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Lecture, week 03The z-transformation — Part I
Week 03, INF3190/4190
Andreas Austeng
Department of Informatics, University of Oslo
September 2019
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 1 / 36
Outline
Outline1 Motivation
z-domain: One of three domains to studyTransfer function (or system function)
2 The z-transformDefinition of the z-transformRational functionsRegion of convergence (ROC)Table of z-transform pairsQuiz
3 The Inverse z-transformApproachesPartial fraction expansion
4 Properties of the z-transform]Properties
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 2 / 36
Motivation
Outline1 Motivation
z-domain: One of three domains to studyTransfer function (or system function)
2 The z-transformDefinition of the z-transformRational functionsRegion of convergence (ROC)Table of z-transform pairsQuiz
3 The Inverse z-transformApproachesPartial fraction expansion
4 Properties of the z-transform]Properties
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 3 / 36
Motivation z-domain: One of three domains to study
3 domains
We can analyze digital signals and systems in three differentdomains:
I n-domain or time domainF The domain of sequences, impulse responses and difference
equationsF Signals are generated and processed in this domain.F Implementation of filters take place in this domain
I !̂-domain or frequency domainF The domain of the frequency responses & spectrum representationF Physical significant when analyzing sound, but seldom used for
implementation (in HW).I z-domain
F The domain of z-transforms, operators, poles & zeros.F Exists primary for its convinience in mathenatical analysis & synthesis
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 4 / 36
Motivation z-domain: One of three domains to study
Why bother with an other domain???
Difficult analysis in one domain may be easier in one other!More domains will result in increased understanding (???)Examples:Cascade combination of LTI-systems:
I In n-domain, this introduced the "new" (and less familiar) techniqueof convolution.
I In z-domain, a cascade combination is performed with polynomialmultiplication.
Stability:I In n-domain: BIBO.I In z-domain, Unit circle within ROC.
Causality:I In n-domain: Only use current or previous samplesI In z-domain, All poles are within the ROC.
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 5 / 36
Motivation z-domain: One of three domains to study
10/26/2007 © 2003, JH McClellan & RW Schafer 5
THREE DOMAINS
al ,bk{ }
Z-TRANSFORM-DOMAIN: poles & zerosPOLYNOMIALS: H(z)
Use H(z) to getFreq. Response
z = e j ˆ ω H(z) =
bkz−k∑
1− alz−l∑
FREQ-DOMAIN
l
ll
ω
ω
ω
ˆ
1
ˆ
0ˆ
1)(
jN
kjM
kk
j
ea
ebeH
−
=
−
=
∑
∑
−=
TIME-DOMAIN
∑∑==
−+−=M
kk
Nknxbnyany
01][][][
ll l
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 6 / 36
Motivation z-domain: One of three domains to study
Three domains of FIR and IIR filters
FIR-filtersI n-domain: y [n] =
PM
k=0 h[k ]x [n � k ] =PM
k=0 bk x [n � k ]I z-domain: H(z) =
PM
k=0 h[k ]z�k =PM
k=0 bk z�k
I !̂-domain: H(e|!k ) = H(!) =PM
k=0 h[k ]e�|!k =PM
k=0 bk e�|!k
IIR-filters: y [n] = �P
N
k=1 aky [n � k ] +P
M
k=0 bkx [n � k ]I Simple example; y [n] = a1y [n � 1] + b0y [n]
F By assuming initial rest, i.e. y [n] = 0 8 n < 0F Use that h[n] is given as response of �[n] ... we get (next slide)F h[n] = b0(a1)
nu[n],
F H(z) =P1
n=0 h[n]z�n = b01�a1z�1 , |a1| < 1
I IIR part is difficult! Could use linearity of the z-transform;Y (z) = a1z�nY (z) + b0X (z) and H(z) = Y (z)
X(z) =b0
1�a1z�1 , |a1| < 1
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 7 / 36
Motivation z-domain: One of three domains to study
10/26/2007 © 2003, JH McClellan & RW Schafer 17
y[n] = a1y[n −1]+ b0x[n]
IMPULSE RESPONSE
u[n] =1, for n ≥ 0
h[n] = a1h[n −1]+ b0δ[n]
][)(][ 10 nuabnh n=
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 8 / 36
Motivation Transfer function (or system function)
Transfer/system function
From last week:I An arbitary sequence can be written: x [n] =
P1k=�1 x [k ]�[n � k ].
