Chapter 5
Z-Transforms.
4.1
Definitions
T
Unilateral Z-transform.
f (n)z −n
SR
Z[f (n)] =
∞
X
M
IS
Let {f (n)} = f (0), f (1), f (2), · · · be any sequence defined for n = 0, 1, 2, · · · .Then the
Z-transform of the given sequence { f(n) }is defined as
r.E
.S
ur
es
where z is arbitrary complex variable.
h,
n=0
The right hand side of the above equation is a function of z and hence it is denoted by
Z[f (n)] = F (z).
Bilateral Z-transform
D
Let {f (n)}be a sequence defined for n = 0, ±1, ±2, · · · .Then the Z-transform the given
sequence { f(n) }is defined as
∞
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 + ...
3. − log(1 − z) = z +
4. ez = 1 +
z2 z3 z4
+
+
+ ...
2
3
4
z
z2 z3 z4
+
+
+
+ ...
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
af (n)z −n +
n=0
∞
X
= a
∞
X
bg(n)z −n
n=0
f (n)z
−n
+b
n=0
∞
X
g(n)z −n
n=0
z
z−1
SR
1. Z[1] =
M
IS
= aF (z) + bG(z)
T
=
n=0
∞
X
h,
Proof.
∞
X
∞
X
n
1
Z[f (n)] =
f (n)z =
f (n)
z
n=0
n=0
2 3
∞ n
X
1
1
1
1
Z[1] =
=1+ +
+
+ ···
z
z
z
z
n=0
−1 −1
1
z−1
z
=
1−
=
=
z
z
z−1
2. Z[n] =
D
r.E
.S
ur
es
−n
z
, |z| > 1
(z − 1)2
Proof.
∞
X
∞
X
n
1
Z[f (n)] =
f (n)z =
f (n)
z
n=0
n=0
2
3
∞
n
X
1
1
1
1
Z[n] =
n
=0+1
+2
+3
+ ···
z
z
z
z
n=0
−2
−2
1
1
1 z−1
1 z2
z
=
1−
=
=
, |z| > 1
2 =
z
z
z
z
z (z − 1)
(z − 1)2
−n
3
3. Z[an ] =
Proof.
z
if |z| > |a|
z−a
∞
X
∞
X
n
1
Z[f (n)] =
f (n)z =
f (n)
z
n=0
n=0
n X
∞
∞ a a 2 a 3
X
a n
n
n 1
Z(a ) =
a
=
=1+
+
+
+ ···
z
z
z
z
z
n=0
n=0
−1
a −1
z−a
z
= 1−
=
=
, |z| > |a|
z
z
z−a
4.Z[(−a)n ] =
−n
z
if |z| > |a|
z+a
Proof.
∞
X
∞
X
h,
SR
M
IS
T
n
1
Z[f (n)] =
f (n)z =
f (n)
z
n=0
n=0
n X
n
∞
∞ a a 2 a 3
X
−a
n 1
n
Z[(−a) ] =
(−a)
=
=1−
+
−
+ ···
z
z
z
z
z
n=0
n=0
−1
a −1
z+a
z
= 1+
=
=
, if |z| > |a|
z
z
z+a
−n
Proof.
r.E
.S
ur
es
1
z
5. Z
= log
, |z| > 1
n
z−1
Z[f (n)] =
∞
X
n=0
f (n)z
−n
∞
X
n
1
=
f (n)
z
n=0
D
1
is n = 1, 2, · · · . Hence we have
n
∞ n
X
1
1
1
Z
=
n
n
z
n=1
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
Here the domain of the sequence
4
1
z
= z log
, |z| > 1
6. Z
n+1
z−1
Proof.
=
=
=
=
=
T
IS
1
Z
n+1
n
1
f (n)z =
f (n)
z
n=0
n=0
∞
n
X
1
1
n+1
z
n=0
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
z
z
−1
z−1
z
z log
= z log
, |z| > 1
z
z−1
M
r.E
.S
ur
es
1
1
z
7. Z
= log
, |z| > 1
n−1
z
z−1
Z[f (n)] =
1
Z
n−1
h,
Proof.
∞
X
−n
SR
Z[f (n)] =
∞
X
=
D
=
=
=
∞
X
∞
X
n
1
f (n)z =
f (n)
z
n=0
n=0
2
3
4
∞
n
X
1
1
1 1
1 1
1 1
=
+
+
+ ···
n
−
1
z
1
z
2
z
3
z
n=2
" #
2
3
1
1
1 1
1 1
+
+
+ ···
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
−n
1
1
8. Z
= ez
n!
Proof.
∞
X
∞
X
n
1
Z[f (n)] =
f (n)z =
f (n)
z
n=0
n=0
n
2
3
4
∞
X
1
1 1
1 1
1 1
1 1
1 1
Z
=
=1+
+
+
+
+ ···
n!
n! z
1! z
2! z
3! z
4! z
n=0
−n
5
2 3 4
1
1
1
1
1
z
z
z
z
= 1+
+
+
+
+ · · · = ez
1!
2!
3!
4!
h 1
i
1
z
9. Z
=z e −1
(n + 1)!
Proof.
=
=
=
=
10. Evaluate Z [ean ]
r.E
