chapter 8 design of infinite impulse response digital filter
TRANSCRIPT
Chapter 8Chapter 8
Design of infinite impulse Design of infinite impulse response digital filterresponse digital filter
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IIR filter– Recursive equation of IIR filter
– Transfer function of IIR filter
1. Basic property of IIR filters1. Basic property of IIR filters
1 20 1 2
1 21 2
0
1
( )1
1
NN
MM
Nk
kk
Mk
kk
a a z a z a zH z
b z b z b z
a z
b z
0
0 1
( ) ( ) ( )
( ) ( )
k
N M
k kk k
y n h k x n k
a x n k b y n k
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– Transfer function of IIR filter• Factored form
1 2
1 2
( )( ) ( )( )
( )( ) ( )N
M
K z z z z z zH z
z p z p z p
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2. Design of IIR filter using analog 2. Design of IIR filter using analog filterfilter
Impulse invariant method– Identical impulse response of discrete filter to that of
analog filter
Analog Filter
Transfer Function
Digital Filter
Transfer Function
Impulse ResponseImpulse Response
Series
( )H s ( )H z
( )h t ( )h nT
Fig. 8-1.
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– Design a LPF using Impulse invariant method
( )a
H ss a
Fig. 8-2.
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• Inverse Laplace transform
• Sampling with T of inverse Laplace transform
• z-transform
( ) , 0ath t ae t
( ) anTnh h nT ae
0
0
1
0
1
( )
( )
1
z
z
z
nn
n
anT n
n
aT n
n
aT
H z h z
ae
a e
a
e
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• Transfer function with single pole
• Inverse Laplace transform
• Sampling with T of inverse Laplace transform
1
( )N
i
i i
CH s
s p
1
( ) i
Np t
ii
h t C e
1
( ) i
Np nT
ii
h nT C e
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• z-transform
• Using commutative law
0
0 1
( ) ( )
i
n
n
Np nT n
in i
H z h nT z
C e z
1 0
1
1 0
( )
( )
i
i
Np nT n
ii n
Np t n
ii n
H z C e z
C e z
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• Using an infinite series
11
0
1( )
1i
i
p T np T
n
e ze z
11
( )1 i
Ni
p Ti
CH z
e z
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– Repeated poles in designing filter• Repeated pole of l order
• z-transform
1 1
1 1
( 1)( )
( 1)! 1i
l li
l aT
a p
CH z
l a e z
( )( )
il
i
CH s
s p
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– Complex number in designing filter (1)
• z-transform
– Complex number in designing filter (2)
• z-transform
( )( )( )
s aH s
s a jb s a jb
1
2 21
1 ( cos )( )
1 (2 cos ) z
aT
aTaT
e bT zH z
e bT z e
( )( )( )
bH s
s a jb s a jb
1
2 21
( sin )( )
1 (2 cos ) z
aT
aTaT
e bT zH z
e bT z e
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– Example 8-1• second order Butterworth filter
• Partial fraction
• Impulse response function (T=1)
2
1( )
1 2H s
s s
2 2( )
( 1 ) 2 ( 1 ) 2
j jH s
s j s j
( 1 ) 2 1 ( 1 ) 2 1
1 2 1
1 2 1 2 2 2
2 2( )
1 1
2 sin(1 2)
1 2 cos(1 2)
j j
j jH z
e z e z
e z
e z e z
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• Magnitude of impulse response function
– Summary of impulse invariance method(1)
(2) Multiply H(z) by T
1( ) ( )sT sz e
n
H z H s jmT
Fig. 8-3.
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Bilinear z transform– Replacing s in the transfer function depending on the
filter required
• Arranging to z variable
2 1
1
zs
T z
12
12
Ts
zT
s
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• Replacing
• Considering frequency scaling
– Replacing ,
(1 )2 2
(1 )2 2
T Tj
zT T
j
s j
2 2
2 2
2 1
1
2
2tan
2
d
d
d d
d d
j T
j T
j T j T
j T j T
d
ej
T e
e e
T e eT
jT
s j dj Tz e
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– Relationship between analog frequency and digital frequency
2tan
2dT
T
tan2 2
dTT
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• Frequency warping
Fig. 8-4.
