Properties of ROC

15
Lecture 4
Properties of ROC
1) ROC is a ring < <
2 1
r z r o
2) The Fourier Transform of x(n) converges if and only if ROC of X(z) includes the unit circle.
(Remember, z = re
j
and if |z| = 1 then r = 1 and then ( ) ( ) ( )
n j j
e n x e X z X

+

= = =
Fourier Transform of x(n). So, if X(z) convergence region includes the unit circle, then
X(e
j
)=X() exists.)
3) ROC cannot contain any poles.
4) If x(n) is finite, then ROC is the entire plane except z = o/
5) If x(n) is right-sided (i.e., x(n) = o for n < N
1
, < ) ROC is the exterior of the largest pole.
6) If x(n) is the left-sided (i.e. x(n) = o for n > N
2
> -) then ROC is the innermost ring of the
smallest pole.
7) If x(n) is two-sided, ROC consists of a ring in z plane, bounded on the interior and exterior by
a pole and not containing any pole.
8) ROC must be a connected region.

Properties of the Z-Transform
Time Shifting: x(n k) Z z
-k
X(z)
Linearity: ax
1
(n) + bx
2
(n) Z aX
1
(z) + bX
2
(z)
But we cannot say ROC = ROC
1
+ ROC
2

Example 1:
( )

=
else
N n
n x
1 0
0
1

Direct Method:

( ) ( )
= =
=
1
1
1
1
1
1
0
1
0
z if
z if
z
z
N
z
z n x z X
N
N
n
n
N

N -1
x(n)
1
0
16
The function
( ) 1
1
1
1
1 1
z z
z
z
z
N
N n
has two poles at 0 and 1, but z = 1 is not a pole for X(z)
because it is defined to be 1 at z = 1. Therefore, ROC the entire plane except z = 0.
Now using Z-Transform properties:
x(n) = u(n) u(n N)
( ) ( ) ( ) ( )
1
1
1
1 1

= =
z
z z U z z X
N N

and ROC of this one is |z| > 1 while that is different from ROC found earlier.
So, if the linear combination of several signals has finite duration, the ROC of its z-transform is
exclusively dictated by the finite duration of this signal and not by the ROC of the individual
transforms.
Scaling ( )
|
.
|
\
|
a
z
X n x a
n
ROC: |a| r
1
< |z| < |a| r
2

Time-Reversal ( ) ( )
1
z X n x ROC:
1 2
r
1
z
r
1
< <
Differentiation ( )
( )
dz
z dX
z n nx same ROC
Example 2:
Determine x(n) if X(z) = log (1+az
-1
) and |z| > || .
( ) ( )
( )
(

=
+
=
+
1
1
1
1
1
2
1
1
1 1 az
az
az
az
dz
z dX
z
az
az
dz
z dX

( ) ( )
( ) ( )
( ) ( ) ( ) n nx n u a a
az
az
InvZ
az
n u a
n n
=
)
`
1
1 1
1
1
1
1
1
Q
( ) ( ) ( ) 1 1
1
=

n u
n
a
n x
n
n

Convolution
( ) ( ) n x n x
2 1
* ( ) ( ) z X z X
2 1

ROC is at least the intersection of that for X
1
(z) and X
2
(z).
Correlation
( ) l
2 1x x
r ( ) ( )
1
2 1

z X z X remember that ( ) ( ) ( ) l l l =
2 1 2 1
* x x r
x x

x*(n) x*(z*)
Z
Z
Z
17
Time Multiplication
( ) ( ) n x n x
2 1
Z ?
( ) ( ) ( ) ( ) ( )
( ) ( )
( ) dv v
v
z
X v X
j
dv v z v n x v X
j
z n x dv v v X
j
z n x n x z X
n n
n
c
n
n
n n
n
1
2 1
1
2 1
2
1
1 2 1
2
1
2
1
2
1
+
+
=
+
=

+
=
|
.
|
\
|
=
(
=
(
= =

Parseval's Theorem
( ) ( ) ( ) dv v
v
X v X
j
n x n x
1 *
2 1
*
2 1
*
1
2
1

|
.
|
\
|
=

It is like evaluating ( ) ( ) { } n x n x Z
*
2 1
at z = 1 circle.
Initial Value Theorem
If x(n) is causal, ) ( lim ) 0 ( z X x
z
=
Proof: ( ) ( ) ( ) ( ) ( ) ... 2 1
2 1
+ + + = =

z x z x o x z n x z X
o
n

if z z
-n
o therefore, ) ( lim ) 0 ( z X x
z
= .
Example:
Using Z-transform properties, find X(z) of the following signal.
( ) ( ) ( ) ( ) 2 2
3
cos 5 . 0 2
2
=

n u n n n x
n

( ) ( ) ( )
( ) ( )
(
)
`
=
)
`
n u
n
Z
dz
d
z z
n u
n
n Z z z X
n
n
3
cos 5 . 0
3
cos 5 . 0
2
2

