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E R . R A H U L A G A R W A L
KEC-403 SIGNAL SYSTEM
Content
 Introduction
 z-Transform
 Zeros and Poles
 Region of Convergence
 Important z-Transform Pairs
 Inverse z-Transform
 z-Transform Theorems and Properties
 System Function
INTRODUCTION
The z-Transform
Why z-Transform?
 A generalization of Fourier transform
 Why generalize it?
 FT does not converge on all sequence
 Notation good for analysis
 Bring the power of complex variable theory deal with the
discrete-time signals and systems
Definition
 The z-transform of sequence x(n) is defined by






n
n
z
n
x
z
X )
(
)
(
 Let z = ej.
( ) ( )
j j n
n
X e x n e
 



 
Fourier
Transform
z-Plane
Re
Im
z = ej







n
n
z
n
x
z
X )
(
)
(
( ) ( )
j j n
n
X e x n e
 



 
Fourier Transform is to evaluate z-transform
on a unit circle.
z-Plane
Re
Im
X(z)
Re
Im
z = ej

Periodic Property of FT
Re
Im
X(z)

 
X(ej)
Can you say why Fourier Transform is
a periodic function with period 2?
ZEROS AND POLES
The z-Transform
Definition
 Give a sequence, the set of values of z for which the z-
transform converges, i.e., |X(z)|<, is called the region
of convergence.



 









n
n
n
n
z
n
x
z
n
x
z
X |
||
)
(
|
)
(
|
)
(
|
ROC is centered on origin and
consists of a set of rings.
Example: Region of Convergence
Re
Im



 









n
n
n
n
z
n
x
z
n
x
z
X |
||
)
(
|
)
(
|
)
(
|
ROC is an annual ring centered
on the origin.

 
 x
x R
z
R |
|
r
}
|
{ 





 x
x
j
R
r
R
re
z
ROC
Stable Systems
 A stable system requires that its Fourier transform is
uniformly convergent.
Re
Im
1
 Fact: Fourier transform is to
evaluate z-transform on a unit
circle.
 A stable system requires the
ROC of z-transform to include
the unit circle.
Example: A right sided Sequence
)
(
)
( n
u
a
n
x n

1 2 3 4 5 6 7 8 9 10
-1
-2
-3
-4
-5
-6
-7
-8
n
x(n)
. . .
Example: A right sided Sequence
)
(
)
( n
u
a
n
x n

n
n
n
z
n
u
a
z
X 




 )
(
)
(





0
n
n
n
z
a





0
1
)
(
n
n
az
For convergence of X(z), we
require that






0
1
|
|
n
az 1
|
| 1


az
|
|
|
| a
z 
a
z
z
az
az
z
X
n
n




 



 1
0
1
1
1
)
(
)
(
|
|
|
| a
z 
a
a
Example: A right sided Sequence ROC for
x(n)=anu(n)
|
|
|
|
,
)
( a
z
a
z
z
z
X 


Re
Im
1
a
a
Re
Im
1
Which one is stable?
Example: A left sided Sequence
)
1
(
)
( 


 n
u
a
n
x n
1 2 3 4 5 6 7 8 9 10
-1
-2
-3
-4
-5
-6
-7
-8
n
x(n)
.
.
.
Example: A left sided Sequence
)
1
(
)
( 


 n
u
a
n
x n
n
n
n
z
n
u
a
z
X 



 


 )
1
(
)
(
For convergence of X(z), we
require that






0
1
|
|
n
z
a 1
|
| 1


z
a
|
|
|
| a
z 
a
z
z
z
a
z
a
z
X
n
n






 



 1
0
1
1
1
1
)
(
1
)
(
|
|
|
| a
z 
n
n
n
z
a 






1
n
n
n
z
a






1
n
n
n
z
a






0
1
a
a
Example: A left sided Sequence ROC for
x(n)=anu( n1)
|
|
|
|
,
)
( a
z
a
z
z
z
X 


Re
Im
1
a
a
Re
Im
1
Which one is stable?
REGION OF
CONVERGENC
E
The z-Transform
Represent z-transform as a Rational Function
)
(
)
(
)
(
z
Q
z
P
z
X 
where P(z) and Q(z) are
polynomials in z.
Zeros: The values of z’s such that X(z) = 0
Poles: The values of z’s such that X(z) = 
Example: A right sided Sequence
)
(
)
( n
u
a
n
x n
 |
|
|
|
,
)
( a
z
a
z
z
z
X 


Re
Im
a
ROC is bounded by the
pole and is the exterior
of a circle.
Example: A left sided Sequence
)
1
(
)
( 


 n
u
a
n
x n
|
|
|
|
,
)
( a
z
a
z
z
z
X 


Re
Im
a
ROC is bounded by the
pole and is the interior
of a circle.
Example: Sum of Two Right Sided Sequences
)
(
)
(
)
(
)
(
)
( 3
1
2
1
n
u
n
u
n
x n
n



3
1
2
1
)
(




z
z
z
z
z
X
Re
Im
1/2
)
)(
(
)
(
2
3
1
2
1
12
1




z
z
z
z
1/3
1/12
ROC is bounded by poles
and is the exterior of a circle.
ROC does not include any pole.
Example: A Two Sided Sequence
)
1
(
)
(
)
(
)
(
)
( 2
1
3
1




 n
u
n
u
n
x n
n
2
1
3
1
)
(




z
z
z
z
z
X
Re
Im
1/2
)
)(
(
)
(
2
2
1
3
1
12
1




z
z
z
z
1/3
1/12
ROC is bounded by poles
and is a ring.
ROC does not include any pole.
Example: A Finite Sequence
1
0
,
)
( 


 N
n
a
n
x n
n
N
n
n
N
n
n
z
a
z
a
z
X )
(
)
( 1
1
0
1
0







 

Re
Im
ROC: 0 < z < 
ROC does not include any pole.
1
1
1
)
(
1





az
az N
a
z
a
z
z
N
N
N


 1
1
N-1 poles
N-1 zeros
Always Stable
Properties of ROC
 A ring or disk in the z-plane centered at the origin.
 The Fourier Transform of x(n) is converge absolutely iff the ROC includes the
unit circle.
 The ROC cannot include any poles
 Finite Duration Sequences: The ROC is the entire z-plane except possibly z=0
or z=.
 Right sided sequences: The ROC extends outward from the outermost finite
pole in X(z) to z=.
 Left sided sequences: The ROC extends inward from the innermost nonzero
pole in X(z) to z=0.
More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform
with the pole pattern:
Find the possible
ROC’s
More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform
with the pole pattern:
Case 1: A right sided Sequence.
More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform
with the pole pattern:
Case 2: A left sided Sequence.
More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform
with the pole pattern:
Case 3: A two sided Sequence.
More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform
with the pole pattern:
Case 4: Another two sided Sequence.
0 5 10 15 20
0
0.2
0.4
0.6
0.8
1
Bounded Signals
-5
0
5
a=0.4
-5
0
5
a=0.9
-5
0
5
a=1.2
0 5 10
-5
0
5
a=-0.4
0 5 10
-5
0
5
a=-0.9
0 5 10
-5
0
5
a=-1.2
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
0 2 4 6 8
-1
-0.5
0
0.5
1
BIBO Stability
 Bounded Input Bounded Output Stability
 If the Input is bounded, we want the Output is bounded, too
 If the Input is unbounded, it’s okay for the Output to be
unbounded
 For some computing systems, the output is
intrinsically bounded (constrained), but limit cycle
may happen
IMPORTANT
Z-TRANSFORM
PAIRS
The z-Transform
Z-Transform Pairs
Sequence z-Transform ROC
)
(n
 1 All z
)
( m
n 
 m
z All z except 0 (if m>0)
or  (if m<0)
)
(n
u 1
1
1

