Transcript 슬라이드 1 - Chonbuk

Nonrecursive Digital Filters
Digital Filters & Filter Equation
General Equation
N
a
k 0
- FIR
- Convolution
k
y[n  k ] 
M
y[n]   bk x[n  k ]
 H ( z) 
M
b z
k 0
b
k 0
k 0
Transfer
function
M
k
k
k
x[n  k ]
M
Y ( z )  X ( z )  bk z k
k 0
H () 
M
b
k 0
k
exp( jk)
Frequency
response
Nonrecursive Filter
• Disadvantage
: takes computation time
• Advantage
: stable (zeros only)
linear phase (no phase distortion)
same phase shift to all frequencies
H ( ) 
M
b
k  M
k
exp( jk)
 b M e jM  .... b 2 e j 2   b1e j  b0  b1e  j  b2 e  j 2   .... bM e  jM
 b0  2b1 cos  2b2 cos 2  ...  2bM cos M
M
 b0  2 bk cos k
k 1
Moving Average Filter
Impulse response of moving average filter
H () 

1
1  2 cos  2 cos2  ...  2 cos M
(2M  1)
M
1
1  2 
cos(k)


k 1

(2M  1) 
2M+1 coefficients, symmetric to n=0
 Smoothness of the signal  correlated to the increment of M
 Width of mainlobe  negatively correlated to M
 increment of M  narrow band lowpass filter
Frequency Response of Moving Average Filter
5-point
(M = 2)
21-point
(M = 10)
No zeros at z=0
since passband around   0
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)
1
 x(n)  x(n  1)  x(n  2)  x(n  3)  x(n  4)
5
1
Y ( z )  X ( z ) 1  z 1  z  2  z 3  z  4
5
y ( n) 

1 z4  z3  z2  z 1
H ( z) 
5
z4

Ideal Lowpass Filter Method
H ( )

1
h[n] 
1
2
1

2
1.0

  H () exp(jn)d

1
1  exp( jn) 
1

exp
(
j

n
)
d



1
2 
jn
 1
1
1
exp( j1n)  exp( j1n)
2jn
1 exp( j1n)  exp( j1n)

n
2j
1

 h[n] 
sin(n1 )  1 sin c(n1 )
n



1
Design of Highpass/Bandpass Filters using Lowpass Filter
h[n] 
Lowpass Filter
H () 
1
sin( n1 ) cos( n 0 ) (bandwidth : 21
n
1


 2 h[k ] cos(k)
k 1
center frequency :  0 )
Replace bk with hn
Limit to 2M+1 terms, and start from n=0
| H () | 
1

M
 2 h[k ] cos(k)
k 1
Center frequency : 0  600
bandwidth : 1  300
Lowpass Filter Design

Sampling rate : T  1 sec f s  106 Hz
Cutoff frequency : 100 KHz


2f
2 105
 c  T 

 0.2
fs
106
H 
 0.2
hLP n  
0
0.2
1
1
sin n1  
sin 0.2n 
n
n



Highpass Filter Design
highpass filter H 
Cutoff frequency 0.5
H 
0
hHP n  


0.5
1
1
n
sin n1  cos n 0    1
sin 0.5n 
n
n
0  1800  1
n
hHP 0  0.5
hHP  1  
1

hHP  2  0
hHP  3 
1

3
Bandpass Filter Design
Cutoff frequency : 160  f1  200Hz
Sampling rate : 800 Hz
Duration of impulse response : 50m sec
0.4  0.5
 0.45
2
Center frequency
0 
Bandwidth
21  0.5  0.4  0.1
hBP n  
 1 
0.1
 0.05
2
1
sin 0.05n   2 cos 0.45n 
n
Frequency Transformation
Recursive Digital Filters
General Form of Filters
M
H ( z) 
b
k 0
N
a
k 0
k
Z k
k
Z k

K ( z  z1 )( z  z 2 )( z  z 3 )     
( z  p1 )(z  p 2 )(z  p 3 )     
a0 y[n]  a1 y[n  1]  ... a N y[n  N ]  b0 x[n]  b1 x[n  1]  ... bM x[n  M ]
 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
Example #1
Find the difference equation of Bandpass Filter
(a) Center frequency :  = /2,
-3dB Bandwidth :  /40,
(b) No frequency component at  = 0,  = 

