Digital Signal Processing
Applications(310253)
UNIT-III
Z-Transform
Prof. Vina M. Lomte
RMDSSOE,Warje
7/13/2017
310253 Digital Signal Processing
Applications
•
Teaching Scheme: Examination Scheme:
•
Theory: 3 Hrs/Week
•
In Semester Assessment: 30 Marks
•
End Semester Assessment: 70 Marks
7/13/2017
Course Objectives:
· Study and understanding of representation of signals and systems.
· To learn and understand different Transforms for Digital Signal Processing
· Design and analysis of Discrete Time signals and systems
· To Generate foundation for understanding of DSP and its applications like audio,
Image, telecommunication and real world
7/13/2017
Syllabus
•
Definition of Z-Transform, ZT and FT, ROC, ZT properties, pole-zero plot,
•
Inverse Z-Transform, Methods, System function H(Z), Analysis of DT LTI
•
systems in Z-domain: DT system representation in time and Z domain.
•
Relationship of FT and ZT
7/13/2017
Teaching Plan
Sr. No.
Topic
01
Definition of Z-Transform, ZT
and FT
Lectures
Required
01
02
ROC, ZT properties & Examples
03(1 + 2
Extra )
03
Pole-zero plot & Examples
03(1 + 2
Extra )
04
Inverse Z-Transform, Methods,
System function H(Z) &
Examples
02(1 + 1
Extra )
05
Analysis of DT LTI
systems in Z-domain: DT system
representation in time and Z
domain
02(1 + 1
Extra )
7/13/2017
References
1. Steven W. Smith, “The
Scientist and Engineer's
Guide to Digital Signal
Processing”
2. P. Ramesh Babu , Fourth
Edition ,” Digital Signal
Processing”
Session 1
•
Introduction
•
Why z-Transform?
•
Definition of Z-Transform,
•
Relationship ZT and FT
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What is ZT
What is Z ?
It is
Real
Z= x + iy
Real
7/13/2017
Why z-Transform?
• It is very simple method for analyzing system(by
ROC properties) ex. LTI system
• 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
• The z-transform is a very important tool in describing and
analyzing digital systems.
• It offers the techniques for digital filter design and frequency
analysis of digital signals.
Definition
•
The z-transform of sequence x(n) is defined by
Frequency
Domain
Time
Domain
Convert

X
(z)
x
(n
)z

n
n


Fourier
Transform
 Let z = ej.

X
(
e)
x
()
n
e

j

n




j

n
Relationship Between FT and ZT
•
The following Eq.(1) and (2) are FT and ZT, respectively.
•
Replacing Z with
7/13/2017
, ZT will become FT
Session 2
•
ROC
•
ZT properties,
•
pole-zero plot
7/13/2017
Definition of ROC
• The region in which Z is valid
• 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
|
X
(
z
)
|

x
(
n
)
z

|
x
(
n
)
||
z
|

 
n


n


ROC is centered on origin and
consists of a set of rings.
Example: Region of Convergence



n

n
|
X
(
z
)
|

x
(
n
)
z

|
x
(
n
)
||
z
|

 
n


Im
n


ROC is annual ring centered an on the
origin.
r
Re
Rx | z|Rx
j

ROC

{
z

re
|R

r

R
}
x

x

Stable Systems
•
A stable system requires that its Fourier transform is uniformly convergent.
Im
 Fact: Fourier transform is to evaluate ztransform on a unit circle.
 A stable system requires the ROC of ztransform to include the unit circle.
1
Re
Example: A right sided Sequence
A right hand sequence x(n) is one for which x(n)>=0 for all n<no where no is
+ve or –ve but finite . If n0>=0 the resulting sequence is causal sequence .
For such type of sequence ROC is entire z-plane except at z=0
x(n)au(n)
n
All positive
values
x(n)
...
-8 -7 -6 -5 -4 -3 -2 -1
1 2 3 4 5 6 7 8 9 10
n
Example: A right sided Sequence
For convergence of X(z), we require that
x(n)au(n)
n


X
(z
)
au
(
n
)
z
n
n



n
1
|
az
 |
n0
| z || a |

anzn
n0

(az1)n
n0
| az1 | 1

1 z
X
(
z
)

(
az
) 