 z
1
|
| 
z
)
1
( 

 n
u 1
1
1

 z
1
|
| 
z
)
(n
u
an 1
1
1

 az
|
|
|
| a
z 
)
1
( 

 n
u
an 1
1
1

 az
|
|
|
| a
z 
Z-Transform Pairs
Sequence z-Transform ROC
)
(
]
[cos 0 n
u
n
 2
1
0
1
0
]
cos
2
[
1
]
[cos
1








z
z
z
1
|
| 
z
)
(
]
[sin 0 n
u
n
 2
1
0
1
0
]
cos
2
[
1
]
[sin







z
z
z
1
|
| 
z
)
(
]
cos
[ 0 n
u
n
rn
 2
2
1
0
1
0
]
cos
2
[
1
]
cos
[
1








z
r
z
r
z
r
r
z 
|
|
)
(
]
sin
[ 0 n
u
n
rn
 2
2
1
0
1
0
]
cos
2
[
1
]
sin
[







z
r
z
r
z
r
r
z 
|
|


 


otherwise
0
1
0 N
n
an
1
1
1




az
z
a N
N
0
|
| 
z
Signal Type ROC
Finite-Duration Signals
Infinite-Duration Signals
Causal
Anticausal
Two-sided
Causal
Anticausal
Two-sided
Entire z-plane
Except z = 0
Entire z-plane
Except z = infinity
Entire z-plane
Except z = 0
And z = infinity
|z| < r1
|z| > r2
r2 < |z| < r1
Some Common z-Transform Pairs
Sequence Transform ROC
1. [n] 1 all z
2. u[n] z/(z-1) |z|>1
3. -u[-n-1] z/(z-1) |z|<1
4. [n-m] z-m all z except 0 if m>0 or ฅ if m<0
5. anu[n] z/(z-a) |z|>|a|
6. -anu[-n-1] z/(z-a) |z|<|a|
7. nanu[n] az/(z-a)2 |z|>|a|
8. -nanu[-n-1] az/(z-a)2 |z|<|a|
9. [cos0n]u[n] (z2-[cos0]z)/(z2-[2cos0]z+1) |z|>1
10. [sin0n]u[n] [sin0]z)/(z2-[2cos0]z+1) |z|>1
11. [rncos0n]u[n] (z2-[rcos0]z)/(z2-[2rcos0]z+r2) |z|>r
12. [rnsin0n]u[n] [rsin0]z)/(z2-[2rcos0]z+r2) |z|>r
13. anu[n] - anu[n-N] (zN-aN)/zN-1(z-a) |z|>0
INVERSE Z-
TRANSFORM
The z-Transform
Inverse Z-Transform by Partial Fraction Expansion
 Assume that a given z-transform can be expressed as
 Apply partial fractional expansion
 First term exist only if M>N
 Br is obtained by long division
 Second term represents all first order poles
 Third term represents an order s pole
 There will be a similar term for every high-order pole
 Each term can be inverse transformed by inspection
 






 N
k
k
k
M
k
k
k
z
a
z
b
z
X
0
0
 
 


 












s
1
m
m
1
i
m
N
i
k
,
1
k
1
k
k
N
M
0
r
r
r
z
d
1
C
z
d
1
A
z
B
z
X
Partial Fractional Expression
 Coefficients are given as
 Easier to understand with examples
 
 


 












s
1
m
m
1
i
m
N
i
k
,
1
k
1
k
k
N
M
0
r
r
r
z
d
1
C
z
d
1
A
z
B
z
X
    k
d
z
1
k
k z
X
z
d
1
A 



   
   
  1
i
d
w
1
s
i
m
s
m
s
m
s
i
m w
X
w
d
1
dw
d
d
!
m
s
1
C
















Example: 2nd Order Z-Transform
 Order of nominator is smaller than denominator (in terms of z-1)
 No higher order pole
 
2
1
z
:
ROC
2
1
1
4
1
1
1
1
1


















z
z
z
X
 

















 1
2
1
1
z
2
1
1
A
z
4
1
1
A
z
X
  1
4
1
2
1
1
1
z
X
z
4
1
1
A 1
4
1
z
1
1 
























 


  2
2
1
4
1
1
1
z
X
z
2
1
1
A 1
2
1
z
1
2 























 


Example Continued
 ROC extends to infinity
 Indicates right sided sequence
 
2
1
z
z
2
1
1
2
z
4
1
1
1
z
X
1
1




















     
n
u
4
1
-
n
u
2
1
2
n
x
n
n













Example #2
 Long division to obtain Bo
   
 
1
z
z
1
z
2
1
1
z
1
z
2
1
z
2
3
1
z
z
2
1
z
X
1
1
2
1
2
1
2
1























1
z
5
2
z
3
z
2
1
z
2
z
1
z
2
3
z
2
1
1
1
2
1
2
1
2














 
 
1
1
1
z
1
z
2
1
1
z
5
1
2
z
X















  1
2
1
1
z
1
A
z
2
1
1
A
2
z
X 
 




  9
z
X
z
2
1
1
A
2
1
z
1
1 











    8
z
X
z
1
A
1
z
1
2 




Example #2 Continued
 ROC extends to infinity
 Indicates right-sides sequence
  1
z
z
1
8
z
2
1
1
9
2
z
X 1
1





 

       
n
8u
-
n
u
2
1
9
n
2
n
x
n









An Example – Complete Solution
3
8
6z
z
14
14z
3z
lim
U(z)
lim
c 2
2
z
z
0 










4
-
z
14
14z
3z
8
6z
z
14
14z
3z
2)
(z
(z)
U
2
2
2
2









2
-
z
14
14z
3z
8
6z
z
14
14z
3z
4)
(z
(z)
U
2
2
2
4









1
4
-
2
14
2
14
2
3
(2)
U
c
2
2
1 






8
6z
z
14
14z
3z
U(z) 2
2





4
z
c
2
z
c
c
U(z) 2
1
0





3
2
-
4
14
4
14
4
3
(4)
U
c
2
4
2 






4
z
3
2
z
1
3
U(z)












 

0
k
,
4
3
2
0
k
3,
u(k) 1
k
1
k
Inverse Z-Transform by Power Series Expansion
 The z-transform is power series
 In expanded form
 Z-transforms of this form can generally be inversed easily
 Especially useful for finite-length series
 Example
   






n
n
z
n
x
z
X
            
 







