chapter 6: properties and the inverse of z-transform 1
TRANSCRIPT
Chapter 6:
Properties and the Inverse of z-Transform
1
Properties of z-Transform
Some of the properties of the z-transform are:
• Linearity• Right shift• Left shift• Final value theorem• Z-differentiation (Multiplication by n)
2
Example: Linearity
The z-transform of a unit ramp function r(nT) is
Find the z-transform of the function 5r(nT).
SolutionUsing the linearity property of z-transform,
3
2)1()(
z
TzzR
.)1(
5)(5)}(5{
2z
TzzRnTrZ
Right-shifting property
• Suppose that the z-transform of f(nT) is F(z).
• Let y(nT) = f(nT-mT), then
assuming that for n<0.
4
)()( zFzzY m
Left-shifting property
• Suppose that z-transform of f(nT) is F(z).• Let y(nT) = f(nT+mT). Then
• However, if for n=0 to m-1, this simplifies to
5)()( zFzzY m
.)()()(1
0
m
i
im ziTfzFzzY
Final value theorem
• Suppose that the z-transform of f (nT ) is F(z). Then the final value of the time response is given by
• Note that this theorem is valid if the poles of (1−z−1)F(z) are inside the unit circle or at z = 1.
6
).()1(lim)(lim 1
1zFznTf
zn
Example 6.5Given the function
find the final value of g(nT).
Solution:Using the final value theorem,
7
,)208.0416.0)(1(
792.0)(
2
zzz
zzG
.1)208.0416.0(
792.0lim
,)208.0416.0)(1(
792.0)1(lim
)()1(lim)(lim
21
21
1
1
1
zz
zzz
zz
zGznTg
z
z
zn
Multiplication by n
• Let
• Then
Example
Given
And
Then, 8
)()}({
)()}({
zFdz
dznTnfZ
zFnTfZ
.)2(
2)()(
2)(
2)(2)(
2
z
zzF
dz
dzzG
nnTg
z
zzFnTf
n
n
Z-differentiation
Inverse z-Transforms
• Given the z-transform, Y(z), of a function, y(n), it is required to find the time-domain function y(n).
• Here, we will study the following methods:power series (long division).expanding Y (z) into partial fractions and using
z-transform tables to find the inverse.9
Method 1: Power series (long division)
This method involves dividing the denominator of Y (z) into the numerator such that a power series of the form
is obtained. Notice that the values of y(n) are the coefficients in the power series.
10
33
22
110)( zyzyzyyzY
Example 6.6Find the inverse z-transform for the polynomial
Solution: Dividing the denominator into the numerator gives
11
.43
)(2
2
zz
zzzY
and the coefficients of the power series arey(0) = 1,y(T) = 4,y(2T) = 8,y(3T) = 8,
:The required sequence is
y(t) = δ(t) + 4δ(t-T) + 8δ(t-2T) + 8δ(t-3T) + …
The first few samples of the time sequence y(nT) is shown below
12
Example 6.7Find the inverse z-transform for Y(z) given by
Solution:Dividing the denominator into the numerator gives
13
.23
)(2
zz
zzY
and the coefficients of the power series arey(0) = 0,y(T) = 1,y(2T) = 3,y(3T) = 7,y(4T) = 15,:
The required sequence is
y(t) = δ(t-1) + 3δ(t-T) + 7 δ(t-2T) + 15 δ(t-3T) + …
The first few samples of the time sequence y(nT) is shown below
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The disadvantage of the power series method is that it does not give a closed form of the resulting
sequence. We often need a closed-form result, and other methods should be used when this is
the case.
15
Method 2: Partial fractions
• A partial fraction expansion of Y (z) is found, and then tables of z-transform can be used to determine the inverse z-transform.
• Looking at the z-transform tables, we see that there is usually a z term in the numerator. It is therefore more convenient to find the partial fractions of Y (z)/z and then multiply the partial fractions by z to obtain a z term in the numerator.
16
Example 6.8
17
Find the inverse z-transform of the function
SolutionThe above expression can be written as
The values of A and B can be found by equating like powers in the numerator, i.e.
We find A = − 1, B = 1, giving
.)2)(1(
)(
zz
zzY
.21)2)(1(
1)(
z
B
z
A
zzz
zY
.1)1()2( zBzA
.2
1
1
1)(
zzz
zY
From the z-transform tables we find that
and the coefficients of the power series arey(0) = 0,y(T) = 1,y(2T) = 3,y(3T) = 7,y(4T) = 15,
:so that the required sequence is
y(t) = δ(t-1) + 3δ(t-T) + 7 δ(t-2T) + 15 δ(t-3T) + …
Which is the same answer as Example 6.7.
