Z transform and Properties of Z Transform

Properties of Z Transform

1. 1
Lecture 7 –10
Z -Transform
June 07, 2005
By
Dr. Mukhtiar Ali Unar

2. 2
The Direct Z-Transform
Thez-transformofadiscretetimesignalisdefinedasthepower
series
(1)
Wherezisacomplexvariable.Forconvenience,thez-transformofa
signalx[n]isdenotedby
X(z)=Z{x[n]}
Sincethez-transformisaninfiniteseries,itexistsonlyforthose
valuesofzforwhichthisseriesconverges.TheRegionof
Convergence(ROC)ofX(z)isthesetofallvaluesofzforwhich
thisseriesconverges.
Weillustratetheconceptsbysomesimpleexamples.




n
n
z]n[x)z(X

3. 3
Example1:Determinethez-transformof
thefollowingsignals
(a)x[n]=[1,2,5,7,0,1]
Solution:X(z)=1+2z
-1
+5z
-2
+7z
-3
+z
-5,
ROC:entirezplaneexceptz=0
(b)y[n]=[1,2,5,7,0,1]
Solution:Y(z)=z
2
+2z+5+7z
-1
+z
-3
ROC:entirez-planeexceptz=0andz=.
(c)z[n]=[0,0,1,2,5,7,0,1]
Solution:z
-2
+2z
-3
+5z
-4
+7z
-5
+z
-7,
ROC:allzexceptz=0

4. 4
(d) p[n] = [n]
Solution: P(z) = 1, ROC: entire z-plane.
(e) q[n] = [n –k], k > 0
Solution: Q(z) = z
-k
, entire z-plane except
z=0.
(f) r[n] = [n+k], k > 0
Solution: R(z) = z
k
,
ROC: entire z-plane except z = .

5. 5
Example 2: Determine the z-transform of
x[n] = (1/2)
n
u[n]
Solution:
ROC: |1/2 z
-1
| < 1, or equivalently |z| > 1/21
2
1
n
0n
1n
n
0n
n
n
z
2
1
1
1
.......z
2
1
z
2
1
1
z
2
1
z
2
1
z]n[x)z(X





































6. 6
Example 3: Determine the z-transform of the
signalx[n] = a
n
u[n]
Solution: 

|a||z:|ROC
az1
1
.......azaz1
azza)z(X
1
2
11
n
0n
1n
0n
n














7. 7
Properties of z-transform
Linearity
Ifx
1[n] X
1(z)
and x
2[[n] X
2(z)
then
a
1x
1[n] + a
2x
2[n] a
1X
1(z) + a
2X
2(z)

8. 8
Example:Determinethez-transformof
thesignalx[n]=[3(2
n
)–4(3
n
)]u[n]
Solution: 
11
nn
1
n
z31
1
4
z21
1
3]34)2(3[z
az1
1
]]n[ua[z








Example4:Determinethez-transformof
thesignal(cosw
0n)u[n] 
  
2
0
1
0
1
1jw1jw0
njwnjw
0
zwcosz21
wcosz1
ze1
1
2
1
ze1
1
2
1
]n[unwcosz
e
2
1
e
2
1
]n[unwcos
00
00













9. 9
Time Shifting Property:
If x[n] X(z) then x[n-k] z
-k
X(z)
Proof:
since
then the change of variable m = n-k
produces




n
n
z]kn[x]]kn[x[z )z(Xzz]m[xz
z]m[x]]kn[x[z
k
m
mk
m
)km(












10. 10
Example: Find the z-transform of a unit step
function. Use time shifting property to find z-
transform of u[n] –u[n-N].
The z-transform of u[n] can be found as
Now the z-transform of u[n]-u[n-N] may be
found as follows:1
21
0n
n
n
n
z1
1
.......zz1
zz]n[u]]n[u[z










  1
N
1
N
1
z1
z1
z1
1
z
z1
1
]]Nn[u]n[u[z













11. 11
Scaling in the z-domain
If x[n] X(z)
Then a
n
x[n] X(a
-1
z)
For any constant a, real or complex.
Proof:
Example 5: Determine the z-transform of the signal
a
n
(cosw
0n)u[n].
Solution: since    zaXza]n[xz]n[xa]n[xaz
1
n
n
1n
n
nn 