40
-
rad   4.50
Maximum gain : 1
r
origin
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
Example #1
M
H ( z) 
b Z
k
k
k 0
N
a Z

k
K ( z  z1 )(z  z2 )(z  z3 )     
( z  p1 )(z  p2 )(z  p3 )     
①
k
k 0
• -3dB band-width : /40
• Maximum gain : 26.15 (28.35dB)
• in equaiton ①, K = (6.15)-1 = 0.03824
Y ( z)
0.03824( z  1)(z  1)
H ( z) 

X ( z)

 
  

z

0
.
961
exp
j
z

0
.
961
exp

j






2
2





0.03824( z 2  1)

z 2  0.9235
• 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 #2
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) -
2(1  r )


H ( z) 
fs = 1.2 kHz
fmax : 600Hz
2 : 1200 = o: 60
o (60Hz) = 0.1 
10

, r  1
 0.97382
600
120
Y ( z)
{ z  exp( j 0.1 ) }{ z  exp( j 0.1 ) }

X ( z ) { z  0.97382exp( j 0.1 ) }{ z  0.97382exp( j 0.1 ) }
z 2  2 z cos(0.1 )  1
z 2 1.9021z  1
 2
 2
z 1.9476cos(0.1 ) z  0.94833 z 1.8523z  0.94833
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]
Types of Filters
Butterworth
Chebyshev – 1st order
Chebyshev – 2nd order
Elliptic
Butterworth, Chebyshev, Elliptic Filters
analog
Butterworth
Chebyshev
Elliptic
ripple
H ( ) 
H ( ) 
H ( ) 
digital
1
| H ( ) | 
1/ 2
2n

   

1    


  1 

1
| H ( ) | 
1/ 2

2   
2
1   Cn  
 1 

1
1/ 2
2n
 
  
tan( )  

 
2  
1  
C  

  tan( )  

2  
 

1
1/ 2

 

t an( ) 



2
2
2 
1   Cn 

 t an( C ) 


2 



1
1/ 2


2 
2


1


R
L

n 

1



  1  (1   2 ) 1/ 2
C0 ( x)  1 및 C1 ( x)  x
Cn ( x)  2 x Cn 1 ( x)  C n  2 ( x)
Example #3
Find the minimum order of Filter
Cutoff frequency 1= 0.2
Frequency response of less than 30dB at  = 0.4

2

H (0.4 ) 
1
1

t an 0.2 2 n 1/ 2 {1  2.2362 n }1/ 2
{1  [
] }
t an 0.1

1  30
log10
(
)  0.03162
20
1
 0.03162
{1  2.2362 n }1/ 2
1  2.2362 n  1000,
1
2
n  4.29, n  5
 30  20log10 x

x  10
30
20
Bilinear Transformation
H ( s) 
K s  z1 s  z 2 s  z3 ...
s  p1 s  p2 s  p3 ...
s
z 1
z 1
s  jw
z  e j
H(s)
H(z)
0   
0   
Bilinear Transformation
e j  1 e j / 2 ( e j / 2  e  j / 2 )
jw  j
 j / 2 j / 2  j / 2
e  1 e (e
e
)
2 jsin( / 2)

 j t an( / 2)
2cos( / 2)
 w  t an( / 2)
0   
0   



t an / 2   t an  1
4
2

Impulse-invariant Filters
 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
K ( s  z1 )s  z2 s  z3  
K3
K1
K2
H ( s) 




( s  p1 )s  p2 s  p3  
s  p1 s  p2 s  p3

Ki
s p
i 1
i
Transfer function of
analog filter
Impulse-invariant
filter
Impulse-invariant Filters
 The impulse response of each analog subfilter takes a simple exponential
form
 For the i-th subfilter
hi (t )  K i exp(pi t )
0
t0
t0
hi [ n]  hi (t ) |t  nT  K i exp(npiT )
0
H i ( z) 

