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z-Transform Properties & LTI System Analysis Presentation
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z-Transform Properties
and LTI System Analysis
EE4015 Digital Signal Processing
Dr. Lai-Man Po
Department of Electrical Engineering
City University of Hong Kong
Overview
• Properties of z-transform
• Transfer Function
• Transfer Function & Difference Equation
• Transfer Function & Impulse Response
• Transfer Function & System Stability
• Difference Equation & System Stability
• Pole-Zero Plot
• Stability Analysis based Pole-Zero Plot
L.M. Po
2
Key Properties of the z-Transform
1. Linearity : π₯! π β· π! π§ π
ππΆ = π
! and π₯" π β· π" π§ π
ππΆ = π
"
ππ₯! π + ππ₯" π β· ππ!(π§) + ππ"(π§) π
ππΆ = π
! ∩ π
"
2. Time Shifting : π₯[π − π# ] β· π§ $%! π(π§)
3. Time Reversal : π₯ −π β· π 1/π§
4. Exponential Scaling : π% π₯ π β· π π§/π
5. Z-domain Differentiation : ππ₯ π β· π§
&'())
&+
6. Convolution : π₯! π ∗ π₯" π β· π!(π§)π"(π§) π
ππΆ = π
! ∩ π
"
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3
Linearity of the z-Transform
• If π₯! π β· π! π§ for π§ in ROC of R1 and π₯" π β· π" π§ for π§ in ROC
of R2 then
ππ₯! π + ππ₯! π β· ππ!(π§) + ππ"(π§) for π§ in ROC of R1 ∩ R2
• Let π¦ π = ππ₯! π + ππ₯" π then
'&
'&
π π§ = , π¦[π]π§ %# = , 0ππ₯! π + ππ₯" π )π§ %#
#$%&
'&
#$%&
'&
= π , π₯! π π§ %# + π , π₯" π π§ %# = ππ! π§ + ππ" (z)
#$%&
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#$%&
π
ππΆ = π
! ∩ π
"
4
ROC Property of the z-Transform
• Note that the ROC of combined sequence may be larger
than either ROC
§ This would happen if some pole/zero cancellation
occurs
• Example: π₯ π = π! π’ π − π! π’[π − π]
§ Both sequences are right-sided
§ Both sequences have a pole π§ = π
§ Both have a ROC defined as π§ > π
§ In the combined sequence the pole at π§ = π cancels
with a zero at π§ = π
§ The combined ROC is the entire z plane except z=0
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This zero
cancelled the
pole at π§ = π
ROC: π§ > 0
π§ ! − π!
π(π§) = !"# /
π§
π§−π
1
5
Linearity Example 1
Find the z-transform of the signal π₯ π defined by
π₯ π = π’ π − 0.5 # π’[π]
• Applying the Linear property of the z-transform, we have
π π§ =π π₯ π
=π π’ π
= π π’ π − 0.5 # π’[π]
− π − 0.5 # π’[π]
π§
π§
=
−
,
π§ − 1 π§ − 0.5
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π
ππΆ = z > 1
6
Linearity Example 2
Find the z-transform of the signal π₯ π defined by
π₯ π = 3 2 #−4 3 # π’ π
• Applying the linearity of the z-transform, we have
§ As we know π π $ π’ π
=
1
1 − ππ§ "#
π
ππΆ: π§ > π
§ Applying the linearity of the z-transform, we have
1
1
π(π§) = 3
−4
1 − 2π§ %!
1 − 3π§ %!
π§
π§
=3
−4
π§−2
π§−3
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π
ππΆ: π§ > 2 ∩ π§ > 3
π
ππΆ: π§ > 3
7
Linearity Example 3
Find the z-transform of the signal π₯ π defined by π₯ π = cos π: π π’ π
• We know
π &'! $ + π "&'! $ 1 &' $ 1 "&' $
!
cos π% π =
= π ! + π
2
2
2
1
π π &'! $ π’[πB =
, π
ππΆ = z > 1
1 − π &'! π§ "#
• Applying the linearity of the z-transform, we have
π π§ =
1
1
1
1
1
1
π π &'! $ π’[πB + π π "&'! $ π’[πB =
+
2
2
2 1 − π &'! π§ "#
2 1 − π "&'! π§ "#
1
1 − π "&'! π§ "#
=
2 1 − π &'! π§ "# 1 − π "&'! π§ "#
1
1 − π &'! π§ "#
+
2 1 − π "&'! π§ "# 1 − π &'! π§ "#
2 − π§ "# π &'! + π "&'!