  2
0
1
0
1
0
zwcosz21
wcosz1
]n[u)nw[cos(z




  
22
0
1
0
1
0
n
zawcosaz21
wcosaz1
]]n[unwcosa[z






12. 12
Time reversal
If x[n] X(z) then x[-n] X(z
-1
)
Proof:
Example 6: Determine the z-transform of
u[-n].
Solution: since z[u[n]] = 1/(1 –z
-1
)
Therefore,
Z[u[-n]] = 1/(1-z) 











m m
1
m
1m
n
n
)z(Xz]m[xz]m[xz]n[x]]n[x[z

13. 13
Differentiation in the z -Domain
x[n] X(z) then nx[n] = -z(dX(z)/dz)
Tutorial4:Q1:Provethedifferentiation
propertyofz–transform.
Example7:Determinethez-transformofthe
signalx[n]=na
n
u[n].
Solution:  
2
1
1
1
n
1
n
az1
az
az1
1
dz
d
z]]n[una[z
az1
1
]]n[ua[z











14. 14
Convolution of two sequences
If x
1[n] X
1(z) and x
2[n] X
2(z) then
x
1[n]*x
2[n] X
1(z)X
2(z)
Proof:
The convolution of x
1[n] and x
2[n] is defined as



k
2121
]kn[x]n[x]n[x*]n[x]n[x
Thez-transformofx[n]is















n
n
n
21
n
n
zknx]k[xz]n[x)z(X
Upon interchanging the order of the summation
and applying the time shifting property, we obtain   zXzXz]k[xzXzknxkx)z(X
1
k
2
k
12
n
n
2
k
1 
















15. 15
Example8:Compute theconvolution
ofthesignalsx
1[n]=[1,-2,1]and
Solution:
X
1(z)=1–2z
-1
+z
-2
X
2(z)=1+z
-1
+z
-2
+z
-3
+z
-4
+z
-5
NowX(z)=X
1(z)X
2(z)=1–z
-1
–z
-6
+z
-7
Hencex[n]=[1,-1,0,0,0,0,-1,1]
Note:Youshouldverifythisresultfromthe
definitionoftheconvolutionsum.

 

elsewhere,0
5n0,1
]n[x
2

16. 16
Correlation of two sequences
If x
1[n] X
1(z) and x
2[n] X
2(z)
then r
x1x2[k] = X
1(z)X
2(z
-1
)
Tutorial 4 Q2: Prove this property.
The Initial Value Theorem:
If x[n] is causal then)z(Xlim]0[x
z

Proof:....z]2[xz]1[x]0[xz]n[x)z(X
21
0n
n






Obviously,asz,z
-n
0sincen>0,this
provesthetheorem.

17. 17
Final Value Theorem
If x[n] X(z), then )z(Xz1lim][x
1
1z



Tutorial 4 Q3: Prove the Final Value
Theorem
Example9:Findthefinalvalueof21
1
z8.0z8.11
z2
)z(X




Solution:   
21
1
11
z8.0z8.11
z2
z1)z(Xz1




  
  
1
1
11
1
1
z5.01
z2
z5.01z1
z2
z1









The final value theorem yields10
2.0
2
z8.01
z2
lim][y
1
1
1z







18. 18
Inverse z-transform
Ingeneral,theinversez-transformmaybe
foundbyusinganyofthefollowing
methods:
Powerseriesmethod
Partialfractionmethod

19. 19
Power Series Method
Example 2: Determine the z-transform of 21
z5.0z5.11
1
)z(X



By dividing the numerator of X(z) by its
denominator, we obtain the power series ...zzzz1
zz1
1
4
16
313
8
152
4
71
2
3
2
2
11
2
3




x[n] = [1, 3/2, 7/2, 15/8, 31/16,…. ]

20. 20
Power Series Method
Example 2:Determine the z-transform of 21
1
zz22
z4
)z(X





By dividing the numerator of X(z) by its
denominator, we obtain the power series
x[n]=[2,1.5,0.5,0.25,…..]

