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EC 2314 Digital Signal Processing
By
Dr. K. Udhayakumar
The z-Transform
Dr. K. Udhayakumar
Content
Introduction z-Transform Zeros and PolesRegion of Convergence Important z-Transform Pairs Inverse z-Transform z-Transform Theorems and Properties System Function
The z-Transform
Introduction
Why z-Transform?
A generalization of Fourier transformWhy 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
z-Transform
Definition The z-transform of sequence x(n) is defined by
n
nznxzX )()(
Let z = ej.
( ) ( )j j n
n
X e x n e
Fourier Transform
z-Plane
Re
Im
z = ej
n
nznxzX )()(
( ) ( )j j n
n
X e x n e
Fourier Transform is to evaluate z-transform on a unit circle.
Fourier Transform is to evaluate z-transform on a unit circle.
z-Plane
Re
Im
X(z)
Re
Im
z = ej
Periodic Property of FT
Re
Im
X(z)
X(ej)
Can you say why Fourier Transform is a periodic function with period 2?
Can you say why Fourier Transform is a periodic function with period 2?
The z-Transform
Zeros and Poles
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 znxznxzX |||)(|)(|)(|
ROC is centered on origin and consists of a set of rings.
ROC is centered on origin and consists of a set of rings.
Example: Region of Convergence
Re
Im
n
n
n
n znxznxzX |||)(|)(|)(|
ROC is an annual ring centered on the origin.
ROC is an annual ring centered on the origin.
xx RzR ||
r
}|{ xx
j RrRrezROC
Stable Systems
Re
Im
1
A stable system requires that its Fourier transform is uniformly convergent.
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
)()( nuanx n )()( nuanx 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
)()( nuanx n )()( nuanx n
n
n
n znuazX
)()(
0n
nn za
0
1)(n
naz
For convergence of X(z), we require that
0
1 ||n
az 1|| 1 az
|||| az
az
z
azazzX
n
n
10
1
1
1)()(
|||| az
aa
Example: A right sided Sequence ROC for x(n)=anu(n)
|||| ,)( azaz
zzX
|||| ,)( az
az
zzX
Re
Im
1aa
Re
Im
1
Which one is stable?Which one is stable?
Example: A left sided Sequence
)1()( nuanx n )1()( nuanx n
1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8n
x(n)
. . .
Example: A left sided Sequence
)1()( nuanx n )1()( nuanx n
n
n
n znuazX
)1()(
For convergence of X(z), we require that
0
1 ||n
za 1|| 1 za
|||| az
az
z
zazazX
n
n
10
1
1
11)(1)(
|||| az
n
n
n za
1
n
n
n za
1
n
n
n za
0
1
aa
Example: A left sided Sequence ROC for x(n)=anu( n1)
|||| ,)( azaz
zzX
|||| ,)( az
az
zzX
Re
Im
1aa
Re
Im
1
Which one is stable?Which one is stable?
The z-Transform
Region of Convergence
Represent z-transform as a Rational Function
)(
)()(
zQ
zPzX 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
)()( nuanx n |||| ,)( azaz
zzX
Re
Im
a
ROC is bounded by the pole and is the exterior of a circle.
Example: A left sided Sequence
)1()( nuanx n|||| ,)( az
az
zzX
Re
Im
a
ROC is bounded by the pole and is the interior of a circle.
Example: Sum of Two Right Sided Sequences
)()()()()( 31
21 nununx nn
31
21
)(
z
z
z
zzX
Re
Im
1/2
))((
)(2
31
21
121
zz
zz
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()()()()( 21
31 nununx nn
21
31
)(
z
z
z
zzX
Re
Im
1/2
))((
)(2
21
31
121
zz
zz
1/3
1/12
ROC is bounded by poles and is a ring.
ROC does not include any pole.
Example: A Finite Sequence
10 ,)( Nnanx n
nN
n
nN
n
n zazazX )()( 11
0
1
0
Re
Im
ROC: 0 < z <
ROC does not include any pole.
