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1
z-Transform
清大電機系林嘉文cwlin@ee.nthu.edu.tw03-5731152
Chapter 3
The Direct z-Transform� A discrete-time version of Laplace Transform� For a given sequence x(n), its (direct) z-transform X(z) is
defined as:
where z = Re(z) + j Im(z) is a complex variable� For convenience, the z-transform of a signal x(n0 is
denoted by
� Whereas the relationship between x(n) and X(z) is indicated by
� The region of convergence (ROC) of X(z) is the set of all values of z of which X(z) attains a finite value
( ) ( ) n
n
X z x n z∞
−
=−∞
≡ ∑
{ }( ) ( )X z Z x n≡
2010/4/7 2Introduction to Digital Signal Processing
( ) ( )zx n X z←→
2
ROC of the z-Transform� Express the complex variable z in polar form as
Where , then X(z) can be expressed as
� In the ROC of X(z), |X(z)| < ∞. But
� Hence |X(z)| is finite if x(n)r−n is absolutely summable� The problem of finding the ROC for X(z) is equivalent to
determining the range of r for which x(n)r−n is absolutely summable
jz re θ=
( ) ( )jn j n
z ren
X z x n r eθθ
∞− −
==−∞
= ∑
2010/4/7 3Introduction to Digital Signal Processing
and r z zθ= =�
( ) ( )
( ) ( )
n j n
n
n j n n
n n
X z x n r e
x n r e x n r
θ
θ
∞− −
=−∞
∞ ∞− − −
=−∞ =−∞
=
≤ =
∑
∑ ∑
ROC of the z-Transform
2010/4/7 4Introduction to Digital Signal Processing
1
0
1 0
( ) ( ) ( )
( )( )
n n
n n
nn
n n
X z x n r x n r
x nx n r
r
− ∞− −
=−∞ =
∞ ∞
= =
≤ +
≤ − +
∑ ∑
∑ ∑
3
The Direct z-Transform
� Example – Determine the z-Transform X(z) of the causal sequence x[n] = αn
u(n) and its ROC
� The above power series converges to
� ROC is the annular region |z| > |α|
0
( ) ( )n n n n
n n
X z u n z zα α∞ ∞
− −
=−∞ =
= =∑ ∑
1for ,1
1)( 1
1<
−= −
− zz
zX αα
2010/4/7 5Introduction to Digital Signal Processing
The Direct z-Transform
� Example – Consider the anti-causal sequence
� Its z-transform is given by
� ROC is the annular region |z| < |α|
1
1
11
10
11
( )
1
1 , for 1
1
n n l l
n l
l l
l
X z z z
zz z
z
zz
α α
αα α
α
αα
− ∞− −
=−∞ =
−∞− −
−=
−−
= − = −
= − = −−
= <−
∑ ∑
∑
2010/4/7 6Introduction to Digital Signal Processing
0, 0( ) ( 1)
, 1n
n
nx n u n
nα
α≥
= − − − = − ≤ −
4
z-Transform
� Note: the z-Transforms of two sequences αnu(n) and −αnu(−n −1) are identical even though the two parent sequences are different
� Only way a unique sequence can be associated with a z-transform is by specifying its ROC
2010/4/7 7Introduction to Digital Signal Processing
z-Transform
� Example – the finite energy sequence
has a DTFT given by
which converges in the mean-square sense� However, hLP[n] does not have a z-transform as it is not
absolutely summable for any value of r
sin( ) ,c
LP
nh n n
n
ωπ
= −∞ < < ∞
( )
≤<
≤≤=
πωωωωω
c
cjLP eH
,0
0,1
2010/4/7 8Introduction to Digital Signal Processing
5
Characteristic Families of Signals with Their Corresponding ROCs
2010/4/7 9Introduction to Digital Signal Processing
The Inverse z-Transform
2010/4/7 10Introduction to Digital Signal Processing
� The z-transform X(z) is given by
� Multiply both side by zn−1 and integrate both sides over a closed contour within the ROC of X(z)
� Interchange the order of integration and summation:
� According to the Cauchy integral theorem
( ) ( ) k
k
X z x k z∞
−
=−∞
= ∑
1 1( ) ( )n n k
C Ck
X z z dz x k z dz∞
− − −
=−∞
= ∑∫ ∫� �
1 1( ) ( )n n k
C Ck
X z z dz x k z dz∞
− − −
=−∞
= ∑∫ ∫� �
1 1,1
0,2n k
C
k nz dz
k njπ− − =
= ≠
∫�11
( ) ( )2
n
Cx n X z z dz
jπ−= ∫�
6
The Inverse z-Transform (from IDTFT)
2010/4/7 11Introduction to Digital Signal Processing
� General Expression : Recall that, for z = rejω, the z-transform X(z) given by
is merely the DTFT of the modified sequence x(n)r−n
� Accordingly, the inverse DTFT is thus given by
� By making a change of variable z = rejω, the previous equation can be converted into a contour integral given by
where C′ is a counterclockwise contour of integration defined by |z| = r
( ) ( ) ( )n n j n
n nX z x n z x n r e ω∞ ∞− − −
=−∞ =−∞= =∑ ∑
( )1( )
2n j j nx n r X re e d
π ω ω
πω
π−
−= ∫
11( ) ( )
2n
Cx n X z z dz
jπ−= ∫�
Properties of the z-Transform
� Linearity - If
and
then
� Example – Determine the z-Transform and the ROC of
Ans:
2010/4/7 12Introduction to Digital Signal Processing
1 1( ) ( )zx n X z←→
2 2( ) ( )zx n X z←→
1 1 2 2 1 1 2 2( ) ( ) ( ) ( ) ( ) ( )zx n a x n a x n X z a X z a X z= + ←→ = +
( ) ( )( ) 3 2 4 3 ( )n nx n u n = −
1 1 1
1( ) 2 ( ) ( ) , ROC: 2
1 2znx n u n X z z
z−= ←→ = >
−
2 2 1
1( ) 3 ( ) ( ) , ROC: 3
1 3znx n u n X z z