I ) LTI-system: y [n] =P1
k=�1 x [k ]h[n � k ] =P1
k=�1 h[k ]x [n � k ].I The impulse response h[n] completely specifies the LTI system!
) In general, any sequence change shape through a LTI system.But: Does a sequences exist that retains its shape?
Yes, sequences being eigenfunctions does!
Lets consider the sequence: x [n] = zn, 8n, z = <(z) + |=(z).
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 9 / 36
Motivation Transfer function (or system function)
Transfer/system function, ... continuesLet x [n] = zn, 8n.
y [n] =P1
k=0 h[k ]x [n � k ] =P1
k=0 h[k ]zn�k =⇣P1
k=0 h[k ]z�k
⌘zn.
Define: H(z) , P1k=0 h[k ]z�k .
Then the output sequence becomes: y [n] = H(z)zn, 8n.I.e. Output sequence is the same as the input sequence,multiplied by a constant H(z).We say that the complex exponential sequences areeigenfunctions to LTI systems, with H(z) being the eigen value.Then, due to linearity:
I If input to LTI can be expressed as: x [n] =P
kck zn
k,
I then output of LTI would be: y [n] =P
kck H(zk )zn
k, 8n.
I Note: If H(zk ) = 0 for some zk , the LTI system would not let thecorresponding complex exponential zk through!
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 10 / 36
The z-transform
Outline1 Motivation
z-domain: One of three domains to studyTransfer function (or system function)
2 The z-transformDefinition of the z-transformRational functionsRegion of convergence (ROC)Table of z-transform pairsQuiz
3 The Inverse z-transformApproachesPartial fraction expansion
4 Properties of the z-transform]Properties
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 11 / 36
The z-transform Definition of the z-transform
Definition of the z-transform
X (z) ⌘ Z{x [n]} =P1
n=�1 x [n]z�n,where z = re|!̂ is a complex variable.Infinite power series;
I it exists only for those values of z for which this serie converges) Region Of Convergence (ROC);
I the set of values for which X (z) attain a finite value.
Notation:x [n]
z ! X (z)
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 12 / 36
The z-transform Definition of the z-transform
Definition of the z-transform ...
The z-transform is a complexvariable. It is convenient to describe itusing the complex z-plane.z = Re(z) + |Im(z) = re|!̂
The z-transform evaluated on the unitcircle corresponds to the DTFT:( ... Chapter 4 ... )
X (e|!̂) = X (z)|z=e|!̂
If the DTFT exist, the unit circle lieswithin the ROC
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 13 / 36
The z-transform Definition of the z-transform
Some elementary sequencesDelta function / impuls:
X (z) =P1
n=�1 �[n]z�n = z0 = 1, ROC: All z
Firkantpuls:
X (z) =P
M
n=0 z�n = 1�z�(M+1)
1�z�1 1, ROC: |z| > 1
Exponential, finit length:
X (z) =P1
n=0 anz�n =P1
n=0(az�1)n
= 1�aM+1z�(M+1)
1�az�1 , ROC: |z| > |a|
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 14 / 36
The z-transform Definition of the z-transform
Drill problem1 Let x [n] = (0.5)nu[n]. Find its z-transformation X (z) (and ROC).2 Let y [n] = (�0.5)nu[n]. Find Y (z) (and ROC).3 Let g[n] = �(0.5)nu[�n � 1]. Find G(z) (and ROC).