1
1 − π§ "# cos π%
=
=
2 1 − π§ "# π &'! + π "&'! + π§ "(
1 − 2π§ "# cos π% + π§ "(
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1 1 − π "&'! π§ "# + 1 − π &'! π§ "#
=
2 1 − π "&'! π§ "# 1 − π &'! π§ "#
π
ππΆ = z > 1
8
Time Shifting of the z-Transform
• A πF -sample delay in the time domain appears in the z domain as a π§ %#! factor.
'&
• More generally, π π§ = , π₯[π]π§ %#
π₯[π]
π) Delay
π₯[π − π) ]
π₯[π]
π§ "$"
π₯[π − π) ]
#$%&
• Let π¦[π] = π₯[π − πF ]
'&
'&
π π§ = , π¦[π]π§ %# = , π₯[π − πF ]π§ %#
#$%&
• Substitute π = π − πF
#$%&
'&
π π§ = , π₯[π ]π§
G$%&
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'&
%G%#!
= π§ %#! , π₯[π ]π§
%G
= π§ %#! π(z)
G$%&
9
Time Shifting Property Example
Find the z-transform of the signal π₯ π using time shifting property
π₯ π = 0.5 #%J π’ π − 5
Solution
π 0.5 " π’ π
=
1
,
1 − 0.5π§ #$
π
ππΆ: π§ > 0.5
• Applying the time shifting property of the z-transform, we have
π π§ = π§ $,π 0.5 % π’ π
$1
π§
= π§ $,
=
$!
1 − 0.5π§
π§ − 0.5
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π
ππΆ = z > 0.5
10
Time Reversal Property
• π₯ −π β· π 1/π§
π
ππΆ = 1/π
2
π = −π
Proof
0/
$/
0/
π΅{π₯[−π]} = < π₯[−π]π§ $% = < π₯[π]π§ 1 = < π₯[π] π§ $! $1
%.$/
1./
1.$/
= π π§ $!
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Time Reversal Property Example
Find the z-transform of the signal π₯ π = πIJ π’ −π
Solution
• The sequence of π$% π’ −π is Time-reversed version of π% π’ π
• Applying the time reversal theorem of the z-transform, we have
1
1
1
−π %! π§ %!
%!
π§
<
π
π π§ =
=
=
,
1 − ππ§
−π π§ − 1 1 − π %! π§ %!
π
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ROC is inverted
12
Exponential Scaling Property
,
-
π π₯ π ↔ π π§/π
π
ππΆ = π π
L
• ROC is scaled by π
• All pole/zero locations are scaled
• If π is a positive real number: z-plane shrinks or expands
• If π is a complex number with unit magnitude it rotates
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13
Exponential Scaling Example
• Example: We know the z-transform pair
π’π
M
N
↔
NIO "#
π
ππΆ = z > 1
• Let’s find the z-transform of
π₯π
= π % cos π# π
1
1
%
%
34
$34
!
!
π’ π = ππ
π’ π + ππ
π’π
2
2
1⁄2
1⁄2
π π§ =
+
1 − ππ LM% π§ %! 1 − ππ %LM% π§ %!
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π
ππΆ = π§ > π
14
Z-Domain Differentiation Property
ππ₯ π
M
ST(O)
↔ −π§
SO
π
ππΆ = π
L
• For example, we want the inverse z-transform of
π π§ = log 1 + ππ§ %!
π§ > π
• Let’s differentiate to obtain rational expression
ππ(π§)
−ππ§ %"
ππ π§
1
%!
=−
⇒ −π§
= ππ§
ππ§
1 + ππ§ %!
ππ§
1 + ππ§ %!