z−= ←→ = >
−
1 1
3 4( ) , ROC: 3
1 2 1 3X z z
z z− −= − >
− −
7
Properties of the z-Transform
� Example - Consider the two-sided sequence x[n] = αnu[n] − bnu[−n −1]
� Let x[n] = αnu[n] and y[n] = − bnu[−n −1] with X1(z) and X2(z) denoting, respectively, their z-transforms
� Now
and
� Using the linearity property we arrive at
� The ROC of X(z) is given by the overlap regions of |z| > |α| and |z| < |b|
1 1
1( ) ,
1X z z
zα
α −= >
−
2 1
1( ) ,
1X z z b
bz−= <
−
1 2 1 1
1 1( ) ( ) ( )
1 1X z X z X z
z bzα − −= + = +− −
2010/4/7 13Introduction to Digital Signal Processing
Properties of the z-Transform
2010/4/7 14Introduction to Digital Signal Processing
8
Properties of the z-Transform
� Example - Determine the z-transform and its ROC of the causal sequence
� We can express x[n] = v[n] + v*[n] where
� The z-transform of v[n] is given by
� Using the conjugation property we obtain the z-transform of v*[n] as
01 1
[ ] [ ] [ ]2 2
j n nv n e u n u nω α= =
01 1
1 1 1 1( ) , 1
2 1 2 1 jV z zz e zω α
α − −= ⋅ = ⋅ > =− −
01 1
1 1 1 1*( *) , 1
2 1 * 2 1 jV z z
z e zω αα −− −= ⋅ = ⋅ > =
− −
2010/4/7 15Introduction to Digital Signal Processing
( )0[ ] cos ( )x n n u nω=
Properties of the z-Transform
� Using the linearity property we get
or,
� Therefore, we obtain
Similarly
0 01 1
1 1 1( ) ( ) *( *)
2 1 1j jX z V z V z
e z e zω ω−− −
= + = + − −
( )( )
10
1 20
1 cos( ) , 1
1 2cos
zX z z
z z
ωω
−
− −
−= >
− +
2010/4/7 16Introduction to Digital Signal Processing
( ) ( )( )
10
0 1 20
1 cos( ) cos ( ) ( ) , ROC: 1
1 2 cosz z
x n n u n X z zz z
ωω
ω
−
− −
−= ←→ = >
− +
( ) ( )( )
10
0 1 20
sin( ) sin ( ) ( ) , ROC: 1
1 2 cosz z
x n n u n X z zz z
ωω
ω
−
− −= ←→ = >− +
9
Properties of the z-Transform
2010/4/7 17Introduction to Digital Signal Processing
� Time Shifting - If
then
And the ROC remains unchanged except for z = 0 if k > 0 and z = ∞ if k < 0
� Example – Determine the z-Transform of the signal
Ans:
( ) ( )zx n X z←→
( ) ( )z kx n k z X z−− ←→
1, 0 1( )
0, elsewhere
n Nx n
≤ ≤ −=
( ) ( ) ( )x n u n u n N= − −
{ } { } ( ) 1
1( ) ( ) ( ) 1 , ROC: 1
1NX z Z u n Z u n N z z
z−
−= − − = − >−
Properties of the z-Transform
2010/4/7 18Introduction to Digital Signal Processing
� Scaling in the z-domain -If
then
Fro any constant a
� Proof
1 2( ) ( ), ROC: zx n X z r z r←→ < <
{ }
( )
( )
1
1
( ) ( )
( )
n n n
n
n
n
Z a x n a x n z
x n a z
X a z
∞−
=−∞
∞ −−
=−∞
−
=
=
=
∑
∑
11 2( ) ( ), ROC: zna x n X a z a r z a r−←→ < <
11 2ROC: r a z r−< < 1 2ROC: a r z a r< <or
10
Properties of the z-Transform
2010/4/7 19Introduction to Digital Signal Processing
� Example - Determine the z-transform and its ROC of the causal sequence
and
� Ans:
By applying the scaling property
and similarly
( ) ( )( )
10
0 1 2 20
1 cos( ) cos ( ) ( ) , ROC:
1 2 coszn r z
x n r n u n X z z rr z r z
ωω
ω
−
− −
−= ←→ = >
− +
( )0( ) cos ( )nx n r n u nω=
( )0( ) sin ( )nx n r n u nω=
( ) ( )( )
10
0 1 2 20
sin( ) sin ( ) ( ) , ROC:
1 2 coszn r z
x n r n u n X z z rr z r z
ωω
ω
−
− −= ←→ = >− +
Properties of the z-Transform
2010/4/7 20Introduction to Digital Signal Processing
� Time Reversal -If
then
� Proof
� Example -
1 2( ) ( ), ROC: zx n X z r z r←→ < <
{ } ( ) ( )1 1( ) ( ) ( )ln
n l
Z x n x n z x l z X z∞ ∞ −− − −
=−∞ =−∞
− = − = =∑ ∑
1
2 1
1 1( ) ( ), ROC: zx n X z z
r r−− ←→ < <
11 2ROC: r z r−< <
2 1
1 1ROC: z
r r< <or
( ) ( )x n u n= −
1
1( ) , ROC: 1
1zu n z
z−←→ >
−1
( ) , ROC: 11
zu n zz
− ←→ <−
11
Properties of the z-Transform
2010/4/7 21Introduction to Digital Signal Processing
� Differentiation in the z-Domain -If
then
ROC remains unchanged
� Example – Find the z-Transform of
� Example – Find the inverse z-transform of
( ) ( )zx n X z←→
( )( ) z dX z
nx n zdz
←→−
( ) ( )nx n na u n=
( )1( ) log 1 , ROC :X z az z a−= + >
Properties of the z-Transform
2010/4/7 22Introduction to Digital Signal Processing
� Convolution of Two Sequences -If
then
The ROC is the intersection of that for X1(z) and X2(z)
� Proof
1 1( ) ( )zx n X z←→
1 2 1 2( ) ( ) ( ) ( ) ( ) ( )zx n x n x n X z X z X z= ∗ ←→ =
1 2( ) ( ) ( )k
x n x k x n k∞
=−∞
= −∑
1 2( ) ( ) ( ) ( )n n
n n k
X z x n z x k x n k z∞ ∞ ∞
− −
=−∞ =−∞ =−∞
= = −
∑ ∑ ∑
2 2( ) ( )zx n X z←→and
Intercahnge the order of the summations and apply the time shifting property
1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( ) ( )n k
k n k
X z x k x n k z x k z X z X z X z∞ ∞ ∞
− −
=−∞ =−∞ =−∞
= − = =
∑ ∑ ∑
12
Properties of the z-Transform
2010/4/7 23Introduction to Digital Signal Processing
