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Signal Processing in the Discrete Time Domain
Microprocessor Applications (MEE4033)
Sogang UniversityDepartment of Mechanical Engineering
Definition of the z-Transform
Overview on Transforms
• The Laplace transform of a function f(t):
0)()( dtetfsF st
• The z-transform of a function x(k):
0
)()(n
kzkxzX
0
)()(k
kTiTi ekxeX
• The Fourier-series of a function x(k):
Example 1: a right sided sequence
1 2 3 4 5 6 7 8 9 10-1-2-3-4-5-6-7-8k
x(k)
. . .
kakx )( 0kfor , is
az
z
azazza
zkxzX
k
k
k
kk
k
k
10
1
0 1
1)(
)()(
For a signal )(zX
Example 2: a lowpass filter
)()1()( kbrkayky Suppose a lowpass filter law is
where1ba
0 0
1
0 0
)1(1
0 00
)()(
)()1(
)()1()(
k k
kk
k k
kk
k k
kk
k
k
zkbrzkyaz
zkbrzkyaz
zkbrzkayzky
1/3
Example 2: a lowpass filter
0 0
1
0
)()()(k k
kk
k
k zkbrzkyazzky
2/3
Rearranging the equation above,
00
1 )()()1(k
k
k
k zkrbzkyaz
)()1(
)()( zRaz
zazR
az
bzzY
Signals
Transfer function
Example 2: a lowpass filter
3/3
Signals
Transfer function
The block-diagram representation:
)(kr )(kyaz
za
)1(
Example 3: a highpass filter
)()()( kykrkh A highpass filter follows:
where)()1()1()( krakayky
1/2
)()1(
)()1(
)(
)()()(
zRaz
za
zRaz
zazR
zYzRzH
Transfer function
z-Transform Pairs
Discrete-time domain signal z-domain signal
otherwise 0,
0for,1)(
kk 1
mz
11
1 z
11
1 az
1/2
otherwise 0,
for,1)(
mkmk
otherwise 0,
0for,1)(
kku
0for, ka k
z-Transform Pairs
2210
10
]cos2[1
]sin[
zrzr
zr
2210
10
]cos2[1
]cos[1
zrzr
zr
210
10
]cos2[1
][sin
zz
z
0for),cos( 0 kk 210
10
]cos2[1
][cos1
zz
z
2/2
Discrete-time domain signal z-domain signal
0for),sin( 0 kk
0for),cos( 0 kkr k
0for),sin( 0 kkr k
Example 4: a decaying signal
Suppose a signal is for . Find .kky 9.0)(
1
0
1
0
9.01
1
)9.0(9.0)()(
z
zzzkyzY kkkk
0k )(zY
19.01
1)(
zzYkky 9.0)( 0kfor
z-transform
Inversez-transform
Example 5: a signal in z-domain
Suppose a signal is given in the z-domain:
25.0)(
2
z
zzY
221
1
5.0)5.0cos(5.021
)5.0sin(5.02)(
zz
zzY
)5.0sin()5.0(2)( kky k 0kfor
z-transform
Inversez-transform
From the z-transform table,
25.0)(
2
z
zzY
The signal is equivalent to
Properties of the z-Transform
Linearity of z-Transform
)()]([ zXkx Z
)()]([ zYky Z
)()()]()([ zbYzaXkbykax Z
where a and b are any scalars.