• Making use of z-transform properties and ROC
ππ₯ π = π −π #%! π’[π − 1]
π₯π =
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&
N
−1 #%! π’[π − 1]
#
15
The z-Transform Convolution Property
• Convolution in time domain is equal to the multiplication in frequency
domain and vice versa.
• If π₯ π β· π π§ and π¦[π] β· π π§ , then
π₯ π ∗ π¦ π β· π π§ π(π§)
π
ππΆ = π
2 ∩ π
5
• Proof
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Convolution Property Example 1
• Consider the two sequences
π₯! π = 3πΏ π + 2πΏ π − 1
π₯" π = 2πΏ π − πΏ π − 1
• Find the z-transform of convolution
π₯ π = π₯! π ∗ π₯" π
• Determine the convolution sum using the z-transform
Solution
π! π§ = Z π₯! π
= 3 + 2z $!
π" π§ = Z π₯" π
= 2 − z $!
π π§ = Z π₯! π ∗ π₯" π
π₯ π = π $! π π§
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= π! π§ π" π§ = 6 + z $! − 2z $"
= 6πΏ π + πΏ π − 1 − 2πΏ π − 2
17
Convolution Property Example 2
• Compute the convolution of the following two sequences using z-transform
π₯! π = 1,
Solution
−2,
1
π₯" π = [
1,
0,
0≤π<5
= 1, 1, 1, 1, 1
πππ ππ€βπππ
π! π§ = Z π₯! π
= 1 − 2z $! + π§ $"
π" π§ = Z π₯" π
= 1 + z $! + π§ $" + z $6 + π§ $-
π π§ = π! π§ π" π§ = 1 − z $! − π§ $, + z $7
π₯ π = π₯! π₯ ∗ π₯" π = π %! π π§
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= 1, −1, 0, 0, 0, −1, 1
18
Properties
of the zTransform
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LTI System Analysis Using the z-Transform
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20
Transfer Function
π₯[π]
π π§
π» π§ =
8+
'+
LTI System
βπ β· π» π§
π¦[π]
π π§
is referred as Transfer Function of the system.
• It is the ratio of the output to the input in the z domain:
§ π π§ is the z transform of the output π¦[π]
§ π π§ is the z-transform of the input π¦[π]
§ π» π§ is the z-transform of the impulse response β[π]
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Impulse Response
• The impulse response β[π] of the discrete-time LTI system π»(π§) can be
obtained by solving its difference equation using a unit impulse input
πΏπ
• With the help of the z-transform and noticing that π π§ = π{πΏ π } = 1
§ β π = π IN {π π§ /π(π§)} = π IN {π» π§ }
π₯ π =πΏ π
π π§ =1
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LTI System
βπ β· π» π§
π¦ π = β[π]
π π§ =π» π§
22
System Outputs in Time and z domains
• The LTI system output can be find using three different ways.
Time Domain
π₯[π]
Difference Equation
π¦[π]
π₯[π]
Impulse Response
β[π]
π¦ π οΌβ π ∗ π₯ π
z Domain
π π§
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Transfer Function
π» π§
π π§ =π» π§ π π§
23
Rational Transfer Function
• Transfer function can be expressed as a rational function consist of numerator
polynomial divided by denominator polynomial.
"*
π π§
π' + π$ π§ #$ + β― + π(#$ π§ (#$ + π( π§ #( ∑,
*+) π* π§
π» π§ =
=
=
π π§
π' + π$ π§ #$ + β― + π)#$ π§ )#$ + π) π§ #) ∑!*+) π* π§ "*
• The highest power in a polynomial is called its degree.
• In a proper rational function, the degree of the numerator is less than or equal to
the degree of the denominator. (π ≤ π)
• In a strictly proper rational function, the degree of the numerator is less than the
degree of the denominator. (π < π)
• In an improper rational function, the degree of the numerator is greater than the
degree of the denominator. (π > π)
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24
Transfer Function and Difference Equation
• A linear constant coefficient difference equation be described by a rational function in
z-transform as a ratio of Polynomials in z.