% S c r i p t som p l o t t e r X( z ) , Y( z ) og G( z )
% Se t te r opp akser og et " meshgrid "ax = �10:1/100:10; ay = �10:1/100:10;[xs ,ys ] = meshgrid (ax ,ay ) ;
% Finner verd ien t i l z , dvs z = R* exp ( j *Omega) , Omega \ i n [�\p i . . \ p i )z = s q r t (xs . ^2 + ys . ^ 2 ) . * exp (j* atan2 (ys ,xs ) ) ;
%% X( z )X = z . / ( z�0.5) ;X (xs . ^2 + ys . ^2 <= 0 . 5 . ^ 2 ) = NaN;
f i g u r e ( 1 ) ; meshz (ax ,ay , abs (X ) ) ; x l a b e l ( 'Re( z ) ' ) ; y l a b e l ( ' Im ( z ) ' ) ;%%zlim ( [ 0 5 ] ) ; cax is ( [ 0 5 ] )
%% Y( z )Y = z . / ( z+0.5) ;Y (xs . ^2 + ys . ^2 <= 0 . 5 . ^ 2 ) = NaN;
f i g u r e ( 2 ) ; mesh(ax ,ay , abs (Y ) ) ; x l a b e l ( 'Re( z ) ' ) ; y l a b e l ( ' Im ( z ) ' ) ;%%zlim ( [ 0 5 ] ) ; cax is ( [ 0 5 ] )
%% G( z )G = z . / (z� 0 .5 ) ;G (xs . ^2 + ys . ^2 >= 0 . 5 . ^ 2 ) = NaN;f i g u r e ( 3 ) ; meshz (ax ,ay , abs (G ) ) ; x l a b e l ( 'Re( z ) ' ) ; y l a b e l ( ' Im ( z ) ' ) ;
%%ax is([�1 1 �1 1 0 5 ] ) ; cax is ( [ 0 5 ] )
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 15 / 36
The z-transform Definition of the z-transform
ExampleTwo-sided exponential sequence
x [n] =
(an, n � 0�bn, n < 0
X (z) = �P�1
n=�1 bnz�n +P1
n=0 anz�n.
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 16 / 36
The z-transform Rational functions
The z-transform of rational functions
Many signals in digital signal processing have z-transform that arerational functions of z:
X (z) = A(z)B(z) =
PM
k=0 bk z�k
PN
l=0 al z�k
= b0z�M
a0z�N ⇥ zM+(b1/b0)z
M�1+···+bM/b0zN+(a1/a0)zN�1+···+aN/a0
= b0z�M
a0z�N ⇥ (z�z1)(z�z2)···(z�zM )(z�p1)(z�p2)···(z�pN )
= b0a0
zN�M ⇧M
k=1(z�zk )⇧N
k=1(z�pk )
I Roots of the numerator polynomial, zk , are referred to as zeros ofX (z).
I Roots of the denominator polynomial, pk , are referred to as poles ofX (z).
I Poles and zeros uniquely defines that functional form of a rationalz-transform to within a constant.
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 17 / 36
The z-transform Rational functions
Rational z-transforms ...
X (z) = b0a0
zN�M ⇧M
k=1(z�zk )
⇧N
k=1(z�pk )
M finite zeros at z = z1, z2, . . . , zM .N finite poles at z = p1, p2, . . . , pM .|N �M| zeros (if N > M) or poles (if N < M) at origin z = 0.Poles or zeros may occur at z =1. A zero exists at z =1 ifX (1) = 0 and a pole exists at z =1 if X (1) =1.If we counts the poles at zero and infinity, we find that X (z) hasexactly the same number of poles as zeros.ROC can not contain any poles.If all zeros/poles known; can determine X (z) to within a scalingfactor (b0
a0).
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 18 / 36
The z-transform Region of convergence (ROC)
ROC
The ROC is, in general, an annulus of the form↵ < |z| < �.
I If ↵ = 0, ROC may also include the point z = 0.I If � = 0, ROC may also include the point z =1.
For a rational X (z), ROC will contain no poles.Finite-duration signals
I Causal: Entire z-plane except z = 0.I Anti-causal: Entire z-plane except z =1.I Two-sided: Entire z-plane except z = 0 and z =1.
