Transcript 2. Linear phase response

Chapter 7
Finite Impulse Response(FIR)
Filter Design
1. Features of FIR filter
 Characteristic
of FIR filter
N
y ( n)   h( k ) x ( n  k )
k 0
N
H ( z )   h( k ) z  k
k 0
– FIR filter is always stable
– FIR filter can have an exactly linear phase response
– FIR filter are very simple to implement. Nonrecursive FIR
filters suffer less from the effect of finite wordlength than IIR
filters
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2. Linear phase response
 Phase
response of FIR filter
H (e
jT
N
)   h(k )e  j kT
(1)
k 0
= H (e jT ) e j ( )
where  ( )  arg  H (e jT ) 
(2)
– Phase delay  p and group delay  g
p  
 ( )

g  
d ( )
d
3/74
– Condition of linear phase response
 ( )  
(3)
 ( )    
(4)
Where  and

is constant
Constant group delay and phase delay response
4/74
• If a filter satisfies the condition given in equation (3)
– From equation (1) and (2)
 ( )  
N
= tan 1
 h(n) sin  nT
n 0
N
 h(n) cos nT
n 0
thus
N
tan =
 h(n) sin  nT
n 0
N
 h(n) cos  nT
n 0
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N
 h(n) cos nT sin   sin nT cos    0
n 0
N
 h(n)sin(  nT )  0
n 0
h(n)  h( N  n  1),
0n N
  NT / 2
– It is represented in Fig 7.1 (a),(b)
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• When the condition given in equation (4) only
– The filter will have a constant group delay only
– It is represented in Fig 7.1 (c),(d)
h(n)  h( N  n)
  NT / 2
  /2
7/74
Center of symmetry
Fig. 7-1.
8/74
Table 7.1 A summary of the key point about the four types of linear phase FIR filters
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 Example
7-1
(1) Symmetric impulse response for linear phase response.
No phase distortion
h(n)  h( N  n) or h(n)  h( N  n)
(2) h(n)  h( N  n), 0  n  N , N  10
h(0)  h(10)
h(1)  h(9)
h(2)  h(8)
h(3)  h(7)
h(4)  h(6)
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• Frequency response
H ( )
H ( )  H (e jT )
10
=  h(k )e  jkT
k 0
=h(0)  h(1)e  jT  h(2)e  j 2T  h(3)e  j 3T  h(4)e  j 4T  h(5)e  j 5T
+h(6)e  j 6T  h(7)e  j 7T  h(8)e  j 8T  h(9)e  j 9T  h(10)e  j10T
=e  j 5T [h(0)e j 5T  h(1)e j 4T  h(2)e j 3T  h(3)e j 2T  h(4)e jT  h(5)
+h(6)e  jT  h(7)e  j 2T  h(8)e  j 3T  h(9)e  j 4T  h(10)e  j 5T ]
H ( )=e j 5T [h(0)(e j 5T  e j 5T )  h(1)(e j 4T  e  j 4T )  h(2)(e j 3T  e  j 3T )
 h(3)(e j 2T  e j 2T )  h(4)(e jT  e  jT )  h(5)]
 e j 5T [2h(0) cos(5T )  2h(1) cos(4T )  2h(2) cos(3T )
 2h(3) cos(2T )  2h(4) cos(T )  h(5)]
5
H ( )= a(k ) cos(kT )e j 5T  H ( ) e j ( )
k 0
5
where
H ( ) = a(k ) cos(kT )
k 0
 ( )  5T
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(3) N  9
h(0)  h(9)
h(1)  h(8)
h(2)  h(7)
h(3)  h(6)
h(4)  h(5)
H ( )=e j 9T /2 [h(0)(e j 9T  e j 9T )  h(1)(e j 7T  e j 7T )  h(2)(e j 5T  e  j 3T )
 h(3)(e j 3T  e j 3T )  h(4)(e jT  e jT )]
 e j 9T /2 [2h(0) cos(9T / 2)  2h(1) cos(7T / 2)  2h(2) cos(5T / 2)
 2h(3) cos(3T / 2)  2h(4) cos(T / 2)]
= H ( ) e j ( )
5
where
H ( )= b(k ) cos[ (k  1/ 2)T ]
k 1
 ()  (9 / 2)T
b( k )  2 h(
N 1
 k ), k  1, 2,
2
,
N 1
2
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3. Zero distribution of FIR filters
 Transfer
function for FIR filter
N
H ( z )   h( k ) z  k
k 0
N
H ( z )   h( N  k ) z  k
k 0
0
=  h( k ) z k z  N
k  N
=z  N H ( z 1 )
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 Four
types of linear phase FIR filters
– H ( z ) have zero at
z  z0
z0  re j
z01  r 1e j
h(n) is real and
z0
is imaginary
z0*  re j
( z0* )1  r 1e j
(1  re j z 1 )(1  re j z 1 )(1  r 1e j z 1 )(1  r 1e j z 1 )
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– If zero on unit circle
r  1, 1/ r  1
z0  e j
z01  e j  z0*
(1  e j z 1 )(1  e j z 1 )
– If zero not exist on the unit circle
(1  rz 1 )(1  r 1 z 1 )
– If zeros on z  1
(1  z 1 )
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Necessary
zero
Necessary
zero
Necessary
zero
Necessary
zero
Fig. 7-2.
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4. FIR filter specifications
 Filter
specifications
 p peak passband deviation (or ripples)
 r stopband deviation
 p passband edge frequency
r stopband edge frequency
s sampling frequency
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ILPF
Satisfies spec’s
Fig. 7-3.
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 Characterization
of FIR filter
N
y ( n)   h( k ) x ( n  k )
k 0
N
H ( z )   h( k ) z  k
k 0
– Most commonly methods for obtaining h(k )
• Window, optimal and frequency sampling methods
19/74
5. Window method
 FIR
filter
– Frequency response of filter H I ( )
– Corresponding impulse response hI (n)
1
hI (n) 
2

