z- Transforms - MetaLab

z- Transforms
• Frequency domain representations of discretetime
signals and LTI discrete-time systems are
made possible with the use of DTFT.
• However not all discrete-time signals ( e.g. unit
step sequence) are guaranteed to have DTFT
because of convergence condition.
• As a result for these cases we are not able to
use such frequency domain characterization.

Definition of z-Transform.
DTFT is given by -
G(e
jω
)
=
Certain function of g[n] will not allow the above summation to converge.
Therefore we need to find a way of overcoming this non - convergence.
Let us modified the signal g[n] to g[n]r
DTFT of
∞
∑
n=−∞
g[n]r
g[
n]
e
-n
− jω
n
is G(e
.
jω
)
=
∞
∑
n=−∞
g[
n]
r
-n
−n
e
− jω
n
=
∞
∑
n=−∞
g[
n](
re
jω
)
−n
.
Letting
re
jω
=
z,
therefore G(z)
=
∞
∑
n=−∞
g[
n]
z
−n
..... Eqn.6.1

where z = re
Region of Convergence .
Another way looking at the transform G(z)
z - transform is the DTFT of modified signal
jω
.
∞ = ∑ g[
n]
z
n=
−∞
g[n]r
By choosing the right value of r,
we can make the summation above to converge.
The values of r that make this convergence possible
is known as Region of Convergence (ROC).
ROC are circles or annular rings in the z - plane.
-n
.
−n
.-

Point z in a
Complex z-
plane
Imaginary Axis
z = re
jω
r
ω
1
Real Axis
Unit Circle

Annular ring
in a
Complex z-
plane
Imaginary Axis
r
ω
z = re
jω
ROC
1
Real Axis
Unit Circle

Example 6.1 z-Transform of Causal
What is
X(z)
∴X(z)
Exponential Sequence.
Using eqn.6.1,
=
the z - transform of
∞
∑
n=−∞
x[
n]
z
G(z)
(The above summation is a sum of
1
=
1- ( αz
only if
The z - transform here is an algebraic expression
but must be accompany with the
convergence or else it is
-1
−n
)
,
=
=
∞
∑
n=−∞
∞
∑
n=−∞
x[n]
g[
n]
z
n
α µ [ n]
z
| αz
not valid.
| < 1.
( αz
geometric series)
condition of
ROC here is outside a circle(annular region)| z | > | α |
-1
n
= α µ [ n]?
−n
−n
=
∞
∑
n=
0
−1
)
n
.

1
Amplitude
0.9
0.8
0.7
0.6
0.5
0.4
0.3
r
−n
= 1. 2
−n
0.2
0.1
0
0 5 10 15 20 25 30 35 40
Time index n
α = 1.1
50
45
40
Amplitude
35
30
25
20
15
n
n
x[ n]
= α µ [ n]
= (1.1) µ [ n].
10
5
0
0 5 10 15 20 25 30 35 40
Time index n

Imaginary Axis
ROC
| α | < 1
1
Real Axis
Unit Circle

n
If α = 1, x[n] = 1
DTFT of
=
∞
∑
n=
0
X(z)
1×
e
only if
|
Unit step sequence
The unit step sequence is
hence it DTFT does
But z - transform of
1
=
1- ( αz
z
µ [ n]
=
− jω
n
-1
⇒ ∞(DTFT does
-1
)
=
× µ [ n]
∞
∑
n=−∞
µ [ n]:
−
1
1- z
µ [ n]
e
not converge.
-1
| < 1i.e.| z | > 1.(ROC).
,
= µ [ n].
− jω
n
not exit)
not absolutely summable,

Imaginary Axis
ROC
1
Real Axis
Unit Circle
Since unit circle is
not in ROC,
DTFT of
unit step does not converge.