.S
ur
es
h,
=
T
IS
1
Z
(n + 1)!
M
∞
X
n
1
f (n)z =
f (n)
z
n=0
n=0
∞
n
X
1
1
(n + 1)! z
n=0
2
3
1
1 1
1 1
1 1
+
+
+
+ ···
1! 2! z
3! z
4! z
" #
2
3
4
1 1
1 1
1 1
1 1
z
+
+
+
+ ···
1! z
2! z
3! z
4! z
"
#
2
3
4
1 1
1 1
1 1
1 1
z 1+
+
+
+
+ ··· − 1
1! z
2! z
3! z
4! z
h 1
i
z ez − 1
−n
SR
Z[f (n)] =
∞
X
w.k.t Z[an ] =
z
z−a
z
z − ea
D
Z [ean ] = Z [(ea )n ] =
11. Evaluate Z a−n
z
z−a
n −n −n −1 n 1
z
az
=
Z a
= Z a
=Z a
=Z
=
1
a
az − 1
z− a
w.k.t Z[an ] =
12. Evaluate Z [(−1)n ]
w.k.t Z[an ] =
z
z
⇒ 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)
(z − 1)2
6
14. Find Z [sin nθ] , Z [cos nθ]
Proof.
Proof. Simply
z 2 − z cosh θ
z 2 − 2z cosh θ + 1
D
15.Z [cosh nθ] =
r.E
.S
ur
es
h,
SR
M
IS
T
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
=
(z − cos θ)2 + sin2 θ
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
(z − cos θ) + sin θ
z(z − cos θ)
z sin θ
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)
Z [cos nθ] = 2
z − 2z cos θ + 1
z sin θ
Z [sin nθ] = 2
z − 2z cos θ + 1
(z 2 − z cos θ)
z 2 − 2z cos θ + 1
(z 2 − z cosh θ)
Z [cosh nθ] = 2
z − 2z cosh θ + 1
Z [cos nθ] =
∴
16.Z [sinh nθ] =
z2
z sinh θ
− 2z cosh θ + 1
Proof. Simply
z sin θ
− 2z cos θ + 1
z sinh θ
Z [sinh nθ] = 2
z − 2z cosh θ + 1
Z [sin nθ] =
∴
z2
7
nπ i
z2
17. Prove that Z cos
= 2
2
z +1
h
Solution. We know that
Z [cos nθ] =
z 2 − z cos θ
z 2 − 2z cos θ + 1
z 2 − z cos π2
nπ i
z 2 − z(0)
z2
Z cos
= 2
=
=
2
z 2 − 2z(0) + 1
z2 + 1
z − 2z cos π2 + 1
h
nπ i
z
18. Prove that Z sin
= 2
2
z +1
h
Solution. We know that
z2
z sin θ
.
− 2z cos θ + 1
IS
Hence
T
Z [sin nθ] =
M
z sin π2
nπ i
z(1)
z
Z sin
= 2
= 2
= 2
π
2
z − 2z(0) + 1
z +1
z − 2z cos 2 + 1
SR
h
r.E
.S
ur
es
h,
h
nπ i
19. Find Z sin2
2
Solution. We know that
1 − cos 2θ
2
h
i
1 h
nπ i
2 nπ
Z sin
=
Z 1 − cos 2
2
2
2
1
=
[Z[1] − Z [cos nπ]]
2
1
=
[Z[1] − Z [(−1)n ]]
2
1
z
z
=
−
2 z−1 z+1
z
=
z2 − 1
D
w.k.tsin2 θ =
8
Property 2.
First shifting (or) Frequency shifting Property (or)
Damping Rule
(i) Z[an f (n)] = Z[f (n)]Z→ z = F
z a
(i)
∞
X
a
∞
X
n
1
f (n)z =
f (n)
z
n=0
n=0
n
∞
X
1
n
a f (n)
z
n=0
∞
∞
a n X
z −n
X
f (n)
=
f (n)
z
a
n=0
n=0
z F
a
Z[f (n)] =
Z[an f (n)] =
=
=
−n
T
If Z[f (n)] = F (z) then
Proof.
M
IS
20. Find the Z-transform of rn cos nθ
w.k.t Z[cos nθ] =
z(z − cos θ)
− 2z cos θ + 1
Z[cos nθ]z→ az
z(z − cos θ)
z 2 − 2z cos θ + 1 z→ z
a
z z
− cos θ
r r
z2
z
− 2 cos θ + 1
2
r
r
z(z − r cos θ)
z 2 − 2rz cos θ + r2
z2
r.E
.S
ur
es
h,
Z[rn cos nθ] =
SR
Solution.
=
D
Z[rn cos nθ] =
=
20. Obtain the Z-transform of rn sin nθ
Solution.
w.k.t Z[sin nθ] =
Z[rn sin nθ] =
=
Z[rn sin nθ] =
=
z sin θ
− 2z cos θ + 1
Z[sin nθ]z→ zr
z sin θ
z 2 − 2z cos θ + 1 z→ z
r
z
sin θ
r
z2
z
− 2 cos θ + 1
2
r
r
rz sin θ
z 2 − 2rz cos θ + r2
z2
9
21. Obtain the Z-transform of nan
Solution.
w.k.t Z[n] =
z
(z − 1)2

