
z[
z + 1
(z − 1)(z2 + 1)
]
u1 = lim
z→∞
z2
[
(1 + 1
z
)
z3(1 − 1
z
)(1 + 1
z2 )
] = lim
z→∞
[
(1 + 1
z
)
z(1 − 1
z
)(1 + 1
z2 )
] = lim
z→∞
(
1
z
)[
(1 + 1
z
)
(1 − 1
z
)(1 + 1
z2 )
]
u0 = (0)[
(1 + 0)
(1 − 0)(1 + 0)
] = (0)(1) = 0
Substituting U(z), u0 and u1 in the above equation
ZT (un+1) = z2
[
z
z − 1
−
z
z2 + 1
− 1 −
(0)
z
]
ZT (un+1) = z2
[
z
z − 1
−
z
z2 + 1
− 1]

Continued Unit - II on Inverse Z - Transforms and Formulas ...
Definition of the Inverse Z - Transforms: The function U(z) is said to be
inverse Z - transforms of the function un. The inverse Z - transforms of the function
U(z) is denoted by Z−1
[U(z)] and is given by
un = Z−1
[U(z)]
where, Z−1
is the inverse Z - transforms operator.
Formulas of the Inverse Z - Transforms of standard functions:
1.
Z−1
[
z
z − 1
] = 1
2.
Z−1
[
z
z + 1
] = −1
3.
Z−1
[
z
z − a
] = an
4.
Z−1
[
z
z + a
] = (−a)n
5.
Z−1
[
1
z − a
] = an−1

Continued Unit - II on Formulas of the Inverse Z - Transforms ...
6.
Z−1
[
1
z + a
] = (−a)n−1
7.
Z−1
[
1
(z − a)2
] = (n − 1)an−2
8.
Z−1
[
1
(z + a)2
] = (n − 1)(−a)n−2
9.
Z−1
[
1
(z − a)3
] =
1
2
(n − 1)(n − 2)an−3
10.
Z−1
[
z2
(z − a)2
] = (n + 1)an
11.
Z−1
[
z3
(z − a)3
] =
1
2!
(n + 1)(n + 2)an
un

Continued Unit - II on Formulas of the Inverse Z - Transforms ...
12.
Z−1
[
z
(z − a)2
] = nan−1
13.
Z−1
[
az
(z − a)2
] = nan
14.
Z−1
[
az2
+ a2
z
(z − a)3
] = n2
an
15.
Z−1
[
z
z − 1
] = H(n)
16.
Z−1
[1] = δ(n)

Continued Unit - II Problems on Inverse Z - Transforms ...
Problem - 01 : Find the inverse Z - transforms of z
(z−1)(z+2)
.
Solution: by data z
(z−1)(z+2)
, Take
U(z) =
z
(z − 1)(z + 2)
U(z)
z
=
1
(z − 1)(z + 2)
(3)
Applying the partial fraction for the RHS expression
1
(z − 1)(z + 2)
=
A
(z − 1)
+
B
(z + 2)
1 = A(z + 2) + B(z − 1)
put, z = 1, we get 1 = A(3) + B(0), A = 1
3
put, z = −2, we get 1 = A(0) + B(−3), B = −1
3
. Substituting the value of A and B
1
(z − 1)(z + 2)
=
1
3(z − 1)
−
1
3(z + 2)
(4)

Using Eqn. (4) in Eqn. (3)
U(z)
z
=
1
3(z − 1)
−
1
3(z + 2)
Multiply throughout by z
U(z) =
1
3
(
z
z − 1
) −
1
3
(
z
z + 2
)
Applying inverse Z - transforms on both sides
Z−1
[U(z)] =
1
3
Z−1
[
z
z − 1
] −
1
3
Z−1
[
z
z + 2
]
Z−1
[
z
(z − 1)(z + 2)
] =
1
3
(1) −
1
3
(−2)n
Z−1
[
z
(z − 1)(z + 2)
] =
1
3
[1 − (−2)n
]