( ) ( )
5 . 0 ;
25 . 0 25 . 0 1
25 . 0 1
25 . 0
3
cos 5 . 0 2 1
3
cos 5 . 0 1
3
cos 5 . 0
2 1
1
2 1
1
>
+

=
+
|
.
|
\
|
|
.
|
\
|
=
)
`
z
z z
z
z z
z
n u
n
Z
n
From Table on page 174

18
( )
)
`
+

=

2 1
1
1
25 . 0 5 . 0 1
5 . 0 1
z z
z
dz
d
z z X
( )
4 3 2 1
5 4 3
0625 . 0 25 . 0 75 . 0 1
0625 . 0 5 . 0 25 . 0

+ +
+
=
z z z z
z z z
z X ; |z| > 0.5
Now, lets use MATLAB to see if weve computed correctly.
b = [0, 0, 0, 0.25, - 0.5, 0.0625];
a = [1, -1, 0.75, -0.25, 0.1625];
n = 0: 20 % checking the fist 21 samples of x(n)
delta = [n =0]; % creating (n)
x = filter (b, a, delta),
plot (n, x), hold
x = [zeros (1, 2) n.* (0.5.
n) * cos(pi * n/3)]; % creating the original signal

n1 = 0:22;
plot (n1, x, r)

Rational Z-Transforms
( )
=
N
k
k
k
M
k
k
k
z a
z b
z X
0
0
if a
o
and b
o
0, then we can rewrite it as: ( )
( )
( )
k
N
k
k
M
k N M
p z
z z
z
a
b
z X

=
=
= +
1
1
0
0

It has M finite zeros at z
1
, z
2
, , z
M
and N finite poles at p
1
, p
2
, ., p
N
as well as N M zeros or
M N poles at origin and a possible zero/pole at . Depending on the location of the poles, the
signal has different behaviors. Read Section 3.3.2.

The System Function of a LTI System
Y(z) = H(z) X(z) H(z) is called the system-function. A system in general can be presented
by a difference equation:

( ) ( ) ( )
( ) ( ) ( ) z X z b z Y z a z Y
k n x b k n y a n y
k
N
k
k
k
N
k
k
M
k
k
N
k
k
=
= =

+ =
+ =
0 1
0 1

19
( )
( )
( )

=
+
= =
N
k
k
k
M
k
k
k
N
k
k
k
M
o k
k
k
z a
z b
z a
z b
z X
z Y
z H
0
0
1
1
, where a
0
= 1
Special Cases:
If a
k
= 0 for 1 < k < N, then ( )
k M
M
k
k M
k
M
k
k
z b
z
z b z H

=
=

= =
0 0
1
, which is an all-zeros system.
The system has M trivial poles at the origin. Such a system has a finite duration impulse response
and therefore is called FIR system.
On the other hand, if b
k
= 0 for M k 1 then ( ) 1 ,
1
0
0
0
1
0
= =
+
=

=
a
z a
z b
z a
b
z H
N
K
k N
k
N
N
k
k
k
.
This system is an all-pole system (has N trivial zeros at origin) and therefore, has an infinite
duration impulse response and thus is called IIR system. A pole-zero system is still IIR because
of the poles.

The Inverse of Z Transform
( ) ( )
k
k
z k x z X

+
=
=
By multiplying both sides of the above formula by z
n-1
and integrating both sides over a closed
contour within ROC of X(z), which encloses the origin, we have:
( ) ( ) dz z n x dz z z X
k n
c
k
n
c

+
=
=
1 1

Since the series converges on this contour, we can interchange and
. Then
( )
43 42 1
=
=

=
k n
k n j
c
k n
k
c
n
dz z k x dz z z X
0
2
1 1
) (
Cauchy Integral Theorem

( ) ( ) dz z z x
j
n x
n
c
1
2
1

One of Caushy Theorems states ( )
=

=
1
1
2 n
n
j
o
dz z z
n
c
o

20
Let z
o
= o. Then, f(z) = z
n
. If n is positive the antiderivitive
1
1
+
+
n
z
n
is analytic every where and
therefore, its contour integral is zero. But only for f(z) = z
-1
it doesnt have an antiderivitive even
in a punctured plane. For 2 n , it is analytic in a punctured plane with origin deleted.
Remember that if f is analytic in a simply connected domain, D, and is any loop (close
contour) in D, then 0 ) ( =
dz z f because in a simply connected domain any loop can be shrunk

to a point. Therefore, the integral of a continuous function over a shrinking loop converges to
zero.

Properties of ROC

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15

From Table on page 174

n) * cos(pi * n/3)]; % creating the original signal

Cauchy Integral Theorem

dz z f because in a simply connected domain any loop can be shrunk

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