 
 2
1
1
2
2
1
0
1
2 z
x
z
x
x
z
x
z
x
z
X
    
1
2
1
1
1
2
z
2
1
1
z
2
1
z
z
1
z
1
z
2
1
1
z
z
X


















         
1
n
2
1
n
1
n
2
1
2
n
n
x 










 



















2
n
0
1
n
2
1
0
n
1
1
n
2
1
2
n
1
n
x
Z-Transform Properties: Linearity
 Notation
 Linearity
 Note that the ROC of combined sequence may be larger than either ROC
 This would happen if some pole/zero cancellation occurs
 Example:
 Both sequences are right-sided
 Both sequences have a pole z=a
 Both have a ROC defined as |z|>|a|
 In the combined sequence the pole at z=a cancels with a zero at z=a
 The combined ROC is the entire z plane except z=0
 We did make use of this property already, where?
    x
Z
R
ROC
z
X
n
x 

 

        2
1 x
x
2
1
Z
2
1 R
R
ROC
z
bX
z
aX
n
bx
n
ax 



 


     
N
-
n
u
a
-
n
u
a
n
x n
n

Z-Transform Properties: Time Shifting
 Here no is an integer
 If positive the sequence is shifted right
 If negative the sequence is shifted left
 The ROC can change the new term may
 Add or remove poles at z=0 or z=
 Example
    x
n
Z
o R
ROC
z
X
z
n
n
x o


 

 
 
4
1
z
z
4
1
1
1
z
z
X
1
1

















   
1
-
n
u
4
1
n
x
1
-
n







Z-Transform Properties: Multiplication by Exponential
 ROC is scaled by |zo|
 All pole/zero locations are scaled
 If zo is a positive real number: z-plane shrinks or expands
 If zo is a complex number with unit magnitude it rotates
 Example: We know the z-transform pair
 Let’s find the z-transform of
    x
o
o
Z
n
o R
z
ROC
z
z
X
n
x
z 

 /
  1
z
:
ROC
z
-
1
1
n
u 1
-
Z


 

             
n
u
re
2
1
n
u
re
2
1
n
u
n
cos
r
n
x
n
j
n
j
o
n o
o 






  r
z
z
re
1
2
/
1
z
re
1
2
/
1
z
X 1
j
1
j o
o




 




Z-Transform Properties: Differentiation
 Example: We want the inverse z-transform of
 Let’s differentiate to obtain rational expression
 Making use of z-transform properties and ROC
   
x
Z
R
ROC
dz
z
dX
z
n
nx 


 

    a
z
az
1
log
z
X 1


 
   
1
1
1
2
az
1
1
az
dz
z
dX
z
az
1
az
dz
z
dX











     
1
n
u
a
a
n
nx
1
n




     
1
n
u
n
a
1
n
x
n
1
n




Z-Transform Properties: Conjugation
 Example
    x
*
*
Z
*
R
ROC
z
X
n
x 

 

   
     
        
 
n
x
Z
z
n
x
z
n
x
z
X
z
n
x
z
n
x
z
X
z
n
x
z
X
n
n
n
n
n
n
n
n
n
n














































Z-Transform Properties: Time Reversal
 ROC is inverted
 Example:
 Time reversed version of
   
x
Z
R
1
ROC
z
/
1
X
n
x 

 


   
n
u
a
n
x n

 
 
n
u
an
  1
1
1
-
1
-1
a
z
z
a
-
1
z
a
-
az
1
1
z
X 






Z-Transform Properties: Convolution
 Convolution in time domain is multiplication in z-domain
 Example:Let’s calculate the convolution of
 Multiplications of z-transforms is
 ROC: if |a|<1 ROC is |z|>1 if |a|>1 ROC is |z|>|a|
 Partial fractional expansion of Y(z)
        2
x
1
x
2
1
Z
2
1 R
R
:
ROC
z
X
z
X
n
x
n
x 

 


       
n
u
n
x
and
n
u
a
n
x 2
n
1 

  a
z
:
ROC
az
1
1
z
X 1
1 

    1
z
:
ROC
z
1
1
z
X 1
2 

 
     
  
1
1
2
1
z
1
az
1
1
z
X
z
X
z
Y 





  1
z
:
ROC
asume
az
1
1
z
1
1
a
1
1
z
Y 1
1











 

     
 
n
u
a
n
u
a
1
1
n
y 1
n



Z - T R A N S F O R M T H E O R E M S
A N D P R O P E R T I E S
The z-Transform
Linearity
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
y
R
z
z
Y
n
y 
 ),
(
)]
(
[
Z
y
x R
R
z
z
bY
z
aX
n
by
n
ax 



 ),
(
)
(
)]
(
)
(
[
Z
Overlay of
the above two
ROC’s
Shift
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
x
n
R
z
z
X
z
n
n
x 

 )
(
)]
(
[ 0
0
Z
Multiplication by an Exponential Sequence



 x
x- R
z
R
z
X
n
x |
|
),
(
)]
(
[
Z
x
n
R
a
z
z
a
X
n
x
a 

 
|
|
)
(
)]
(
[ 1
Z
Differentiation of X(z)
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
x
R
z
dz
z
dX
z
n
nx 


)
(
)]
(
[
Z
Conjugation
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
x
R
z
z
X
n
x 
 *)
(
*
)]
(
*
[
Z
Reversal
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
x
R
z
z
X
n
x /
1
)
(
)]
(
[ 1


 
Z
Real and Imaginary Parts
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
x
R
z
z
X
z
X
n
x
e 

 *)]
(
*
)
(
[
)]
(
[ 2
1
R
x
j R
z
z
X
z
X
n
x 

 *)]
(
*
)
(
[
)]
(
[ 2
1
Im
Initial Value Theorem
0
for
,
0
)
( 
 n
n
x
)
(
lim
)
0
( z
X
x
z 


Convolution of Sequences
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
y
R
z
z
Y
n
y 
 ),
(
)]
(
[
Z
y
x R
R
z
z
Y
z
X
n
y
n
x 

 )
(
)
(
)]
(
*
)
(
[
Z
Convolution of Sequences






k
k
n
y
k
x
n
y
n
x )
(
)
(
)
(
*
)
(
 















n
n
k
z
k
n
y
k
x
n
y
n
x )
(
)
(
)]
(
*
)
(
[
Z
 









k
n
n
z
k
n
y
k
x )
(
)
(  









k
n
n
k
z
n
y
z
k
x )
(
)
(
)
(
)
( z
Y
z
X

SYSTEM
FUNCTION
The z-Transform
Signal Characteristics from Z-Transform
 If U(z) is a rational function, and
 Then Y(z) is a rational function, too
 Poles are more important – determine key
characteristics of y(k)
m)
u(k
b
...
1)
u(k
b
n)
y(k
a
...
1)
y(k
a
y(k) m
1
n
1 
















 m
1
j
j
n
1
i
i
)
p
(z
)
z
(z
D(z)
N(z)
Y(z)
zeros
poles
Why are poles important?













