Post on 03-Jun-2018
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2013-10-02Dan Ellis 1
ELEN E4810: Digital Signal Processing
Topic 4: The Z Transform
1. The Z Transform
2. Inverse Z Transform
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2013-10-02Dan Ellis 2
1.The Z Transform! Powerful tool for analyzing & designing
DT systems
! Generalization of the DTFT:
!
zis complex...! z= ej!DTFT
! z= rej!
G(z) = Z g n[ ]{ }= g[n]z!n
n=!"
"
# Z Transform
g n[ ]r!n
e!j"n
n#
DTFT ofr-ng[n]
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2013-10-02Dan Ellis 3
Region of Convergence (ROC)! Critical question:
Does summation
converge(to a finite value)?
! In general, depends on the value of z
! !Region of Convergence:
Portion of complexz-plane
for which a particularG(z)
will converge
G(z) = x[n]z!n
n=!"
"
#
z-plane
Re{z}
Im{z}
ROC|z| > "
"
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2013-10-02Dan Ellis 4
ROC Example! e.g.x[n] = n[n]
! converges onlyfor |z-1| < 1i.e. ROC is |z| > ||
! || < 1(e.g. 0.8) - finite energy sequence
! || > 1(e.g. 1.2) - divergent sequence,
infinite energy, DTFT does notexistbut still has ZTwhen |z| > 1.2 (in ROC)
! X(z) = "nz#n
n=0
$
% =1
1# "z#1
(previous slide)
n-1 1 2 3 4-2
M
when|z-1| < 1
closed form
Re{z}
Im{z}
"
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2013-10-02Dan Ellis 5
About ROCs! ROCs always defined in terms of |z|
!circularregions onz-plane
(inside circles/outside circles/rings)
! If ROC includesunit circle(|z| = 1),
!g[n]has a DTFT
(finite energysequence)
z-plane
Re{z}
Im{z}
Unit circlelies in ROC
!DTFT OK
1
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2013-10-02Dan Ellis 6
Another ROC example! Anticausal (left-sided) sequence:
! SameZT as n[n], differentsequence?
x n[ ] = !"n!n !1[ ] n-1
1 2 3 4
-2-3-4-5
X z( )=
!"
n
!n !1
[ ]( )z!n
n
#= ! "
nz
!n
n =!$
!1
# = ! "!m
zm
m=1
$
#
=
!"!1
z
1
1! "!1z=
1
1! "z!1
ROC:|"| > |z|
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2013-10-02Dan Ellis 7
ROC is necessary!! A closed-form expression for ZT
mustspecify the ROC:
x[n] = n[n]
ROC|z| > ||
x[n] = -n[-n-1]
ROC|z| < ||! A single G(z)expression can match
several sequences with different ROCs
!X(z) =1
1" #z"1
!X(z) =1
1" #z
"1
n-1
1 2 3 4-2-3-4 z-plane
Re
Im
|"|
n-1 1 2 3 4-2-3-4
DTFTs?
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2013-10-02Dan Ellis 8
Rational Z-transforms! G(z)expression can be any function;
rational polynomialsare importantclass:
!
By convention, expressed in terms ofz-1
matches ZT definition
! (Reminiscent of LCCDE expression...)
G z( ) = P z( )D z( )
=p0 + p1z!1 +
+ pM!1z!(M!1) + pMz!M
d0 + d1z!1++ dN!1z
!(N!1)+ dNz
!N
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2013-10-02Dan Ellis 9
Factored rational ZTs! Numerator, denominator can be
factored:
! {!}are roots of numerator
!G(z) = 0!{!}are thezerosofG(z)
! {!}are roots of denominator
!G(z) =#!{!}are the polesofG(z)
G z( )=
p0!
!=1
M1"#
!z"1( )
d0!
!=1
N 1" $!z"1( )
=
zM
p0!
!=1
Mz "#
!( )
zNd0!
!=1N z " $
!( )
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2013-10-02Dan Ellis 10
Pole-zero diagram! Can plot poles and zeros on
complexz-plane:
! (Value of) expression determinedby roots
z-plane
Re{z}
Im{z}
1! o
o
o
o
!
!
poles !
(cpx conj for real g[n])
zeros !
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2013-10-02Dan Ellis 11
Z-plane surface! G(z): cpx functionof a cpx variable
! Can calculate value over entire z-plane
M
ROCnot
shown!!
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2013-10-02Dan Ellis
1
0.5
0
0.5
1
1
0.5
0
0.5
1
0
2
4
6
8
10
12
ROCs and sidedness! Two sequences have:
! Each ZT pole!region in ROCoutsideor inside||for R/Lsided term in g[n]
! Overall ROC is intersection of each terms
G(z) = 11! "z
!1
z-plane
|"|
ROC |z| >||!g[n] = n[n]
n
ROC |z|
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2013-10-02Dan Ellis 13
ZT is Linear!
! Thus, if
then
G(z) = Z g n[ ]{ }= g[n]z
!n
"n# Z Transform
y[n] =!1"1n
[n]+!2"2n
[n]
Y(z) =!