Problem - 02 : Find the inverse Z - transforms of 5z
(2−z)(3z−1)
.
Solution: by data 5z
(2−z)(3z−1)
, Take
U(z) =
5z
(2 − z)(3z − 1)
U(z)
z
=
5
(2 − z)(3z − 1)
(5)
5
(2 − z)(3z − 1)
=
A
(2 − z)
+
B
(3z − 1)
5 = A(3z − 1) + B(2 − z)
put, z = 2, we get 5 = A(5) + B(0), A = 1
put, z = 1
3
, we get 5 = A(0) + B(5
3
), B = 3. Substituting the value of A and B in the
above equation
5
(2 − z)(3z − 1)
=
1
(2 − z)
+
3
(3z − 1)
(6)

U(z)
z
=
1
(2 − z)
+
3
(3z − 1)
U(z) =
z
(2 − z)
+ 3
z
(3z − 1)
Z−1
[U(z)] = Z−1
[
z
−(z − 2)
] + 3Z−1
[
z
3(z − 1
3
)
]
Z−1
[
5z
(2 − z)(3z − 1)
] = (−1)(2)2
+
3
3
(
1
3
)n
Z−1
[
5z
(2 − z)(3z − 1)
] = −(2)2
+ (
1
3
)n

Problem - 03 : Find the inverse Z - transforms of 2z2
+3z
(z+2)(z−4)
.
Solution: by data 2z2
+3z
(z+2)(z−4)
, Take
U(z) =
2z2
+ 3z
(z + 2)(z − 4)
U(z)
z
=
2z + 3
(z + 2)(z − 4)
(7)
2z + 3
(z + 2)(z − 4)
=
A
(z + 2)
+
B
(z − 4)
2z + 3 = A(z − 4) + B(z + 2)
put, z = −2, we get −1 = A(−6) + B(0), A = 1
6
put, z = 4, we get 11 = A(0) + B(6), B = 11
6
. Substituting the value of A and B in
the above equation
2z + 3
(z + 2)(z − 4)
=
1
6(z + 2)
+
11
6(z − 4)
(8)

U(z)
z
=
1
6(z + 2)
+
11
6(z − 4)
U(z) =
1
6
[
z
(z + 2)
] +
11
6
[
z
(z − 4)
]
Z−1
[U(z)] =
1
6
Z−1
[
z
(z + 2)
] +
11
6
Z−1
[
z
(z − 4)
]
Z−1
[
2z2
+ 3z
(z + 2)(z − 4)
] =
1
6
(−2)n
+
11
6
(4)n
Z−1
[
2z2
+ 3z
(z + 2)(z − 4)
] =
1
6
[(−2)n
+ 11(4)n
]

Problem - 04 : Find the inverse Z - transforms of 4z2
−2z
z3−5z2+8z−4
.
Solution: by data 4z2
−2z
z3−5z2+8z−4
, Take
U(z) =
4z2
− 2z
z3 − 5z2 + 8z − 4
Factorize the denominator expression z3
− 5z2
+ 8z − 4 = (z − 1)(z − 2)2
U(z) =
4z2
− 2z
(z − 1)(z − 2)2
U(z)
z
=
4z − 2
(z − 1)(z − 2)2
(9)
4z − 2
(z − 1)(z − 2)2
=
A
(z − 1)
+
B
(z − 2)
+
C
(z − 2)2
4z − 2 = A(z − 2)2
+ B(z − 1)(z − 2) + C(z − 1)
put, z = 1, we get 2 = A(1) + B(0) + C(0), A = 2,
put, z = 2, we get 6 = A(0) + B(0) + C(1), C = 6,
put, z = 0, we get −2 = A(4) + B(2) + C(−1), −2 = 8 + 2B − 6, B = 0. Substituting
the value of A, B and C in the above equation

4z − 2
(z − 1)(z − 2)2
=
2
(z − 1)
+
0
(z − 2)
+
6
(z − 2)2
4z − 2
(z − 1)(z − 2)2
=
2
(z − 1)
+
6
(z − 2)2
(10)
U(z)
z
=
2
(z − 1)
+
6
(z − 2)2
U(z) = 2
z
(z − 1)
+ 6
z
(z − 2)2
Z−1
[U(z)] = 2Z−1
[
z
(z − 1)
] + 6Z−1
[
z
(z − 2)2
]
Z−1
[
4z2
− 2z
z3 − 5z2 + 8z − 4
] = 2(1) + 6n(2)n−1