Infinite-duration signalsI Causal: |z| > r2I Anti-causal: |z| < r1I Two-sided: r2 < |z| < r1
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 19 / 36
The z-transform Table of z-transform pairs
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 20 / 36
The z-transform Quiz
Quiz
1 La x [n] = (�2)nu[n]. ROC til X (z) er da(a) |z| > 2, (b) |z| < 2, (c) |z| > 1/2, (d) |z| < 2, (e) |z| > 0.
2 Hvilket type system beskriver H[z] = z2+2z+1
z?
(a) Kausalt, (b) HP, (c) FIR, (d) Rekursivt, (e) Lineær fase.3 Et kausalt filter har poler z = 0.3, �0.5, 0.7.
Hva er ROC og er systemet stabilt?(a) |z| > 0.7 og stb, (b) |z| > 0.7 og ustb,(c) |z| > 0.3 og stb, (d) |z| > 0.3 og ustb.
4 Hva er impulsresponsen h[n] til systemet H(z) = z�1z+1 , |z| > 1?
5 La y [n] + 0.5y [n � 1] = 2x [n � 1].Finn transferfunksjonen HI(z) til det inverse systemet.
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 21 / 36
The Inverse z-transform
Outline1 Motivation
z-domain: One of three domains to studyTransfer function (or system function)
2 The z-transformDefinition of the z-transformRational functionsRegion of convergence (ROC)Table of z-transform pairsQuiz
3 The Inverse z-transformApproachesPartial fraction expansion
4 Properties of the z-transform]Properties
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 22 / 36
The Inverse z-transform Approaches
The Inverse z-transform
Three possible approachesContour integral: x [n] ⌘ Z�1{X (z)} = 1
2⇡j
Hc
X (z)zn�1dz.Power series: The z-transform is a power series expansionX (z) =
P1n=�1 x [n]z�n = . . . x [�1]z + x [0] + x [1]z�1 . . ..
I Then we can read x [n] directly!
Partial fraction expansionFor z-transforms that are rational functions of z:
X (z) =P
M
k=0 bk z�k
1+P
N
k=1 ak z�k,
where a0 = 1.I Turn it into a sum of simple parts and invert each part.
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 23 / 36
The Inverse z-transform Partial fraction expansion
Partial fraction expansion
X (z) =P
M
k=0 bk z�k
1+P
N
k=1 ak z�k, where a0 = 1.
Proper if aN 6= 0 and M < N.Proper and distinct poles:X(z)
z= A1
z�p1+ A2
z�p2+ · · ·+ AN
z�pN
I Ak = (z�pk )X(z)z
���z=pk
, k = 1, 2, . . .N
I x [n] =PN
k=1 Ak (pk )nu[n]
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 24 / 36
The Inverse z-transform Partial fraction expansion
10/26/2007 © 2003, JH McClellan & RW Schafer 37
GENERAL INVERSE Z
(pole)n
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 25 / 36
The Inverse z-transform Partial fraction expansion
Ekample (Manolakis 3.8)Real and distinct poles
Consider x [n] with z-transform X (z) = 1+z�1
(1�z�1)(1�0.5z�1).
X (z) is proper and rational, with poles p1 = 1 and p2 = 0.5 )X (z) = 1+z�1
(1�z�1)(1�0.5z�1)= A1
1�z�1 + A21�0.5z�1
To find A1, multiply with (1� z�1) and set z = p1 = 1.) A1 = 4
To find A2, multiply with (1� 0.5z�1) and set z = p2 = 2.) A2 = �3) X (z) = 4
1�z�1 � 31�0.5z�1
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 26 / 36
The Inverse z-transform Partial fraction expansion
Ekample (Manolakis 3.8) continues ...Real and distinct poles
To find sequence x [n], we need the ROCI If ROC: |z| < .5, both sequences are anti-causal and
x [n] = �4u[�n � 1] + 3( 12 )
nu[�n � 1].I If ROC: |z| > 1, both sequences are causal and
x [n] = 4u[n]� 3( 12 )
nu[n].I If ROC: 0.5 < |z| < 1, x [n] becomes two-sided:
x [n] = �4u[�n � 1]� 3( 12 )
nu[n].