� Correlation of Two Sequences -If
then
The ROC is at least the intersection of that for X1(z) andX2(z-1)
� Proof
� Example -
1 1( ) ( )zx n X z←→
1 2 1 2
11 2 1 2( ) ( ) ( ) ( ) ( ) ( )z
x x x xn
r l x n x n l R z X z X z∞
−
=−∞
= − ←→ =∑
1 2 1 2( ) ( ) ( )x xr l x l x l= ∗ −
{ } { } 11 2 1 2( ) ( ) ( ) ( ) ( )X z Z x l Z x l X z X z−= − =
2 2( ) ( )zx n X z←→and
( ) ( )nx n a u n=
( )1 2 1 1 2
1 1 1 1( ) , ROC:
1 1 1x xR z a z
az az aa z z a− −= = < <
− − − + +
Properties of the z-Transform
2010/4/7 24Introduction to Digital Signal Processing
� Multiplication of Two Sequences -If
then
where C is the closed contour that encloses the origina and lies withing the ROC common to both X1(v) and X2(v-1)
� Proof
1 1( ) ( )zx n X z←→
11 2 1 2
1( ) ( ) ( ) ( ) ( )
2z
C
zx n x n x n X z X v X v dv
j vπ− = ←→ =
∫�
1 2( ) ( ) ( ) n
n
X z x n x n z∞
−
=−∞
= ∑
11 1
1( ) ( )
2n
Cx n X v v dv
jπ−= ∫�
2 2( ) ( )zx n X z←→and
11 2
1( ) ( ) ( )
2
n
Cn
zX z X v x n v dv
j vπ
−∞−
=−∞
=
∑∫�
Since
2
zX
v
13
Properties of the z-Transform
2010/4/7 25Introduction to Digital Signal Processing
� Parseval’s relation -If
then
� The Initial Value Theorem -
� If x(n) is causal, then
� Proof -
1 1( ) ( )zx n X z←→
* * 11 2 1 2 *
1 1( ) ( ) ( )
2 Cn
x n x n X v X v dvj vπ
∞−
=−∞
=
∑ ∫�
2 2( ) ( )zx n X z←→and
(0) lim ( )z
x X z→∞
=
1 2
0
( ) ( ) (0) (1) (2)n
n
x z x n z x x z x z∞
− − −
=
= = + + +∑ �
Properties of the z-Transform
2010/4/7 26Introduction to Digital Signal Processing
14
Common z-Transform Paris
2010/4/7 27Introduction to Digital Signal Processing
Rational z-Transforms
� In the case of LTI discrete-time systems we are concerned with in this course, all pertinent z-transforms are rational functions of z−1
� That is, they are ratios of two polynomials in z−1
� The degree of the numerator polynomial B(z) is M and the degree of the denominator polynomial A(z) is N
� An alternate representation of a rational z-transform is as a ratio of two polynomials in z:
10 1 0
10 1
0
( )( )
( )
M kMkM k
NN kN kk
b zb b z b zB zX z
A z a a z a z a z
−− −=
− − −
=
+ + += = =
+ + +∑∑
�
�
10 1 0 0
10 1 0 0
( / ) /( )
( / ) /
M M MM
N N NN
b z z b b z b bX z
a z z a a z a a
− −
− −
+ + +=
+ + +
�
�
2010/4/7 28Introduction to Digital Signal Processing
15
Rational z-Transform
� A rational z-transform can be alternately written in factored form as
� At a root z = zk of the numerator polynomial X(zk) = 0. These values of z are known as the zeros of G(z)
� At a root z = pk of the denominator polynomial X(pk) → ∞. These values of z are known as the poles of G(z)
� There are |N – M| zeros (if N > M) or poles (if N < M) at z = 0
( )( ) ( )( )( ) ( )
( )
( )
1 20
0 1 2
1
1
( ) MM N
N
M
kM N k
N
kk
z z z z z zbX z z
a z p z p z p
z zGz
z p
− +
− + =
=
− − −=
− − −
−=
−
∏
∏
�
�
2010/4/7 29Introduction to Digital Signal Processing
0
0
bG
a=
Rational z-Transform
� Example – the z-transform
has a zero at z1 = 0 and a pole at p1 = a
( ) 1
1, ROC:
1
zX z z a
az z a−= = >
− −
2010/4/7 30Introduction to Digital Signal Processing
Pole-zero plot
16
Rational z-Transform
� Example – Determine the pole-zero plot for the signal
Since a > 0 zM = aM has M roots at
The zero z0 = a cancels the pole at z = a. Thus
( ) ( ) ( )( )
111
1 10
1
1
MM MM n
Mn
az z aX z az
az z z a
−−−
− −=
− −= = =
− −∑
2010/4/7 31Introduction to Digital Signal Processing
Pole-zero plot
, 0 1( )
0, elsewhere
na n Mx n
≤ ≤ −=
0a >
2 / 0,1, , 1j k Mkz ae k Mπ= = −…
( ) ( )( ) ( )1 2 11
MM
z z z z z zX z
z−
−
− − −=
�
Rational z-Transform
� Example – Determine the z-transform and the signal corresponds to the pole-zero plot
� There are 2 zeros (M = 2) at z1 =0 and z2 = rcosω0 and 2 poles (N = 2) at and
2010/4/7 32Introduction to Digital Signal Processing
Pole-zero plot
( )1
01 2 2
0
1 cos, ROC:
1 2 cos
rzX z G z r
rz r z
ωω
−
− −
−= >
− +
( )0( ) cos ( )nx n G r n u nω=
( ) ( )( )( )( )
( )( )( )0 0
1 2
1 2
0cos, ROC:
j j
z z z zX z G
z p z p
z zG z r
z re z reω ω
ω−
− −=
− −
−= >
− −
01
jp re ω= 02
jp re ω−=
17
Rational z-Transform
� A physical interpretation of the concepts of poles and zeros can be given by plotting the log-magnitude 20log10|X(z)| for
• The magnitude plot exhibits very large peaks around the poles of X(z) (z = 0.4 ± j 0.6928)
• It also exhibits very narrow and deep wells around the location of the zeros (z = 1.2 ± j 1.2)
1 2
1 2
1 2.4 2.88( )
1 0.8 0.64
z zX Z
z z
− −
− −
− +=
− +
2010/4/7 33Introduction to Digital Signal Processing