Example 6.2 z-Transform of Anticausal
Exponential Sequence.
n
What is the z - transform of x[n] = −α
µ [ −n
−1]?{left hand sequence}
Using eqn.6.1, X(z) =
= −
−∞
∑
n=−1
( αz
−1
1
= 1−
−
1- ( α
1
)
n
,
z)
= −
∞
∑
m=
1
∞
( αz
∑
n=−∞
−1
x[
n]
z
)
−m
−n
= −
=
∞
∑
m=
0
∞
∑
n=−∞
( αz
−
provided that | ( α
1
n
−α
µ [ −n
−1]
z
−1
)
−m
z) | < 1.
+ 1 = 1−
−n
∞
∑
m=
0
= −
−
( α
1
−1
∑
n=−∞
z)
m
( αz
−1
)
n
.
−1
- ( α z)
X(z) = =
−1
1- ( α z)
1 1
=
1
+ 1
1−
( αz
−1
− ( α z)
The end algebraic expression is exactly the same as in example 6.1,
the only that is different here is ROC
,
)
−
| ( α
z) | < 1.
∴ROC is inside a circle(annular region) | z | < | α |
-1
1
−
provided that | ( α
1
z) | < 1.

Imaginary Axis
| α | < 1
ROC
1
Real Axis
Unit Circle

z-Transform of Finite Sequences
G(z) =
If g[n]is finite e.g.g[n] = [1,-2,3,-1,0,3,7].
G(z) =
ROC for this G(z) is the entire z - plane except for z =
If g[n]is finite e.g.g[n] = [-1,0,3,7].
G(z) =
ROC for this G(z) is the entire z - plane except for z =
If g[n]is finite e.g.g[n] = [1,-2,3,-1]
G(z) =
∑
3
∑
n=−3
3
∞
n=−∞
∑
n=
0
0
∑
n=−3
g[
n]
z
g[
n]
z
g[
n]
z
g[
n]
z
−n
−n
−n
−n
= 1z
= 1z
3
= ( −1)
z
3
+ ( −2)
z
0
↑
+ 0z
+ ( −2)
z
+ 3z
+ 3z
+ 3z
+ ( −1)
z
+ 7z
−1.
ROC for this G(z) is the entire z - plane except for z = ∞.
2
−1
2
↑
↑
1
1
−2
0
−3
.
+ 0z
−1
+ 3z
−2
+ 7z
0 and z = ∞
0.
−3
.

Rational z-transforms
Ratio of
P(
z)
H ( z)
= =
D(
z)
Alternate Representation as ratio of polynomials in z : -
H ( z)
= z
Factoring above equations : -
H ( z)
=
l
two polynomials P(z) and D(z) in z
( N −M
)
p
d
0
0
Π
Π
M
l=
1
N
l=
1
p0
+ p1z
d + d z
0
0
p0z
d z
M
N
(1 −ξl
z
(1 − λ z
1
−1
−1
−1
−1
+ p1z
+ d z
l
1
+ p
+ d
M −1
N −1
)
= z
)
2
z
z
2
−2
−2
+ p
+ d
2
( N −M
)
....... + p
....... + d
2
z
z
M −2
....... + p
....... + d
....... Eqn.6.13
.......... Eqn.6.14
( z −ξl
)
................... Eqn.6.15
( z − λ )
If N > M, there are additionl (N - M) zeros at z = 0(origin of z - plane.
If N < M, there are additionl (M - N) poles at z = 0(origin of z - plane.
N −2
p
d
0
0
Π
Π
M
l=
1
N
l=
1
-1
M −1
N −1
: −
−(
M −1)
−(
N −1)
M −1
N −1
l
+ p
+ d
z + p
z + d
z = ξl
are M zeros of H(z) on z - plane. i.e.obtained by taking H(z) = 0.
z = λ are N poles of H(z) on z - plane. i.e.obtained by taking H(z) = ∞.
z
z
N
M
N
M
z
z
−M
−N

Examples of Rational z-Transforms
(1)
z - transform of
µ
[ n]:
−
µ
( z)
=
1
1- z
having a zero at z = 0, and a pole at z = 1.
(2) H(z) =
-1
z
= , for | z | > 1.(ROC).
z −1
1- 2.4z
1−
0.8z
+ 2.88z
+ 0.64z
−2
Conjugate poles at z = 0.4 ± j0.6928,
-1
−1
−2
Conjugate zeros at z
=
1.2
±
j1.2

ROC for right-sided sequence

Methods of computing the Inverse
z-transform
• Using Cauchy’s Residue Theorem
• Table look-up
• Partial Fraction Expansion
• Power Series Expansion or Long Division.