21. 21
Partial Fraction Method:
Example 1: Find the signal corresponding to
the z-transform21
3
zz32
z
)z(X




Solution: 5.0z1zz
5.0
z5.0z5.1z
5.0
zz32
z
)z(X
2321
3







   5.0z
4
1z
1
z
1
z
3
5.0z1zz
5.0
z
)z(X
22






 5.0z
z
)4(
1z
z
z
1
3)z(X




or11
1
z5.01
1
4
z1
1
z3)z(X





 ]n[u5.04]n[u]1n[]n[3]n[x
n


22. 22
Partial Fraction Method:
Example 2: Find the signal corresponding to the z-
transform  
2
11
z2.01z2.01
1
)z(Y



Solution:   
2
3
2.0z2.0z
z
)z(Y

     
22
2
2.0z
1.0
2.0z
75.0
2.0z
25.0
2.0z2.0z
z
z
)z(Y







  
2
2.0z
z1.0
2.0z
z75.0
1z
z25.0
)z(Y





  
2
1
1
2.0
1.0
11
z2.01
z2.0
z2.01
1
75.0
z2.01
1
25.0








   ]n[u2.0n5.0]n[u2.075.0]n[u2.025.0]n[y
nnn


23. 23
The One-Sided z-Transform
The one-sided or unilateral z-transform of a signal
x[n] is defined by 




0n
n
z]n[x)z(X
Characteristics:
•It does not contain information about the
signal x[n] for negative values of time.
•It is unique only for causal signals.

24. 24
Pulse Transfer Function
Itisdefinedastheratioofthez-transformofthe
outputtothez-transformoftheinputwhenall
initialconditionsareassumedtobezero.
Mathematically,











N
0k
k
k
M
0k
k
k
N
N
2
2
1
10
M
M
2
2
1
10
za
zb
za.....zazaa
zb.....zbzbb
)z(H
The roots of the denominator of a pulse transfer function
are called Polesand those of the denominator are called
Zeros.
The above pulse transfer function has M zeros and N poles.
We can represent X(z) graphically by a pole-zero plotin the
Complex plane, which shows the location of poles by crosses
() and the location of zeros by circles (o).

25. 25
Stability:
A discrete time system is said to be stable if,
and only if, all of its poles lie inside a unit
circle.
If any pole(s) lie(s) on the unit circle, the
system is said to be marginally stable.
Im(z)
Re(z)




Unit circle
Stable System


Unstable System

Marginally stable

26. 26
Example:Asystemischaracterizedby
by the difference equation
y[n]–0.1y[n-1]–0.02y[n-2]=2x[n]–x[n-1].
Findthesystemtransferfunctionandunit
impulseresponse.
Solution:Takingthez-transformofbothsidesofthe
differenceequation(ignoringtheinitial
conditions)wehave
Y(z)–0.1z
-1
Y(z)–0.02z
-2
Y(z)=2X(z)–z
-1
X(z)
Y(z)[1–0.1z
-1
–0.02z
-2
]=X(z[2–z
-1
]21
1
z02.0z1.01
z2
)z(X
)z(Y





Thistransferfunctionhastwopoles(i.e.z=-0.1,0.2)and
Twozerosatz=0,0.5.Thisshowsthatthesystemis
BIBOstable.ThePole-zeroplotisgivenonthenextslide.

27. 27
-1.5-1 -0.5 0 0.5 1 1.5
-1
0
1
RealPart
Imaginary
Part
Pole-zero plot

28. 28
Tofindtheunitimpulseresponse,we
computetheinversez-transformofH(z)by
usingpartialfractionexpansion:  1.0z2.0z
2z2
02.0z1.0z
zz2
z02.0z1.01
z2
)z(H
2
2
2
21
1










    1.0z
4
2.0z
2
1.0z2.0z
1z2
z
)z(H







 




















 11
z1.01
1
4
z2.01
1
2
1.0z
z4
2.0z
z2
)z(H
and, the unit impulse response is
H[n] = -2(0.2)
n
u[n] + 4(-0.1)
n
u[n]

29. 29
Response of systems with non-zero initial
conditions
Wecanusez-transformtosolvethedifference
equationthatcharacterizesacausal,linear,time
invariantsystem.Thefollowingexpressionsare
especiallyusefultosolvethedifference
equations:
z[y[(n-1)T]=z
-1
Y(z)+y[-T]
Z[y(n-2)T]=z
-2
Y(z)+z
-1
y[-T]+y[-2T]
Z[y(n-3)T]=z
-3
Y(z)+z
-2
y[-T]+z
-1
y[-2T]+
y[-3T]

30. 30
Tutorial 5 Q1: Determine the step response of
the system
y[n]=ay[n-1] + x[n], -1 < a < 1
when the initial condition is y[-1] = 1.