Z−1
[
4z2
− 2z
z3 − 5z2 + 8z − 4
] = 2 + 6n(2)n−1
Problem - 05 : Find the inverse Z - transforms of 8z−z3
(4−z)3 .
Solution: by data 8z−z3
(4−z)3 , Take
U(z) =
8z − z3
(−1)3(z − 4)3
=
8z − z3
−(z − 4)3
U(z) =
z3
− 8z
(z − 4)3
We can observe in the denominator expression has repeated linear factors and we
have Z−1
[ z
(z−4)
] = 4n
, Z−1
[ 4z
(z−4)2 ] = n4n
and Z−1
[4z2
+16z
(z−4)3 ] = n2
4n
We have, apply
the partial fractions for the RHS expression for the above equation
z3
− 8z
(z − 4)3
= A
z
z − 4
+ B
4z
(z − 4)2
+ C
4z2
+ 16z
(z − 4)3
z3
− 8z = Az(z − 4)2
+ 4Bz(z − 4) + 4Cz(z + 4)

Canceling z on both sides
z2
− 8 = A(z − 4)2
+ 4B(z − 4) + 4C(z + 4)
put, z = 4, we get −8 = A(0) + 4B(0) + 4C(8), C = 1
4
.
Equating the coefficient of z2
on both sides, we get, A = 1.
Again by equating the coefficient of z on both sides, we have, 0 = −8A + 4B + 4C,
we get, B = 7
4
. Substituting the value of A, B and C in the above equation
z3
− 8z
(z − 4)3
= (1)
z
z − 4
+ (
7
4
)
4z
(z − 4)2
+ (
1
4
)
4z2
+ 16z
(z − 4)3
Applying the inverse Z - transforms on both sides
Z−1
[
z3
− 8z
(z − 4)3
] = Z−1
[
z
z − 4
] +
7
4
Z−1
[
4z
(z − 4)2
] +
1
4
Z−1
[
4z2
+ 16z
(z − 4)3
]
Using the above results, we get
Z−1
[
z3
− 8z
(z − 4)3
] = 4n
+
7
4
n4n
+
1
4
n2
4n

Problem - 06 : Find the inverse Z - transforms of z3
−20z
(z−2)3(z−4)
.
Solution: by data z3
−20z
(z−2)3(z−4)
, Take
U(z) =
z3
− 20z
(z − 2)3(z − 4)
We can observe in the denominator expression has repeated linear factors in first
factor and we have Z−1
[ z
(z−2)
] = 2n
, Z−1
[ 2z
(z−2)2 ] = n2n
, Z−1
[2z2
+4z
(z−2)3 ] = n2
4n
and
Z−1
[ z
(z−4)
] = 4n
We have, apply the partial fractions for the RHS expression for the
above equation
z3
− 20z
(z − 2)3(z − 4)
= A
z
z − 2
+ B
2z
(z − 2)2
+ C
2z2
+ 4z
(z − 2)3
+ D
z
z − 4
z3
− 20z = Az(z − 2)2
(z − 4) + 2Bz(z − 2)(z − 4) + C(2z2
+ 4z)(z − 4) + Dz(z − 2)3

Canceling z on both sides
z2
− 20 = A(z − 2)2
(z − 4) + 2B(z − 2)(z − 4) + C(2z + 4)(z − 4) + D(z − 2)3
put, z = 2, we get −16 = A(0) + 2B(0) + C(−16) + D(0), C = 1.
put, z = 4, we get −4 = A(0) + 2B(0) + C(0) + D(8), D = −1
2
.
on both sides, we get, A + D = 0, A = 1
2
.
put, z = 0, we get −20 = A(4)(−4) + 2B(8) + C(−16) + D(−8), B = 0. Substituting
the value of A, B, C and D in the above equation
z3
− 20z
(z − 2)3(z − 4)
= (
1
2
)
z
z − 2
+ (0)
2z
(z − 2)2
+ (1)
2z2
+ 4z
(z − 2)3
+ (
−1
2
)
z
z − 4
Z−1
[
z3
− 20z
(z − 2)3(z − 4)
] =
1
2
Z−1
[
z
z − 2
] + Z−1
[
2z2
+ 4z
(z − 2)3
] −
1
2
Z−1
[
z
z − 4
]
Using the above results, we get
Z−1
[
z3
− 20z
(z − 2)3(z − 4)
] =
1
2
2n
+ n2
2n
−
1
2
4n
= 2n−1
+ n2
2n
− 22n−1