H

I
( )e j n d 
– Ideal lowpass response
1 
1 c jn
j n
hI (n) 
1 e d  
e d
2 
2 c
 2 f c sin(nc )
, n  0, -  n  

n

=
c

2 fc ,
n0

20/74
Fig. 7-4.
21/74
 Truncation
to FIR
h(n)  hI (n)w(n)
"windowing"
H(e j )  21 H I (e j ) W(e j )

1
2



H I (e j )W(e j (  ) )d
– Rectangular Window
1 , n  0,1,..., N
w(n)  
0 , elsewhere
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Fig. 7-5.
23/74
Fig. 7-6.
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Fig. 7-7.
25/74
Table 7.2 summary of ideal impulse responses for standard frequency selective filters
hI (n)
hI (n)
hI (n)
fc , f1 and f 2 are the normalized passband or stopband edge frequencies; N is the length of filter
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 Common
window function types
– Hamming window
0.54  0.46cos  2 n / N  , 0  n  N
w n  
0
, elsewhere

F  3.32 / N
where N is filter length and
 F is normalized transition width
27/74
– Characteristics of common window functions
Fig. 7-8.
28/74
Table 7.3 summary of important features of common window functions
29/74
– Kaiser window
2 
 
2
n

N


 I0   1  
 
N
 

 
wn   
, 0n N
I0   


0
, elsewhere

where I 0 ( x) is the zero-order modified Bessel function of the first kind
 ( x / 2) k 
I 0 ( x)  1   
k ! 
k 1 
L
2
where typically L  25
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• Kaiser Formulas – for LPF design
A  20log10 ( )
  min( p ,  s )
if A  21dB
0,

  0.5842( A  21)0.4  0.07886( A  21), if 21dB  A  50dB
0.1102( A  8.7)
if A  50dB

A  7.95
N
14.36F
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 Example
7-2
– Obtain coefficients of FIR lowpass using hamming
window
Passband cutoff frequency
f p : 1.5kHz
Transition width
f : 0.5kHz
Stopband attenuation
 50dB
Sampling frequency
f s : 8kHz
• Lowpass filter
sin  nc 