Example 6: a signal in z-domain
Suppose a signal is given in the z-domain:
25.0)(
2
2
z
zzY
11 5.01
5.0
5.01
5.0)(
zz
zY
kkky )5.0(5.0)5.0(5.0)(
0kfor
z-transform
Inversez-transform
Since the z-transform is a linear map,
11 5.01
5.0
5.01
5.0)(
zz
zY
Arranging the right hand side,
Shift
)()]([ zXkx Z
)()]([ zXzmkx mZ
Example 7: arbitrary signals
z-transform
Inversez-transform
Any signals can be represented in the z-domain:
1 2 3 4 5 6 7 8 9 10-1-2-3-4k
y(k)5
55)( 0 zzY
z-transform
Inversez-transform
1 2 3 4 5 6 7 8 9 10-1-2-3-4k
y(k)3
321 213)( zzzzY2
1
Discrete-Time Approximation
Backward approximation
)( kTty )(1 1
zYT
z
Forward approximation
)( kTty )(1
zYT
z
Trapezoid approximation
)( kTty )(1
12zY
z
z
T
Multiplication by an Exponential Sequence
)()]([ zXkx Z
)()]([ 1zaXkxa k Z
Initial Value Theorem
0for ,0)( nnx
)(lim)0( zXxz
Convolution of Sequences
)()]([ zXkx Z
)()]([ zYky Z
)()(
)()()]()([0
zYzX
iyikxkykxk
i
ZZ
1/2
0
)()()()(i
ikyixkykx
0 0
)()()]()([k
k
i
zikyixkykxZ
0 0
)()(i
k
k
zikyix
0 0
)()(i
k
k
i zkyzix
)()( zYzX
2/2
Convolution of Sequences
Proof:
z-Transform of Linear Systems
Linear Time-Invariant System
)(kx
)(zX
)(kg
)(zG
k
i
ixikgky0
)()()(
)()()( zXzGzY
Nth-Order Difference Equation
M
r
rr
N
i
ii zbzXzazY
00
)()(
M
rr
N
ii rkxbikya
00
)()(
N
i
ii
M
r
rr zazbzG
00)(
z-Transform
Stable and Causal Systems
Re
Im
1
N
ii
M
rr
dz
czc
zG
1
10
)(
)(
)(
The system G(z) is stable if all the roots (i.e., di) of the denominator are in the unit circle of the complex plane.
Stable and Causal Systems
Re
Im
1
The system G(z) is causal if the number of poles is greater than that of zeros (i.e., M N).
N
ii
M
rr
dz
czc
zG
1
10
)(
)(
)(
Example 8: a non-causal filter
011
1
011
1
...
...
)(
)(
azazaz
bzbzbzb
zR
zYn
nn
mm
mm
Suppose a transfer function is given
By applying the inverse z-Transform
Therefore, the system is causal if
)()1(...)1()(
)()1(...)1()(
011
011
nkrbnkrbnmkrbnmkrb
nkyankyakyaky
mm
n
nm
Example 9: open-loop controller
ukyycym Suppose the dynamic equation of a system is
Approximating the dynamic equation byT
kykyy
)()1(
)()(112
2
2
zUzYkT
zc
T
zzm
The transfer function from U(z) to Y(z) is
012
2
)(
)(
azaz
b
zU
zY
1/2
Example 9: open-loop controller
A promising control algorithm is
2/2
012
2
azaz
b
)(zU )(zY
012
2
azaz
b
)(zU
)(zY2
012
b
azaz )(zR
However, the control algorithm is non-causal.
Frequency Response of H(z)
• The z-transform of a function x(k):
k
kzkxzX )()(
k
kTiTj ekxeX )()(
• The Fourier-transform of a function x(k):
(Recall: Similarity of the z-Transform and Fourier Transform)
• The frequency response is obtained by settingTjez
where T is the sampling period.
Example 10: frequency response of a low pass filter
Suppose a lowpass filter
1/2
)(1
)1()(
1zR
az
azY
By substituting for z,Tie
)(1
)1()( Ti
TiTi eR
ae
aeY
The magnitude is
)(1
)1()( Ti
TiTi eR
ae
aeY
2/2
Since ,
)(1
)1()( Ti
TiTi eR
ae
aeY
)sin()cos( TiTe Ti
)cos(21
1
)(sin))cos(1(
1
))sin()(cos(1
)1(
1
)1(
2
22
Taa
a
TTa
a
TiTa
a
ae
aTi
Example 10: frequency response of a low pass filter
IIR Filters and FIR Filters
An IIR (Infinite Impulse Response) filter is
)(
)(
)(
)(
i
i
pz
zzk
zR
zY
A FIR (Finite Impulse Response) filter is
)()(
)(izzk
zR
zY
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