m
1
j j
j
0
m
1
j
j
n
1
i
i
p
z
c
c
)
p
(z
)
z
(z
D(z)
N(z)
Y(z)






m
1
j
1
-
k
j
j
impulse
0 p
c
(k)
u
c
Y(k)
Z-
1
Z domain
Time domain
poles
componen
ts
Various pole values (1)
-1 0 1 2 3 4 5 6 7 8 9
0
0.5
1
1.5
2
2.5
-1 0 1 2 3 4 5 6 7 8 9
0
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 9
0
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 9
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
-1 0 1 2 3 4 5 6 7 8 9
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 9
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
p=1.1
p=1
p=0.9
p=-1.1
p=-1
p=-0.9
Various pole values (2)
-1 0 1 2 3 4 5 6 7 8 9
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1 0 1 2 3 4 5 6 7 8 9
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1 0 1 2 3 4 5 6 7 8 9
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
p=0.9
p=0.6
p=0.3
-1 0 1 2 3 4 5 6 7 8 9
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 9
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 9
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
p=-0.9
p=-0.6
p=-0.3
Conclusion for Real Poles
 If and only if all poles’ absolute values are smaller
than 1, y(k) converges to 0
 The smaller the poles are, the faster the
corresponding component in y(k) converges
 A negative pole’s corresponding component is
oscillating, while a positive pole’s corresponding
component is monotonous
How fast does it converge?
 U(k)=ak, consider u(k)≈0 when the absolute value of
u(k) is smaller than or equal to 2% of u(0)’s absolute
value
|
a
|
ln
4
k
3.912
ln0.02
|
a
|
kln
0.02
|
a
| k






11
0.36
4
|
0.7
|
ln
4
k
0.7
a







Rememb
er
This!
0 2 4 6 8 10 12
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
y(k)=0.7k
y(11)=0.0198
When There Are Complex Poles …
U(z)
z
a
...
z
a
1
z
b
...
z
b
Y(z) n
n
1
1
m
m
1
1









 c)...
bz
(az2


2a
4ac
b
b
z
2




0,
4ac
b2


)
2a
4ac
b
b
)(z
2a
4ac
b
b
a(z
c
bz
az
2
2
2 










0,
4ac
b2

 )
2a
b
4ac
i
b
)(z
2a
b
4ac
i
b
a(z
c
bz
az
2
2
2 










If
If
Or in polar coordinates,
)
ir
r
)(z
ir
r
a(z
c
bz
az2
θ
θ
θ
θ sin
cos
sin
cos 






What If Poles Are Complex
 If Y(z)=N(z)/D(z), and coefficients of both D(z) and N(z) are all real
numbers, if p is a pole, then p’s complex conjugate must also be a
pole
 Complex poles appear in pairs





















l
1
j
2
2
j
j
0
l
1
j j
j
0
r
)z
(2r
z
)
r
dz(z
bzr
p
z
c
c
ir
r
z
c'
ir
r
z
c
p
z
c
c
Y(z)
θ
θ
θ
θ
θ
θ
θ
cos
cos
sin
sin
cos
sin
cos
coskθ
dr
sinkθ
br
p
c
(k)
u
c
y(k) k
k
m
1
j
1
-
k
j
j
impulse
0 




 

Z-
1
Time domain
An Example
0 2 4 6 8 10 12 14 16 18 20
-1
-0.5
0
0.5
1
1.5
2
)
3
kπ
cos(
0.8
)
3
kπ
sin(
0.8
2
y(k)
0.64
0.8z
z
z
z
Y(z)
k
k
2
2









Z-Domain: Complex Poles
Time-Domain:
Exponentially Modulated Sin/C
Poles Everywhere
Observations
 Using poles to characterize a signal
 The smaller is |r|, the faster converges the signal
 |r| < 1, converge
 |r| > 1, does not converge, unbounded
 |r|=1?
 When the angle increase from 0 to pi, the frequency of oscillation
increases
 Extremes – 0, does not oscillate, pi, oscillate at the maximum frequency
Change Angles
0.9
-0.9 Re
Im
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 2 4 6 8 10 12 14
-6
-4
-2
0
2
4
6
8
10
12
Changing Absolute Value
Im
Re
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 2 4 6 8 10 12 14
-3
-2
-1
0
1
2
3
4
Conclusion for Complex Poles
 A complex pole appears in pair with its complex
conjugate
 The Z-1-transform generates a combination of
exponentially modulated sin and cos terms
 The exponential base is the absolute value of the
complex pole
 The frequency of the sinusoid is the angle of the
complex pole (divided by 2π)
Steady-State Analysis
 If a signal finally converges, what value does it converge to?
 When it does not converge
 Any |pj| is greater than 1
 Any |r| is greater than or equal to 1
 When it does converge
 If all |pj|’s and |r|’s are smaller than 1, it converges to 0
 If only one pj is 1, then the signal converges to cj
 If more than one real pole is 1, the signal does not converge … (e.g. the ramp signal)

 k
dr
k
br k
k
cos
sin 




 

m
1
j
1
-
k
j
j
impulse
0 p
c
(k)
u
c
y(k) 2
1
-1
)
z
(1
z


An Example
k
k
0.9)
(
3
0.5
2
u(k)
0.9
z
3z
0.5
z
z
1
z
2z
U(z)











0 10 20 30 40 50 60
-1
0
1
2
3
4
5
6
converge to 2
Final Value Theorem
 Enable us to decide whether a system has a steady
state error (yss-rss)
Final Value Theorem
1
Theorem: If all of the poles of (1 ) ( ) lie within the unit circle, then
lim ( ) lim ( 1) ( )
k z
z Y z
y k z Y z


 
2
1 1
0.11 0.11
( )
1.6 0.6 ( 1)( 0.6)
0.11
( 1) ( ) | | 0.275
0.6
z z
z z
Y z
z z z z
z
z Y z
z
 
 
 
   

   
 0 5 10 15
-0.35
-0.3
-0.25
-0.2
-0.15
-0.1
-0.05
0
k
y(k)
If any pole of (1-z)Y(z) lies out of or ON the
unit circle, y(k) does not converge!
What Can We Infer from TF?
 Almost everything we want to know
 Stability
 Steady-State
 Transients
 Settling time
 Overshoot
 …
Shift-Invariant System
h(n)
x(n) y(n)=x(n)*h(n)
X(z) Y(z)=X(z)H(z)
H(z)
Shift-Invariant System
H(z)
X(z) Y(z)
)
(
)
(
)
(
z
X
z
Y
z
H 
Nth-Order Difference Equation

 




M
r
r
N
k
k r
n
x
b
k
n
y
a
0
0
)
(
)
(

 




M
r
r
r
N
k
k
k z
b
z
X
z
a
z
Y
0
0
)
(
)
(







N
k
k
k
M
r
r
r z
a
z
b
z
H
0
0
)
(
Representation in Factored Form








 N
k
r
M
r
r
z
d
z
c
A
z
H
1
1
1
1
)
1
(
)
1
(
)
(
Contributes poles at 0 and zeros at cr
Contributes zeros at 0 and poles at dr
Stable and Causal Systems








 N
k
r
M
r
r
z
d
z
c
A
z
H
1
1
1
1
)
1
(
)
1
(
)
( Re
Im
Causal Systems : ROC extends outward from the outermost pole.
Stable and Causal Systems