Z[nan ] = Z[n]z→ az = 
z
a

z
az
a =
2
(z − a)2
−1
h
nπ i
az
22.Show that Z an sin
= 2
2
z + a2
Solution.
z sin θ
− 2z cosθ +
1
π
h nπ i
z sin
z(1)
z
2π Z sin
=
= 2
= 2
2
z − 2z(0) + 1
z +1
z 2 − 2z cos
+1
2
By Scaling Property
z Z[an f (n)] = F
a
z
h
i
nπ
az
n
a
Z a sin
=
= 2
2
z
2
z + a2
+1
a2
w.k.t Z [sin nθ] =
r.E
.S
ur
es
h,
SR
M
IS
T
z2
nπ i
z2
23. Show that Z a cos
= 2
2
z + a2
h
n
Solution. We know that
D
Z [cos nθ] =
z 2 − z cos θ
z 2 − 2z cos θ + 1
π 2
h
z − z cos
nπ i
z 2 − z(0)
z2
π 2
Z cos
=
= 2
= 2
2
z − 2z(0) + 1
z +1
z 2 − 2z cos
+1
2
2
h
Z an cos
nπ i
2
=
z
a2
z2
a2
+1
=
z2
z 2 + a2
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
k
Z[n ] =
Z[nk−1 ] =
∞
X
n=0
∞
X
nk z −n
(1)
nk−1 z −n
(2)
n=0
Differentiating (2) with respect to z we get
k−1 −n
n
z
=
n=0
n
k−1
(−n)z
−n−1
∞
X
nk z −n
M
SR
h,
d
Z(nk−1 )
dz
= −z
−1
d −n z
dz
n=0
1
= − Z[nk ]
z
Z(nk ) = −z
nk−1
n=0
n=0
=⇒
∞
X
T
=
∞
X
!
IS
d
d
Z[nk−1 ] =
dz
dz
∞
X
24. Z[n] =
z
(z − 1)2
r.E
.S
ur
es
Using the above the theorem we have the following
d
Z[nk−1 ]
dz
d
d
z
Z[n] = −z Z(1) = −z
dz
dz z − 1
(z − 1)(1) − z(1)
−1
z
= −z
=
−z
=
(z − 1)2
(z − 1)2
(z − 1)2
D
w.k.t Z[nk ] = −z
25. Z[n2 ] =
z2 + z
(z − 1)3
d
Z[nk−1 ]
dz
d
Z[n2 ] = −z Z(n)
dz d
z
(z − 1)2 (1) − z 2(z − 1)
= −z
= −z
dz (z − 1)2
(z − 1)4
z − 1 − 2z
−z − 1
z2 + z
= −z
=
−z
=
(z − 1)3
(z − 1)3
(z − 1)3
w.k.t Z[nk ] = −z
11
26. Z[n(n − 1)] =
2z
(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
27. Z[n3 ] =
z 3 + 4z 2 + z
(z − 1)4
d
Z[nk−1 ]
dz
d
−z Z(n2 )
dz d
z2 + z
−z
dz (z − 1)3
(z − 1)3 (2z + 1) − (z 2 + z)3(z − 1)2
−z
(z − 1)6
(z − 1)(2z + 1) − 3(z 2 + z)
−z
(z − 1)4
2
2z − z − 1 − 3z 2 − 3z
−z
(z − 1)4
2
−z − 4z − 1
z 3 + 4z 2 + z
−z
=
(z − 1)4
(z − 1)4
w.k.t Z[nk ] = −z
=
=
D
28. Find the Z transform of
Solution.
M
r.E
.S
ur
es
=
SR
=
h,
=
IS
T
Z[n3 ] =
1
n(n + 1)
1
A
B
=
+
n(n + 1)
n n+1
A(n + 1) + Bn = 1
n=0
n = −1
1
Z
n(n + 1)
=⇒ A = 1
=⇒ B = −1
1
1
= Z
−Z
n
n+1
z
z
= log
− z log
z−1
z−1
z
= (1 − z) log
z−1
12
29. Find the Z transform of
1
(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
n = −2
1
Z
(n + 1)(n + 2)
=⇒ A = 1
=⇒ B = −1
=
=
M
r.E
.S
ur
es
=
SR
=
h,
=
IS
T
=
1
1
Z
−Z
n+1
n+2
∞
∞
X 1 1
X 1 1
−
n + 1 z n n=0 n + 2 z n
n=0
11 1 1
1 11 1 1
1+
+
+ ··· −
+
+
+ ···
2 z 3 z2
2 3 z 4 z2
1 1 1
1 1
1 1
1 1
2 1 1
z
+
+
+ ··· − z
+
+
+ ···
z 2 z2 3 z3
2 z2 3 z3 4 z4
1
1
1
2
−z log 1 −
− z − log 1 −
−
z
z
z
1
1
−z log 1 −
+ z 2 log 1 −
+z
z
z
z−1
z + (z 2 − z) log
z
=
D
Property 4 Second Shifting Property (or) Translation theorem
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)
then decompose F (z)
into
z
z
partial fractions, multiply this with z then find the inverse transform.
Examples
10z
(z − 1)(z − 2)
T
IS
Example 1. Find Z
−1
10z
(z − 1)(z − 2)