31. 31
Example:Considerthefollowingdifference
equation:
y[nT]–0.1y[(n-1)T]–0.02y[(n-2)T]=2x[nT]–
x[(n-1)T]
wheretheinitialconditionsarey[-T]=-10andy[-
2T]=20.Findtheoutputy[nT]whenx[nT]isthe
unitstepinput.
Solution:
Computing the z-transform of the difference
equation gives
Y(z) –0.1[z
-1
Y(z) + y[-T]] –0.02[z
-2
Y(z) + z
-1
y[-T]
+ y[-2T]] = 2X(z) –z
-1
X(z)
Substituting the initial conditions we get
Y(z)–0.1z
-1
Y(z)+1–0.02z
-2
Y(z)+0.2z
-1
–0.4=
(2–z
-1
)X(z)

32. 32    6.0z2.0
z1
1
z2)z(Yz02.0z1.01
1
1
121





   6.0z2.0
z1
z2
z02.0z2.01)z(Y
1
1
1
21







      
111
21
211
21
z1.01z2.01z1
z2.0z6.04.1
z02.0z1.01z1
z2.0z6.04.1
)z(Y









   1.0z2.0z1z
z2.0z6.0z4.1
23


 1.0z
830.0
2.0z
567.0
1z
136.1
z
)z(Y






 111
z1.01
1
830.0
z2.01
1
567.0
z1
1
136.1)z(Y







andtheoutputsignaly[nT]is]nT[u)1.0(830.0]nT[u)2.0(567.0]nT[u136.1]nT[y
nn


33. 33
Pole-Zero Cancellation
Whenaz-transformhasapolethatisatthe
samelocationasazero,thepoleiscancelledby
thezeroand,consequently,thetermcontaining
thatpoleintheinversez-transformvanishes.
Pole-zerocancellationmayoccureitherinthe
systemfunctionitselforintheproductofthe
systemfunctionwiththez-transformofthe
inputsignal.
Whenthezeroislocatedverynearthepolebut
notexactlyatthesamelocation,theterminthe
responsehasaverysmallamplitude.
Pole-zerocancellationmaycauseinstability
andshouldbeavoidedinmostcases.

34. 34
Example:Determinetheunitimpulseresponseof
thesystemcharacterizedbythedifference
equation
y[n]=2.5y[n-1]–y[n-2]+x[n]–5x[n-1]+6x[n-2]
Solution: Taking the z-transform of both sides of
the above difference equation and ignoring all
initial conditions, we obtain the following pulse
transfer function:  
  
1
2
1
1
1
2
1
1
11
2
1
11
21
21
z1
z5.2
1
z1
z31
z21z1
z31z21
zz5.21
z6z51
)z(H



















andtherefore ]1n[u5.2]n[]n[h
1n
2
1



35. 35
System Frequency response
The frequency response of a BIBO stable
system is defined as
H(w) = H(z)|
z = e
jw
The frequency response function is usually
expressed in terms of its magnitude |H(w)|
and phase (w), where
H(w) = |H(w)|e
j(w)
Usually, the magnitude is plotted on a
logarithmic scale as
|H(w)|
dB= 20log
10|H(w)|

36. 36
Example: Determine the frequency response
function H(w) and the magnitude |H(w)|
dBfor the
LTI system characterized by the difference
equation
y[n] = 1.8y[n-1] –0.81y[n-2] + x[n] + 0.95x[n-1]
Solution: The pulse transfer function is 
2
1
1
21
1
z9.01
z95.01
z81.0z8.11
z95.01
)z(H










Nowthefrequencyresponsefunctionmaybeobtainedas
 
2
jw
jw
e9.01
e95.01
wH





The magnitude response iswcos8.181.1
wcos9.19025.1
|)w(H|




37. 37
Wenotethat|H(w)|hasitsmaximumatw=0,
where|H(0)|=195.Itiscustomarytonormalize
|H(w)|byitspeakvalueandplot
20log|H(w)|/|H|
max.Aplotofthenormalized
magnitudeofthefrequencyresponseisshown
below:
0
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
- 

38. 38
Example:Findthefrequencyresponseforthe
system
y[n]=-0.1y[n-1]+0.2y[n-2]+x[n]+x[n-1]
Solution: The system function is21
1
z2.0z1.01
z1
)z(H





Now )zz(2.0)zz(08.005.1
zz2
z2.0z1.01
z1
z2.0z1.01
z1
)z(H)z(H
221
1
221
1
1












 wcos8.0wcos16.045.1
wcos22
|)z(H)z(H||)w(H|
2ez
12
jw




 wcos8.0wcos16.045.1
wcos22
|)w(H|
2