 N
k
r
M
r
r
z
d
z
c
A
z
H
1
1
1
1
)
1
(
)
1
(
)
( Re
Im
Stable Systems : ROC includes the unit circle.
1
Example
Consider the causal system characterized by
)
(
)
1
(
)
( n
x
n
ay
n
y 


1
1
1
)
( 


az
z
H
Re
Im
1
a
)
(
)
( n
u
a
n
h n

Determination of Frequency Response from pole-
zero pattern
 A LTI system is completely characterized by its
pole-zero pattern.
)
)(
(
)
(
2
1
1
p
z
p
z
z
z
z
H




Example:
)
)(
(
)
(
2
1
1
0
0
0
0
p
e
p
e
z
e
e
H j
j
j
j



 



0

j
e
Re
Im
z1
p1
p2
Determination of Frequency Response from pole-
zero pattern
 A LTI system is completely characterized by its
pole-zero pattern.
)
)(
(
)
(
2
1
1
p
z
p
z
z
z
z
H




Example:
)
)(
(
)
(
2
1
1
0
0
0
0
p
e
p
e
z
e
e
H j
j
j
j



 



0

j
e
Re
Im
z1
p1
p2
|H(ej)|=? H(ej)=?
Determination of Frequency Response from pole-
zero pattern
 A LTI system is completely characterized by its
pole-zero pattern.
Example:
0

j
e
Re
Im
z1
p1
p2
|H(ej)|=? H(ej)=?
|H(ej)| =
| |
| | | | 1
2
3
H(ej) = 1(2+ 3 )
Example
1
1
1
)
( 


az
z
H
Re
Im
a
0 2 4 6 8
-10
0
10
20
0 2 4 6 8
-2
-1
0
1
2
dB

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E R . R A H U L A G A R W A L KEC-403 SIGNAL SYSTEM