= Z −1 [F (z)]
SR
Z
−1
M
Solution.
10z
(z − 1)(z − 2)
F (z)
10
=
z
(z − 1)(z − 2)
10
A
B
=
+
(z − 1)(z − 2)
z−1 z−2
=⇒ A(z − 2) + B(z − 1) = 10
r.E
.S
ur
es
h,
F (z) =
A = −10 ,
D
Put z = 1
z=2
B = 10
F (z)
−10
10
=
+
z
z−1 z−2
−10z
10z
F (z) =
+
z − 1 z − 2 z
z
−1
−1
−1
Z [F (z)] = −10Z
+ 10Z
z−1
z−2
−1
n
n
n
n
Z [F (z)] = 10(−1 + 2 ) = 10(2 − 1 )
Example 2. Find the inverse transform of
z2
z
− 7z + 12
Solution.
Z
−1
z
2
z − 7z + 12
= Z −1 [F (z)]
F (z) =
z2
z
z
=
− 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
B = −1 ,
Put z = 3
z(z 2 − z + 2)
(z + 1)(z − 1)2
M
SR
Example 3. Find Z
−1
IS
T
z=4
A=1
F (z)
1
1
=
−
z
z−4 z−3
z
z
F (z) =
−
z − 4 z − 3
z
z
−1
−1
−1
Z [F (z)] = Z
−Z
z−4
z−3
−1
n
n
Z [F (z)] = 4 − 3
Solution.
z(z 2 − z + 2)
(z + 1)(z − 1)2
h,
= Z −1 [F (z)]
r.E
.S
ur
es
Z
−1
z(z 2 − z + 2)
(z + 1)(z − 1)2
F (z)
z2 − z + 2
=
z
(z + 1)(z − 1)2
z2 − z + 2
A
B
C
=
+
+
2
(z + 1)(z − 1)
z + 1 (z − 1) (z − 1)2
D
F (z) =
=⇒ A(z − 1)2 + B(z + 1)(z − 1) + C(z + 1) = z 2 − z + 2
Substituting z = 1
z = −1
z=0
F (z)
=
z
F (z) =
Z −1 [F (z)] =
Z −1 [F (z)] =
2C = 2 =⇒ C = 1 ,
4A = 4 =⇒ A = 1
A − B + C = 2,
1 − B + 1 = 2 =⇒ B = 0
1
1
+0+
z+1
(z − 1)2
z
z
+
z + 1 (z − 1)2
z
z
−1
−1
Z
−Z
z+1
(z − 1)2
(−1)n + n
15
Example 4. Find Z
−1
z2 − z
(z + 1)(z 2 + 1)
Solution.
Z
−1
z2 − z
(z + 1)(z 2 + 1)
= Z −1 [F (z)]
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
2A = −2 =⇒ A = −1 ,
A + C = −1
−1 + C = −1 =⇒ C = 0
2A + 2B + 2C = 0
−2 + 2B + 0 = 0 =⇒ B = 1
1
z
−
+ 2
z+1 z +1
z
z2
−
+ 2
z + 1 z +
1
2 z
z
−1
−1
−Z
+Z
2
z+1
z +1
nπ
−(−1)n + cos
2
z=1
h,
F (z)
=
z
SR
M
IS
T
Substituting z = −1
z=0
r.E
.S
ur
es
F (z) =
Z −1 [F (z)] =
Z −1 [F (z)] =
D
Example 5. Find the inverse z-transform of
z(z + 1)
.
(z − 1)3
Solution.
F (z) =
F (z)
=
z
F (z)
=
z
F (z) =
Z −1 [F (z)] =
Z −1 [F (z)] =
z(z + 1)
(z − 1)3
(z + 1)
(z − 1 + 2)
(z − 1)
2
=
=
+
3
(z − 1)3
(z − 1)3
(z − 1)
(z − 1)3
1
2
2 +
(z − 1)
(z − 1)3
z
2z
2 +
(z − 1)
(z − 1)3
z
2z
−1
−1
Z
+Z
(z − 1)3
(z − 1)3
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 constants.
Solution. Given yn = a(2)n + b(−2)n ,
Solving Difference equations using Z-Transform
IS
F (z)
zF (z) − zy(0)
z 2 F (z) − z 2 y(0) − zy(1)
z 3 F (z) − z 3 y(0) − z 2 y(1) − zy(2)
M
=
=
=
=
SR
Z[y(n)]
Z [y(n + 1)]
Z [y(n + 2)]
Z [y(n + 3)]
T
Points to Remember
r.E
.S
ur
es
h,
Examples
Example 1. Solve un+2 − 5un+1 + 6un = 4n with u0 = 0, u1 = 1.
Solution. The given difference equation is
un+2 − 5un+1 + 6un = 4n , u0 = 0, u1 = 1.
D
Consider the difference equation as
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
F (z) =
z(z − 3)
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
⇒
−2A = 1
⇒
A=−
1
2
Put z = 4,
⇒
2B = 1
⇒
B=
1
2
T
Put z = 2
r.E
.S
ur
es
h,
SR
M
IS
F (z)
1 A
1 B