1
1# "1z#1+
!2
1# "2z#1
y[n] = g[n]+ h[n]
Y(z) = (g[n]+h[n])z-n
=g[n]z-n+ h[n]z-n= G(z)+H(z)
ROC:
|z|>|"1|,|"2|
Linear!
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2013-10-02Dan Ellis 14
G(z) = 11! "
1z!1
+ 11! "
2z!1
ROC intersections
! Consider
with |1| < 1, |2| > 1...
!
Two possible sequences for
1term...
! Similarly for 2 ...
!4 possible g[n]seqs and ROCs ...
no ROC specified
1n[n]
n-
1n[-n-1]
nor
n-2n[-n-1]
n
2n[n]or
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2013-10-02Dan Ellis 15
ROC intersections: Case 1
ROC:|z| >|1| and|z| >|2|
ImRe
|"1| |"2|
Im
G(z) =1
1! "1z!1+
1
1! "2z!1
n
g[n] = 1n[n]+ 2
n[n]
both right-sided:
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2013-10-02Dan Ellis 16
ROC intersections: Case 2
ImRe
|"1| |"2|
Im
G(z) =1
1! "1z!1+
1
1! "2z!1
n
g[n] = -1n[-n-1]- 2
n[-n-1]
both left-sided:
ROC:|z|
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ROC intersections: Case 3
ROC:|z| >|1| and|z|
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ROC intersections: Case 4
ROC:|z| |2| ?
Re
|"1| |"2|
Im
G(z) =1
1! "1z!1+
1
1! "2z!1
n
g[n] = -1n[-n-1]+ 2
n[n]
two-sided:
no ROC...
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2013-10-02Dan Ellis 19
ROC intersections! Note: Two-sided exponential
! Nooverlap in ROCs!ZT does not exist(does not converge for anyz)
g n[ ] =!n
=!
n
n[ ]+!n
"
n"1[ ]
!"< n
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2013-10-02Dan Ellis 20
ZT of LCCDEs! LCCDEs have solutions of form:
! Hence ZT
! Each terminin g[n] corresponds to a
poleiof G(z) ... and vice versa
! LCCDE solns are right-sided
ROCs are |z| > |i|
yc[n]=!i"inn[ ]+...
Yc z( ) =!
i
1" #iz"1 +!
(sames)
outsidecircles
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2013-10-02Dan Ellis 21
Z-plane and DTFT! Slice between surfaceand unit cylinder
(|z| = 1z= ej) is G(ej), the DTFT
|G(ej
)|z = ej
0 / rad/samp
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2013-10-02Dan Ellis 22
Some common Z transforms
g[n] G(z) ROC [n] 1 !z
[n] |z|> 1
n[n] |z|> ||
rncos(0n)[n] |z|> r
rnsin(0n)[n] |z|> r
1
1!z!1
11!"z
!1
1!rcos "0( )z
!1
1!2rcos "0( )z
!1+r
2z!2
rsin !0( )z"1
1"2rcos !0( )z
"1+r
2z"2
sum ofrnej0n + rne-j0n
poles atz = rej0!
!
conjugate polepair
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2013-10-02Dan Ellis 23
Z Transform properties
g[n] G(z) w/ROCRg
Conjugation g*[n] G*(z*) Rg
Time reversal g[-n]
G(1/z)
1/Rg
Time shift g[n-n0] z-n0G(z) Rg(0/#?)
Exp. scaling n
g[n]
G(z/) ||Rg
Diff. wrtz ng[n] Rg(0/#?)!zdG(z)
dz
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2013-10-02Dan Ellis 24
Z Transform properties
g[n] G(z) ROC
Convolution g[n] h[n] G(z)H(z)
Modulation g[n]h[n]
Parseval:
"at least
Rg#Rh
1
2!j G v( )H z v( )v
"1dvC#
g n[ ]h* n[ ]n=!"
"
# = 12$j G v( )H* 1
v( )v!1
dvC%
at least
RgRh
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2013-10-02Dan Ellis 25
ZT Example! ; can express as
! Hence,
x n[ ] = rn cos !0n( ) n[ ]
1
2 n[ ] rej
!0( )
n
+ re"j!
0( )n
$%
'( =v n[ ]+ v* n[ ]
v[n] = 1/2[n]n; = rej0
!V(z) = 1/(2(1- rej0z-1))ROC:|z| > r
X z( ) = V z( )+ V* z*( )= 1
21
1!re j"0z
!1+ 1
1!re! j"0z
!1( )=
1!rcos "0( )z
!1
1!2rcos "0( )z
!1+r
2z!2
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2013-10-02Dan Ellis 26
Another ZT example
y n[ ] = n +1( )!n n[ ]= x[n]+ nx[n] wherex[n] = n[n]
X z( ) = 11!"z
!1
! "zdX z( )
dz
="z d
dz
1
1"#z"1
$
%&
'
() =
#z"1
(1"#z"1
)
2
!Y z( ) =1
1"#z"1+
#z"1
(1"#z"1)2=
1
(1"#z"1)2
repeatedroot - IZT
( |z| > || )
ROC |z| > ||
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2013-10-02Dan Ellis 27
2.Inverse Z Transform (IZT)! Forwardztransform was defined as:
! 3 approaches to invertingG(z)to g[n]:! Generalization of inverse DTFT! Power series inz(long division)
! Manipulate into recognizablepieces (partial fractions)
G(z) = Z g n[ ]{ }= g[n]z!n
n=!"