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– Example 8-2
• Using bilinear z transform
• in transfer function
1( )
11 ( )
1
H zk z
z
1( )
1 ( )c
H ss
k
11 1( )
2 2H z z
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1) Impulse response of analog filter :
2) Frequency response of analog filter:
3) Impulse response of digital filter :
4) Frequency response of digital filter:
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1
1 ( )c 1( )c - tan
cos2dT
1 sin-tan
1 cosd
d
T
T
cTce
Magnitude =
phase =
1 1, , 0, 0,
2 2
Magnitude =
phase =
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5) Relationship between and
tan2d
c
T
dT
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– Example 8-3• Specification of the desired filter
– Filter response : -3dB at 1000Hz
: -10dB at 3000Hz
– Sampling frequency : 10kHz
– Monotonic decrease in transition region(1000~3000Hz)
• Digital parameter from specification
–
–
–
1/100000[sec]T
2 1000[rad/sec], 0.2dp dpT
2 3000[rad/sec], 0.6dr drT
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• Considering Frequency warp
– Prewarp
• Determining order of Butterworth filter
tan( ) 0.32492 2p dpT T
tan( ) 1.37642 2
drr TT
21.376410 log 1 ( ) 10
0.3249N
1N
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• Using bilinear z transform
1( )
1 ( / )p
H ss
0.3949( )
10.3249
10.2452( 1)
0.5095
H zzz
z
z
2 1
1
zs
T z
Fig. 8-5.
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Two transformation method– Impulse invariant method
– Bilinear z transform
3. Comparing two transformation 3. Comparing two transformation methodmethod
sTz e
2 1
1
zs
T z
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– Example 8-4• Transfer function of analog filter
• Frequency response
• Using impulse invariant method (1)
1.333( )
1.333
aH s
s a s
1
( , 0.25 sec )0.75
a T
2 2
1.333( ) ( )
1.333aH j H j
1( ) tan1.333
j
1 1
1.333( )
1 1 0.716aT
aH z
e z z
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• Frequency response with
• Using bilinear z transform (2)
cos( ) sin( )
cos(0.25 ) sin(0.25 )
j Tz e T j T
j
2 2
1.333( ) ( )
1 0.716cos(0.25 ) 0.716sin(0.25 )
j TIIH e H j
1 0.716sin(0.25 )( ) tan
1 0.716cos(0.25 )j
2 1
1
( ) ( )
( 2) ( 2)
0.333 0.333
2.333 1.667
zs
T z
H z H s
aTz aT
aT z aT
z
z
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• Frequency response
2 22
2 2
( ) ( )
0.777cos (0.25 ) 0.222cos(0.25 ) 0.555 0.777sin(0.25 ) 1.332sin(0.25 )
2.333cos(0.25 ) 1.667 2.333sin(0.25 )
j TBH e H j
12 2
1.332sin(0.25 )( ) tan
0.777cos (0.25 ) 0.222cos(0.25 ) 0.555 0.777sin (0.25 )j
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Analog
Constant Impulse Response
Bilinear z Transform
Fig. 8-6.
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– Example 8-5
• Partial fraction of impulse response
• Inverse Laplace transform
2
1( )
2 1H s
s s
/ 2 / 2( )
1 1
2 2
j jH s
j js s
1
/ 2
( ) ( )
2 sin2
t
h t H s
te
L
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• Frequency response
• Analog filter with -3db at
2
4
1( )
( ) 2( ) 1
1
1
H jj j
2
2 2
7
2 3 7
( ) ( )
2
3.948 10
8.886 10 3.948 10
c
c
c c
H s H s
s s
s s
2 1000[rad/sec]c
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• Impulse invariant method
– Partial fraction using sampling period ( ) 410 [sec]T
( 2 )
1 ( 2 ) 2( 2 )2
2 sin( 2)( )
2 cos( 2)
c
c c
Tc c
T Tc
e T zH z
z ze T e
0.20.4443
2 2cT
3
1 2
2.449 10( )
1.158 0.4112
zH z
z z
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• Bilinear z transform
2 1
1
2
22
2 2 2
2 2 2 2 2
2
2
( ) ( )
2 1 2 12
1 1
( 1)
4( 1) 2 2 ( 1) ( 1)
0.064( 2 1)
1.168 0.424
zB sT z
c
c c
c
c c
H z H s
z zT z T z
T z
z T z T z
z z
z z
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Design of various filters using frequency transformation
4. Frequency transformation4. Frequency transformation
Analog low pass filter
(normalization filter)
Analogfrequency transform Low pass Low pass High pass Band pass Band reject
Desired digital filter
Bilinear z transform
or
Impulse invariant method
Fig. 8-7.