18
21)(
z
z
z
zzY
nnTy 21)(
Example 6.9
Find the inverse z-transform of the function
Answer:The above expression can be written as
19
.)2)(1(
1)(
zzzY
.)2()1()2)(1(
1)(
z
C
z
B
z
A
zzzz
zY
• The values of A, B, C can be found as follows
20
5.0)2)(1(
1)2(
)()2(lim
1)2)(1(
1)1(
)()1(lim
5.0)2)(1(
1)(lim
2
1
0
zzzz
z
zYzC
zzzz
z
zYzB
zzzz
z
zYzA
z
z
z
.)2(
5.0
)1(
15.0)(
zzzz
zY
• Using the inverse z-transform, we find
• The coefficients of the power series are
and the required sequence is
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.)2(
5.0
)1(5.0)(
z
z
z
zzY
.21)(5.0)2(5.01)(5.0)( 1 nn nnnTy
15)5(
7)4(
3)3(
1)2(
0)(
0)0(
Ty
Ty
Ty
Ty
Ty
y
.)5(15)4(7)3(3)2()( TtTtTtTtty
Example 6.11
Using the partial expansion method described above, find the inverse z-transform of
SolutionRewriting the function as
22
.)1)(8.0)(5.0(
)(2
zzz
zzzY
.18.05.0)1)(8.0)(5.0(
1)(
z
C
z
B
z
A
zzz
z
z
zY
we find that
Thus
23
.20)1)(8.0)(5.0(
1)1(
,30)1)(8.0)(5.0(
1)8.0(
,10)1)(8.0)(5.0(
1)5.0(
1
8.0
5.0
z
z
z
zzz
zzC
zzz
zzB
zzz
zzA
.1
20
8.0
30
5.0
10)(
z
z
z
z
z
zzY
The inverse transform is found from the tables as
and the coefficients of the power series arey(0) = 0,y(T) = 1,y(2T) = 3.3,y(3T) = 5.89,
:so that the required sequence is
y(t) = δ(t-T) + 3.3δ(t-2T) + 5.89δ(t-3T) + … 24
20)8.0(30)5.0(10)( nnnTy
Example 6.12
Find the inverse z-transform of
SolutionRewriting the function as
25
.)2)(8.0)(5(
23)(
2
2
zzz
zzzY
.)2()2(8.05)2)(8.0)(5(
23)(22
2
z
E
z
D
z
C
z
B
z
A
zzzz
zz
z
zY
We obtain
2629.0
)]8.0)(5([
)44.83)(23()8.0)(5()32(
)8.0)(5(
23
,48.0)2)(8.0)(5(
23)2(
,16.0)2)(8.0)(5(
23)8.0(
,0056.0)2)(8.0)(5(
23)5(
,125.0)2)(8.0)(5(
23
2
2
22
2
2
2
2
22
8.0
2
2
5
2
2
0
2
2
z
z
z
z
z
z
zzz
zzzzzzzz
zzz
zz
dz
dD
zzzz
zzzE
zzzz
zzzC
zzzz
zzzB
zzzz
zzzA
• We can now write Y(z) as
• The inverse transform is found from the tables as
• Note that for the last term, we used the multiplication by n property which is equivalent to a z-differentiation.
27
.)2(
48.0
)2(
29.0
8.0
016.0
5
0056.0125.0)(
2
z
z
z
z
z
z
z
zzY
.)2(24.0)2(29.0)8.0(016.0)5(0056.0)(125.0)( nnnn nnny
Pulse Transfer Function and Manipulation of Block Diagrams
• Given the following system:
• Suppose we are interested in finding the relationship between the sampled output y*(s) and the sampled input u*(s).
• The continuous output y(s) is given by:
• Then, sampling the output signal gives:
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).()()( * sGsesy
)35.6()()()(
)()()]()([)( *****
zezGzy
sGsesGsesy
• Note that if at least one of the continuous functions has been sampled, then the z-transform of the product is equal to the product of the z-transforms of each function (note that [e*(s)]* = [e*(s)], since sampling an already sampled signal has no further effect).
• G(z) is the ratio of the z-transform of the sampled output and the sampled input at the sampling instants, and is called the pulse transfer function.
• Notice from (6.35) that we have no information about the output y(z) between the sampling instants.
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Example 6.17Derive an expression for the z-transform of the output of the system.
Solution The following expressions can be written for the system:
Which gives
or
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),()()(),()()(
),()()(),()()(**
2**
2
**1
**1
sXsGsYsXsGsY
sesGsXsesGsX
),()()()( **2
*1
* sEsGsGsY
).()()()( 21 zEzGzGzY
• For example, if
• Then
and the output function is given by
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,)(,1
)( 21 as
asG
ssG
,)}({,1
)}({ 21 aTez
azsGZ
z
zsGZ
),(1
)( zEez
az
z
zzY
aT
).())(1(
)(2
zEezz
azzY
aT
Example 6.16
The figure shows an open-loop sampled data system. Derive an expression for the z-transform of the output of the system.
32
SolutionThe following expression can be written for the system:
or
And
where
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),()()()( 21* sGsGsesy
)(])[()]()()([)( *21
**21
** sGGsesGsGsesy
),()()( 21 zezGGzy
).()()}()({)( 212121 zGzGsGsGZzGG
• For example, if
• Then, from the z-transform tables,
and the output is given by
34
,)(,1
)( 21 as
asG
ssG
,))(1(
)1(
)()}()({ 21 aT
aT
ezz
ez
ass
aZsGsGZ
).())(1(
)1()( zE
ezz
ezzY
aT
aT
Note about using Matlab• In Matlab, you can use the function c2d to convert a continuous
function into discrete. For example, to convert the transfer function
using a sample period T=1, you can use the following commands:>> G = tf([1],[1 5 6]);>> c2d(G,1,’impulse’)
• Matlab output:
35
.65
1)(
2
sssG