= −
+
z
2 (z − 2) 2 (z − 4)
1
z
1
z
F (z) = −
+
2 z−2
2 z−4
1 −1
z
1 −1
z
−1
Z [F (z)] = − Z
+ Z
2
z−2
2
z−4
1
1
y(n) = − (2)n + (4)n
2
2
Example 2. Solve yn+2 + 6yn+1 + 9yn = 2n with y0 = y1 = 0.
Solution. The given difference equation is
D
yn+2 + 6yn+1 + 9yn = 2n , y0 = y1 = 0
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
z
2
(z + 6z + 9)F (z) =
z−2
18
F (z) = Z [y(n)] =
z
(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
Put z = −3
⇒
−5C = 1
⇒
C=−
Put z = 2,
⇒
25A = 1
⇒
A=
1
5
1
25
T
Equating the coefficient of z 2 on both sides we get
IS
1
25
SR
M
A + B = 0 ⇒ B = −A ⇒ B = −
D
r.E
.S
ur
es
h,
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
F (z) =
−
−
25 z − 2
25 z + 3
5 (z + 3)2
1 −1
z
1 −1
z
1
z
−1
Z [F (z)] =
Z
− Z
−
25
z−2
25
z+3
5 (z + 3)2
(2)n (−3)n 1 1
(−3)z
−1
=
−
−
Z
25
25
5 (−3)
(z + 3)2
az
−1
∵Z
= n(a)n
(z − a)2
=
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
z
2
(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)
(z − 3)(z + 1)(z + 3)
F (z)
(z − 2)
A
B
C
=
=
+
+
z
(z − 3)(z + 1)(z + 3)
(z − 3) (z + 1) (z + 3)
M
IS
T
F (z) =
⇒
−8A = −3
h,
Put z = −1
SR
⇒ A(z + 3)(z − 3) + B(z + 1)(z − 3) + C(z + 1)(z + 3) = z − 2
⇒
A=
3
8
⇒
12B = −5
⇒
B=−
5
12
Put z = 3,
⇒
24C = 1
⇒
C=
1
24
r.E
.S
ur
es
Put z = −3,
D
F (z)
3
1
5
1
1
1
∴
=
−
+
z
8 z+1
12 z + 3
24 z − 3
3 −1
z
5 −1
z
1 −1
z
−1
Z [F (z)] =
Z
− Z
+ Z
8
z+1
12
z+3
24
z−3
1 −1
z
1 −1
z
1
z
−1
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)]
z [F (z) − y(0) − y(1)z −1 ] − 4z[F (z) − y(0)] + 4F (z)
z 2 [F (z) − 1 − 0] − 4z[F (z) − 1] + 4F (z)
(z 2 − 4z + 4)F (z) − z 2 + 4z
(z − 2)2 F (z)
2
F (z) =
T
IS
M
r.E
.S
ur
es
Z −1 [F (z)] =
z (z − 4)
(z − 2)2
(z − 4)
z−2−2
2 =
(z − 2)
(z − 2)2
(z − 2)
2
1
2
−
2 −
2 =
z − 2 (z − 2)2
(z − 2)
(z − 2)
z
2z
−
z−2
(z − 2)2
z
2z
−1
−1
Z
−Z
z−2
(z − 2)2
n
n
n
(2) − n(2) = (2) (1 − n)
SR
F (z)
=
z
F (z)
=
z
0
0
0
0
z 2 − 4z
h,
F (z) =
=
=
=
=
=
y(n) =
D
Example 5. Solve the difference equation y(n + 3) − 3y(n + 1) + 2y(n) = 0 given that
y(0) = 4, y(1) = 0 and y(2) = 8, by the method of Z - transform.
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)]
Z [y(n + 1)]
Z [y(n + 2)]
Z [y(n + 3)]
=
=
=
=
F (z)
zF (z) − zy(0)
z 2 F (z) − z 2 y(0) − zy(1)
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
3
Z[y(n + 3)] − 3Z[y(n + 1)] + 2Z[y(n)] = 0
F (z) − y(0) − y(1)z − y(2)z −2 − 3z[F (z) − y(0)] + 2F (z) = 0
using y(0) = 4, y(1) = 0, y(2) = 8
3
z F (z) − 4 − 8z −2 − 3z[F (z) − 4] + 2F (z) = 0
−1
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)
(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)
F (z) = Z [y(n)] =
⇒ 4(z + 1) = A(z + 2) + B(z − 1)
⇒
3A = 8
⇒
Put z = −2,
⇒
−3B = −4
⇒
A=
T
Put z = 1
4
3
M
IS
B=
8
3
r.E
.S
ur
es
h,
SR
F (z)
8
1
4
1
=
+
z
3 z−1
3 z+2
8
z
4