Inverse z-Transform via Table
Example 6.12.
H(z) =
H(z) =
n
nα
µ [ n]
⇔
∴h[n]
=
Look-Up
0.5z
, | z | > 5.
2
z − z + 0.25
-1
0.5z 0.5z
=
2
−1
(z − 0.5) (1−
0.5z
)
From Table 6.1: -
ZT
n(0.5)
−
αz
(1 −αz
n
1
−1
µ [ n]
)
2
2
, | z | > | α | .
.

Inverse z-Transform via Partial Fraction Expansion
Example 6.13.
-1 -2 -3
2 + 0.8z + 0.5z + 0.3z
H(z) =
.
-1 -2
1+
0.8z + 0.2z
This is an improper function since M = 3 > N = 2.
Perform long division first by reversing order of both polynomial s
-1
(0.3/0.2)z − (0.7 / 0.2)
H(z) =
0.2z
-2
+ 0.8z
-1
+ 1
0.3z
-3
+ 0.5z
-2
+ 0.8z
-1
+ 2
to get at the
proper function.
∴ H(z)
= 1.5z
−1
− 3.5 +
-3 -2 -1
0.3z + 1.2z + 1.5z
-2 -1
- 0.7z − 0.7z + 2
-2 -1
- 0.7z − 2.8z − 3.5
-1
− 2.1z + 5.5
-1
− 2.1z + 5.5
.
-2 -1
0.2z + 0.8z + 1

Continue Example 6.13
-1
-1
− 2.1z + 5.5 −10.5z
+ 27.5
Now from the proper function
=
and
-2 -1
-2 -1
0.2(z + 4z + 5) (z + 4z + 5)
2
- b ± b − 4ac
quadratic formula
,
2a
- 4 ± 16 − 20
Roots of denominator of proper function =
= −2
± j
2
Performining the partial fraction expansion for the proper function : -
-1
− 2.1z + 5.5
=
-2 -1
0.2z + 0.8z + 1
( z
−1
-1
−10.5z
+ 27.5
−1
− ( −2
+ j))(
z − ( −2
− j))

Continue Example 6.13
]
[
j)
(-2
)
2
(
6.875
16.375
]
[
j)
(-2
)
2
(
6.875
16.375
1]
[
1.5
]
[
3.5
h[n]
)
2
(
)
2
(
.
|
|
z |
|
,
)
(1
1
]
[
)
)
2
(
)(1
2
(
6.875
16.375
)
)
2
(
)(1
2
(
6.875
16.375
))
2
(
(
6.875
16.375
))
2
(
(
6.875
16.375
))
2
(
4)(
(
27.5
65.5
4)
))(
2
(
(
27.5
65.5
))
2
(
))(
2
)(
2
(
(1
)
2
27.5(
10.5
))
2
)(
2
(
))((1
2
(
(
)
2
27.5(
10.5
))
2
(
))(
2
)(
2
(
(1
)
2
27.5(
10.5
))
2
)(
2
(
))((1
2
(
(
)
2
27.5(
10.5
))
2
(
))(
2
(
)
2
((
27.5
)
2
10.5(
))
2
(
)
2
))((
2
(
(
27.5
)
2
10.5(
-n
-n
1
1
1
n
1
1
1
1
1
1
1
1
1
1
1
1
1
1
-1
1
1
-1
n
j
j
n
j
j
n
n
j
and
j
z
n
z
j
j
j
z
j
j
j
j
z
j
j
z
j
j
z
j
j
z
j
j
z
j
j
j
j
j
j
z
j
j
z
j
j
j
j
j
j
z
j
j
z
j
j
j
j
j
j
z
j
ZT
µ
µ
δ
δ
α
α
α
µ
α
−
−
−
−
+
+
+
+
−
−
−
+
−
+
= −
∴
−
−
+
−
=
>
−
⇔
−
−
−
−
−
−
+
+
+
−
−
+
−
−
−
=
−
−
−
+
+
+
−
−
−
=
−
−
−
−
−
−
+
−
+
−
−
+
−
=
−
−
−
−
−
+
−
−
−
−
+
−
+
+
−
−
−
−
+
−
−
+
−
+
−
=
−
−
−
−
−
+
−
−
−
−
+
−
+
+
−
−
−
−
+
−
−
+
−
+
−
=
−
−
−
+
−
−
−
−
+
−
−
−
+
−
−
−
+
−
+
−
−
+
+
−
−
=
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−