  • 1. E R . R A H U L A G A R W A L KEC-403 SIGNAL SYSTEM
  • 2. Content  Introduction  z-Transform  Zeros and Poles  Region of Convergence  Important z-Transform Pairs  Inverse z-Transform  z-Transform Theorems and Properties  System Function
  • 4. Why z-Transform?  A generalization of Fourier transform  Why generalize it?  FT does not converge on all sequence  Notation good for analysis  Bring the power of complex variable theory deal with the discrete-time signals and systems
  • 5. Definition  The z-transform of sequence x(n) is defined by       n n z n x z X ) ( ) (  Let z = ej. ( ) ( ) j j n n X e x n e        Fourier Transform
  • 6. z-Plane Re Im z = ej        n n z n x z X ) ( ) ( ( ) ( ) j j n n X e x n e        Fourier Transform is to evaluate z-transform on a unit circle.
  • 8. Periodic Property of FT Re Im X(z)    X(ej) Can you say why Fourier Transform is a periodic function with period 2?
  • 9. ZEROS AND POLES The z-Transform
  • 10. Definition  Give a sequence, the set of values of z for which the z- transform converges, i.e., |X(z)|<, is called the region of convergence.               n n n n z n x z n x z X | || ) ( | ) ( | ) ( | ROC is centered on origin and consists of a set of rings.
  • 11. Example: Region of Convergence Re Im               n n n n z n x z n x z X | || ) ( | ) ( | ) ( | ROC is an annual ring centered on the origin.     x x R z R | | r } | {        x x j R r R re z ROC
  • 12. Stable Systems  A stable system requires that its Fourier transform is uniformly convergent. Re Im 1  Fact: Fourier transform is to evaluate z-transform on a unit circle.  A stable system requires the ROC of z-transform to include the unit circle.
  • 13. Example: A right sided Sequence ) ( ) ( n u a n x n  1 2 3 4 5 6 7 8 9 10 -1 -2 -3 -4 -5 -6 -7 -8 n x(n) . . .
  • 14. Example: A right sided Sequence ) ( ) ( n u a n x n  n n n z n u a z X       ) ( ) (      0 n n n z a      0 1 ) ( n n az For convergence of X(z), we require that       0 1 | | n az 1 | | 1   az | | | | a z  a z z az az z X n n           1 0 1 1 1 ) ( ) ( | | | | a z 
  • 15. a a Example: A right sided Sequence ROC for x(n)=anu(n) | | | | , ) ( a z a z z z X    Re Im 1 a a Re Im 1 Which one is stable?
  • 16. Example: A left sided Sequence ) 1 ( ) (     n u a n x n 1 2 3 4 5 6 7 8 9 10 -1 -2 -3 -4 -5 -6 -7 -8 n x(n) . . .
  • 17. Example: A left sided Sequence ) 1 ( ) (     n u a n x n n n n z n u a z X          ) 1 ( ) ( For convergence of X(z), we require that       0 1 | | n z a 1 | | 1   z a | | | | a z  a z z z a z a z X n n             1 0 1 1 1 1 ) ( 1 ) ( | | | | a z  n n n z a        1 n n n z a       1 n n n z a       0 1
  • 18. a a Example: A left sided Sequence ROC for x(n)=anu( n1) | | | | , ) ( a z a z z z X    Re Im 1 a a Re Im 1 Which one is stable?
  • 20. Represent z-transform as a Rational Function ) ( ) ( ) ( z Q z P z X  where P(z) and Q(z) are polynomials in z. Zeros: The values of z’s such that X(z) = 0 Poles: The values of z’s such that X(z) = 
  • 21. Example: A right sided Sequence ) ( ) ( n u a n x n  | | | | , ) ( a z a z z z X    Re Im a ROC is bounded by the pole and is the exterior of a circle.
  • 22. Example: A left sided Sequence ) 1 ( ) (     n u a n x n | | | | , ) ( a z a z z z X    Re Im a ROC is bounded by the pole and is the interior of a circle.
  • 23. Example: Sum of Two Right Sided Sequences ) ( ) ( ) ( ) ( ) ( 3 1 2 1 n u n u n x n n    3 1 2 1 ) (     z z z z z X Re Im 1/2 ) )( ( ) ( 2 3 1 2 1 12 1     z z z z 1/3 1/12 ROC is bounded by poles and is the exterior of a circle. ROC does not include any pole.
  • 24. Example: A Two Sided Sequence ) 1 ( ) ( ) ( ) ( ) ( 2 1 3 1      n u n u n x n n 2 1 3 1 ) (     z z z z z X Re Im 1/2 ) )( ( ) ( 2 2 1 3 1 12 1     z z z z 1/3 1/12 ROC is bounded by poles and is a ring. ROC does not include any pole.
  • 25. Example: A Finite Sequence 1 0 , ) (     N n a n x n n N n n N n n z a z a z X ) ( ) ( 1 1 0 1 0           Re Im ROC: 0 < z <  ROC does not include any pole. 1 1 1 ) ( 1      az az N a z a z z N N N    1 1 N-1 poles N-1 zeros Always Stable
  • 26. Properties of ROC  A ring or disk in the z-plane centered at the origin.  The Fourier Transform of x(n) is converge absolutely iff the ROC includes the unit circle.  The ROC cannot include any poles  Finite Duration Sequences: The ROC is the entire z-plane except possibly z=0 or z=.  Right sided sequences: The ROC extends outward from the outermost finite pole in X(z) to z=.  Left sided sequences: The ROC extends inward from the innermost nonzero pole in X(z) to z=0.
  • 27. More on Rational z-Transform Re Im a b c Consider the rational z-transform with the pole pattern: Find the possible ROC’s
  • 28. More on Rational z-Transform Re Im a b c Consider the rational z-transform with the pole pattern: Case 1: A right sided Sequence.
  • 29. More on Rational z-Transform Re Im a b c Consider the rational z-transform with the pole pattern: Case 2: A left sided Sequence.
  • 30. More on Rational z-Transform Re Im a b c Consider the rational z-transform with the pole pattern: Case 3: A two sided Sequence.
  • 31. More on Rational z-Transform Re Im a b c Consider the rational z-transform with the pole pattern: Case 4: Another two sided Sequence.
  • 32. 0 5 10 15 20 0 0.2 0.4 0.6 0.8 1 Bounded Signals -5 0 5 a=0.4 -5 0 5 a=0.9 -5 0 5 a=1.2 0 5 10 -5 0 5 a=-0.4 0 5 10 -5 0 5 a=-0.9 0 5 10 -5 0 5 a=-1.2 0 10 20 30 40 50 60 70 -1 -0.5 0 0.5 1 0 2 4 6 8 -1 -0.5 0 0.5 1
  • 33. BIBO Stability  Bounded Input Bounded Output Stability  If the Input is bounded, we want the Output is bounded, too  If the Input is unbounded, it’s okay for the Output to be unbounded  For some computing systems, the output is intrinsically bounded (constrained), but limit cycle may happen
  • 35. Z-Transform Pairs Sequence z-Transform ROC ) (n  1 All z ) ( m n   m z All z except 0 (if m>0) or  (if m<0) ) (n u 1 1 1   z 1 | |  z ) 1 (    n u 1 1 1   z 1 | |  z ) (n u an 1 1 1   az | | | | a z  ) 1 (    n u an 1 1 1   az | | | | a z 
  • 36. Z-Transform Pairs Sequence z-Transform ROC ) ( ] [cos 0 n u n  2 1 0 1 0 ] cos 2 [ 1 ] [cos 1         z z z 1 | |  z ) ( ] [sin 0 n u n  2 1 0 1 0 ] cos 2 [ 1 ] [sin        z z z 1 | |  z ) ( ] cos [ 0 n u n rn  2 2 1 0 1 0 ] cos 2 [ 1 ] cos [ 1         z r z r z r r z  | | ) ( ] sin [ 0 n u n rn  2 2 1 0 1 0 ] cos 2 [ 1 ] sin [        z r z r z r r z  | |       otherwise 0 1 0 N n an 1 1 1     az z a N N 0 | |  z
  • 37. Signal Type ROC Finite-Duration Signals Infinite-Duration Signals Causal Anticausal Two-sided Causal Anticausal Two-sided Entire z-plane Except z = 0 Entire z-plane Except z = infinity Entire z-plane Except z = 0 And z = infinity |z| < r1 |z| > r2 r2 < |z| < r1
  • 38. Some Common z-Transform Pairs Sequence Transform ROC 1. [n] 1 all z 2. u[n] z/(z-1) |z|>1 3. -u[-n-1] z/(z-1) |z|<1 4. [n-m] z-m all z except 0 if m>0 or ฅ if m<0 5. anu[n] z/(z-a) |z|>|a| 6. -anu[-n-1] z/(z-a) |z|<|a| 7. nanu[n] az/(z-a)2 |z|>|a| 8. -nanu[-n-1] az/(z-a)2 |z|<|a| 9. [cos0n]u[n] (z2-[cos0]z)/(z2-[2cos0]z+1) |z|>1 10. [sin0n]u[n] [sin0]z)/(z2-[2cos0]z+1) |z|>1 11. [rncos0n]u[n] (z2-[rcos0]z)/(z2-[2rcos0]z+r2) |z|>r 12. [rnsin0n]u[n] [rsin0]z)/(z2-[2rcos0]z+r2) |z|>r 13. anu[n] - anu[n-N] (zN-aN)/zN-1(z-a) |z|>0