πF π¦[π] + π! π¦[π − 1] + β― + πW π¦[π − π] = πF π₯[π] + π! π₯[π − 1] + β― + πX π₯[π − π]
• Taking the z-transform of both sides
πF π(π§) + π! π§ %! π(π§) + β― + πX π§ %X π(π§) = πF π(π§) + π! π§ %! π(π§) + β― + πW π§ %W π(π§)
π(π§) πF + π! π§ %! + β― + πX π§ %X = π(π§) πF + π! π§ %! + β― + πW π§ %W
• Taking π(π§) and π(π§) common and then cross multiply to get Transfer Function π»(π§)
π(π§) πF + π! π§ %! + β― + πW%! π§ W%! + πW π§ %W
π»(π§) =
=
π(π§) πF + π! π§ %! + β― + πX%! π§ X%! + πX π§ %X
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25
Difference Equation Example 1
Find the transfer function described by the following difference equation.
2π¦ π + π¦ π − 1 + 0.9π¦ π − 2 = π₯ π − 1 + π₯ π − 4
Solution: Taking the z-transforms term by term we get,
2π π§ + π§ %! π π§ + 0.9π§ %" π π§ = π§ %! π π§ + π§ %Y π π§
Factoring out π π§ on the left side and π π§ on the right side:
π π§ 2 + π§ %! + 0.9π§ %" = π π§ π§ %! + π§ %Y
The transfer function is
π(π§)
π§ %! + π§ %Y
π» π§ =
=
π(π§) 2 + π§ %! + 0.9π§ %"
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Difference Equation Example 2
Find the transfer function described by the following difference equation.
π¦ π − 0.2π¦ π − 1 = π₯ π + 0.8π₯ π − 1
Solution: Taking z transforms term by term we get,
π π§ − 0.2π§ %! π π§ = π π§ + 0.8π§ %! π π§
Factoring out π π§ on the left side and π π§ on the right side:
π π§ 1 − 0.2π§ %! = π π§ 1 + 0.8π§ %!
The transfer function is
π(π§) 1 + 0.8π§ %!
π» π§ =
=
π(π§) 1 − 0.2π§ %!
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27
Difference Equation Example 3
Find the transfer function described by the following difference equation.
π¦ π = 0.75π₯ π − 0.3π₯ π − 2 + 0.01π₯ π − 3
Solution: Taking z transforms term by term we get,
π π§ = 0.75π π§ − 0.3π§ %" π π§ + 0.01π§ %Z π π§
Factoring out π π§ on the left side and π π§ on the right side:
π π§ = π π§ 0.75 − 0.3π§ %" + 0.01π§ %Z
The transfer function is
π(π§)
π» π§ =
= 0.75 − 0.3π %" − 0.01π %Z
π(π§)
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28
Difference Equation Example 4
Find the difference equation that correspond to transfer function.
1 + 0.5π§ %!
π» π§ =
1 − 0.5π§ %!
Solution: Since π» π§ = π π§ /π π§ , do the cross multiply to get
π π§ 1 − 0.5π§ $! = π π§ 1 + 0.5π§ $!
then
π π§ − 0.5π§ $!π π§ = π π§ + 0.5π§ $!π π§
Finally taking the inverse z-transform term by term to get
⇒
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π¦ π − 0.5π¦ π − 1 = π₯ π + 0.5π₯ π − 1
π¦ π = π₯ π + 0.5π₯ π − 1 + 0.5π¦ π − 1
29
Difference Equation Example 5
Find the difference equation that correspond to transfer function.
1 + 0.8π§ %!
π» π§ =
1 − 0.2π§ %! + 0.7π§ %"
Solution: Since π» π§ = π π§ /π π§ , do the cross multiply to get
π π§ 1 − 0.2π§ $! + 0.7π§ $" = π π§ 1 + 0.8π§ $!
then
π π§ − 0.2π§ $!π π§ + 0.7π§ $"π π§ = π π§ + 0.8π§ $!π π§
Finally taking the inverse z-transform term by term to get
π¦ π − 0.2π¦ π − 1 + 0.7π¦ π − 2 = π₯ π + 0.8π₯ π − 1
⇒ π¦ π = π₯ π + 0.8π₯ π − 1 + 0.2π¦ π − 1 − 0.7π¦ π − 2
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30
Difference Equation Example 6
Find the difference equation that correspond to transfer function.