, n0
2 f c
hI  n   
nc

2 fc
, n0

F f / f s  0.5 / 8  0.0625
N  3.32 / F  3.32 / 0.0625  53.12
N  54
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• Using Hamming window
0  n  54
h(n)  hI (n)w(n)
w(n)  0.54  0.46cos(2 n / 54)
0  n  54
fc'  f p  f / 2  (1.5  0.25) 1.75[kHz]
fc  fc' / f s  1.75 / 8  0.21875

N


2
f
sin
(
n

)

c
 c

2
,


N
hI (n)  
(n  )c
2


2 fc ,

n
N
, 0 n N
2
n
N
2
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2  0.21875
sin(27  2  0.21875)
27  2  0.21875
 0.00655
n  0 : hI (0) 
w(0)  0.54  0.46 cos(0)  0.08
h(0)  hI (0) w(0)   0.00052398
2  0.21875
sin(26  2  0.21875)
26  2  0.21875
 0.011311
n 1 : hI (1) 
w(1)  0.54  0.46 cos(2 / 54)
 0.08311
h(1)  hI (1) w(1)   0.00094054
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2  0.21875
sin(25  2  0.21875)
25  2  0.21875
 0.00248397
n  2 : hI (2) 
w(2)  0.54  0.46 cos(2  2 / 54)
 0.092399
h(2)  hI (2) w(2)  0.000229516
2  0.21875
sin(1 2  0.21875)
1 2  0.21875
 0.312936
n  26 : hI (26) 
35/74
w(26)  0.54  0.46cos(2  26 / 54)
 0.9968896
h(26)  hI (26) w(26)  0.3112226
n  27 : hI (27)  2 f c  2  0.21875
 0.4375
w(27)  0.54  0.46 cos(2  27 / 54)
1
h(27)  hI (27) w(27)  0.4375
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Fig. 7-9.
37/74
 Example
7-3
– Obtain coefficients using Kaiser or Blackman window
Stopband attenuation
: 40dB
passband attenuation
: 0.01dB
f : 500 Hz
Transition region
Sampling frequency
f s : 10kHz
Passband cutoff frequency
f p : 1200Hz
20log(1   p )  0.01dB,
20log( r )  40dB,
 p  0.00115
 r  0.01
  r   p  0.00115
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• Using Kaiser window
N
A  7.95
58.8  7.95

 70.82
14.36F 14.36(500 /10000)
N  71
  0.1102(58.8  8.7)  5.52
fc  fc' / f s  1, 450 /10,000  0.145

N 
2 f c sin  n   c 
2 

hI (n) 
, 0n N
N

n


 c
2

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h(n)  hI (n)w(n)
n  0 : hI (0) 
2  0.145
sin(35.5  2  0.145)
35.5  2  0.145
 0.00717
2

2n  N  


I0   1  


 N  
I (0)

w(0) 
 0
 0.023
I0 ( )
I 0 (5.52)
h(0)  hI (0)w(0)  0.000164935
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n  1: hI (1) 
2  0.145
sin(34.5  2  0.145)
34.5  2  0.145
 0.0001449
2

69  


I 0  5.52 1  


 71   I (1.3)

w(1) 
 0
 0.0337975
I 0 (5.52)
I 0 (5.52)
h(1)  h(1)  hI (1)w(1)  0.000004897
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n  2 : hI (2) 
2  0.145
sin(34.5  2  0.145)
33.5  2  0.145
 0.007415484
2

69  


I 0  5.52 1  


 71   I (1.8266)

w(2) 
 0
 0.04657999
I 0 (5.52)
I 0 (5.52)
h(2)  h(2)  hI (2) w(2)  0.000345413
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n  35 : hI (35) 
2  0.145
sin(0.5  2  0.145)
0.5  2  0.145
 0.280073974
2

