lecture 07 - z-transform
TRANSCRIPT
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Lecture 7: Z-transform
Instructor:
Dr. Gleb V. Tcheslavski
Contact:
Office Hours:
Room 2030
Class web site:
http://ee.lamar.edu/gleb/dsp/index.htm
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Definitions
Z-transform converts a discrete-time signal into a complex
frequency-domain representation. It is similar to the Laplace
transform for continuous signals.
( ) nnn
X z x z| If (where) it exists!
(7.2.1)
n is an integer time index;
j
z re[
! is a complex number; [ - angular freq.
When the magnitude r=1,jz e [!
( ) j j nnn
z e x e[ [p ! If it exists!
(7.2.2)
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Region of Convergence (ROC)
j z re[! (7.3.2)Since z is complex:
( ) n j n n j nn nn n
X z x r e x r e[ [ ! !
The Region of convergence (ROC) is the set of points zin the complex plane,
for which the summation is bounded (converges):
n
n
n
x z g (7.3.1)
In general, z-transform exists for r r r
r z r
(7.3.3)
(7.3.4)
(7.3.5)
r- r+ Re
Im
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Region of Convergence (ROC)
Examples of ROCs
from Mitra
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Region of Convergence (ROC)
Example 7.1: Let xn = an _ a2 1 2..., , , , , ,...n x a a a a !
n
n
n
x z
! gThere are no values ofzsatisfying:
Example 7.2: Let xn = an un a causal sequence
1 10
1( )
1
nn n
n
n n
X z aaz
u z azg
!
! !
!
1 1o af r az z "
a Re
Im
(7.5.1)
ROC
We can modify (7.5.1) as
( )( )
( )
z N zz
z a D z! !
(7.5.2)
( ) 0 0
( ) 0
( )
( )
N z z
D
zero s
pol z z a e s
! !
! !
x
( )n
n
n
X z x z|
roots of numerator: X(z) = 0
roots of denominator: X(z) p g
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Region of Convergence (ROC)
Example 7.3: Let xn = -an u-n-1 an anticausal sequence
_ a
1
11
1
1 1
1 11
1 0
1
1( )n
n n n
n
n n
m m
m m
a u z n n a z
az a z
X z
a z z
a z z a
!g
g g
!
!
! ! u e !
! !
!
!
(7.6.1)
a Re
Imx1 1for a z z a
Conclusion 1: z-transform exists only within the ROC!
Conclusion 3: poles cannot exist in the ROC; only on its boundary.
Note: if the ROC contains the unit circle (|z|=
1), the system is stable.
Conclusion 2: z-transform and ROC uniquely specify the signal.
( )n
n
n
X z x z|
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The transfer function
-i n i j n j
i j
a y b x2 1
! !
! Consider an LCCDE:
Time shift:( ) { } ( )n
l m
n m n m lz
n l
x x z z l n m z x z z z p ! ! ! !
and take z-transform
utilizing time shift( ) ( )i j
i j
i j
a z Y z b z z2 1
! !
!
LTI:( )
( )
( ) ( ) ( )n n m mn n n m n m m n mz
n m m n
z
y h x h x z h xY z Hz zz z ! p ! ! ! 1 44 2 4 43
The system transfer function0
0
( )( )
( )
M
j
j
j n
n
i ni
i
b zY z
z h zX z
a z
!
!
! ! !
(7.7.1)
(7.7.2)
(7.7.3)
(7.7.4)
(7.7.5)
( )n
n
n
X z x z|
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Rational z-transform
Frequently, a z-transform can be described as a rational function, i.e. a
ratio of two polynomials in z-1:1 ( 1)
0 1 1
1 ( 1)
0 1 1
...( )( )
( ) ...
M M
M M
N N
N N
n n z n z n zP zz
z d d z d z d z
! !
Here Mand Nare the degrees of the numerators and denominatorspolynomials. An alternative representation is a ratio of two polynomials in z:
( 1)
0 1 1
1
0 1 1
...( )( )
( ) ...
M M
N M M M
N N
N N
n z n z n z nP zz z
z d z d z d z d
! !
(7.8.1)
(7.8.2)
Finally, a rational z-transform can be written in a factorized form:
1
0 0
1 1( )
1
0 0
1 1
(1 ) ( )
( )
(1 ) ( )
M M
j j
j jN M
N N
i i
i i
n z z n z z
z z
d p z d z p
! !
! !
! !
(7.8.3)
zeros: numerator= 0
Poles: denominator= 0
( )n
n
n
X z x z|
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Notes on poles of a system function
Positions of poles of a transfer function are used to evaluate system
stability. Let assume a single real pole at z=E. Therefore:
( ) 1( )( )
Y z z zY z Y z X z
X z zE
E! ! !
The difference equation is:
1 1 1n n n n n n y y x y y xE E ! !
Therefore, the impulse response is:
1 1n n nh hE H !
for 20,1, 2, 3,... 0,1, , ,...nn h E E ! ! Iff |E| < 1, hn decays as npg and the system is BIBO stable; otherwise, hngrows without limits. Therefore, poles of a stable system (and signals
in fact) must be inside the unit circle.