1
1

az
z

a
n

0

1
n
| z || a |
Example: A right sided Sequence ROC
for x(n)=anu(n)
z
X
(
z
)
 ,
z

a
a
|z
|
|a
|
Which one is stable?
Im
Im
1
1
a
a
Re
ROC
includes
unit
circle
a
Re
Example: A left sided Sequence
A left hand sequence x(n) is one for which x(n)>=0 for all n<no where no is
+ve or –ve but finite . If n0<=0 the resulting sequence is anticausal
sequence . For such type of sequence ROC is entire z-plane except at z=∞
n
x
(n
)

a
u
(
n

1
)
-8 -7 -6 -5 -4 -3 -2 -1
n
...
Negative
Values
1 2 3 4 5 6 7 8 9 10
x(n)
Example: A left sided Sequence
x
(n
)

au
(
n

1
)
n

X
(
z
)


a
u
(

n

1
)
z

n
n


1

anzn

n
For convergence of X(z), we require that

1
|
a
 z|
n0
| z || a |
n

anzn
n1

1
anzn
n0
| a1z | 1


1
n
1 z
X
(
z
)

1

(
a
z
)

1



1
1

a
zz

a
n

0
| z || a |
Example: A left sided Sequence ROC
for x(n)=anu( n1)
z
X
(
z
)
 ,
z

a
a
|z
|
|a
|
Which one is stable?
Im
Im
1
1
a
a
Re
a
Re
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.
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if you need stability then the ROC must contain the unit
circle.
If you need a causal system then the ROC must contain
infinity and the system function will be a right-sided
sequence.
If you need an anticausal system then the ROC must
contain the origin and the system function will be a leftsided sequence.
If you need both, stability and causality, all the poles of
the system function must be inside the unit circle.
7/13/2017
Pole and Zeros
Represent z-transform as a Rational
Function
P(z)
X(z)
Q(z)
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
z
X
(
z
)
 ,
z

a
x(n)anu(n)
|z
|
|a
|
Im
a
Re
ROC is bounded by
the pole and is the
exterior of a circle.
Example: A left sided Sequence
n
x
(n
)

a
u
(
n

1
)
z
X
(
z
)
 ,
z

a
|z
|
|a
|
Im
a
Re
ROC is bounded by
the pole and is the
interior of a circle.
Example: Sum of Two Right Sided Sequences
Im
1/12
1/3
1/2
Re
ROC is bounded by poles
and is the exterior of a
circle.
ROC does not include any pole
Example: A Two Sided Sequence
n
n
1
1
x
(
n
)

(

)
u
(
n
)

(
)
(

n

1
)
3
2u
1
z
z
2z(z12
)
X
(z
) 1 1 
z

(z1
z1
3 z
2
3)(
2)
Im
ROC is bounded by poles
and is a ring.
1/12
1/3
1/2
Re
ROC does not include any pole
Example: A Finite Sequence
n
x
(
n
)

a
,
0

n

N

1
N

1
N

1
n
n
1 N
1(az
)
1 zNaN
X
(
z
)

a
z
(
a
z) 
 N1


1
1az
z
za
n

0
n

0

1
n
Im
N-1 zeros
ROC: 0 < z < 
ROC does not include any pole
N-1 poles
Re
Always Stable
BIBO Stability
• Bounded
Input Bounded Output Stability
 If the input is bounded, we want the output is bounded
too
For limited input sequence its output should respectively
limited
7/13/2017
Z-Transform Pairs
Sequence
z-Transform
(n)
1
(n  m)
z m
ROC
All z
All z except 0 (if m>0)
or  (if m<0)
u (n)
1
1  z 1
| z | 1
u(n1)
1
1  z 1
| z | 1
a u (n)
1
1  az 1
| z || a |
anu(n1)
1
1  az 1
| z || a |
n
Z-Transform Pairs
Sequence
z-Transform
ROC
[cos

]u(n)
0n

1
1

[cos

]
z
0

1

2
1

[
2
cos

]
z

z
0
| z | 1
[sin
0n]u(n)

1
[sin

]
z
0

1

2
1

[
2
cos

]
z

z
0
| z | 1
[rncos

]u(n
)
0n

1
1

[
r
cos

]
z
0

1 2
1

[
2
r
cos

]
z

rz2
0
| z | r
[rnsin

]u(n)
0n

1
[
r
sin

]
z
0

1 2
1

[
2
r
cos

]
z

rz2
0
| z | r
n

a
0
nN

1
1aN zN

1
0 otherwise 1az

| z | 0
Signal Type
ROC
Finite-Duration Signals
Causal
Entire z-plane
Except z = 0
Anticausal
Entire z-plane
Except z = infinity
Two-sided
Entire z-plane
Except z = 0
And z = infinity
Causa
l
Anticausal
Infinite-Duration Signals
|z| > r2
|z| < r1
Two-sided
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]
10. [sin0n]u[n]
11. [rncos0n]u[n]
(z2-[cos0]z)/(z2-[2cos0]z+1)
[sin0]z)/(z2-[2cos0]z+1)
(z2-[rcos0]z)/(z2-[2rcos0]z+r2)
|z|>1
|z|>1
|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
Z-Transform Properties:
1.Linearity
•
Notation
•
Linearity
Z