π» π§ =
*
π§
(2π§ − 1)(4π§ − 1)
. *
Solution: π» π§ = +* ! #,*-$ = / *
Do the cross multiply to get
8π§ 0 − 6π§ + 1 π π§ = π§ π π§ , then 8π§ 0 π π§ − 6π§π π§ + π π§ = π§π π§
⇒
8π π§ − 6π§ #$ π π§ + π§ #0 π π§ = π§ #$ π π§
Finally taking the inverse z-transform term by term to get
8π¦ π − 6π¦ π − 1 + π¦ π − 2 = π₯ π − 1
⇒
⇒
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π¦ π − 0.75π¦ π − 1 + 0.125π¦ π − 2 = 0.125π₯ π − 1
π¦ π = 0.125π₯ π − 1 + 0.75π¦ π − 1 − 0.125π¦ π − 2
31
Pole-Zero Description of Discrete-Time System
• The zeros of a z-transform π»(π§) are the values of z for which π» π§ = 0.
• The poles of a z-transform are the values of z for which π» π§ = ∞ .
• If H(z) is a rational function , then
π» π§
8+
=
'+
9+ 09, + -, 0β―09.-, + .-, 09. + -.
=
;+ 0;, + -, 0β―0;/-, + /-, 0;/ + -/
• After factoring the rational transfer function, the roots π½[ of the numerator
polynomial are zeros and roots πΌ[ of denominator polynomial are poles.
#1
πΎ(π§ − π½$ )(π§ − π½0 ) β― (π§ − π½( ) πΎ ∏(
12$ 1 − π½1 π§
π»(π§) =
=
#1
(π§ − πΌ$ )(π§ − πΌ$ ) β― (π§ − πΌ) )
∏)
12$ 1 − πΌ1 π§
The poles and zeros of a system can provide a great deal of information about the
behavior of the system.
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32
Identify Poles and Zeros (1)
• It is easiest to identify the poles and zeros if the rational transfer
function
"#
%"#
"%
π» π§
k$ lk# O lβ―lk%"# O
lk% O
=
m$ lm# O "# lβ―lm&"# O &"# lm& O "&
is converted to the form
π» π§
k$ O % lk# O %"# lβ―lk%
nIo
=π§
m$ O & lm# O &"# lβ―lm&
which has only positive exponents.
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33
Identify Poles and Zeros (2)
= + π π§ =$! + β― + π
π
π§
>
!
=
<$=
π» π§ =π§
π>π§ < + π!π§ <$! + β― + π<
• The roots of the numerator polynomial are the zeros of the system.
• The roots of the denominator polynomial are the poles of the system.
• In general, numerator and denominator polynomials can always be factored
-'
∏*
(π§ − π½( )(π§ − π½) ) β― (π§ − π½* )
',( 1 − π½' π§
π»(π§) = πΎ
=πΎ +
(π§ − πΌ( )(π§ − πΌ) ) β― (π§ − πΌ+ )
∏',( 1 − πΌ' π§ -'
Where the are π½' the zeros, are πΌ' the poles, and πΎ is called the gain
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34
Effects of Poles and Zeros
• Poles are the values of π§ that make the denominator of a transfer
function zero.
§ Poles have the biggest effect on the behavior of discrete-time LTI
system
• Zeros are the values of π§ that make the numerator of a transfer function
zero.
§ Zeros tend to modulate, to a greater or lesser degree depending on
their position relative to the poles.
• The poles of the system can be found if its transfer function is known.
• Both zeros and poles are in general complex numbers.
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35
Pole-Zero Plot
• A very powerful tool for the discrete-time system
analysis and design is a complex plane called z-plane,
on which poles and zeros of the transfer function are
plotted.
• On the z plane,
§ poles are plotted as crosses (X)
z-plane
Im
unit circle
o
x
a
1
x
b
Re
§ zeros are plotted as circles (O)
• A plot showing pole and zero locations is called a
pole-zero plot.
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Pole-Zero Plot
36
Pole-Zero Plot Example 1
For a first order system the poles and zeros are
z-plane
2
π» π§ =
1 + 0.4π§ %!