1
1
1
)(1
az
az N
az
az
z
NN
N
1
1
N-1 poles
N-1 zeros
Always StableAlways 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
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 15 200
0.2
0.4
0.6
0.8
1
Bounded Signals
-5
0
5a=0.4
-5
0
5a=0.9
-5
0
5a=1.2
0 5 10-5
0
5a=-0.4
0 5 10-5
0
5a=-0.9
0 5 10-5
0
5a=-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
The z-Transform
Important
z-Transform Pairs
Z-Transform Pairs
Sequence z-Transform ROC
)(n 1 All z
)( mn mz All z except 0 (if m>0)or (if m<0)
)(nu 11
1 z
1|| z
)1( nu 11
1 z
1|| z
)(nuan 11
1 az
|||| az
)1( nuan 11
1 az
|||| az
Z-Transform Pairs
Sequence z-Transform ROC
)(][cos 0 nun 210
10
]cos2[1
][cos1
zz
z1|| z
)(][sin 0 nun 210
10
]cos2[1
][sin
zz
z1|| z
)(]cos[ 0 nunr n 2210
10
]cos2[1
]cos[1
zrzr
zrrz ||
)(]sin[ 0 nunr n 2210
10
]cos2[1
]sin[
zrzr
zrrz ||
otherwise0
10 Nnan
11
1
az
za NN
0|| z
Signal Type ROCFinite-Duration Signals
Infinite-Duration Signals
Causal
Anticausal
Two-sided
Causal
Anticausal
Two-sided
Entire z-planeExcept z = 0
Entire z-planeExcept z = infinity
Entire z-planeExcept z = 0And z = infinity
|z| < r1
|z| > r2
r2 < |z| < r1
Some Common z-Transform Pairs
Sequence Sequence Transform Transform ROCROC
1. 1. [n] [n] 1 1 all zall z
2. u[n] 2. u[n] z/(z-1) z/(z-1) |z|>1|z|>1
3. -u[-n-1] 3. -u[-n-1] z/(z-1) z/(z-1) |z|<1|z|<1
4. 4. [n-m] [n-m] zz-m-m all z except 0 if m>0 or all z except 0 if m>0 or ฅฅif m<0if m<0
5. a5. annu[n] u[n] z/(z-a) z/(z-a) |z|>|a||z|>|a|
6. -a6. -annu[-n-1] u[-n-1] z/(z-a) z/(z-a) |z|<|a||z|<|a|
7. na7. nannu[n] u[n] az/(z-a)az/(z-a)22 |z|>|a||z|>|a|
8. -na8. -nannu[-n-1] u[-n-1] az/(z-a)az/(z-a)22 |z|<|a||z|<|a|
9. [cos9. [cos00n]u[n] n]u[n] (z(z22-[cos-[cos00]z)/(z]z)/(z22-[2cos-[2cos00]z+1) ]z+1) |z|>1|z|>1
10. [sin10. [sin00n]u[n] n]u[n] [sin[sin00]z)/(z]z)/(z22-[2cos-[2cos00]z+1) ]z+1) |z|>1|z|>1
11. [r11. [rnncoscos00n]u[n] n]u[n] (z(z22-[rcos-[rcos00]z)/(z]z)/(z22-[2rcos-[2rcos00]z+r]z+r22) ) |z|>r|z|>r
12. [r12. [rnnsinsin00n]u[n] n]u[n] [rsin[rsin00]z)/(z]z)/(z22-[2rcos-[2rcos00]z+r]z+r22) ) |z|>r|z|>r
13. a13. annu[n] - au[n] - annu[n-N] u[n-N] (z(zNN-a-aNN)/z)/zN-1N-1(z-a) (z-a) |z|>0|z|>0
The z-Transform
Inverse 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
kk
M
k
kk
za
zbzX
0
0
s
1mm1
i
mN
ik,1k1
k
kNM
0r
rr
zd1
Czd1
AzBzX
Partial Fractional Expression
Coefficients are given as
Easier to understand with examples
s
1mm1
i
mN
ik,1k1
k
kNM
0r
rr
zd1
Czd1
AzBzX
kdz
1kk zXzd1A
1idw
1sims
ms
msi
m wXwd1dwd
d!ms
1C
Example: 2nd Order Z-Transform
– Order of nominator is smaller than denominator (in terms of z-1)– No higher order pole
2
1z :ROC
2
11
4
11
1
11
zz
zX
1
2
1
1
z21
1
A
z41
1
AzX
1
41
21
1
1zXz
41
1A1
41
z
11
2
21
41
1
1zXz
21
1A1
21
z
12
Example Continued
ROC extends to infinity – Indicates right sided sequence
21
z z
21
1
2
z41
1
1zX
11
nu41
-nu21
2nxnn
Example #2
Long division to obtain Bo
1z z1z
21
1
z1
z21
z23
1
zz21zX
11
21
21
21
1z5
2z3z
21z2z1z
23
z21
1
12
1212
11
1
z1z21
1
z512zX
1
2
1
1
z1A
z21
1
A2zX
9zXz21
1A
21
z
11
8zXz1A1z
12
Example #2 Continued
ROC extends to infinity– Indicates right-sides sequence
1z z1
8
z21
1
92zX 1
1
n8u-nu21
9n2nxn
An Example – Complete Solution
386zz1414z3z
limU(z)limc 2
2
zz0
4-z1414z3z
86zz1414z3z
2)(z(z)U
2
2
2
2
2-z1414z3z
86zz1414z3z
4)(z(z)U
2
2
2
4
86zz1414z3z
U(z) 2
2
4z
c2z
ccU(z) 21
0
14-2
1421423(2)Uc
2
21
32-4
1441443(4)Uc
2
42
4z3
2z1
3U(z)
0k,432
0k3,u(k) 1k1k
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
nz nxzX
2112 2 1 0 1 2 zxzxxzxzxzX
12
1112
z21
1z21
z
z1z1z21
1z zX
1n21
n1n21
2nnx
2n0
1n21
0n1
1n21
2n1
nx
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?