n0
n0

K
n 0
i
exp(npiT ) z
n


 K [exp(np T ) z
n 0
i
Ki
Ki z

1  exp(piT ) z 1
z  exp(piT )
 A zero at the origin of the z-plane
 A polse at z  exp piT 
i
1 n
]
Design of Recursive Digital Filters
Butterworth LP Analog Filter Design (prototype)
| H LP ( jw) | 2 
1
1  ( w / wc ) 2 N
Prototype : when
| H LP ( jw) |2 
or
| H LP ( jw) |
1
(1  [ w / wc ]2 N )1/ 2
c  1
1
1  w2 N
or
frequency responses at N = 1, 2, 3
| H LP ( jw) |
1
(1  w2 N )1/ 2
Determination of Poles
| H LP ( jw) |2  H LP ( s) H LP (s) 
1 s
When N : odd
N=2;
s0  e
s0  e
0s

2N
1 e
j 2k
1  s 2 N  0  s 2 N  1  e j 2k  
When N : even
N=1;
2N
1
1
1
|


   js
2N
1  w2 N
1   js 
1  s2
j
j
0
1
20  
22
s1  e
1
e
j

4
j
1
1
se
j
2k
2N

N
 sk  e
 sk  e
j
2k  
 1
 0.707 j 0.707
s1  0.707 j 0.707, s2  0.707 j 0.707, s3  0.707 j 0.707
N=3;
only 3 effective terms
2N
j
k
N
Determination of Poles
H s  
1
1 s
1st order
H s  
1
s  0.707 j 0.707s  0.707 j 0.707
2nd order
Example
  1rad / sec  2f
 f  0.159Hz
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
20log10 0.1  20dB
Order of filter


log10 10M dB /10   1
N
2 log10  a


log10 1020 /10   1
N
 3.31
2 log10 2
H LP s  
N  4
1
s 4  2.613 s 3  3.414 s 2  2.613 s  1
Chebyshev LP Analog Filter Design (prototype)
| H LP ( jw) |2 
N : order,
1
1   2C N2 ( w)
or
| H LP ( jw) |
1  
1
2
C N2 ( w)

1
2
 : cutoff frequency, r : ripple amplitude (  : ripple parameter)
Order of filter
CN 1 (w)  2CN    CN 1  
CN (w)  cosN cos1  
rdB  10log10 1   2 
Chebyshev Prototype Denominator Polynomials
Example
Maximum passband ripple : 1dB,
Cutoff frequency : less than 1.3 rad/sec
Attenuation in stopband : 40dB for greater than 5 rad/sec
rdB  10log10 1   2 
ripple parameter
 2  0.2589
cutoff frequency : -3dB point is half the magnitude
CN 1 (w)  2CN    CN 1  
H LP  j   0.5
2
H LP  j   0.707
C1 ( w)  
| H LP ( jw) |2 
1
| 1.3  0.69  0.5
1   2C12 ( w)
C2 ( w)  2 2  1
| H LP ( jw) |2 
1
| 1.3  0.41  0.5
1   2C22 ( w)
Passband characteristic
1
| 5  1.6 103  10 4
2
1   C2 ( w)
1
| H LP ( jw) |2 
| 5  1.6 105  10 4
2
2
1   C3 ( w)
| H LP ( jw) |2 
2
 H LP s  
3rd order
K
s 3  0.988 s 2  1.238 s  0.491
2nd order
Analog Filter Frequency Transformation
 form
 LP 
P
 LP
c
 LP 
P
 LP
P
 2 BP   0 2

B BP
 LP 
P
c
 HP
B BS
2
  2 BS   0
s form
s
c
c
s
s 2  0
Bs
2
Bs
2
s 2  0
Example
Butterworth bandpass filter
f u  900Hz
u  2 900  5655rad / sec
fl  600Hz
l  2 600  3770rad / sec
fo  ?
02  u l  21.32106 rad / sec
 / sec
B  u  l  1885rad
Maximum attenuation of 0.2dB for
f o  f  800Hz
Minimum attenuation of 50dB for
0  f  200Hz
800Hz  5026rad / sec
200Hz  1256rad / sec
Prototype equivalent frequency
 LP
P
 2 BP   0 2
1  2 BP  21.32106 




B BP
1885
 BP

Filter order


log10 10M dB /10   1
N 
2 log10  a
a  0.418 M dB  0.2  N  1.75
a  8.35  M dB  50  N  2.71 3rd order
 f0 
0
 735 Hz
2