Continued Unit - II on Application of the Z - Transforms to Solve the
Difference Equations ...
Introduction:
Definition of Difference equation: An equation which involving the set of values
of the dependent variable is known as a difference equation.
Let us consider the second order forward difference of y0 is equal to zero, that is
∆2
y0 = 0
∆(∆y0) = 0
we have, ∆y0 = y1 − y0
∆(y1 − y0) = 0
∆y1 − ∆y0 = 0
(y2 − y1) − (y1 − y0) = 0
y2 − 2y1 + y0 = 0
This kind of equation is called as a difference equation Examples:
1. yn+2 + 5yn+1 + 6yn = 3n
.
2. un+2 − 3yn+1 + 2yn = n2
.

Continued Unit - II on Application of the Z - Transforms to Solve the
Difference Equations ...
Working procedure to solve the difference equations:
Step - 01: Applying the Z - transforms on both sides.
Step - 02: Substituting the shifting property of Z - transforms, that is
Z[un+1] = z[U(z) − u0]
Z[un+2] = z2
[U(z) − u0 − u1
z
]
Z[un+3] = z3
[U(z) − u0 − u1
z
− u2
z2 ]
Step - 03: Substituting the given initial conditions and simplify U(z) in terms of z.
Step - 04: Applying the inverse Z - transforms on both sides, we get the solution of
the given difference equation.

Continued Unit - II Problems on Application of the Z - Transforms to
Solve the Difference Equations ...
Problem - 01: Solve the difference equation un+2 − 5un+1 + 6un = 2, given that
u0 = 3, u1 = 7, by using Z - transforms.
Solution: By data un+2 − 5un+1 + 6un = 2, and u0 = 3, u1 = 7
Z[un+2] − 5Z[un+1] + 6Z[un] = Z[2]
Substituting the shifting property of the Z - transforms
z2
[U(z) − u0 −
u1
z
] − 5(z[U(z) − u0]) + 6(U(z)) = (
2z
z − 1
)
Substituting the given initial conditions u0 = 3 and u1 = 7
z2
[U(z) − (3) −
7
z
] − 5(z[U(z) − (3)]) + 6(U(z)) = (
2z
z − 1
)
Simplify
z2
U(z) − 3z2
− 7z − 5zU(z) + 15z + 6U(z) =
2z
z − 1
z2
U(z) − 5zU(z) + 6U(z) =
2z
z − 1
+ 3z2
+ 7z − 15z

(z2
− 5z + 6)U(z) =
2z
z − 1
+ 3z2
− 8z
(z − 2)(z − 3)U(z) =
2z + 3z2
(z − 1) − 8z(z − 1)
z − 1
U(z) =
3z3
− 11z2
+ 10z
(z − 1)(z − 2)(z − 3)
U(z)
z
=
3z2
− 11z + 10
(z − 1)(z − 2)(z − 3)
(11)
Now, applying the partial fractions for the RHS expression
3z2
− 11z + 10
(z − 1)(z − 2)(z − 3)
=
A
z − 1
+
B
z − 2
+
C
z − 3
3z2
− 11z + 10 = A(z − 2)(z − 3) + B(z − 1)(z − 3) + C(z − 1)(z − 2)
put, z = 1, we get, 2 = A(−1)(−2) + B(0) + C(0), A = 1.
put, z = 2, we get, 0 = A(0) + B(1)(−1) + C(0), B = 0.
put, z = 3, we get, 4 = A(0) + B(0) + C(2)(1), C = 2. Substituting the values of the
A, B and C in the above equation

3z2
− 11z + 10
(z − 1)(z − 2)(z − 3)
=
1
z − 1
+
0
z − 2
+
2
z − 3
3z2
− 11z + 10
(z − 1)(z − 2)(z − 3)
=
1
z − 1
+
2
z − 3
U(z)
z
=
1
z − 1
+
2
z − 3
(12)
U(z) =
z
z − 1
+ 2
z
z − 3
Z−1
[U(z)] = Z−1
[
z
z − 1
] + 2Z−1
[
z
z − 3
]
but, Z−1
[U(z)] = un
un = (1) + 2(3n
) = 1 + 23n
This is required solution for the given difference equation.