Zeros may be placed anywhere. Zeros at the origin produce a time delay.
(7.9.1)
(7.9.2)
(7.9.3)
(7.9.4)
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The transfer function and the
Frequency response
( )1 jn z en
jh z RO HC e H z [[
! g ! !BIBO:
0
0 1
0
0 1
( )
( )
( )
MM
Mj
jj
j jN N
i N
i i
i i
b z z zb z
H z
a z a z z p
! !
! !
! !
where zj are zeros andpiare poles of the transfer function.
BIBO:
10
0
1
0
( )
1
0
1
0
0
1
1
( 1
( ( ) ( )
)
( )
)
M
M
j M N j
j
jj
N
jj
jj
N
j
i
i
M N
j
j
i
i j j
j i
j i
b e e z
e zb
ea
e pe
a e
e b a M N e z e
p
p
[
[
[
[
[ [
[
[
[ [[
!
!
!
!
! !
!
!
!
(7.10.1)
(7.10.2)
(7.10.3)
( )n
n
n
X z x z|
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The transfer function and the
Frequency response
A good way to evaluate the systems frequency response:
_ a
_ a
0
2
0k
M
mj
Nkp
k
P FT be
P FT a
[T
[ !
!
When the frequency approaches a pole, the frequency response has a local
maximum, a zero forces the response to a local minimum.
For real systems, poles and zeros are symmetrical with respect to the real axis.
(7.11.1)
0 0.1 0.2 0.3 0.4 0.5 0
0. 5
1
1. 5
MagnitudeofH
([)
Fractional f requency, [
pole zero
Zero-padded
( )n
n
n
X z x z|
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The transfer function and the SFG
1 ( ) ( ) ( )
( ) ( ) ( )
n n n
T Tzn n n
S AS bx zS z AS z bX z
y c S dx Y z c S z dX z
! ! p
! !
1
1
1
( ) ( ) ( ) ( )
( ) ( ) ( )
( )
T
T
zI S z b z S z zI
Y z c zI b z d z
H z c zI b d
b z
!
! !
!
Poles ofH(z) correspond to the eigenvalues of the
system matrix.
(7.12.1)
(7.12.2)
(7.12.3)
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More on Transfer function
0
1 ( )0 0 1
00
0 1 1
( ) ( )( )
( )( )
( ) ( )
M MMMm
j mmj M Nm m
N N N
k N
k i k
k i k
b z z z z zb zbP z
z zz a
a z a z z p z p
! ! !
! ! !
! ! ! !
| 1 ( )j
j
j
jz e
e BIB z z e
e[
[
[
[!
! p p
1) N> M: zeros at z= 0 of multiplicity N-M
2) M> N: poles at z= 0 of multiplicity M-N
zeros
poles
( )
( )
( )
1
( ) .
j M N
j M N
j M N
e const magnitudee distortionless
e M N lin phase
[[
[ [
! p
!
(7.13.1)
( ) 0 ( ) 0; ( ) ( )i i i i
H z Y z H p Y pp p p g p g (7.13.2)
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Types of digital filters
0( ) 0
M
nm
m n
m
b n MH z b z h otherwise
!
e e! p !
1. FIR (all-zero) filter:
(7.14.1)
All poles are at z= 0: a nest of poles
ROC: the entire z-plane except of the origin (z= 0).
FIR filters are stable.
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Types of digital filters
2. IIR (all-pole) filter:
0
0
( )N
k
k
k
bH z
a z
!
!
All zeros are at z= 0: a nest of zeros
(7.15.1)
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Types of digital filters
3. General IIR (zero-pole) filter:
0
0
( )
M
m
mm
N
k
k
k
b zH z
a z
!
!
!
(7.16.1)
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On test signals
( ) ( ) ( ) n n nz
Y z H z X z y h x! p !
n nhH p
n k n k n
k
x x yH ! p %
Calculate and compare ton k n kk
y x h! ny
%
(7.17.1)
(7.17.2)
(7.17.3)
We dont need any other that a delta function test signals since a unit-pulse
response is a complete systems description.
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Types of sequences and convergence
1. Two-sided:
( )N
n
n M
nG z g z
!
!
Converges everywhere except ofz= 0 and z=g
(7.18.1)
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Types of sequences and convergence
2. Right-sided:
1
0
( ) n n nn n nn nM Mn
G z g z g z g z g
g
! !!
! ! (7.19.1)
Assume: if converges at z = z0, converges for |z| > | z0|
Blows up at z=g
ROC: r-
< |z| < g - exterior ROC
For a causal sequence: |z| > r- = max|pk| - a max pole ofG(z)
To be causal, a sequence must be right-sided (necessary but not sufficient)
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Types of sequences and convergence
3. Left-sided:
1
0
( ) n n nn n nn n n
N N
G z g z g z g z
! !g !
g
! ! (7.20.1)
Blows up at z= 0Converges at z0
ROC: 0 < |z| < r+
- interior ROC
When encountering an interior ROC, we need to check convergence at
z= 0. If the sequence blows up at zero its an anti-causal sequence
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Properties
from Mitra
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Common pairs
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Inverse z-transform
11 ( )2
n
n
c
x z z dzjT
! i (7.23.1)
WhereC
is a counterclockwise closed path encircling the origin and is entirely inthe ROC. ContourCmust encircle all the poles ofX(z).
