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Transcript of Signal Processing in the Discrete Time Domain Microprocessor Applications (MEE4033) Sogang...
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