  • 40. Inverse Z-Transform by Partial Fraction Expansion  Assume that a given z-transform can be expressed as  Apply partial fractional expansion  First term exist only if M>N  Br is obtained by long division  Second term represents all first order poles  Third term represents an order s pole  There will be a similar term for every high-order pole  Each term can be inverse transformed by inspection          N k k k M k k k z a z b z X 0 0                     s 1 m m 1 i m N i k , 1 k 1 k k N M 0 r r r z d 1 C z d 1 A z B z X
  • 41. Partial Fractional Expression  Coefficients are given as  Easier to understand with examples                     s 1 m m 1 i m N i k , 1 k 1 k k N M 0 r r r z d 1 C z d 1 A z B z X     k d z 1 k k z X z d 1 A               1 i d w 1 s i m s m s m s i m w X w d 1 dw d d ! m s 1 C                
  • 42. Example: 2nd Order Z-Transform  Order of nominator is smaller than denominator (in terms of z-1)  No higher order pole   2 1 z : ROC 2 1 1 4 1 1 1 1 1                   z z z X                     1 2 1 1 z 2 1 1 A z 4 1 1 A z X   1 4 1 2 1 1 1 z X z 4 1 1 A 1 4 1 z 1 1                                2 2 1 4 1 1 1 z X z 2 1 1 A 1 2 1 z 1 2                            
  • 43. Example Continued  ROC extends to infinity  Indicates right sided sequence   2 1 z z 2 1 1 2 z 4 1 1 1 z X 1 1                           n u 4 1 - n u 2 1 2 n x n n             
  • 44. Example #2  Long division to obtain Bo       1 z z 1 z 2 1 1 z 1 z 2 1 z 2 3 1 z z 2 1 z X 1 1 2 1 2 1 2 1                        1 z 5 2 z 3 z 2 1 z 2 z 1 z 2 3 z 2 1 1 1 2 1 2 1 2                   1 1 1 z 1 z 2 1 1 z 5 1 2 z X                  1 2 1 1 z 1 A z 2 1 1 A 2 z X          9 z X z 2 1 1 A 2 1 z 1 1                 8 z X z 1 A 1 z 1 2     
  • 45. Example #2 Continued  ROC extends to infinity  Indicates right-sides sequence   1 z z 1 8 z 2 1 1 9 2 z X 1 1                 n 8u - n u 2 1 9 n 2 n x n         
  • 46. An Example – Complete Solution 3 8 6z z 14 14z 3z lim U(z) lim c 2 2 z z 0            4 - z 14 14z 3z 8 6z z 14 14z 3z 2) (z (z) U 2 2 2 2          2 - z 14 14z 3z 8 6z z 14 14z 3z 4) (z (z) U 2 2 2 4          1 4 - 2 14 2 14 2 3 (2) U c 2 2 1        8 6z z 14 14z 3z U(z) 2 2      4 z c 2 z c c U(z) 2 1 0      3 2 - 4 14 4 14 4 3 (4) U c 2 4 2        4 z 3 2 z 1 3 U(z)                0 k , 4 3 2 0 k 3, u(k) 1 k 1 k
  • 47. Inverse Z-Transform by Power Series Expansion  The z-transform is power series  In expanded form  Z-transforms of this form can generally be inversed easily  Especially useful for finite-length series  Example           n n z n x z X                          2 1 1 2 2 1 0 1 2 z x z x x z x z x z X      1 2 1 1 1 2 z 2 1 1 z 2 1 z z 1 z 1 z 2 1 1 z z X                             1 n 2 1 n 1 n 2 1 2 n n x                                 2 n 0 1 n 2 1 0 n 1 1 n 2 1 2 n 1 n x
  • 48. Z-Transform Properties: Linearity  Notation  Linearity  Note that the ROC of combined sequence may be larger than either ROC  This would happen if some pole/zero cancellation occurs  Example:  Both sequences are right-sided  Both sequences have a pole z=a  Both have a ROC defined as |z|>|a|  In the combined sequence the pole at z=a cancels with a zero at z=a  The combined ROC is the entire z plane except z=0  We did make use of this property already, where?     x Z R ROC z X n x              2 1 x x 2 1 Z 2 1 R R ROC z bX z aX n bx n ax               N - n u a - n u a n x n n 
  • 49. Z-Transform Properties: Time Shifting  Here no is an integer  If positive the sequence is shifted right  If negative the sequence is shifted left  The ROC can change the new term may  Add or remove poles at z=0 or z=  Example     x n Z o R ROC z X z n n x o          4 1 z z 4 1 1 1 z z X 1 1                      1 - n u 4 1 n x 1 - n       
  • 50. Z-Transform Properties: Multiplication by Exponential  ROC is scaled by |zo|  All pole/zero locations are scaled  If zo is a positive real number: z-plane shrinks or expands  If zo is a complex number with unit magnitude it rotates  Example: We know the z-transform pair  Let’s find the z-transform of     x o o Z n o R z ROC z z X n x z    /   1 z : ROC z - 1 1 n u 1 - Z                    n u re 2 1 n u re 2 1 n u n cos r n x n j n j o n o o          r z z re 1 2 / 1 z re 1 2 / 1 z X 1 j 1 j o o          
  • 51. Z-Transform Properties: Differentiation  Example: We want the inverse z-transform of  Let’s differentiate to obtain rational expression  Making use of z-transform properties and ROC     x Z R ROC dz z dX z n nx           a z az 1 log z X 1         1 1 1 2 az 1 1 az dz z dX z az 1 az dz z dX                  1 n u a a n nx 1 n           1 n u n a 1 n x n 1 n    
  • 52. Z-Transform Properties: Conjugation  Example     x * * Z * R ROC z X n x                           n x Z z n x z n x z X z n x z n x z X z n x z X n n n n n n n n n n                                              
  • 53. Z-Transform Properties: Time Reversal  ROC is inverted  Example:  Time reversed version of     x Z R 1 ROC z / 1 X n x           n u a n x n      n u an   1 1 1 - 1 -1 a z z a - 1 z a - az 1 1 z X       
  • 54. Z-Transform Properties: Convolution  Convolution in time domain is multiplication in z-domain  Example:Let’s calculate the convolution of  Multiplications of z-transforms is  ROC: if |a|<1 ROC is |z|>1 if |a|>1 ROC is |z|>|a|  Partial fractional expansion of Y(z)         2 x 1 x 2 1 Z 2 1 R R : ROC z X z X n x n x               n u n x and n u a n x 2 n 1     a z : ROC az 1 1 z X 1 1       1 z : ROC z 1 1 z X 1 2              1 1 2 1 z 1 az 1 1 z X z X z Y         1 z : ROC asume az 1 1 z 1 1 a 1 1 z Y 1 1                       n u a n u a 1 1 n y 1 n   
  • 55. Z - T R A N S F O R M T H E O R E M S A N D P R O P E R T I E S The z-Transform
  • 56. Linearity x R z z X n x   ), ( )] ( [ Z y R z z Y n y   ), ( )] ( [ Z y x R R z z bY z aX n by n ax      ), ( ) ( )] ( ) ( [ Z Overlay of the above two ROC’s
  • 58. Multiplication by an Exponential Sequence     x x- R z R z X n x | | ), ( )] ( [ Z x n R a z z a X n x a     | | ) ( )] ( [ 1 Z
  • 59. Differentiation of X(z) x R z z X n x   ), ( )] ( [ Z x R z dz z dX z n nx    ) ( )] ( [ Z
  • 62. Real and Imaginary Parts x R z z X n x   ), ( )] ( [ Z x R z z X z X n x e    *)] ( * ) ( [ )] ( [ 2 1 R x j R z z X z X n x    *)] ( * ) ( [ )] ( [ 2 1 Im
  • 63. Initial Value Theorem 0 for , 0 ) (   n n x ) ( lim ) 0 ( z X x z   
  • 64. Convolution of Sequences x R z z X n x   ), ( )] ( [ Z y R z z Y n y   ), ( )] ( [ Z y x R R z z Y z X n y n x    ) ( ) ( )] ( * ) ( [ Z
  • 65. Convolution of Sequences       k k n y k x n y n x ) ( ) ( ) ( * ) (                  n n k z k n y k x n y n x ) ( ) ( )] ( * ) ( [ Z            k n n z k n y k x ) ( ) (            k n n k z n y z k x ) ( ) ( ) ( ) ( z Y z X 