"
#
the usefulone
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2013-10-02Dan Ellis 28
IZT #1: Generalize IDTFT! If then
!
so
! Any closed contour around origin will do! Cauchy: g[n] = [residues of G(z)zn-1]
z =re j!
IDTFTz = rejd= dz/jz
Counterclockwise
closed contour at |z| = rwithin ROC
Re
Im
g[n]rn = 12
G rej
ejnd
= 12j
CG (z) zn1rndz
G(z) =G(rej) =
g[n]rnejn = DTFTg[n]rn
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IZT #2: Long division! Since if we could express G(z)as a simple
power series G(z) = a+ bz-1+ cz-2...
then can just read off g[n] = {a, b, c, ...}! Typically G(z)is right-sided (causal)
and a rational polynomial
! Can expand as power series throughlong division of polynomials
G(z) = g[n]z!nn=!"
"
#
G z( ) =P(z)
D(z)
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2013-10-02Dan Ellis 30
IZT #2: Long division! Procedure:
! Express numerator, denominator in
descending powers ofz(for a causal fn)
!
Find constant to cancel highest term!first term in result
! Subtract & repeat !lower terms in result
!
Just like long division for base-10numbers
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2013-10-02Dan Ellis 31
IZT #2: Long division
! e.g.
H z( ) = 1+ 2z!1
1+ 0.4z!1! 0.12z
!2
1+ 0.4z!1!0.12z!2
1+ 2z!1)
1+1.6z!1! 0.52z
!2+ 0.4z
!3...
1+ 0.4z!1! 0.12z
!2
1.6z!1
+ 0.12z!2
1.6z!1
+ 0.64z!2
!0.192z
!3
!0.52z!2+ 0.192z
!3
...
Result
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2013-10-02Dan Ellis 32
IZT#3: Partial Fractions! Basic idea: Rearrange G(z)as sumof
terms recognizedas simple ZTs
! especially
or sin/cos forms
! i.e. given products
rearrange to sums
1
1!"z
!1 # "
n
n[ ]
P(z)
1!
"z!1
( )1!
#z!1
( )!
A
1!"z!1+
B
1! #z!1+!
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2013-10-02Dan Ellis 33
Partial Fractions!
Note that:
! Can do the reversei.e.
go from to
! if orderof P(z)is less thanD(z)
A
1!"z!1+
B
1! #z!1+
C
1! $z!1=
A 1! #z!1( ) 1! $z!1( )+ B 1!"z!1( ) 1! $z!1( )+C1!"z!1( ) 1! #z!1( )1!"z!1( ) 1! #z!1( ) 1! $z!1( )order 3 polynomial
order 2 polynomial
u + vz-1 + wz-2
P(z)
!!=1N (1" #
!z"1)
!!
1" #!z"
1!=1
N
$else cancelw/ long div.
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2013-10-02Dan Ellis 34
Partial Fractions! Procedure:
! where
i.e. evaluateF(z)at the pole but multipliedby the pole term !dominates = residueof pole
F(z) =P(z)
!!=1N
(1" #!z"1)=
$!
1" #!z"1!=1
N
%
!!= 1" #
!z"1( )F z( ) z=#
!
! f n[ ] = "! #!( )n n[ ]
!=1
N
$
orderN-1
(cancels term indenominator)
no repeatedpoles!
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2013-10-02Dan Ellis 35
Partial Fractions Example! Given (again)
factor:
! where:
H z( ) =1+ 2
z
!1
1+ 0.4z!1! 0.12z
!2
=1+ 2z
!1
1+ 0.6z!1( ) 1! 0.2z!1( )=
"1
1+ 0.6z!1+
"2
1!0.2z!1
!1 = 1+ 0.6z
"1
( )H z( ) z="0.6 =1+ 2z
"1
1" 0.2z"1
z="0.6
="1.75
!2= 1+ 2z
"1
1+ 0.6z"1
z=0.2
=2.75
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2013 10 02Dan Ellis 36
Partial Fractions Example
! Hence
! If we know ROC |z| > || i.e. h[n]causal:
= 1.75{ 1 -0.6 0.36 -0.216 ...}
+2.75{ 1 0.2 0.04 0.008 ...}
= {1 1.6 -0.52 0.4 ...}
H z( ) = !1.751+ 0.6z
!1+ 2.75
1! 0.2z!1
! h n[ ] = "1.75( ) "0.6( )n n[ ]+ 2.75( ) 0.2( )
n n[ ]
same aslong division!