z
F (z) =
+
3 z−1
3 z+2
8 −1
z
4 −1
z
−1
Z [F (z)] =
Z
+ Z
3
z−1
3
z+2
8 n 4
y(n) =
(1) + (−2)n
3
3
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.
D
Solution. The given difference equation is
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)
z 2 + 3z − 4
F (z)
(3z + 7)
A
B
=
=
+
z
(z + 4)(z − 1)
(z + 4) (z − 1)
F (z) =
⇒ 3z + 7 = A(z − 1) + B(z + 4)
⇒
5B = 10
⇒
B=2
Put z = −4,
⇒
−5A = −5
⇒
A=1
T
Put z = 1
r.E
.S
ur
es
h,
SR
M
IS
F (z)
1
1
= 1
+2
z
z+4
z−1
z
z
F (z) = 1
+2
z+4
z−1
z
z
−1
−1
−1
Z [F (z)] = Z
+ 2Z
z+4
z−1
n
n
y(n) = (−4) + 2(1)
Inverse Z-Transform by Residue Theory
Z −1 [F (z)] = sum of residues of f (z) at its poles lies inside C.
D
where f (z) = z n−1 F (z) and C is the contour which include all the poles of f (z) .
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)
m−1
z→a (m − 1)! dz
R(a) = lim
Examples
Example 1. Find the inverse z-transform of
10z
by residue method.
(z − 1)(z − 2)
23
10z
(z − 1)(z − 2)
−1
Z [F (z)] = sum of the residues , z = 1, 2 are the simple poles (or) pole of order 1
Solution. Let F (z) =
R(1) = lim(z − 1)z n−1 F (z)
z→1
10z
n−1
= 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 )
h,
SR
M
IS
T
R(2) = lim(z − 2)z n−1 F (z)
z→2
10z
n−1
= lim (z − 2)z
z→2
(z − 1)(z − 2)
10z n
= lim (z − 2)
z→1
(z − 1)(z − 2)
n
10z
R(1) = lim
= 10(2)n
z→2 (z − 1)
r.E
.S
ur
es
=⇒ Z −1 [F (z)] = R(1) + R(2) = 10(2n − 1n )
Example 2. Find the inverse z-transform of
z(z + 1)
(z − 1)3
D
Solution. Let F (z) =
z(z + 1)
by residue method.
(z − 1)3
Z −1 [F (z)] = sum of the residues , z = 1 is a pole of order 3.
R(1) =
=
=
=
=
=
n−1 2
1 d2
(z + z)
3z
lim
(z − 1)
z→1 2! dz 2
(z − 1)3
1 d2 n+1
lim
z
+ zn
2
z→1 2! dz
1 d lim
(n + 1)nz n + nz n−1
z→1 2! dz
1 lim
(n + 1)nz n−1 + n(n − 1)z n−2
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
f (n) ∗ g(n) =
n
X
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)]
n=0
" n
∞
X
X
IS
#
f (r)g(n − r) z −n
r.E
.S
ur
es
=
[f (0)g(n) + f (1)g(n − 1) + f (2)g(n − 2) + · · · + f (n)g(0)] z −n
h,
=
n=0
SR
n=0
∞
X
M
F (z) G(z) = Z[f (n)] Z[g(n)]
"∞
#" ∞
#
X
X
=
f (n)z −n
g(n)z −n
T
Proof. From the definition of Z-transform we have
w.k.t f (n) ∗ g(n) =
n=0
n
X
r=0
f (r)g(n − r)
r=0
=
∞
X
[f (n) ∗ g(n)] z −n = Z [f (n) ∗ g(n)]
Formula
D
n=0
1 + a + a2 + a3 + · · · + an =
1 − an+1
1−a
25
Inverse Z-Transform by Convolution
Examples
z2
Example 1. Find Z
.
(z − a)(z − b)
z
z
−1
n
−1
Solution. We rnow that Z
= a and Z
= bn
z−a
z−b
−1
Z
−1
z2
(z − a)(z − b)
= Z
−1
= Z
−1
z
z
(z − a) (z − b)
z
z
−1
∗Z
(z − a)
(z − b)
By Convolution Theorem
= an ∗ b n
n
a ∗b
n
=
IS
f (r)g(n − r)
M
r=0
n
X
n
X
SR
f (n) ∗ g(n) =
n
X
T
By Defn. Convolution
r n−r
ab
=
r n −r
ab b
=
n
X
b
n
a r
b
a a 2
a n = bn 1 +
+
+ ··· +
b
b
b
r=0
r=0
r.E
.S
ur
es
h,
r=0