Behavior of a Single Real-Pole Causal Signal
2010/4/7 34Introduction to Digital Signal Processing
1
1( ) ( ) ( ) , ROC:
1znx n a u n X z z a
az−= ←→ = >
−
18
Behavior of a Double Real-Pole Causal Signal
2010/4/7 35Introduction to Digital Signal Processing
( )21
1( ) ( ) ( ) , ROC:
1
znx n na u n X z z aaz−
= ←→ = >−
Behavior of a Causal Signal with a Pair of Complex-Conjugate Poles
2010/4/7 36Introduction to Digital Signal Processing
( )0( ) cos ( )nx n r n u nω=
19
Behavior of a Causal Signal with a Double Pair of Complex-Conjugate Poles
2010/4/7 37Introduction to Digital Signal Processing
( )0( ) cos ( )x n n n u nω=
The System Function of an LTI System
� According to the convolution property, the I/O-relationship can be expressed as
� The system function (impulse/unit-sample response) H(z) is
� If an LTI system is described by the following constant-coefficient difference equation
� Taking the z-Transform at both sides
2010/4/7 38Introduction to Digital Signal Processing
( ) ( ) ( )Y z H z X z=
( )( )
( )
Y zH z
X z=
( ) ( ) ( )1 0
N M
k kk k
y n a y n k b x n k= =
= − − + −∑ ∑
1 0
( ) ( ) ( )N M
k kk k
k k
Y z a Y z z b X z z− −
= =
= − +∑ ∑
20
The System Function of an LTI System
� Therefore an LTI system described by a constant-coefficient difference equation has a rational system function
� Two important special forms of pole-zero systems:� If ak = 0 for 1 ≤ k ≤ N
� If bk = 0 for 1 ≤ k ≤ M
2010/4/7 39Introduction to Digital Signal Processing
1 0
( ) 1 ( )N M
k kk k
k k
Y z a z X z b z− −
= =
+ =
∑ ∑
0
1
( )( )
( ) 1
Mk
kk
Nk
kk
b zY z
H zX z
a z
−
=
−
=
= =+
∑
∑
0 0
1( )
M Mk M k
k kMk k
H z b z b zz
− −
= =
= =∑ ∑
0
1 0
1( )
1
N
N Nk N k
k kk k
b zH z
a z a z− −
= =
= =+∑ ∑
(all-zero system)
(all-pole system)
(pole-zero systems)
2010/4/7 40Introduction to Digital Signal Processing
The System Function of an LTI System
� Example – A causal LTI IIR filter is described by a constant coefficient difference equation given by
� y(n) = x(n −1) −1.2 x(n − 2) + x(n − 3) +1.3 y(n −1) −1.04 y(n − 2) + 0.222 y(n − 3)
� Its transfer function is therefore given by
� Alternate forms:
� Note: Poles farthest from z = 0 have a magnitude� ROC:
321
321
222.004.13.11
2.1)( −−−
−−−
−+−+−
=zzz
zzzzH
( )( )( )( )( )7.05.07.05.03.0
8.06.08.06.0222.004.13.1
12.1)(
23
2
jzjzz
jzjzzzz
zzzH
−−+−−−−+−
=
−+−+−
=
0.74
0.74z >
21
MATLAB Functions for Rational z-Transform
� In general, if the rational z-transform has N poles with R distinct magnitudes, then it has R + 1 ROCs, and R + 1 distinct sequences having the same z-transform
� Hence, a rational z-transform with a specified ROC has a unique sequence as its inverse z-transform
� MATLAB [z,p,k] = tf2zp(num,den) determines the zeros, poles, and the gain constant of a rational z-transform with the numerator coefficients specified by num and the denominator coefficients specified by den
� [num,den] = zp2tf(z,p,k) implements the reverse process� The factored form of the z-transform can be obtained using
sos = zp2sos(z,p,k)
2010/4/7 41Introduction to Digital Signal Processing
MATLAB Functions for Rational z-Transform
� The pole-zero plot is determined using the function zplane� The z-transform can be either described in terms of its
zeros and poles: zplane(zeros,poles) or, in terms of its numerator and denominator coefficients zplane(num,den)
� Example – The pole-zero plot of
obtained using MATLAB
4 3 2
4 3 2
2 16 44 56 32( )
3 3 15 18 12
z z z zX z
z z z z
+ + + +=
+ − + −
2010/4/7 42Introduction to Digital Signal Processing
22
Inversion of the z-Transform
2010/4/7 43Introduction to Digital Signal Processing
� The inverse z-transform of X(z) is given by
� There are three methods for the evaluation of the inverse z-transform in practice:1. Direct evaluation by the contour integration2. Power series expansion3. Partial-fraction expansion and table lookup
11( ) ( )
2n
Cx n X z z dz
jπ−= ∫�
Inverse z-Transform by Contour Integration
2010/4/7 44Introduction to Digital Signal Processing
� Cauchy’s integral theorem . Let f(z) be a function of the complex variable z and C be a closed path in the z-plane. If the derivative df(z)/dz exists on and inside the contour C and if f(z) has no poles at z = z0, then
� More generally, if the (k+1)-order derivative of f(z) exist and f(z) has no poles at z = z0, then
0 0
00
( ), if is inside 1 ( )0, if is outside 2 C
f z z Cf zdz
z Cj z zπ
= −
∫�
( )( )
0
1
01
0
0
1 ( ), if is inside 1 ( ) 1 !