Im
unit circle
• One Pole at π§ = -0.4
• One Zero at π§ = 0
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x o
-0.4
1
Re
37
LTI System Analysis using the z-Transform
• π¦ π = π₯ π ∗ β[π]
π π§
LTI System
π»(π§)
π π§
• π π§ = π π§ π»(π§)
• π»(π§) is the z-transform of the impulse response
β[π], which is called transfer function
• Stable system <=> Unit circle in the ROC
• Causal system => β[π] is right-sided sequence
z-plane
Im
unit circle
o x x
1 Re
=> ROC outside outermost pole
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38
System Stability based Pole Locations
• The position of the poles and zeros on the z-plane can give clue about the way a
discrete-time system will behave.
• One reason the poles of a system are so useful is that they determine whether the
system is stable or not.
• The system is stable if the poles lie inside the unit circle, which is a circle of unit
radius on the z-plane.
• Since poles are complex numbers, this requires that their magnitudes be less than
one.
• Mathematically, the region of stability can be described as
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39
z-plane
Stable Causal LTI System
Im
unit circle
• If the magnitude of each pole is less than one,
the poles are less than one unit’s distance from
the center of the unit circle, and the system is
stable. The ROC includes unit circle.
• If any of the poles of a system lie outside the
unit circle, the system is unstable.
• If the outermost pole lies on the unit circle, the
filter is described as being marginally stable
o
x
x
x
1 Re
Stable System
Im
x
unit
circle
x
o
x
1 Re
Unstable System
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40
ROC of π» π§ =
7
?
@
78 9 A?
ROC: |z|>1/2
ROC: |z|<1/2
• Stable system (unit circle in ROC)
• Unstable system (unit circle not in ROC)
• Casual system (h[n] is right-sided
sequence)
• Non-casual system (h[n] is left-sided
sequence)
• βπ =
! #
"
Im
π’π
unit circle
o
z-plane
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x
1/2 1 Re
• β π =−
! #
"
Im
π’ −π − 1
unit circle
o
x
1/2 1 Re
z-plane
41
System Stability Example 1
Find the poles and zeros and stability for the causal discrete-time system
whose transfer function is
4π§ %!
π» π§ =
4 − 9π§ %! + 2π§ %"
Solution
Eliminating negative exponents yields
π» π§
FG -,
G -,
G -,
= FHIG -,JKG -0 = LHK.KMG -,JN.MG -0 = LHN.KMG , LHKG -,
G0
G
$ G 0 = GHN.KM GHK
• Two Poles at π§ = 0.25 and π§ = 2
• One Zeros at π§ = 0
• As one pole lie outside the unit circle at z = 2, hence the system is unstable.
L.M. Po
42
System Stability Example 2
Find the poles and zeros and stability for the causal discrete-time system
whose transfer function is
z-plane
1 − π§ %"
π» π§ =
1 + 0.7π§ %! + 0.9π§ %"
Im
Solution
unit circle
Two zeros at 0.
Two poles are located at −0.35 ± π0.88
• For these poles the distance from the center of the unit circle is
§ π§ =
−0.35 1 + 0.88 1 = 0.9487 < 1
x
o
0.88
o1
-0.35
x
Re
-0.88
As both poles lie inside the unit circle,
Therefore, the system is stable.
L.M. Po
43
System Stability Example 3
z-plane
Determine the stability of the following causal system.
π» π§ =
unit circle
1 + 1.2π§ %! + 0.45π§ %"
Solution: Eliminating negative exponents yields
π» π§
Im
π§ %! − 0.5π§ %"
+ -, $>.,+ -0
+0
+$>.,
=
T = 0
!0!."+ -, 0>.-,+ -0 + 0
+ 0!."+0>.-,
x
0.3
-0.6
0.5
-0.3
x
o
1
Re
Poles: at π§ = −0.6 + π0.3 and π§ = −0.6 − π0.3
Zero: at π§ = 0.5
As all poles lie inside the unit circle, hence the system is stable.
L.M. Po
44
System Stability Example 4
Find the stability of the filter if the difference equation of the filter is
π¦ π + 0.8π¦ π − 1 − 0.9π¦ π − 2 = π₯ π − 2
Solution
• Poles are found most easily from the transfer function.