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 27 / 36
The Inverse z-transform Partial fraction expansion
10/26/2007 © 2003, JH McClellan & RW Schafer 36
CALCULATE the RESPONSE
H(z)
Use the Z-Transform MethodAnd PARTIAL FRACTIONS
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 28 / 36
The Inverse z-transform Partial fraction expansion
10/26/2007 © 2003, JH McClellan & RW Schafer 38
SPLIT Y(z) to INVERT
� Need SUM of Terms:
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 29 / 36
The Inverse z-transform Partial fraction expansion
10/26/2007 © 2003, JH McClellan & RW Schafer 39
INVERT Y(z) to y[n]
� Use the Z-Transform Table
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 30 / 36
The Inverse z-transform Partial fraction expansion
10/26/2007 © 2003, JH McClellan & RW Schafer 40
TWO PARTS of y[n]
�� TRANSIENTTRANSIENT� Acts Like (pole)n
� Dies out ?� IF |a1|<1
b0a1
a1 −e j ˆ ω 0
⎛ ⎝ ⎜
⎞ ⎠ ⎟ a1nu[n]
b0
1− a1e− j ˆ ω 0
⎛ ⎝ ⎜
⎞ ⎠ ⎟ e j ˆ ω 0n u[n]
�� STEADYSTEADY--STATESTATE� Depends on the input� e.g., Sinusoidal
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 31 / 36
The Inverse z-transform Partial fraction expansion
10/26/2007 © 2003, JH McClellan & RW Schafer 41
STEADY STATE HAPPENS
� When Transient dies out� Limit as “n” approaches infinity� Use Frequency Response to get
Magnitude & Phase for sinusoid
yss[n]→ b0
1 −a1e−j ˆ ω 0
⎛ ⎝ ⎜
⎞ ⎠ ⎟ e j ˆ ω 0n = H(e j ˆ ω 0 )e j ˆ ω 0n
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 32 / 36
Properties of the z-transform
Outline1 Motivation
z-domain: One of three domains to studyTransfer function (or system function)
2 The z-transformDefinition of the z-transformRational functionsRegion of convergence (ROC)Table of z-transform pairsQuiz
3 The Inverse z-transformApproachesPartial fraction expansion
4 Properties of the z-transform]Properties
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 33 / 36
Properties of the z-transform Properties
Properties of the z-transform
Linearity:Z{a1x1[n] + a2x2[n]} = a1X1(z) + a2X2(z);ROC: at least ROCx1 \ ROCx2.Time sample shifting:Z{x [n � n0]} = z�n0X (z);ROC = ROCx , with the possible exception of adding or deletingthe points z = 0 and z =1.Frequency shifting:Z{anx [n]} = X ( z
a);
ROC = ROCx scaled by |a|.
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 34 / 36
Properties of the z-transform Properties
Properties of the z-transform ...
Time reversal:If Z{x [n]} = X (z); ROC : r1 < |z| < r2,then Z{x [�n]} = X (z�1);ROC : 1/r2 < |z| < 1/r1.Differentiation in the z-domain:Z{nx [n]} = �z
dX(z)dz
;ROC = ROCx .Convolution of two sequences:Z{x1[n] ⇤ x2[n]} = X1(z)X2(z);ROC : ROCx1 \ ROCx2.ROC may be larger if there is a pole-zero cancellation in theproduct X1(z)X2(z).
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 35 / 36
Properties of the z-transform Properties
Properties of the z-transform ....
Complex conjugation:Z{x⇤[n]} = X ⇤(z⇤);ROC = ROCx .The initial value theorem:If x [n] causal, then x [0] = limz!1 X (z).
AA, IN3190/4190 (Ifi/UiO) Lecture, week 03 Sept. 2019 36 / 36