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The Z-transform
1
M. Sami Fadali
Professor of Electrical Engineering
UNR
Z-transform Definition
Definition 2.1 Given the causal sequence{u0, u1, u2, ... , uk,
... } then its z-transform is
defined as
2
operatordelaytime
...)(
1
0
2
2
1
10
=
=
+++++=
=
z
zu
zuzuzuuzU
k
k
k
k
k L
Example
Obtain the z-transform of the sequence
{ } { }0,0,0,4,0,2,3,1)( =ku
3
Solution Definition 2.1 gives
U(z) = 1+3z1 +2z2 + 4z4
Z-transform Definition
Definition 2.2 Laplace transform the impulsetrain representation
of sampled signal
k kTtuTtuTtututu +++++=
...)(...)2()()()( 210*
4
sT
k
k
k
k
skT
k
skT
k
sT
k
k
ezzu
eu
eueuuU*(s)
tu
==
=
++++=
=
=
=
=
,0
0
10
0
LL
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Identities Used Repeatedly
1,1
1 1
0
=
+
= a
a
aa
nn
k
k
5
1,1
1
0
=
=
=
=
14
Y z i j z
i z j z F z F z
j i
ji
i j
ji
( ) ( ) ( )
( ) ( ) ( ) ( )
( )=
=
=
+
=
=
=
=
f f
f f
1 2
00
1 2
00
1 2
Example
Find the z-transform of the convolution oftwo sampled step
sequences.
Solution:By the convolution theorem,
15
z- rans orm = pro uc o e z- rans orms otwo step sequences.
F zz
z
z
z
z
z
( ) =
=
1 1
1
2
Multiplication by Exponential
Proof
{ } )()(f zaFka k =Z
16
( )
LHS a k z
k a z
F a z
k
k
k
=
=
=
=
=
f
f
( )
( )
( )
0
0
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Example
Find the z-transform of the exponentialsequence
f( ) =k e kkT = , , , , ...0 1 2 1 1
17
( ){ }( ) TT
kT
ez
z
zeke
=
=
11
1)(fZ
( )f( ) =k e kTk =1 0 1 2, , , ,...
- z z=
(same as earlier example ak )
Complex Differentiation
Proof(Induction)(i) Establish validity for m = 1.
{ } )()(f zFzd
dzkk
m
m
=Z
18
m m .For m = 1, we have
{ }
)()(f)(f
)(f)(f
00
0
zFzd
dzzk
zd
dzz
zd
dzk
zkkkk
k
k
k
k
k
k
=
=
=
=
=
=
=
Z
Proof (Cont.) For any m f fm
mk k k k ( ) ( ), , , , ...= = 0 1 2
{ } )(f)(f0
d
zkkkk
k
k
k
mm
=
=
=
Z
19
)()(f0
0
zFzd
dzzk
zd
dz
zd
m
k
k
m
k
m
=
=
=
=
Substitute for Fm(z) { } )()(f1
zFzd
dzkk
m
m
+
=Z
ExampleFind the z-transform of the sampled ramp sequence
f(k) = k, k= 0, 1, 2,
Solution: z-transform of a sampled step F zz
z( ) =
1
20
r te f(k) as: f(k) = k 1,k= 0, 1, 2,
Apply the complex differentiation property
{ } ( )
( )
( ) ( )22 11
1
11 =
=
= z
z
z
zz
zz
z
zd
d
zkZ
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Inversion of the z-Transform
1. Long division: gives as many termsof series as desired.
21
.look-up: similar to Laplace transforminversion.
Long Division
(i) Using long division, expand F(z) as a series
F z f f z f z
z
t i
i
ik
( ) ...= + + +
=
0 1
1
22
k=0
(ii) Write the inverse transform as the sequence
{ }f f fi0 1, ,..., ,...
ExampleF z
z
z z( )
. .=
+
+ +
1
0 2 012
Solution:(i) Long Division ......26.08.0
1.02.0
11.02.0
321
1-
2++
++
+++
zzz
zz
zzz
Inverse z-transform
23
...26.0
08.016.08.0
10.08.0
1
21-
1-
++
z
zz
z
321
)26.0(8.00)(
+++= zzzzFt
(ii) Inverse Transformation { } { }fk = 0 108 0 26, , . , .
,...
Partial Fraction Expansion
(i) Find the partial fraction expansion of F(z)/z.
(ii) Obtain the inverse transform f(k) using the
z-transform tables.
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Three types of z-domain functions F(z):
1. F(z) with simple (non-repeated) real poles.
2. F(z) with complex conjugate & real poles.
3. F(z) with repeated poles.
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I: Simple Real RootsResidue of a complex function F(z) at a
simple polezi
]A z z F zi i z zi= ( ) ( )
Residue = partial fraction coefficient of the ith
25
term o t e expans on
( )F zA
z z
i
ii
n
==
1
Example
Obtain the inverse z-transform of the function
F zz
z z( )
. .=
+
+ +
1
0 3 0 022
Solution:Solve using two different methods.
26
(i) Partial Fraction Expansion (dividing by z)
F z
z
z
z z z
A
z
B
z
C
z
( )
( . . )
. .
=+
+ +
= ++
++
1
0 3 0 02
01 0 2
2
Example (cont.)
A zF z
zF
B zF z
z
z
z
=
= = =
= +
=
=
=
=
( )( )
.
( . )( ) .
( . )( . ).
0
0 1
01
0 0250
011 01
01 0190
27
C zz
z z= +
=
=
=
( . )( ) .
( . )( . ).0 20 2
1 0 2
0 2 0140
Partial fraction expansion
F zz
z
z
z
z
z( ) . .= + + +
50 90
01
40
0 2
Example (cont.)
(ii) Table Lookup
f kk k
k
k k
( )( ) ( . ) ( . ) ,
,=
+