Inverse z-Transform Via Power Series Expansion/Long Division
........
4]
[
0.2224
3]
[
0.4
2]
[
0.52
1]
[
1.6
]
[
1
]
[
........
0.2224
0.4z
0.52z
1.6z
1
H(z)
...............................
0.0480
0.2224
0.0480
0.1600
0.400
0.0624
0.400
0.0624
0.208
0.52
0.192
0.52
0.192
0.64
1.6
0.12
1.6
0.12
0.4z
1
........
0.2224
0.4z
0.52z
1.6z
1
2.0z
1
0.12
0.4z
1
.
0.12
0.4z
1
2.0z
1
H(z)
Example 6.19
4
-3
-2
-1
5
4
5
4
3
4
3
4
3
2
3
2
3
2
1
2
1
2
-1
4
-3
-2
-1
-1
2
-1
2
-1
-1
+
−
−
−
+
−
−
−
+
=
∴
+
−
+
−
+
=
+
−
−
+
−
+
−
−
+
−
−
+
+
−
+
+
−
+
−
+
+
−
+
−
+
+
=
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
n
n
n
n
n
n
h
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
z
δ
δ
δ
δ
δ

Properties of z-Transform
• Linearity
• Time Shifting
• Scaling in the z-domain
• Time Reversal
• Time Expansion
• Conjugation
• Convolution
• Differentiation in the z-domain
• The initial-value Theorem.

z-T Property associated with
Linearity
z−T
x [ n]
←⎯ →
1
X
1
( z)with ROC denoted by R
1
x
2
z−T
[ n]
←⎯ →
X
2
( z)with ROC denoted by R
2
ax [ n]
+ bx
I
1
is
2
LT
[ n]
←⎯→ aX
( z)
+ bX
the symbol for intersect with.
1
2
( z)
with ROC containing R
1
I
R
2
.

z-T Properties continued
zT
x[
n]
←⎯→ X ( z),
with ROC = R,
Time Shifting : -
then x[
n − n
0
zT
] ←⎯→ z
−n
0
X ( z),
with ROC = R,except for possible addition
or deletion of the origin or infinity.
Scaling in the z - domain : -
n zT z
then z0x[
n]
←⎯→ X (
z
0
),
with ROC = | z
0
| R.
Time Reversal: -
zT
then x[
−n]
←⎯→ X
1
( ),
z
with ROC =
1
.
R
Time Expansion : -
x
( k )
[ n]
= x[
n ] if n is a multiple of k,
k
= 0.if n is not a multiple of k.
then x
( k )
zT
[ n]
←⎯→ X ( z
k
),
with ROC = R
1
k
Conjugation : -
then x
*
zT
[ n]
←⎯→ X
*
( z
*
),
with ROC = R.

z-T Property associated with
Convolution
x
1
zT
[ n]
←⎯→ X
1
( z)with ROC denoted by R
1
x
2
zT
[ n]
←⎯→
X
2
( z)with ROC denoted by R
2
x
1
I
[ n]*
x
is
2
zT
[ n]
←⎯→ X
( z)
X
( z)
the symbol for intersect with.
1
2
with ROC containing R
1
I
R
2
.