xZ RROC zXnx
21 xx21
Z21 RRROC zbXzaXnbxnax
N-nua-nuanx nn
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
xnZ
o RROC zXznnx o
41
z z
41
1
1z zX
1
1
1-nu41
nx1-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
xooZn
o RzROCzzXnxz /
1z:ROC z-1
1nu 1-
Z
nure21
nure21
nuncosrnxnjnj
on oo
rz zre1
2/1zre1
2/1zX
1j1j oo
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 RROC dz
zdXznnx
az az1logzX 1
1
11
2
az11
azdz
zdXz
az1az
dzzdX
1nuaannx 1n
1nuna
1nxn
1n
Z-Transform Properties: Conjugation
Example
x**Z* RROC zXnx
nxZz nxz nxzX
z nxz nxzX
z nxzX
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
R1
ROC z/1Xnx
nuanx n
nuan
111-
1-1
az za-1
za-az1
1zX
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)
2x1x21
Z21 RR:ROC zXzXnxnx
nunx and nuanx 2n
1
az:ROC az11
zX 11
1z:ROC z1
1zX 12
1121 z1az11
zXzXzY
1z :ROC asume az11
z11
a11
zY 11
nuanua1
1ny 1n
The z-Transform
z-Transform Theorems and Properties
Linearity
xRzzXnx ),()]([Z
yRzzYny ),()]([Z
yx RRzzbYzaXnbynax ),()()]()([Z
Overlay of the above two
ROC’s
Shift
xRzzXnx ),()]([Z
xn RzzXznnx )()]([ 0
0Z
Multiplication by an Exponential Sequence
xx- RzRzXnx || ),()]([Z
xn RazzaXnxa || )()]([ 1Z
Differentiation of X(z)
xRzzXnx ),()]([Z
xRzdz
zdXznnx
)()]([Z
Conjugation
xRzzXnx ),()]([Z
xRzzXnx *)(*)](*[Z
Reversal
xRzzXnx ),()]([Z
xRzzXnx /1 )()]([ 1 Z
Real and Imaginary Parts
xRzzXnx ),()]([Z
xRzzXzXnxe *)](*)([)]([ 21R
xj RzzXzXnx *)](*)([)]([ 21Im
Initial Value Theorem
0for ,0)( nnx
)(lim)0( zXxz
Convolution of Sequences
xRzzXnx ),()]([Z
yRzzYny ),()]([Z
yx RRzzYzXnynx )()()](*)([Z
Convolution of Sequences
k
knykxnynx )()()(*)(
n
n
k
zknykxnynx )()()](*)([Z
k
n
n
zknykx )()(
k
n
n
k znyzkx )()(
)()( zYzX
The z-Transform
System Function
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(kb...1)u(kbn)y(ka...1)y(kay(k) m1n1
m
1jj
n
1ii
)p(z
)z(z
D(z)N(z)
Y(z)
zeros
poles
Why are poles important?
m
1j j
j0m
1jj
n
1ii
pz
cc
)p(z
)z(z
D(z)N(z)
Y(z)
m
1j
1-kjjimpulse0 pc(k)ucY(k)
Z-1
Z domain
Time domain
poles
components
Various pole values (1)
-1 0 1 2 3 4 5 6 7 8 90
0.5
1
1.5
2
2.5
-1 0 1 2 3 4 5 6 7 8 90
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 90
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 90
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 90
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 90
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|ln4
k
3.912ln0.02|a|kln
0.02|a| k
110.36
4|0.7|ln
4k
0.7a
Remember
This!