20, if is outside
k
k
k z zC
d f zz Cf z k dzdz
j z zz C
π
−
−
=
−= −
∫�
23
Inverse z-Transform by Contour Integration
2010/4/7 45Introduction to Digital Signal Processing
� Suppose the integrand of the contour integral is a proper fraction f(z)/g(z), where f(z) has no poles inside the contour C and g(z) is a polynomial with distinct roots inside C. Then
� In the case of the inverse z-transform, we have
1
1 1
1 ( ) 1
2 ( ) 2
1
2
ni
C Ci i
n ni
iCi ii
Af zdz dz
j g z j z z
Adz A
j z z
π π
π
=
= =
= −
= =−
∑∫ ∫
∑ ∑∫
� �
� ( ) ( )
( ) ii i z z
f zA z z
g z == −
{ }
( )
1
1
all poles inside
1
1( ) ( )
2
residue of ( ) at
( )
i
i
n
C
ni
z C
ni z z
i
x n X z z dzj
X z z z z
z z X z z
π−
−
−
=
=
= =
= −
∫
∑
∑
�
Inverse z-Transform by Contour Integration
2010/4/7 46Introduction to Digital Signal Processing
� Example – Evaluate the inverse z-transform of
Using the complex inversion integral� Sol:
(1) If n ≥ 0, f(z) has only zeros and hence no poles inside C. The only pole inside C is z = a. Hence
(2) If n < 0, f(z) has an nth-order poles at z = 0 which is also inside C. If n = 1, we have
If n = 2, we have
x(n) = 0 for n < 0
1
1( ) ,
1X z z a
az−= >
−
0( ) ( ) , 0nx n f z a n= = ≥
1
1
1 1( )
2 1 2
n n
C C
z zx n dz dz
j az j z aπ π
−
−= =
− −∫ ∫� �
0
1 1 1 1( 1) 0
2 ( )Cz z a
x dzj z z a z a zπ = =
− = = + =− −∫�
2 20
1 1 1 1( 2) 0
2 ( )Cz az
dx dz
j z z a dz z a zπ ==
− = = + = − − ∫�
( ) ( )nx n a u n=
24
Inverse z-Transform by Power Series Expansion
2010/4/7 47Introduction to Digital Signal Processing
� Expand X(z) into a power series as
which converges in the given ROC. Then, by the uniqueness of the z-transform, x(n) = cn for all n.
� Example : Determine the inverse z-transform of
If (a) ROC: |z| > 1, (b) ROC: |z| < 0.5Sol: (a)
( ) nn
n
X z c z∞
−
=−∞
= ∑
1 2
1( )
1 1.5 0.5X z
z z− −=
− +
1 2 3 4
1 2
1 3 7 15 31( ) 1
3 1 2 4 8 1612 2
X z z z z zz z
− − − −
− −= = + + + + +
− +�
3 7 15 31( ) 1, , , , ,
2 4 8 16X n
=
…
Inverse z-Transform by Partial-Fraction Expansion
� A rational X(z) can be expressed as
� X(z) is called improper if M ≥ N. It can be re-expressed as
where the degree of B1(z) is less than N� The rational function B1(z)/A(z) is called a proper fraction� Example – Consider
By long division we arrive at
10 0 1
110
( )( )
( ) 1
M i Mii M
N NiNii
b z b b z b zB zX z
B z a z a za z
− − −=
− −−
=
+ + += = =
+ + +∑∑
�
�
11 10 1
0
( ) ( )( )
( ) ( )
M NM N
M N
B z B zX z c z c c z c z
A z A z
−− − −
−=
= + = + + + +∑ �
�
�
�
1 2 3
1 2
2 0.8 0.5 0.3( )
1 0.8 0.2
z z zX z
z z
− − −
− −
+ + +=
+ +
11
1 2
5.5 2.1( ) 3.5 1.5
1 0.8 0.2
zX z z
z z
−−
− −
+= − + +
+ +2010/4/7 48Introduction to Digital Signal Processing
25
Inverse z-Transform by Partial-Fraction Expansion
� Simple Poles : In most practical cases, the rational z-transform of interest X(z) is a proper fraction with simple poles
� Let the poles of X(z) be at z = pk 1 ≤ k ≤ N� A partial-fraction expansion of X(z) is then of the form
� The constants in the partial-fraction expansion are called the residues and are given by
1 21 1 1 1
1 1 2
( )1 1 1 1
NN
N
AA A AX z
p z p z p z p z− − − −=
= = + + − − − − ∑ �
� �
�
( )11 ( ) z pA p z X z−== −�� �
2010/4/7 49Introduction to Digital Signal Processing
Inverse z-Transform by Partial-Fraction Expansion
� x(n) = Z-1{X(z)} can be obtained by inverting each term as:
� If x(n) is causal, the ROC is |z| > pmax where pmax = max{|p1|, |p2|, …, |pN|}, and x(n) becomes
� The above approach with a slight modification can also be used to determine the inverse of a rational z-transform of a non-causal sequence
( )
( )
11
if ROC: ( ),
(causal sequence)1
1 if ROC: ( 1)
(anticausal sequence)
n kk
nk kk
z pp u n
Zp z z p
p u n
−−
>
=
− < − − −
( )1 1 2 2( ) ( )n n nN Nx n A p A p A p u n= + + +�
2010/4/7 50Introduction to Digital Signal Processing
26
Inverse z-Transform by Partial-Fraction Expansion
� Example – Determine the inverse z-transform of
If (a) ROC: |z| > 1, (b) ROC: |z| < 0.5, and (c) 0.5 < |z| < 1� Ans:
(a) ROC: |z| > 1
(b) ROC: |z| < 0.5
(b) ROC: 0.5 < |z| < 1
( )( )1 2 1 1
1 1( )
1 1.5 0.5 1 1 0.5X z
z z z z− − − −= =
− + − −
1 21 1
( )1 1 0.5
A AX z
z z− −= +
− −
2010/4/7 51Introduction to Digital Signal Processing
( ) ( )1 11 1 2 0.51 ( ) 2, 1 0.5 ( ) 1z zA z H z A z H z− −
= == − = = − = −
( )( ) 2 0.5 ( )nx n u n= −
( )( ) 2 0.5 ( 1)nx n u n= − + − −
( )( ) 2 ( 1) 0.5 ( )n
x n u n u n= − − − −
Inverse z-Transform by Partial-Fraction Expansion
� Multiple Poles : If X(z) has multiple poles, the partial-fraction expansion is of slightly different form
� Let the pole at z = ν be of multiplicity L and the remaining N − L poles be simple and at z = λ, 1 ≤ l ≤ N − L
� Then the partial-fraction expansion of G(z) is of the form
where the constants are computed using
� The residues ρl are calculated as before
( )∑ ∑∑−
= =−−
−
=
−
−+
−+=
LN L
ii
iNM
vzzzzG
1 111
0 11)(
� �
�
�
�
�
γλρ
η
( ) ( )( ) ( )[ ] LizGvz
zd
d
viLvz
L
iL
iL
iLi ≤≤−−−
= =−
−−
−
− 1,1)(!