Problem - 02: Solve yn+2 + 2yn+1 + yn = n, given that y0 = 0, y1 = 0, by using Z -
transforms.
Solution: Given yn+2 + 2yn+1 + yn = n and y0 = 0, y1 = 0
Z[yn+2] + 2Z[yn+1] + Z[yn] = Z[n]
z2
[Y (z) − y0 −
y1
z
] + 2(z[Y (z) − y0]) + (Y (z)) = (
z
(z − 1)2
)
Substituting the given initial conditions y0 = 0 and y1 = 0
z2
[Y (z) − (0) −
0
z
] + 2(z[Y (z) − (0)]) + (Y (z)) =
z
(z − 1)2
Simplify
z2
Y (z) − (0) − (0) + 2zY (z) − (0) + Y (z) =
z
(z − 1)2
(z2
+ 2z + 1)Y (z) =
z
(z − 1)2

(z + 1)2
Y (z) =
z
(z − 1)2
Y (z) =
z
(z − 1)2(z + 1)2
Y (z)
z
=
1
(z − 1)2(z + 1)2
(13)
Applying the partial fractions for the RHS expression
1
(z − 1)2(z + 1)2
=
A
(z − 1)
+
B
(z − 1)2
+
C
(z + 1)
+
D
(z + 1)2
1 = A(z − 1)(z + 1)2
+ B(z + 1)2
+ C(z + 1)(z − 1)2
+ D(z − 1)2
put, z = 1, we get, 1 = A(0) + B(4) + C(0) + D(0), B = 1
4
.
put, z = −1, we get, 1 = A(0) + B(0) + C(0) + D(4), D = 1
4
.
Rewrite the above equation
1 = A(z3
+ z2
− z − 1) + B(z2
+ 2z + 1) + C(z3
− z2
− z + 1) + D(z2
− 2z + 1)
on both sides A + C = 0 Equating the coefficient of z2
on both sides A + B − C + D = 0, A − C = −1
2
,

solving these two equations, we get A = −1
4
and C = 1
4
. Substituting the values of A,
B, C and D in the above equation.
1
(z − 1)2(z + 1)2
=
−1
4(z − 1)
+
1
4(z − 1)2
+
1
4(z + 1)
+
1
4(z + 1)2
(14)
Using Eqn. (14) in (13)
Y (z)
z
=
−1
4(z − 1)
+
1
4(z − 1)2
+
1
4(z + 1)
+
1
4(z + 1)2
Y (z) = −
1
4
z
z − 1
+
1
4
z
(z − 1)2
+
1
4
z
z + 1
+
1
4
z
(z + 1)2
Z−1
[Y (z)] = −
1
4
[Z−1
[
z
z − 1
] + Z−1
[
z
(z − 1)2
] + Z−1
[
z
z + 1
] + Z−1
[
z
(z + 1)2
]]

but, Z−1
[Y (z)] = yn
yn =
1
4
[−(1) + n(1)n−1
+ (−1)n
+ n(−1)n−1
] =
1
4
[−1 + n + (−1)n
+ n(−1)n−1
]
Problem - 03: Solve yn+2 − 4yn+1 + 3yn = H(n), given that y0 = 0, y1 = 0, by
using Z - transforms.
Solution: Given yn+2 − 4yn+1 + 3yn = H(n) and y0 = 0, y1 = 0
Z[yn+2] − 4Z[yn+1] + 3Z[yn] = Z[H(n)]
z2
[Y (z) − y0 −
y1
z
] − 4(z[Y (z) − y0]) + 3(Y (z)) = (
z
(z − 1)
)
Substituting the given initial conditions y0 = 0 and y1 = 0
z2
[Y (z) − (0) −
0
z
] − 4(z[Y (z) − (0)]) + 3(Y (z)) =
z
(z − 1)