0 2 4 6 8 10 120
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)za...za1
zb...zbY(z)
nn
11
mm
11
c)...bz(az2
2a
4acbbz
2
0,4acb2 )
2a4acbb
)(z2a
4acbba(zcbzaz
222
0,4acb2 )2a
b4acib)(z
2ab4acib
a(zcbzaz22
2
If
If
Or in polar coordinates,
)irr)(zirra(zcbzaz2 θθθθ sincossincos
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
1j22
j
j0
l
1j j
j0
r)z(2rz)rdz(zbzr
pz
cc
irrzc'
irrzc
pz
ccY(z)
θ
θθ
θθθθ
cos
cossin
sincossincos
coskθdrsinkθbrpc(k)ucy(k) kkm
1j
1-kjjimpulse0
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
kπcos(0.8)
3kπ
sin(0.82y(k)
0.640.8zzzz
Y(z)
kk
2
2
Z-Domain: Complex Poles
Time-Domain:Exponentially Modulated Sin/Cos
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
Re1
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)
kdrkbr kk cossin
m
1j
1-kjjimpulse0 pc(k)ucy(k) 21
-1
)z(1z
An Example
kk 0.9)(30.52u(k)0.9z
3z0.5zz
1z2z
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
AAAAAAAAAAAAAAAAAAAAAAAAAAAA
2
1 1
0.11 0.11( )
1.6 0.6 ( 1)( 0.6)
0.11( 1) ( ) | | 0.275
0.6z z
z zY z
z z z z
zz 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 timeOvershoot
– …
Shift-Invariant System
h(n)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)H(z)X(z) Y(z)
)(
)()(
zX
zYzH
)(
)()(
zX
zYzH
Nth-Order Difference Equation
M
rr
N
kk rnxbknya
00
)()(
M
rr
N
kk rnxbknya
00
)()(
M
r
rr
N
k
kk zbzXzazY
00
)()(
N
k
kk
M
r
rr zazbzH
00)(
N
k
kk
M
r
rr zazbzH
00)(
Representation in Factored Form
N
kr
M
rr
zd
zcAzH
1
1
1
1
)1(
)1()(
N
kr
M
rr
zd
zcAzH
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
kr
M
rr
zd
zcAzH
1
1
1
1
)1(
)1()(
N
kr
M
rr
zd
zcAzH
1
1
1
1
)1(
)1()( Re
Im
Causal Systems : ROC extends outward from the outermost pole.
Stable and Causal Systems
N
kr
M
rr
zd
zcAzH
1
1
1
1
)1(
)1()(
N
kr
M
rr
zd
zcAzH
1
1
1
1
)1(
)1()( Re
ImStable Systems : ROC includes the unit circle.
1
Example
Consider the causal system characterized by
)()1()( nxnayny
11
1)(
azzH 11
1)(
azzH
Re
Im
1
a
)()( nuanh n
Determination of Frequency Response from pole-zero pattern
A LTI system is completely characterized by its pole-zero pattern.
))(()(
21
1
pzpz
zzzH
))(()(
21
1
pzpz
zzzH
Example:
))(()(
21
1
00
0
0
pepe
zeeH jj
jj
))(()(
21
1
00
0
0
pepe
zeeH jj
jj
0je
Re
Im
z1
p1
p2
Determination of Frequency Response from pole-zero pattern
A LTI system is completely characterized by its pole-zero pattern.
))(()(
21
1
pzpz
zzzH
))(()(
21
1
pzpz
zzzH
Example:
))(()(
21
1
00
0
0
pepe
zeeH jj
jj
))(()(
21
1
00
0
0
pepe
zeeH jj
jj
0je
Re
Im
z1
p1
p2
|H(ej)|=?|H(ej)|=? H(ej)=?H(ej)=?
Determination of Frequency Response from pole-zero pattern
A LTI system is completely characterized by its pole-zero pattern.
Example:
0je
Re
Im
z1
p1
p2
|H(ej)|=?|H(ej)|=? H(ej)=?H(ej)=?
|H(ej)| =| |
| | | | 1
2
3
H(ej) = 1(2+ 3 )
Example
11
1)(
azzH 11
1)(
azzH
Re
Im
a
0 2 4 6 8-10
0
10
20
0 2 4 6 8-2
-1
0
1
2d
B