bn+1 − an+1
1−


n
b 
bn+1
= bn 

a =b 
b−a
1−
b
b
n+1
n+1
b
−
a
= bn+1 n
b b (b − a)
2
n+1
z
b
− an+1
−1
Z
=
(z − a)(z − b)
b−a
a n+1 

D

Example 2. Using convolution theorem evaluate Z
−1
z2
(z − 1)(z − 3)
Solution.
Derive the solution of Example 1
Z
Put a = 1, b = 3 we get
−1
Z
−1
z2
bn+1 − an+1
=
(z − a)(z − b)
b−a
z2
3n+1 − 1n+1
3n+1 − 1
=
=
(z − 1)(z − 3)
3−1
2
26
Example 3. Using convolution theorem evaluate Z
−1
8z 2
(2z − 1)(4z − 1)
Solution.

Z
−1
8z 2
(2z − 1)(4z − 1)



z2

= Z −1 

1
1 
z−
z−
2
4
Derive the solution of Example 1
M


z2

 =

1
1 
z−
z−
2
4
n+1 n+1
n+1 n+1
1
1
1
1
−
−
4
2
4
2
=
1
1
1
−
−
4
2
4
" 2(n+1) #
n+1
n+1 n+1
1
1
1
1
−
−
2
2
2
4
=
2
2
1
1
2
2
" #
n+1−2
2n+2−2
1
1
−
2
2
" 2n #
n−1
1
1
−
2
2
SR

r.E
.S
ur
es
h,
Z
z2
bn+1 − an+1
=
(z − a)(z − b)
b−a
IS
1
1
Put a = , b = we get
2
4

−1
T
Z
−1
=
=
8z 2
(2z − 1)(4z − 1)
D
Z
−1
=
Example 4. Using convolution theorem evaluate Z
−1
8z 2
(2z − 1)(4z + 1)
Solution.
Z
−1
8z 2
(2z − 1)(4z + 1)
z2
= Z
(z − 12 )(z + 14 )
n "
n+1 #
2 1
1
=
1− −
3 2
2
−1
Derive the solution of Example 1
Z
−1
z2
bn+1 − an+1
=
(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 n n+1
4
1
1
4
1
1
1
1
=
− −
=
− −
−
3
2
4
3
2
2
4
4
n
n
n
n
8z 2
2
1
1
1
2 1
2
1
−1
Z
=
+
−
=
+
−
(2z − 1)(4z + 1)
3
2
2
4
3 2
3
4