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Analog
low pass filter
(normalization)
Digital
low pass filter
(normalization)
Analog
Low pass
High pass
Band pass
Band reject
Digital
Low pass
High pass
Band pass
Band reject
Bilinear transform
Bilinear transform
Analog frequency transform
Digital frequency transform
Fig. 8-8.
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– Low pass filter• -3dB at
– High pass filter• By replacing to
• -3dB at
2
1( )
2 1H s
s s
2
1( )
( / ) 2( / ) 1c c
H ss s
1[rad/sec]c
( )1
sH s
s
( )c
sH s
s
s1
s
1[rad/sec]c
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– Band pass filter• -3dB at
– Band reject filter• -3dB at
2 20s
ss
2 20
ss
s
1[rad/sec]c
1[rad/sec]c
Table. 8-1 Analog Frequency Transform
Low pass filter (cutoff frequency )
Low pass filter (cutoff frequency : )
High pass filter (cutoff frequency : )
Band pass filter(Upper cufoff frequency : ,Lower cufoff
frequency: , Band pass frequency : )
Band reject filter (Upper cufoff frequency : ,Lower cufoff frequency :
, Band reject frequency : )
1[rad sec]c ( )H s
c
ss
c
css
20 ,h l h l
h l2 2
0
( )h l
ss
s
20 ,h l h l
h l2 2
0
( )h lss
s
c
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– Example 8-6• Specification of filter design
– -3dB at 10Hz
– Sampling frequency ( )
– Bilinear z transform
– Transfer function :
• Considering prewarp
2
1( )
1.414 1H s
s s
'tan tan 0.325
2 10c
c
T
100[Hz]sf
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• Transfer function of analog LPF
• Bilinear z transform
2
2
1( )
1.414 10.325 0.325
0.105
0.46 0.105
H ss s
s s
11
2
2
2
1 2
1 2
0.105( ) ( )
1 10.46 0.105
1 1
0.105 0.21 0.105
1.56 1.79 0.645
0.067 0.135 0.067
1 1.147 0.413
zzs
H z H sz zz z
z z
z z
z z
z z
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• For computational efficiency
• For accurate frequency
'2tan
2c
c
T
T
2 1
1
zs
T z
'tan
2c
c
T
1
1
zs
z
'2tan
2
2tan 65[rad / sec]
0.01 10
cc
T
T
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Fig. 8-9.
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– Example 8-7• Specification of filter design
– cutoff frequency ( )
– Sampling frequency ( )
– Bilinear z transform
– Transfer function :
• Cutoff frequency using prewarp
1( )
1LH ss
'tan tan 0.7265
2 5c
c
T
150[Hz]sf
30[Hz]sf
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• Analog HPF using table 8-1
• Bilinear z transform
( ) ( )
1
0.72651
cs
L s
c
H s H s
s
ss
11
1
1
( ) ( )
0.5792 0.5792
1 0.1584
zzs
H z H s
z
z
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Fig. 8-10.
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– Example 8-8• Specification of filter design
– Band pass frequency ( )
– Sampling frequency ( )
– Order of filter : 2
– Bilinear z transform
• Using table 8-1
200 ~ 300[Hz]sf
2[kHz]sf
2 20
( )h l
ss
s
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• Analog bandpass filter
1( )
1LH ss
20
'tan tan 0.3249
2 10
' 3tan tan 0.5095
2 20
0.1655
0.1846
hh
ll
h l
h l
T
T
2 20
2
( ) ( )
0.1846
0.1846 0.0274
s
s
Ls
H s H s
s
s s
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• Bilinear z transform
11
2
1 2
( ) ( )
0.1367 0.1367
1 1.2362 0.7265
zzs
H z H s
z
z z