nonrecursive digital filters. digital filters & filter equation general equation transfer...
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Nonrecursive Digital Filters
Digital Filters & Filter Equation
General Equation
M
kk
N
kk knxbknya
00
][][
M
kk
M
k
kk
M
k
kk
M
kk
jkbHzbzH
zbzXzYknxbny
00
00
)exp()()(
)()(][][
Transfer function
Frequency response
- FIR- Convolution
• Disadvantage : takes computation time
• Advantage : stable (zeros only)
linear phase (no phase distortion)
same phase shift to all frequencies
M
kk
M
jMM
jjjjjMM
M
Mkk
kbb
Mbbbb
ebebebbebebeb
jkbH
10
210
22101
22
cos2
cos2...2cos2cos2
........
)exp()(
Nonrecursive Filter
M
kk
M
MM
H
1)cos(21
)12(
1
cos2...2cos2cos21)12(
1)(
Smoothness of the signal correlated to the increment of M
Width of mainlobe negatively correlated to M
increment of M narrow band lowpass filter
2M+1 coefficients, symmetric to n=0
Moving Average Filter
Impulse response of moving average filter
5-point(M = 2)
No zeros at z=0
since passband around 0
21-point(M = 10)
Frequency Response of Moving Average Filter
= 0 peak value = 1
unwanted side lobe first side lobe 22% of main lobe
5 terms 4 zeros missing zero at z = 1
21 terms 20 zeros passband contains at = 0
Zeros lie actually on the unit circle true nulls in the corresponding
frequency
ex)
4
234
4321
1
5
1)(
1)(5
1)(
)4()3()2()1()(5
1)(
z
zzzzzH
zzzzzXzY
nxnxnxnxnxny
Frequency Response of Moving Average Filter
11
)(H1.0
)(sin)sin(1
][
2
)exp()exp(1
)exp()exp(2
1
)exp(
2
1)(exp1
2
1
)exp()(2
1][
11
1
11
11
1
1
1
1
ncnn
nh
j
njnj
n
njnjjn
jn
njdnj
dnjHnh
Ideal Lowpass Filter Method
)cos()sin(1
][ 01 nnn
nh
1
1 )cos(][2)(k
kkhH
Limit to 2M+1 terms, and start from n=0
M
k
kkhH1
1 )cos(][2|)(|
(bandwidth : center frequency : )12 0
Center frequency :
bandwidth :
00 60
01 30
Lowpass Filter
Replace withkb nh
Design of Highpass/Bandpass Filters using Lowpass Filter
Cutoff frequency : Sampling rate : KHz100 HzfT s610sec1
2.010
10226
5
s
c f
fT
H
0 2.02.0
nn
nn
nhLP
2.0sin1
sin1
1
Lowpass Filter Design
Cutoff frequency highpass filter 5.0 H
nn
nnn
nh nHP
5.0sin
11cossin
101
n118000
5.00 HPh 1
1 HPh 02 HPh 31
3HPh
H
0 5.0
Highpass Filter Design
Cutoff frequency : Sampling rate :
Duration of impulse response :
Hzf 200160 1 Hz800
sec50m
45.0
2
5.04.00
Center frequency
Bandwidth
1.04.05.02 1 05.0
2
1.01
nnn
nhBP
45.0cos205.0sin1
Bandpass Filter Design
Frequency Transformation