1

−1 
z
z
1−
1+
2
4
−1
IS
T

Example 7. Using convolution theorem and evaluate Z −1 




h,



1
1
 = Z −1  

1
1 
z −1
z −1 
1−
1+
1−
1+
2z
4z
2
4


r.E
.S
ur
es

Z −1 


SR
M
Solution.
= Z
−1
2


1
8z
−1


 2z − 1 4z + 1  = Z
(2z − 1) (4z + 1)
2z
4z
D
FromExample 8. we get the Solution.
z2
Example 9. Find Z
.
(z − a)2
z
−1
Solution. We know that Z
= an
z−a
−1
Z
−1
z2
(z − a)2
z
z
= Z
(z − a) (z − a)
z
z
−1
−1
= Z
∗Z
(z − a)
(z − a)
n
n
X
X
= an ∗ an =
ar an−r =
ar an a−r
−1
r=0
= an + an + · · · + an
= (n + 1)an
r=0
(n + 1) times
28
z2
Example 10. Find Z
.
(z + a)2
z
−1
Solution. We know that Z
= (−a)n
z+a
−1
Z
−1
z2
(z + a)2
z
z
= Z
(z + a) (z + a)
z
z
−1
−1
= Z
∗Z
(z + a)
(z + a)
n
X
= (−a)n ∗ (−a)n =
(−a)r (−a)n−r
−1
r=0
n
n
= (−a) + (−a) + · · · + (−a)n
= (n + 1)(−a)n
z2
Example 11. Find Z
.
(z − 1)3
z
−1
Solution. We rnow that Z
= 1n
z−1
z2
(z − 1)3
=
T
IS
M
z
z
Z
(z − 1) (z − 1)2
z
z
−1
−1
Z
∗Z
(z − 1)
(z − 1)2
1n ∗ n = n ∗ 1n
n
X
r1n−r = 1 + 2 + · · · + n
−1
r.E
.S
ur
es
=
SR
Z
−1
h,
−1
(n + 1) times
=
=
r=0
n(n + 1)
2
D
=
z3
z3
−1
Example 12. Find Z
and hence deduce Z
.
(z − a)3
(z − 1)3
z
z2
−1
n
−1
Solution. We know that Z
= a ,Z
= (n + 1)an
z−a
(z − a)2
z3
z2
z
−1
−1
Z
= Z
(z − a)3
(z − a)2 (z − a)
z2
z
−1
−1
= Z
∗Z
(z − a)
(z − a)2
n
n
= (n + 1)a ∗ a
n
n
n
X
X
X
r
n−r
r
n −r
n
=
a (r + 1)a
=
a (r + 1)a a = a
(r + 1)
−1
r=0
n
Z −1
3
z
(z − a)3
r=0
= a [1 + 2 + · · · + (n + 1)]
(n + 1)(n + 2)
= an
2
r=0
29
Z
−1
z3
(z − 1)3
= 1n
(n + 1)(n + 2)
2
z3
Example 13. Find Z
using Convolution Theorem.
(z − 2)2 (z − 3)
z
z2
−1
n
−1
Solution. We know that Z
= a ,Z
= (n + 1) an
z−a
(z − a)2
z3
z2
z
−1
−1
Z
= Z
.
(z − 2)2 (z − 3)
(z − 2)2 (z − 3)
z2
z
−1
−1
= Z
∗Z
(z − 3)
(z − 2)2
n
n
= (n + 1) 2 ∗ 3
r
n
n
n
X
X
X
2
r n−r
r n −r
n
=
(r + 1) 2 .3
=
(r + 1) 2 .3 .3 = 3
(r + 1)
3
r=0
r=0
r=0
"
#
2
3
n
2
2
2
2
n
= 3 1+2
+3
+4
+ · · · + (n + 1)
3
3
3
3
2
Let x =
3
S = 1 + 2x + 3x2 + 4x3 + · · · + (n + 1)xn
xS = x + 2x2 + 3x3 + 4x4 + · · · + nxn + (n + 1)xn+1
S − xS = 1 + x + x2 + x3 + · · · + xn − (n + 1)xn+1
1 − xn+1
(1 − x) S =
− (n + 1)xn+1
1−x
(1 − xn+1 ) (n + 1)xn+1
S =
−
(1 − x)
(1 − x)2
n+1
n+1
1 − 23
(n + 1) 23
=
2 −
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
D
r.E
.S
ur
es
h,
SR
M
IS
T
−1
Z
−1
z3
(z − 2)2 (z − 3)
= 3n .S
30
"
n+1 #
2
= 3n 9 − 3 (n + 4)
3
n+1
2
n+2
n+1
= 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
M
IS
T
F (z) = Z[f (n)]
∞
X
=
f (n)z −n
n=0
h,
SR
= f (0) + f (1)z −1 + f (2)z −2 + · · · +
f (1) f (2)
= f (0) +
+ 2 + ···
z
z
r.E
.S
ur
es
As z → ∞ we get
f (1) f (2)
lim F (z) = lim f (0) +
+ 2 + ···
z→∞
z→∞
z
z
= f (0)
Final value Theorem
D
If Z[f (n)] = F (z) then lim f (n) = lim(z − 1)F (z).
n→∞
z→1
Proof. By the definition we have
Z[f (n)] =
∞
X
f (n)z
−n
, Z[f (n + 1)] =
n=0
∞
X
f (n + 1)z −n
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
z[F (z) − f (0)] − F (z) =
n=0
∞
X
[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
=
∞
X
n=0
[f (n + 1) − f (n)]
n=0
=
lim (z − 1)F (z) − f (0) =
z→1
lim (z − 1)F (z) =
z→1
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 [f (n + 1)] − f (0)
n→∞
lim f (n + 1)
T
=
n→∞
n→∞
IS
=
n
X
lim
[f (n + 1) − f (n)]
Since lim f (n + 1) = lim f (n) we have
M
n→∞
SR
lim f (n) = lim(z − 1)F (z)
z→1
r.E
.S
ur
es
h,
n→∞
D
n→∞