x
n




X
z
ROC

R
x
Z






ax
n

bx
n




aX
z

bX
z
RO

R

R
1
2
1
2
x
1x
2
– Note that the ROC of combined sequence may be larger than either ROC
– This would happen if some pole/zero cancellation occurs
n
n
– Example:




x
n

a
u
n
a
u
n
N
Proof:
According to defination of ZT


n
X
(
z
)
x
(
n
)
z

Here x(n)=a1x1(n) + a2x2(n)
n


Writing two terms separately we get,
Here a1 & a2 are constants se we can take it outside the summation sign
By comparing eqn 1 & 3 we get
X(z) =a1X1(z)+a2X2(z) Hence proved
ROC : the combined ROC is overlapped or intersection of the individual ROCs
of X1(z) & X2(z)
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2. Time Shifting
n
Z
o




x
n

n




z
X
z
o
ROC

R
x
• 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


 1

1



X
z

z
1

1


1
z


 4
1
z

4
n
1
1





x
n

n
1
 u
4

Here x(n-no) indicates that the sequence is shifted in the time domain
by (-no) samples corresponds to multiplication by
in the frequency
domain
Proof
Statement : if X(n)
Then x(n-k) )
z
z
Z(z)
Z-k X(z) ----- 1
Then Z{x(n-k)} =
-----2
Now Z-n can be written as Z-(n-k)
Z(x(n-k) =
Since the limits of summation are in terms
Z(x(n-k) =
Now put
At
of n we can take Z-k outside of the summation
--------3
n-k=m on RHS the limit will change as follows
n=-∞ , -∞-k = m m=-∞
At n=
+∞, ∞-k=m
Z{x(n-k)} =
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, m= ∞
-----------4
Compare eqn 1 & 4
Z{x(n-k) = Z-k X(z)
hence X(n)
z
Z(z)
Similarly it can be shown that
x(n-k) z Z-k X(z)
= x(n-k)
z z
+k
X(z)
Here x(n-k) indicates that the sequence is shifted in time domain by (-k) samples
corresponding to multiplication by z-k in the frequency domain
ROC of
z-k
is same
as that X(Z) except z=0 if k>0
and z=∞ if k>0
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Example
Find ZT of x( n) =
That means
(n-k)
(n) Z 1
x(n-k) ) Z Z-1 X(z)
hence
(n-k)
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Z
Z-k . 1
3. Scaling in Z-domain (Multiplication by Exponential)
Z
n




z
x
n



X
z
/
z
o
o
•
•
•
•
•
ROC

z
R
o
x
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
Z 1


u
n




ROC
:
z

1
1
1
z
n 1
n
1
j


j

n
o
o










x
n

r
cos

n
u
n

re
u
n

re
u
n
o
2
2

•
Let’s find the z-transform of

1
/
2 1
/
2


X
z
j

z

r


j


1

1
o
o
1

re
z
1

re
z
4. Differentiation


dX
z
Z


nx
n





z
ROC

R
x
dz
•
Example: We want the inverse z-transform of



1


X
z

log
1

az
z

a
•
Let’s differentiate to obtain rational expression
•
Making use of z-transform properties and ROC

2




dX
z

az
dX
z

11



z

az
1

1
dz
dz
1

az
1

az

u



nx
n

a

a
n

1
n

1
n
n

1
a

 u




x
n

1
n

1
n
Multiplying the sequence in time
domain by n is equivalent to
multiplying the sequence the
derivation of its ZT by –Z in the
Z-domain
5. Conjugation

Z *
*
*


x
n




X
z
ROC

R
x
• Example

Xz
x
nzn
n





n
Xz
x
nz  
x
nzn
n
 n






 
X z 
x
n z 
x
nznZx
n
 

n

n
n

6. Time Reversal
Z




x

n




X
1
/
z
• ROC is inverted
• Example:
• Time reversed version of
1
ROC

R
x

n

n
x
na
u
anun
-1

1
1a
z


X
z
 1

1
1

az
1
a
z

1
z

a
Here x(-n) is the folded version of x(n) so,x(-n) is the time reverse signal
thus the folding of signal in time domain is equivalent to replacing z by z-1
in the z-domain
Replacing z by z-1 in the z-domain is called as inversion hence folding in
the time domain is equivalent to the inversion in z-domain
7. Convolution
Z