Recursive Digital Filters
))()((
))()(()(
321
321
0
0
pzpzpz
zzzzzzK
Za
ZbzH
N
k
kk
M
k
kk
][...]1[][][...]1[][ 1010 MnxbnxbnxbNnyanyanya MN
Recursive filter
powerful : separate control over the numerator and denominator of H(z)
If the magnitude of the denominator becomes small at the appropriate frequency
produce sharp response peaks by arranging
General Form of Filters
Find the difference equation of Bandpass Filter
(a) Center frequency : = /2, -3dB Bandwidth : /40, Maximum gain : 1(b) No frequency component at = 0, =
- Assume BC is straight line- d = 1 - r (r > 0.9)- 2d = 2 (1-r)- 2 (1-r) [rad] = /40 = 3.14/40, r = 0.961- No frequency component at = 0 and = - two zeroes at z = +1
and -1
05.440
rad r
origin
Example #1
))()((
))()(()(
321
321
0
0
pzpzpz
zzzzzzK
Za
ZbzH N
k
k
k
M
k
k
k
• -3dB band-width : /40
• Maximum gain : 26.15 (28.35dB)
• in equaiton , K = (6.15)① -1 = 0.03824
①
2exp961.0
2exp961.0
)1)(1(03824.0
)(
)()(
jzjz
zz
zX
zYzH
9235.0
)1(03824.02
2
z
z
• The corresponding difference equaion is :
y[n+2] + 0.9235y[n] = 0.03824{x[n+2] - x[n]}
subtracting 2 from each term in brackets
y[n] = -0.9235y[n-2] + 0.03824{x[n] - x[n-2]}
Example #1
Design a band-reject filter which stops 60Hz powerline noise from ECG signal- 10Hz cutoff bandwidth at -3dB point- Poles and zeros as in the picture
(solution) - fs = 1.2 kHz
- fmax : 600Hz
- 2 : 1200 = o: 60
- o (60Hz) = 0.1 97382.0
1201,
600
10)1(2
rr
94833.08523.1
19021.1
94833.0)1.0cos(9476.1
1)1.0cos(2
})1.0(exp97382.0{})1.0(exp97382.0{
})1.0(exp{})1.0(exp{
)(
)()(
2
2
2
2
zz
zz
zz
zz
jzjz
jzjz
zX
zYzH
y[n+2] - 1.8523 y[n+1] + 0.94833 y[n] = x[n+2] - 1.9021 x[n+1] + x[n]
y[n] = 1.8523 y[n-1] - 0.94833 y[n-2] + x[n] - 1.9021 x[n-1] + x[n-2]
Example #2
Butterworth
Chebyshev – 1st order
Elliptic
Types of Filters
Chebyshev – 2nd order
2/122/12
1
)2
tan(
)2
tan(1
1|)(|
1
1)(
n
C
nHH
2/1
22
2/1
1
22
)2
tan(
)2
tan(1
1|)(|
1
1)(
Cn
nC
H
C
H
2/1
1
221
1)(
LR
H
n
Butterworth
Chebyshev
Elliptic
analog digital
)()(2)(
)( 1)(
)1(1
21
10
2/12
xCxCxxC
xxCxC
nnn
및
ripple
Butterworth, Chebyshev, Elliptic Filters
Find the minimum order of Filter
Cutoff frequency 1= 0.2
Frequency response of less than 30dB at = 0.4
5,29.4,1000236.21
03162.0}236.21{
1
03162.0)20
30(log
}236.21{
1
}]1.0tan2.0tan
[1{
1)4.0(
2
2/12
110
2/122/12
nn
H
n
n
nn
x10log2030
20
30
10
x
2
21
Example #3
jezjwsz
zs
1
1
H(s) H(z)
0 0
...