  • 67. Signal Characteristics from Z-Transform  If U(z) is a rational function, and  Then Y(z) is a rational function, too  Poles are more important – determine key characteristics of y(k) m) u(k b ... 1) u(k b n) y(k a ... 1) y(k a y(k) m 1 n 1                   m 1 j j n 1 i i ) p (z ) z (z D(z) N(z) Y(z) zeros poles
  • 68. Why are poles important?              m 1 j j j 0 m 1 j j n 1 i i p z c c ) p (z ) z (z D(z) N(z) Y(z)       m 1 j 1 - k j j impulse 0 p c (k) u c Y(k) Z- 1 Z domain Time domain poles componen ts
  • 69. Various pole values (1) -1 0 1 2 3 4 5 6 7 8 9 0 0.5 1 1.5 2 2.5 -1 0 1 2 3 4 5 6 7 8 9 0 0.2 0.4 0.6 0.8 1 -1 0 1 2 3 4 5 6 7 8 9 0 0.2 0.4 0.6 0.8 1 -1 0 1 2 3 4 5 6 7 8 9 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 -1 0 1 2 3 4 5 6 7 8 9 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 0 1 2 3 4 5 6 7 8 9 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 p=1.1 p=1 p=0.9 p=-1.1 p=-1 p=-0.9
  • 70. Various pole values (2) -1 0 1 2 3 4 5 6 7 8 9 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1 0 1 2 3 4 5 6 7 8 9 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -1 0 1 2 3 4 5 6 7 8 9 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 p=0.9 p=0.6 p=0.3 -1 0 1 2 3 4 5 6 7 8 9 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 0 1 2 3 4 5 6 7 8 9 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 0 1 2 3 4 5 6 7 8 9 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 p=-0.9 p=-0.6 p=-0.3
  • 71. Conclusion for Real Poles  If and only if all poles’ absolute values are smaller than 1, y(k) converges to 0  The smaller the poles are, the faster the corresponding component in y(k) converges  A negative pole’s corresponding component is oscillating, while a positive pole’s corresponding component is monotonous
  • 72. How fast does it converge?  U(k)=ak, consider u(k)≈0 when the absolute value of u(k) is smaller than or equal to 2% of u(0)’s absolute value | a | ln 4 k 3.912 ln0.02 | a | kln 0.02 | a | k       11 0.36 4 | 0.7 | ln 4 k 0.7 a        Rememb er This! 0 2 4 6 8 10 12 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 y(k)=0.7k y(11)=0.0198
  • 73. When There Are Complex Poles … U(z) z a ... z a 1 z b ... z b Y(z) n n 1 1 m m 1 1           c)... bz (az2   2a 4ac b b z 2     0, 4ac b2   ) 2a 4ac b b )(z 2a 4ac b b a(z c bz az 2 2 2            0, 4ac b2   ) 2a b 4ac i b )(z 2a b 4ac i b a(z c bz az 2 2 2            If If Or in polar coordinates, ) ir r )(z ir r a(z c bz az2 θ θ θ θ sin cos sin cos       
  • 74. What If Poles Are Complex  If Y(z)=N(z)/D(z), and coefficients of both D(z) and N(z) are all real numbers, if p is a pole, then p’s complex conjugate must also be a pole  Complex poles appear in pairs                      l 1 j 2 2 j j 0 l 1 j j j 0 r )z (2r z ) r dz(z bzr p z c c ir r z c' ir r z c p z c c Y(z) θ θ θ θ θ θ θ cos cos sin sin cos sin cos coskθ dr sinkθ br p c (k) u c y(k) k k m 1 j 1 - k j j impulse 0         Z- 1 Time domain
  • 75. An Example 0 2 4 6 8 10 12 14 16 18 20 -1 -0.5 0 0.5 1 1.5 2 ) 3 kπ cos( 0.8 ) 3 kπ sin( 0.8 2 y(k) 0.64 0.8z z z z Y(z) k k 2 2          Z-Domain: Complex Poles Time-Domain: Exponentially Modulated Sin/C
  • 77. Observations  Using poles to characterize a signal  The smaller is |r|, the faster converges the signal  |r| < 1, converge  |r| > 1, does not converge, unbounded  |r|=1?  When the angle increase from 0 to pi, the frequency of oscillation increases  Extremes – 0, does not oscillate, pi, oscillate at the maximum frequency
  • 78. Change Angles 0.9 -0.9 Re Im 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
  • 79. 0 2 4 6 8 10 12 14 -6 -4 -2 0 2 4 6 8 10 12 Changing Absolute Value Im Re 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 5 10 15 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 2 4 6 8 10 12 14 -3 -2 -1 0 1 2 3 4
  • 80. Conclusion for Complex Poles  A complex pole appears in pair with its complex conjugate  The Z-1-transform generates a combination of exponentially modulated sin and cos terms  The exponential base is the absolute value of the complex pole  The frequency of the sinusoid is the angle of the complex pole (divided by 2π)
  • 81. Steady-State Analysis  If a signal finally converges, what value does it converge to?  When it does not converge  Any |pj| is greater than 1  Any |r| is greater than or equal to 1  When it does converge  If all |pj|’s and |r|’s are smaller than 1, it converges to 0  If only one pj is 1, then the signal converges to cj  If more than one real pole is 1, the signal does not converge … (e.g. the ramp signal)   k dr k br k k cos sin         m 1 j 1 - k j j impulse 0 p c (k) u c y(k) 2 1 -1 ) z (1 z  
  • 83. Final Value Theorem  Enable us to decide whether a system has a steady state error (yss-rss)
  • 84. Final Value Theorem 1 Theorem: If all of the poles of (1 ) ( ) lie within the unit circle, then lim ( ) lim ( 1) ( ) k z z Y z y k z Y z     2 1 1 0.11 0.11 ( ) 1.6 0.6 ( 1)( 0.6) 0.11 ( 1) ( ) | | 0.275 0.6 z z z z Y z z z z z z z Y z z                 0 5 10 15 -0.35 -0.3 -0.25 -0.2 -0.15 -0.1 -0.05 0 k y(k) If any pole of (1-z)Y(z) lies out of or ON the unit circle, y(k) does not converge!
  • 85. What Can We Infer from TF?  Almost everything we want to know  Stability  Steady-State  Transients  Settling time  Overshoot  …
  • 88. Nth-Order Difference Equation        M r r N k k r n x b k n y a 0 0 ) ( ) (        M r r r N k k k z b z X z a z Y 0 0 ) ( ) (        N k k k M r r r z a z b z H 0 0 ) (
  • 89. Representation in Factored Form          N k r M r r z d z c A z H 1 1 1 1 ) 1 ( ) 1 ( ) ( Contributes poles at 0 and zeros at cr Contributes zeros at 0 and poles at dr
  • 90. Stable and Causal Systems          N k r M r r z d z c A z H 1 1 1 1 ) 1 ( ) 1 ( ) ( Re Im Causal Systems : ROC extends outward from the outermost pole.
  • 91. Stable and Causal Systems          N k r M r r z d z c A z H 1 1 1 1 ) 1 ( ) 1 ( ) ( Re Im Stable Systems : ROC includes the unit circle. 1
  • 92. Example Consider the causal system characterized by ) ( ) 1 ( ) ( n x n ay n y    1 1 1 ) (    az z H Re Im 1 a ) ( ) ( n u a n h n 
  • 93. Determination of Frequency Response from pole- zero pattern  A LTI system is completely characterized by its pole-zero pattern. ) )( ( ) ( 2 1 1 p z p z z z z H     Example: ) )( ( ) ( 2 1 1 0 0 0 0 p e p e z e e H j j j j         0  j e Re Im z1 p1 p2
  • 94. Determination of Frequency Response from pole- zero pattern  A LTI system is completely characterized by its pole-zero pattern. ) )( ( ) ( 2 1 1 p z p z z z z H     Example: ) )( ( ) ( 2 1 1 0 0 0 0 p e p e z e e H j j j j         0  j e Re Im z1 p1 p2 |H(ej)|=? H(ej)=?
  • 95. Determination of Frequency Response from pole- zero pattern  A LTI system is completely characterized by its pole-zero pattern. Example: 0  j e Re Im z1 p1 p2 |H(ej)|=? H(ej)=? |H(ej)| = | | | | | | 1 2 3 H(ej) = 1(2+ 3 )
  • 96. Example 1 1 1 ) (    az z H Re Im a 0 2 4 6 8 -10 0 10 20 0 2 4 6 8 -2 -1 0 1 2 dB