1 11
γ
2010/4/7 52Introduction to Digital Signal Processing
27
2010/4/7 53Introduction to Digital Signal Processing
Decomposition of Rational z-Transforms
� Suppose we have a rational z-transform X(z) expressed as
� If M ≥ N (i.e., X(z) is improper), we convert X(z) to a sum of a polynomial and a proper function
� If the roots are distinct, it can be expanded in partial fractions as
( )
( )
1
0 10
11
1
1( )
1 1
M
M k kkk k
N Nkkk k
k
z zb z
X z ba z p z
−−
= =
−−
=
=
−= =
+ −
∏∑∑ ∏
( ) ( ) ( )1 21 1 11 2
1 1 1( )
1 1 1N
N
X z A A Ap z p z p z− − −
= + + +− − −
�
pr0
( ) ( )M N
kk
k
X z c z X z−
−
=
= +∑
2010/4/7 54Introduction to Digital Signal Processing
Decomposition of Rational z-Transforms
� If there some complex conjugate pairs of poles, we can group and combine the pairs as
� The partial fraction expansion becomes
where N = K1 + 2K2
( ) ( )( )
1 10 1
1 1 1 21 21 21 1 11 ( )
A A Ap A p z b b zA A
pz p z a z a zp p z pp z
∗ ∗ ∗ − −∗
− ∗ − − −∗ − ∗ −
+ + − − ++ = =
− − + ++ − − +
1 2 10 1
1 1 20 1 1 1 2
( )1 1
K KM Nk k k k
kk k kk k k
b b b zX z c z
a z a z a z
−−−
− − −= = =
+= + +
+ + +∑ ∑ ∑
28
2010/4/7 55Introduction to Digital Signal Processing
Decomposition of Rational z-Transforms
� An alternative form is obtained by expressing X(z) as a product of simple terms. To avoid complex coefficients in the decomposition, the complex-conjugate poles and zeros should be combined as
� Assuming for simplicity that M = N, X(z) can be expressed as
where N = K1 + 2K2
1 21 1 21 2
0 1 1 21 1 1 2
1 1( )
1 1
K Kk k k
k kk k k
b z b z b zX z b
a z a z a z
− − −
− − −= =
+ + += ⋅ ⋅
+ + +∏ ∏
( )( )( )( )
1 1 1 21 2
1 21 11 2
1 1 1
11 1
k k k k
k kk k
z z z z b z b z
a z a zp z p z
− ∗ − − −
− −− ∗ −
− − + +=
+ +− −
2010/4/7 56Introduction to Digital Signal Processing
LTI Discrete-Time Systems in the Transform Domain
� An LTI discrete-time system is completely characterized in the time-domain by its impulse response sequence {h(n)}
� Thus, the transform-domain representation of a discrete-time signal can also be equally applied to the transform-domain representation of an LTI discrete-time system
� Besides providing additional insight into the behavior of LTI systems, it is easier to design and implement these systems in the transform-domain for certain applications
� We consider now the use of the DTFT and the z-transform in developing the transform-domain representations of an LTI system
29
2010/4/7 57Introduction to Digital Signal Processing
LTI Discrete-Time Systems in the Transform Domain
� Consider LTI discrete-time systems characterized by linear constant coefficient difference equations of the form
� Applying the z-transform to both sides of the difference equation and making use of the linearity and the time-invariance properties we arrive at
� A more convenient form of the z-domain representation of the difference equation is given by
( ) ( )0 0
N M
k kk k
a y n k b x n k= =
− = −∑ ∑
0 0
( ) ( )N M
k kk k
k k
a z Y z b z X z− −
= =
=∑ ∑
0 0
( ) ( )N M
k kk k
k k
a z Y z b z X z− −
= =
=
∑ ∑
2010/4/7 58Introduction to Digital Signal Processing
Frequency Response from Transfer Function
� If the ROC of the transfer function H(z) includes the unit circle, then the frequency response H(ejω) of the LTI digital filter can be obtained simply as follows:
� For a real coefficient transfer function H(z) it can be shown that
� For a stable rational transfer function in the form
� the factored form of the frequency response is given
ωω
jez
j zHeH=
= )()(
( ) ( ) ( ) ( ) ( ) ( ) ( )2
1* j
j j j j j
z eH e H e H e H e H e H z H z ω
ω ω ω ω ω− −
== = =
( )( )
0 1
01( )
M
kN M kN
kk
z zbH z z
a z p
− =
=
−=
−
∏∏
( ) ( ) ( )( )
10
01
M jkj N Mj k
N jkk
e zbH e e
a e p
ωωω
ω
− =
=
−=
−
∏∏
30
2010/4/7 59Introduction to Digital Signal Processing
Geometric Interpretation of Frequency Response Computation
� It is convenient to visualize the contributions of the zero factor (z − zk) and the pole factor (z − pk) from the factored form of the frequency response
� The magnitude function is given by
which reduces to
� The phase response for a rational transfer function is of the form
( ) ( ) 10
01
M jkj N Mj k
N jkk
e zbH e e
a e p
ωωω
ω
− =
=
−=
−
∏∏
( ) 10
01
M jkj k
N jkk
e zbH e
a e p
ωω
ω
=
=
−=
−
∏∏
( ) ( ) ( ) ( ) ( )0 01 1
arg arg / arg argM N
j j jk k
k k
H e b a N M e z e pω ω ωω= =
= + − + − − −∑ ∑
2010/4/7 60Introduction to Digital Signal Processing
Geometric Interpretation of Frequency Response Computation
� The magnitude-squared function of a real-coefficient transfer function can be computed using
� The factored form of the frequency response
is convenient to develop a geometric interpretation of the frequency response computation from the pole-zero plot as ω varies from 0 to 2π on the unit circle
� The geometric interpretation can be used to obtain a sketch of the response as a function of the frequency
( ) ( )( )( )( )
2 *2
10
*0
1
M j jk kj k