Simplify
z2
Y (z) − (0) − (0) − 4zY (z) − (0) + 3Y (z) =
z
(z − 1)
(z2
− 4z + 3)Y (z) =
z
(z − 1)
(z − 1)(z − 3)Y (z) =
z
(z − 1)
Y (z) =
z
(z − 1)2(z − 3)
Y (z)
z
=
1
(z − 1)2(z − 3)
(15)
Applying the partial fraction for RHS expression in the above equation
1
(z − 1)2(z − 3)
=
A
(z − 1)
+
B
(z − 1)2
+
C
(z − 3)
1 = A(z − 1)(z − 3) + B(z − 3) + C(z − 1)2

put, z = 1, we get, 1 = A(0) + B(−2) + C(0), B = −1
2
.
put, z = 3, we get, 1 = A(0) + B(0) + C(4), C = 1
4
.
put, z = 0, we get, 1 = A(3) + B(−3) + C(1), A = −1
4
. Substituting the values of the
A, B and C in the above equation
1
(z − 1)2(z − 3)
=
−1
4(z − 1)
+
−1
2(z − 1)2
+
1
4(z − 3)
(16)
Using Eqn. (16) IN Eqn.. (15)
Y (z)
z
=
−1
4(z − 1)
+
−1
2(z − 1)2
+
1
4(z − 3)
Y (z) = −
1
4
z
(z − 1)
−
1
2
z
2(z − 1)2
+
1
4
z
(z − 3)
Z−1
[Y (z)] = −
1
4
Z−1
[
z
(z − 1)
] −
1
2
Z−1
[
z
(z − 1)2
] +
1
4
Z−1
[
z
(z − 3)
]

But, Z−1
[Y (z)] = yn
yn = −
1
4
(1) −
1
2
(n) +
1
4
(3)n
=
−1
4
−
n
2
+
(3)n
4
Problem - 04: Solve the difference equation un+2 − 5un+1 + 6un = Hn, given that
u0 = 0, u1 = 1, where Hn is a unit step sequence, by using Z - transforms.
Solution: Given un+2 − 5un+1 + 6un = Hn and u0 = 0, u1 = 1
Z[un+2] − 5Z[un+1] + 6Z[un] = Z[Hn]
z2
[U(z) − u0 −
u1
z
] − 5(z[U(z) − u0]) + 6(U(z)) = (
z
(z − 1)
)
Substituting the given initial conditions u0 = 0 and u1 = 1
z2
[U(z) − (0) −
1
z
] − 5(z[U(z) − (0)]) + 6(U(z)) =
z
(z − 1)

z2
U(z) − 0 − z − 5zU(z) − 0 + 6U(z) =
z
(z − 1)
(z2
− 5z + 6)U(z) =
z
(z − 1)
+ z =
z + z2
− z
(z − 1)
(z − 2)(z − 3)U(z) =
z2
(z − 1)
U(z) =
z2
(z − 1)(z − 2)(z − 3)
U(z)
z
=
z
(z − 1)(z − 2)(z − 3)
(17)
Applying the partial fractions for the RHS expression
z
(z − 1)(z − 2)(z − 3)
=
A
(z − 1)
+
B
(z − 2)
+
C
(z − 3)
z = A(z − 2)(z − 3) + B(z − 1)(z − 3) + C(z − 1)(z − 2)

put, z = 1, we get, 1 = A(2) + B(0) + C(0), A = 1
2
.
put, z = 2, we get, 2 = A(0) + B(−2) + C(0), B = −1
2
.
put, z = 3, we get, 3 = A(0) + B(0) + C(2), C = 3
2
. Substituting the values of the A,
B and C in the above equation
z
(z − 1)(z − 2)(z − 3)
=
1
2(z − 1)
+
−1
2(z − 2)
+
3
2(z − 3)
U(z)
z
=
1
2(z − 1)
+
−1
2(z − 2)
+
3
2(z − 3)
U(z) =
1
2
z
z − 1
−
1
2
z
z − 2
+
3
2
z
z − 3

Z−1
[U(z)] =
1
2
Z−1
[
z
z − 1
] −
1
2
Z−1
[
z
z − 2
] +
3
2
Z−1
[
z
z − 3
]
But, Z−1
[U(z)] = un
un =
1
2
(1) −
1
2
(2)n
+
3
2
(3)n
=
1
2
− (2)n−1
+
(3)n+1
2

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