zT Properties continued
zT
x[
n]
←⎯→ X
( z),
with ROC
=
R,
Differentiation in the z - Domain : -
zT dX ( z)
then nx[
n]
←⎯→− z , with ROC
dz
=
R.
Initial - value theorem :
If x[n] = 0 for n < 0 ,
then x[0]
= Lim X ( z).
z→∞
-

Analysis & Characterization of LTI systems using
z-Transforms.
If
x[n] =
n
where z
n
z
, y [ n]
=
is eigenfunction of
H(z) is eigenvalue of
H ( z).
z
n
,
LTI system
LTI system,
x[n]
X(z)
h[n]
H(z)
y[n]=x[n]*h[n]
Y(z)=X(z).H(z)
H(z) is known as System Function or
Transfer Function
Frequency response = H(z)
with z
=
e
jω
( unit
circle in
ROC)

Causality.
(1) A LTI system is causal if : -
Its impulse response h[n] = 0 for n < 0 ie. right -sided.
or in other words : -
ROC of its system function H(z) is the exterior of the circle including infinity.
(2) A discrete - time LTI system with rational system function H(z)
is causal if and only if : -
(a) the ROC is the exterior of a circle outside the outermost pole;
and (b) with H(z) expressed as a ratio of polynomials in z,
the order of the numerator cannot be greater than the order
of the denominator.

Stability.
• An LTI system is stable if and only if:-
Its Impulse response h[n] is absolutely
summable.
Or Fourier Transform of h[n] converges.
Or the ROC of its system function H(z) includes
the unit circle, |z|=1
• For causal LTI system, all poles of H(z)
must be in the unit circle, |z|=1

Example 6.14 Inverse z-transform , Causality & Stability.
H ( z)
=
Determine all the possible impulse responses h[n] for the given H(z) above.
Associate each one of
Using Partial - Fraction Expansion : -
H ( z)
=
z(
z + 2.0)
,
( z − 0.2)( z + 0.6)
z(
z + 2.0) 2.75
=
( z − 0.2)( z + 0.6) 1−
0.2z
Poles are z = 0.2 and z = -0.6
the above impulse reponses with the ROC, stability, and causality.
−1
1.75
−
1+
0.6z
−1
.
ROC1: - Right hand impulse reponse h[n] =
2.75 (0.2)
n
µ [ n]
−1.75(
−0.6)
n
µ [ n].
Causal & Stable.
ROC 2 : - R.H & L.H impulse response h[n] = 2.75(0.2)
ROC3 : - Left hand impulse response h[n] = 2.75(-1)(0.2)
n
µ [ n]
−1.75(
−1)(
−0.6)
n
µ [ −n
−1]
−1.75(1)(
−0.6)
n
µ [ −n
−1].
Not Causal & Not Stable.
n
µ [ −n
−1].
Not Causal & Not Stable.
ROC 1:- |z|>0.6
ROC 2:- 0.2

System Function for
Interconnections of LTI Systems
H(z) = H ( z).H
2(
1
z
)
( H 1
z)
H 2
( z)
H 1
( z)
+
H 2
( z)
H(z) = H ( z ) + H ( z)
1 2

System Function for
Interconnections of LTI Systems
X(
z )
Y(
z)
+ H 1
( z)
H 2
( z)
Y(z)
X(z)
=
H(z)
=
1+
H
1(z)
H (z) H(z)
1
2

Block Diagram of Causal LTI systems described by
Difference Equations and Rational System Functions.
Example10.28 A Causal LTIsystem with system function : -
1
H(z) =
1 -1
x[n]
1−
z
4
Y ( z)
1
=
X ( z)
1 -1
1−
z
4
1 -1
Y ( z){1
− z } = X ( z)
4
Taking the inverse z - transform of
y[
n]
−
1
4
y[
n −1]
=
x[
n]
or
y[
n]
=
+
the above equation,
x[
n]
+
1
4
1
4
z
y[
n −1]
-1
y[n-1]
y[n]