...)(
321
321
pspsps
zszszsKsH
Bilinear Transformation
)2/tan(
)2/tan()2/cos(2
)2/sin(2)(
)(
1
12/2/2/
2/2/2/
w
jj
eee
eee
e
ejw
jjj
jjj
j
j
14
tan2/2
tan
0
0
Bilinear Transformation
Another method of deriving a digital filter from an analog filter
A sampled version of that of the reference analog filter
Impulse-invariant Filters
Impulse-invariant Filters
13
3
2
2
1
1
321
321
)(
)()(
i i
i
ps
K
ps
K
ps
K
ps
K
pspsps
zszszsKsH
Transfer function of
analog filter
Impulse-invariant
filter
Impulse-invariant Filters
00
0)exp(|)(][
00
0)exp()(
n
nTnpKthnh
t
ttpKth
iinTtii
iii
)exp()exp(1
])[exp()exp()(
1
0
1
0
Tpz
zK
zTp
K
zTnpKzTnpKzH
i
i
i
i
n
nii
n
niii
A zero at the origin of the z-plane
A polse at Tpz iexp
The impulse response of each analog subfilter takes a simple exponential form
For the i-th subfilter
Impulse-invariant Filters
Design of Recursive Digital Filters
Nc
LP wwjwH
22
)/(1
1|)(|
2/12 )]/[1(
1|)(|
Nc
LP wwjwH
NLP wjwH
22
1
1|)(|
2/12 )1(
1|)(|
NLP wjwH
or
or
Prototype : when 1c
frequency responses at N = 1, 2, 3
Butterworth LP Analog Filter Design (prototype)
NNjsNLPLPLPsjsw
sHsHjwH222
2
1
1
1
1|
1
1)()(|)(|
When N : odd
When N : even
kjNN ess 222 101 N
kj
es 2
2
N
kj
k es
kjNN ess 222 101
N
kj
k es 2
2
11
0
0
jes 11
1
1
jes
707.0707.0422
02
0 jeesjj
N = 1 ;
N = 2 ;
N = 3 ; only 3 effective terms
707.0707.0,707.0707.0,707.0707.0 321 jsjsjs
Determination of Poles
1st order
s
sH
1
1 707.0707.0707.0707.0
1
jsjssH
2nd order
Determination of Poles
frad 2sec/1
Order of filter
Design a lowpass Butterworth filter : -3dB at 1 rad/sec (prototype filter)
gain of less than 0.1 for the frequency greater than 2 rad/sec
Hzf 159.0
dB201.0log20 10
31.3
2log2
110log
10
10/2010
N 4N
1613.2414.3613.2
1234
ssss
sH LP
a
M dB
N10
10/10
log2
110log
Example
)(1
1|)(|
222
wCjwH
NLP
or 2
122 )(1
1|)(|
wCjwH
N
LP
N : order, : cutoff frequency, r : ripple amplitude ( : ripple parameter)
11 2)( NNN CCwC
1coscos)( NwCN
210 1log10 dBr
Chebyshev LP Analog Filter Design (prototype)
Order of filter
Chebyshev Prototype Denominator Polynomials
Maximum passband ripple : 1dB, Cutoff frequency : less than 1.3 rad/sec
Attenuation in stopband : 40dB for greater than 5 rad/sec
ripple parameter 210 1log10 dBr 2589.02
cutoff frequency : -3dB point is half the magnitude 5.02 jH LP
707.0jH LP
11 2)( NNN CCwC
)(1 wC
12)( 22 wC
5.069.0|)(1
1|)(| 3.12
12
2
wCjwH LP
5.041.0|)(1
1|)(| 3.12
22
2
wCjwH LP 2nd order
Passband characteristic
4352
22
2 10106.1|)(1
1|)(|
wCjwH LP
4552
32
2 10106.1|)(1
1|)(|
wCjwH LP 3rd order
491.0238.1988.0 23
sss
KsH LP
Example
c
LPLPP
HP
cLPP
BP
BP
LP BP
20
2
20
2
BS
BSLP
BP
c
s
sc
Bs
s 20
2
20
2 s
Bs
form form s
Analog Filter Frequency Transformation
Butterworth bandpass filter
Hzfu 900
?of
Hzfl 600
sec/56559002 radu
sec/37706002 radl
sec/1032.21 620 radlu Hzf 735
20
0
sec/1885 radB lu
sec/5026800 radHz sec/1256200 radHz
Prototype equivalent frequency
BP
BP
BP
BP
LP BP
6220
2 1032.21
1885
1
a
M dB
N10
10/10
log2
110log
Filter
order
75.12.0418.0 NM dBa
71.25035.8 NM dBa 3rd order
Maximum attenuation of 0.2dB for
Minimum attenuation of 50dB for
Hzffo 800
Hzf 2000
Example
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