x
n

x
n




X
z
X
z
12
1
2
•
•
RO
:
R

R
x
x
1
2
Convolution in time domain is multiplication in z-domain
Example:Let’s calculate the convolution of
n



and




x
n

a
u
n
x
n

u
n
1
2
1
1


X
z

ROC
:
z

a


X
z

ROC
:
z

1
1
2

1

1
1

az
1

z
•
Multiplications of z-transforms is
1





Y
z

X
z
X
z

1 2

1

1
1

az
1

z

•
•

 
ROC: if |a|<1 ROC is |z|>1 if |a|>1 ROC is |z|>|a|
Partial fractional expansion of Y(z)
1
11




Y
z


asum
ROC
:
z

1


1

1
1

a
1

z
1

az




n

1

1u




y
n
n
a
u
n
1

a
Linearity
Z
[
x
(
n
)]

X
(
z
), z

R
x
Z
[
y
(
n
)]

Y
(
z
), z

R
y
Z
[
ax
(
n
)

by
(
n
)]

aX
(
z
)

bY
(
z
),
z

R

R
x
y
Overlay of
the above two
ROC’s
Shift
Z
[
x
(
n
)]

X
(
z
), z

R
x
Z
[
x
(
n

n
)]

zX
(
z
) z

R
0
x
n
0
Multiplication by an Exponential Sequence
Z
[
x
(
n
)]

X
(
z
),R

|z
|

R
xx


1
Z
[
a
x
(
n
)]

X
(
a
z
) z

|a
|

R
x
n
Differentiation of X(z)
Z
[
x
(
n
)]

X
(
z
), z

R
x
dX
(
z
)
Z
[
nx
(
n
)]


z
z

R
x
dz
Conjugation
Z
[
x
(
n
)]

X
(
z
), z

R
x
Z
[
x
*
(
n
)]

X
*
(
z
*)
z

R
x
Reversal
Z
[
x
(
n
)]

X
(
z
), z

R
x

1
Z
[
x
(

n
)]

X
(
z) z

1
/
R
x
Initial Value Theorem
x
(
n
)
0
,
Initial
Value
for
n

0
x(0
)lim
X(z)
z

Convolution of Sequences
Z
[
x
(
n
)]

X
(
z
), z

R
x
Z
[
y
(
n
)]

Y
(
z
), z

R
y
Z
[
x
(
n
)
*
y
(
n
)]

X
(
z
)
Y
(
z
)z

R

R
x
y
Z- Transform Properties Examples:
Linearity:
a and b are arbitrary constants.
Example
Problem:
Find z- transform of
Using z- transform table:
Therefore, we get
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Shift Theorem:
Verification:
Since x(n) is assumed to be causal:
Then we achieve,
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Example
Problem:
Find z- transform of
Solution:
Using shift theorem,
Using z- transform table
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Convolution
In time domain
In z- transform domain,
Verification:
Using z- transform in Eq. (1)
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Eq. (1)
Example
Problem:
Given the sequences,
Find the z-transform of their convolution.
Solution:
Applying z-transform on the two sequences,
From the table
Therefore we get,
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System Function
x(n)
h(n)
y(n)
By using convolution property & system function
Y(z)=H(z)X(z)
Proof:
x(n) * h(n) =y(n)
X(z)H(z)=Y(z)
H(z) = Y(z) / x(z)
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ROC: at least the intersection of the ROCs of H(z) & X(z)
Session 3
•
Inverse Z-Transform,
•
Methods to find IZT
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Inverse z- Transform:
The Procedure of obtaining x(n) from its ZT X(Z) is
called Inverse ZT
Methods to find Inverse z- Transform:
1.Power series expansion
2.Partial fraction expansion
3.Residue method
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1. Inverse Z-Transform by Power Series
Expansion


n


X
z
x
nz
•
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


2
1

1

2












X
z



x

2
z

x

1
z

x
0

x
1
z

x
2
z


X(z)=a0+a1z-1+a2Z-2 + ------- + an Z-n


n


X
z
x
nz
n


If the sequence is causal then the limits of n will be n=0 to n=∞
Expanding the above expression we get,
X(Z) = x(0) z0 + x(1) z-1+x(2) z-2+ ……..x(n) z-n = x(0) + x(1) z-1+x(2) z-2 ……..x(n) z-n
By comparing two equs of X(z) we can write
X(0) =a0
X(1)= a1
X(2)=a2
.
.
.
X(n)=an
Thus the general expression of discrete time causal sequence x(n) is,
X(n) = an n>=0
Expansion of ZT for matching standard pair of ZT to get original
sequence back
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2. Inverse z-Transform by Partial Fraction
•
Assume that a given z-transform can be expressed as
M
b z
Xz  kN0
k
k
a z
k 0
•
•
Apply partial fractional expansion
First term exist only if M>N
– Br is obtained by long division
•
•
k
k
s
A
C
k
m