In general, there is no simple way to compute (7.23.1)
A special case:C is the unit circle (can be used when the ROC includes the unit circle).
The inverse z-transform reduces to the IDTFT.
12
j j n
n x X e e d
T[ [
T
[T
! (7.23.2)
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Inverse z-transform
A. Via Cauchy residue theorem
n i
i
x V! For all poles ofX(z)zn-1 inside C (contour of integration)
WhereViare the residues ofX(z)zn-1
1
1
( )1
( 1)! i
k
i
i k
z p
d z
kdz
JV
!
!
for a pole of multiplicity k:
Residue function: 1( ) ( ) kni i
z X z z z pJ !
(7.24.1)
(7.24.2)
(7.24.3)
( )n
n
n
X z x z|
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Inverse z-transform: Example
( ) ;z
z a zz a
!
a Re
Im
xC
1
0
01 1
02 2
nan
n
aC C
nz z x z dz dz
nj z a j z a
V
V VT T
u! ! !
i i
Example:
V0 is a residue ofX(z)z-n-1 at z=0 involves pole of
multiplicity n wnen n < 0.1
1
1
1( )
( 1)!
k nk n n
a k z ak
d z z a z a
k d z z aV
!!
! ! !
multiplicity1 1
0 1 0
1 ( ) 1( 0) ( 1)( 2)...( ( 1))( )
( 1)! ( 1)!
k
k k k n
k k z
d z aLet n k n z k z a a a
k dz z kV
!
! ! ! ! !
0
0 0nn
n n
n
nn
a nx
a aa u
nx
u !
!
!
( )n
n
n
X z x z|
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Inverse z-transform
B. Via recognition (table look-up)
Example:
( )a
zz e!
0 0
1( ) ; 1
!!
na
z
n
nn
n
a a X z e z a
z n
az
n z
g
!
g
!
! ! ! "
!
n
n n
ax u
n!Therefore:
Sometimes, the z-transform can be modified such way that it can be
found in a table
( )n
n
n
X z x z|
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Inverse z-transform
C. Via long division
1. Right-sided z-transform sequences can be expanded into a power series in
z-1. The coefficient multiplying z-n is the nth sample of the inverse z-transform.
Example:2 1
1
2 2( ) ; 1
1
z z X z z
z
! "
Lower powers first:
1 2
1
2 2( ) ; 1
1
z zz z
z
!
and long division:11 |z
2 3 42 ... z z z 1 2
1
2 2
2 2
z z
z
2
2 3
z
z z
{2, 0,1, 1,1,...}nx !
x0x1x2 x3 x4
( )n
n
n
X z x z|
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Inverse z-transform
2. Left-sided z-transform sequences into a power series in z1
Example: 2 1
1
2 2( ) ; 1
1
z z X z z
z
!
Multiply both numerator and denominator by z2
2 2 1 2
2 1 2
2 2 1 2 2( )
1
z z z z zz
z z z z
! !
Long division
1 2 3 4( ) 1 ...X z z z z z z!
x0 x-1 x-2 x-3 x-4x1 Non-causal
( )n
n
n
X z x z|
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Inverse z-transform
Example:
2 1
1 1
2 2 1( ) ; 1
21 2
z zX z z
z z
!
not suitable for
long division!
Example: 1
1
1 1
( ) 1 ; 1
1
( 1)nn n n n
zX z z z
zz
x uH H
!
q q
!
( )n
n
n
X z x z|
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Inverse z-transform
D. Via partial fraction expansion (PFE)
( )( ) ;
( )
N zLet G z r z r
D z
!
If the degree of the numerator is equal or greater than the degree of the
denominator: Mu N, G(z) is an improper polynomial. Then:
1
0
(( )(
( ))
)
( )
M Nl
l
l
N zG z
D
N zc z
zz D
!
! !
A proper fraction: M1 < N
(7.30.1)
(7.30.2)
Then:1
0 1
( )1
M N N
l ll
l l l
cp
G z zz
V
! ! !
Simple poles: multiplicity of 1.
(7.30.3)
( )n
n
n
X z x z|
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Inverse z-transform
Here Vl is a residue
( ) ( )l
n l
l l z pz p G z zV
!!
poles
Therefore:
0 1
:
:
n M N N
l n l
n l n l l nl l l n l l
p u ifexternal ROC p rg c
p u ifinternal ROC p rH V
! !
e!
u
(7.31.1)
(7.31.2)
This method is suitable for complex poles.
Problem: large polynomials are hard to manipulate
??QUESTIONS??
( )n
n
n
X z x z|