π» π§
+ -0
+0
!
=
T
=
!0>.C+ -, $>.D+ -0 + 0
+ 0 0>.C+$>.D
• The quadratic formula gives the pole locations as
−0.8 ± 0.80 − 4 1 −0.9
−0.8 ± 2.059
π§=
=
= 0.630 πππ − 1.430
2 1
2
• The poles in this case are purely real, without any imaginary component. Clearly the
pole at z = −1.430 lies outside the unit circle, so the system is unstable.
L.M. Po
45
Example 5: LTI System Analysis in z Domain
• Given the following system function:
π» π§
Nlr.stO "#
= Nlr.uO "# Ir.uvO ".
(a) Plot the pole-zero diagram of π» π§ .
(b) Find a stable impulse response β π .
(c) Find a causal impulse response exist that is both stable and causal?
L.M. Po
46
Solution (a)
Plot the pole-zero diagram of π» π§ .
1 + 0.25π§ %!
1 + 0.25π§ %!
π§"
π§ π§ + 0.25
π» π§ =
=
w "=
%!
%"
%!
%!
1 + 0.8π§ − 0.84π§
(1 + 1.4π§ )(1 − 0.6π§ ) π§
(π§ + 1.4)(π§ − 0.6)
Im
• Two zeros at z = 0 and z = -0.25
z-plane
unit circle
• Two poles at z = -1.4 and z = 0.6
x
-1.4
L.M. Po
o o
-0.25 0
x
0.6
1
Re
47
Im
Solution (b)
unit circle
Find a stable impulse response β π .
• For a stable LTI system, the ROC must include the
unit circle. Thus, the ROC is
x
o o
-1.4
-0.25
x
0.6
1
Re
§ π
ππΆ: 0.6 < z < 1.4
• The impulse response β π can be obtained by
inverse z-transform of H(z) with this ROC
1 + 0.25π§ %!
0.575
0.425
π» π§ =
=
+
%!
%!
%!
(1 + 1.4π§ )(1 − 0.6π§ ) 1 − −1.4 z
1 − 0.6z %!
βπ
L.M. Po
Left-sided sequence of
ROC |z|<1.4
= −0.575 −1.4 # π’ −π − 1
Right-sided sequence of
ROC |z|>0.6
+ 0.425 0.6 # π’[π]
48
Partial Fraction
1 + 0.25π§ "#
π΄
π΅
π» π§ =
=
+
(1 + 1.4π§ "# )(1 − 0.6π§ "# ) 1 + 1.4π§ "# 1 − 0.6z "#
1
"#
1 + 0.25
1
+
0.25π§
−1.4 = 0.5750
π΄ = 1 + 1.4π§ "# π» π§ b
=
b
=
"#
1
1 − 0.6π§
-+"#./
-+"#./
1 − 0.6
−1.4
1
"#
1 + 0.25
1
+
0.25π§
0.6 = 0.425
π΅ = 1 − 0.6π§ "# π» π§ b
=
b
=
"#
1
1 + 1.4π§
-+).0
-+).0
1 + 1.4
0.6
1 + 0.25π§ #$
0.575
0.425
π» π§ =
=
+
(1 + 1.4π§ #$ )(1 − 0.6π§ #$ ) 1 + 1.4π§ #$ 1 − 0.6z #$
Im
Solution (c)
Find a causal impulse response exist that is
both stable and causal?
• For a causal LTI system, the impulse response β π
is right-sided sequence ROC. Then, the ROC is
unit circle
x
-1.4
o o
-0.25
x
0.6
1
§ π
ππΆ: 1.4 < π§
• The impulse response β π can be obtained by
inverse z-transform of H(z) with this ROC
0.5750
0.425
π» π§ =
+
1 − −1.4 z %! 1 − 0.6z %!
π
ππΆ: 1.4 < z
β π = 0.5750 −1.4 # π’ π + 0.425 0.6 # π’[π]
L.M. Po
For this transfer function, we cannot achieve both stable and causal as the system is causal, the ROC
cannot include the unit circle.
50
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