N j jk kk
e z e zbH e
a e p e p
ω ωω
ω ω
−
=
−
=
− −=
− −
∏∏
( ) ( ) ( )( )
10
01
M jkj N Mj k
N jkk
e zbH e e
a e p
ωωω
ω
− =
=
−=
−
∏∏
31
Geometric Interpretation of Frequency Response Computation
� A typical factor in the factored form of the frequency response is given by
(ejω − ρejϕ)where ρejϕ is a zero (pole) if it is zero (pole) factor
� As shown below in the z-plane the factor (ejω − ρejϕ)represents a vector starting at the point z = ρejϕ and ending on the unit circle at z = ejω
� As ω is varied from 0 to 2π, the tip of the vector moves counter-clockwise from the point z = 1 tracing the unit circle and back to the point z = 1
2010/4/7 61Introduction to Digital Signal Processing
2010/4/7 62Introduction to Digital Signal Processing
Geometric Interpretation of Frequency Response Computation
� As indicated by
the magnitude response |H(ejω)|at a specific value of ω is given by the product of the magnitudes of all zero vectors divided by the product of the magnitudes of all pole vectors
� Likewise, from
we observe that the phase response at a specific value of ω is obtained by adding the phase of the term b0/a0 and the linear-phase term ω(N − M) to the sum of the angles of the zero vectors minus the angles of the pole vectors
( ) 10
01
M jkj k
N jkk
e zbH e
a e p
ωω
ω
=
=
−=
−
∏∏
( ) ( ) ( ) ( ) ( )0 0 1 1arg arg / arg arg
M Nj j jk kk k
H e b a N M e z e pω ω ωω= =
= + − + − − −∑ ∑
32
3-D View of Transfer Function & Frequency Response
2010/4/7 63Introduction to Digital Signal Processing
UNIT CIRCLE
EVALUATE H(z)EVERYWHERE 1− z −2
1 +0.7225z−2
Flying Thru z-Plane
2010/4/7 64Introduction to Digital Signal Processing
POLES CAUSEPEAKS in H(z)
H(ejωωωω)
H(z)
33
2010/4/7 65Introduction to Digital Signal Processing
Response of Systems with Rational System function
� Assume that the input signal x(n) has a rational z-transform
� For a initially relaxed system, the output becomes
� The output y(n) can be subdivided into two parts. The first part is a function of {pk}, called the natural response of the system. The second is a function of {qk}, called the forced response
( )( )
( )
N zX z
Q z=
( ) ( )
nr fr
1 1
the natral response ( ) the forced response ( )
( ) ( ) ( )N L
n n
k k k kk k
y n y n
y n A p u n Q q u n= =
= +∑ ∑������ ������
1 11 1
( ) ( )( ) ( ) ( )
( ) ( )
1 1
N Lk k
k kk k
B z N zY z H z X z
A z Q z
A Q
p z q z− −= =
= =
= +− −∑ ∑
Transient & Steady-State Responses
� The zero-state response of a system to a given input can be separate into two components: the natural responseand the forced response
where {pk}, k = 1, 2, …, N are the poles of the system.� If |pk| < 1 for all k, then ynr(n) decays to zero as n
approaches infinity. In such a case, we refer to the natural response of the system as the transient response
� The forced response of the system has the form
� where {pk}, k = 1, 2, …, N are the poles in the forcing function
2010/4/7 66Introduction to Digital Signal Processing
( )nr1
( ) ( )N
n
k kk
y n A p u n=
=∑
( )fr1
( ) ( )L
n
k kk
y n Q q u n=
=∑
34
Transient & Steady-State Responses
� When the causal input signal is a sinusoid, the forced response is also a sinusoid. In such case the forced output is called steady-state response of the system
� Example – Determine the transient and steady-state response of the following system (assuming initially rest)
where the input is � Sol
2010/4/7 67Introduction to Digital Signal Processing
( ) 0.5 ( 1) ( )y n y n x n= − +
( ) 10cos( / 4) ( )x n n u nπ=
1
1( )
1 0.5H z
z−=
−( )( )1
1 2
10 1 1/ 2( )
1 2
zX z
z z
−
− −
−=
− +
( )( )( )( )( ) ( ) ( ) ( )
128.7 28.7
1 /4 1 /4 1 1 /4 1 /4 1
10 1 1/ 2 6.3 6.78 6.78( )
1 0.5 1 1 1 0.5 1 1
j j
j j j j
z e eY z
z e z e z z e z e zπ π π π
−−
− − − − − − − −
−= = + +
− − − − − −
nr ( ) 6.3(0.5) ( )ny n u n= ( )fr ( ) 13.56cos / 4 28.7 ( )y n n u nπ= −
Causality and Stability
� As defined previously, a causal LTI system is one whose unit sample response h(n) satisfies the condition
h(n) = 0, n < 0 � The ROC of a causal sequence is the exterior of a circle� A linear LTI system is causal if and only if the ROC of the
system function is the exterior of a circle of radius r < ∞, including the point z = ∞
� Recall that a necessary and sufficient condition for a LTI system to be BIBO stable is
� In turn, this condition implies that H(z) must contain the unit circle within its ROC
2010/4/7 68Introduction to Digital Signal Processing
( )n
S h n∞
=−∞
= < ∞∑
35
Causality and Stability
� Since
� When evaluated on the unit circle (i.e., |z| = 1)
� Hence, if the system is BIBO stable, the unit circle is contained in the ROC of H(z), vice versa
� Therefore, a LTI system is BIBO stable if and only if the ROC of the system function includes the unit circle
� Note that causality and stability are different and that one does not imply the other
� A causal LTI system is BIBO stable if and only if all the poles of H(z) are inside the unit circle (why?)