X
z

B
z






1
m

1
1

d
z
m

1
1

d
z
k
i
Second term represents all first order poles
Third term represents an order s pole
M

N
N

r
r
r

0
k

1
,
k

i


– There will be a similar term for every high-order pole
•
•
•
•
•
Each term can be inverse transformed by inspection
It is expressed as ratios of two polynomials
X(z) = N(z)/D(z) = bo+b1z-1 +b1z-2 + ---+bMz-M / a0+a1z-1 +a2z-2 + --- + aNz-N
N(z)- numerator polynomial , D(z) - denominator polynomial
bo…,bM –coefficient numerator polynomial , ao..aN - coefficient denominator
polynomial , M – Degree of numerator & N- Degree of denominator
Session 4
•
System function H(Z)
7/13/2017
System Function
Signal Characteristics from Z-Transform
• If U(z) is a rational function, and
y(k)

a
y(k

1)

...

a
y(k

n)

b
u(k

1)

...

b
u(

m
1
n
1
m
• Then Y(z) is a rational function, too
n
zeros
(zzi)
N(z)
i1
Y(z)

m
D(z)
(zpj)

j1
poles
• Poles are more important – determine key
characteristics of y(k)
Why are poles important?
Z domain
n
(z

z

i)
mc
N(z)
j
i
1
Y(z)
 m

c


0
D(z)
z

p
j
1
j
(z

p
)
j
j
1
Time domain
poles
Z-1
m
k
1
Y(k)

c

u
(k)

c

p

0 impulse
j
j
j

1
components
Shift-Invariant System
y(n)=x(n)*h(n)
x(n)
h(n)
X(z)
H(z)
Y(z)=X(z)H(z)
Shift-Invariant System
X(z)
Y(z)
H(z)
Y(z)
H(z)
X(z)
Session 5
Analysis of DT LTI systems in Time
domain
7/13/2017
Time-Domain Representation
•
Signals represented as sequences of numbers, called samples
•
Sample value of a typical signal or sequence denoted as x[n] with n being an
integer in the range
•
x[n] defined only for integer values of n and undefined for non integer values
of n
•
Discrete-time signal represented by {x[n]}
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•
Discrete-time signal may also be written as
•
a sequence of numbers inside braces:
•
{ [x n]} ={K,− 0.2,2.2,1.1,0.2,− 3.7,2.9,K}
↑
•
In the above, x[−1] = −0.2, x[0] = 2.2, x[1] =1.1,
etc.
•
The arrow is placed under the sample at time index n = 0
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•
Graphical representation of a discrete-time signal with real-valued samples is as
shown below:
7/13/2017
•
In some applications, a discrete-time sequence {x[n]} may be generated by
periodically sampling a continuous-time signal x a ( t ) at uniform intervals of
time
7/13/2017
•
Here, n-th sample is given by
x[n] = xa (t) t=nT = xa (nT), n =K,− 2,−1,0,1,K
•
The spacing T between two consecutive samples is called the sampling interval
or sampling period
•
Reciprocal of sampling interval T, denoted as FT, is called the sampling
frequency:
FT= 1/T
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•
Two types of discrete-time signals:
- Sampled-data signals in which samples are continuous-valued
- Digital signals in which samples are discrete-valued
•
Signals in a practical digital signal processing system are digital signals
obtained by quantizing the sample values either by rounding or truncation
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•
A discrete-time signal may be a finitelength or an infinite-length sequence
•
Finite-length (also called finite-duration or finite-extent) sequence is defined
only for a finite time interval:N1 ≤ n ≤ N2
•
where −∞ < N1 and N2 < ∞ with N1 ≤ N2
•
Length or duration of the above finitelength
•
sequence is N = N2 − N1 +1
7/13/2017
Session 6
Analysis of DT LTI systems in Zdomain
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LTI System description
Previous basis function: unit sample or DT impulse
 The input sequence is represented as a linear combination
of shifted DT impulses.
 The response is given by a convolution sum of the input
and the reflected and shifted impulse response
Now, use eigenfunctions of all LTI systems as basis function
Relation between DFT & ZT
This means if Z-T is evaluated on the unit
circle at evenly spaced points only; then it
become DFT
7/13/2017