2010/4/7 69Introduction to Digital Signal Processing
( ) ( ) ( ) ( )n n n
n n n
H z h n z h n z h n z∞ ∞ ∞
− − −
=−∞ =−∞ =−∞
= ≤ =∑ ∑ ∑
( ) ( )n
H z h n∞
=−∞
≤ ∑
Causality and Stability
� Example – A linear LTI system is characterized by the system function
Specify the ROC of H(z) and determine h(n) for the following conditions:
(a) The system is stable.
(b) The system is causal.
(c) The system is anticausal.
2010/4/7 70Introduction to Digital Signal Processing
1
1 2 1 1
3 4 1 2( )
1 3.5 1.5 1 0.5 1 3
zH z
z z z z
−
− − − −
−= = +
− + − −
36
Pole-Zero Cancelation
� Pole-zero cancelation: When a z-transform has pole that is the same location as a zero, the pole is canceled by the zero and, consequently, the term containing that pole in the inverse z-transform vanishes
� Pole-zero cancelations can occur either in the system function H(z) itself (a reduced-order system) or in the product of H(z)X(z)
� Properly selecting the position of the zeros of the input signal is possible to suppress one or more modes of the system function, or vice versa
� When the zero is located very near the pole but not exactly at the same location, the term in the response has very small amplitude. The nonexact pole-zero cancellation can occur in practice due to insufficient numerical precision
2010/4/7 71Introduction to Digital Signal Processing
Pole-Zero Cancelation
� Example – Determine the response of the system
to the input signal Sol:
The system function is
� And the z-transform of the input is
2010/4/7 72Introduction to Digital Signal Processing
1 2 1 1
1 1( )
5 1 1 11 1 16 6 2 3
H zz z z z− − − −
= = − + − −
1( ) ( 1) ( ) 3 ( 1)
2y n y n x n x n= − + − −
1( ) ( ) ( 1)
3x n n nδ δ= − −
11( ) 1
3X z z−= −
1
1( ) ( ) ( )
11
2
Y z H z X zz−
= =−
1( ) ( )
2
n
y n u n =
37
Stability Condition in Terms of the Pole Locations
� In addition, for a stable and causal digital filter for which h[n] is a right-sided sequence, the ROC will include the unit circle and entire z-plane including the point z = ∞
� An FIR digital filter with bounded impulse response is always stable
� On the other hand, an IIR filter may be unstable if not designed properly
� In addition, an originally stable IIR filter characterized by infinite precision coefficients may become unstable when coefficients get quantized due to implementation
2010/4/7 73Introduction to Digital Signal Processing
Stability of Second-Order Systems
� Consider a causal two-pole system described by the second-order difference equation,
� The system function is
� This system has two zeros at the origin and poles at
� The system is BIBO stable is the poles lie inside the unit circle, that is, if |p1| < 1 and |p2| < 1
� Thus, if the ROC includes the unit circle |z| = 1, then the system is stable, and vice versa
1 2 0( ) ( 1) ( 2) ( )y n a y n a y n b x n= − − − − +
2010/4/7 74Introduction to Digital Signal Processing
20 0
1 2 2 11 2 1 2
( )( )
( ) 1
b b zY zH z
X z a z a z z a z a− −= = =
+ + + +
21 1 2
1 2
4,
2 4
a a ap p
−= − ±
38
Stability of Second-Order Systems
� The coefficients are related to the roots by
� For stability, the following conditions must be satisfied
� Therefore a two-pole system is stable if and only if the two coefficients satisfy the two conditions.
( )1 1 2 2 1 2, a p p a p p= − + =
2010/4/7 75Introduction to Digital Signal Processing
2 1 2 1 21, 1a p p a a= < < +
Stability of Second-Order Systems
2010/4/7 76Introduction to Digital Signal Processing
39
Stability of Second-Order Systems
2010/4/7 77Introduction to Digital Signal Processing
Stability of Second-Order Systems
2010/4/7 78Introduction to Digital Signal Processing
40
Stability of Second-Order Systems
� Example – Consider the causal LTI IIR transfer function:
� The plot of the impulse response is shown below
� As can be seen from the above plot, the impulse response coefficient h(n) decays rapidly to zero value as n increases
� The absolute summability condition of h(n) is satisfied, ⇒ H(z) is a stable transfer function
21 850586.0845.11
1)( −− +−=
zzzH
2010/4/7 79Introduction to Digital Signal Processing
Stability of Second-Order Systems� Now, consider the case when the transfer function coef. are
rounded to values with 2 digits after the decimal point:
� A plot of the impulse response of is shown below
� In this case, the impulse response coefficient increases rapidly to a constant value as n increases
� Hence, is an unstable transfer function
21 85.085.11
1)( −−
∧
+−=
zzzH
2010/4/7 80Introduction to Digital Signal Processing
�[ ]h n
� ( )h n
� ( )h n
41
Stability Condition in Terms of the Pole Locations
� Example : The factored form of
is
which has a real pole at z = 0.902 and a pole at z = 0.943� Since both poles are inside the unit circle H(z) is BIBO stable� Example : The factored form of
is
which has a pole at z = 1 and the other inside the unit circle� Since one pole is not inside the unit circle, H(z) is not BIBO
stable
21 850586.0845.01
1)(
−− +−=
zzzH
( )( )11 943.01902.01
1)( −− −−=
zzzH
( )21 85.085.11
1−−
∧
+−=
zzzH
( ) ( )( )11 85.011
1−−
∧
−−=
zzzH
2010/4/7 81Introduction to Digital Signal Processing
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