1  

I 0  5.52 1    

 71   I (5.51945)

w(35) 
 0
 0.999503146
I 0 (5.52)
I 0 (5.52)
h(35)  h(35)  hI (35)w(35)  0.279934818
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Fig. 7-10.
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 Summary
of window method
1. Specify the ‘ideal’ or desired frequency response of filter,
H I ( )
2. Obtain the impulse response, hI (n), of the desired filter by
evaluating the inverse Fourier transform
3. Select a window function that satisfies the passband or
attenuation specifications and then determine the number of filter
coefficients
4. Obtain values of w(n) for the chosen window function and the
values of the actual FIR coefficients, h(n), by multiplying
hI (n) by w(n)
h(n)  hI (n)w(n)
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 Advantages
and disadvantages
– Simplicity
– Lack of flexibility
– The passband and stopband edge frequencies cannot be
precisely specified
– For a given window(except the Kaiser), the maximum
ripple amplitude in filter response is fixed regardless of how
large we make N
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6. The optimal method
 Basic
concepts
– Equiripple passband and stopband
E()  W ()[ H I ()  H ()]
Weighted
Approx. error
Weighting
function
Ideal desired
response
Practical
response
min[max E() ]
• For linear phase lowpass filters
– m+1 or m+2 extrema(minima and maxima)
where m=(N+1)/2 (for type1 filters) or m =N/2 (for type2 filters)
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Practical response
Ideal response
Fig. 7-11.
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Fig. 7-12.
49/74
– Optimal method involves the following steps
• Use the Remez exchange algorithm to find the optimum set of
extremal frequencies
• Determine the frequency response using the extremal
frequencies
• Obtain the impulse response coefficients
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 Optimal
FIR filer design
N
H ( z )   h( k ) z  k
k 0
where
h(n)  h(n)
N /2
N /2
k 1
k 0
H ( )  h(0)   2h(k ) cos kT   a(k ) cos kT
where a(0)  h(0) and a(k )  2h(k ),
k  1, 2,
,N /2
Let f   / 2 f s  T / 2
1 ,
H1  f   
0 ,
1
,

W  f   k

1 ,
0  f  fp
f r  f  0.5
0  f  fp
f r  f  0.5
This weighting function permits different peak
error in the two band
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H ( f )  H (e
j 2 f
N /2
)   a(k ) cos k 2 f
k 0
E( f )  W ( f )[ H ( f )  H I ( f )]
where f are  0, f p  and  fr ,0.5
min[max E( f ) ]
Find
a(k )
52/74
– Alternation theorem
Let F  0, f p    f r , 0.5


If E ( f ) has equiripple inside bands F and more than m+2 extremal point
then
H ( f )  HI ( f )
E( fi )  E( fi 1 )  e,
f0  f1 
 fl
where e 
i  0,1,
, l 1
(l  m  1)
max
f 0, f p  &  f r ,0.5 
E( f )
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• From equation (7-33) and (7-34)
W ( fi )  H ( fi )  HI ( fi )  (1)i e,
i  0,1,2,
, m 1
• Equation (7-35) is substituted equation (7-32)
m
 a(k ) cos 2 kf
k 0
i
 H I ( fi )  (1)i
e
,
W ( fi )
i  0,1, 2,
, m 1
• Matrix form
cos 4 f 0
1 cos 2 f 0
1 cos 2 f
cos 4 f1
1



1 cos 2 f m cos 4 f m
1 cos 2 f m1 cos 4 f m1
cos 2 mf 0
cos 2 mf1
cos 2 mf m
cos 2 mf m1
  a(0)   H I ( f 0 ) 
1/ W ( f1 )   a(1)   H I ( f1 ) 







 
(1) m / W ( f m )   a(m)   H I ( f m ) 
( 1) m! / W ( f m1 )   e   H I ( f m1 ) 
1/ W ( f 0 )
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– Summary
Step 1. Select filter length as 2m+1
Step 2. Select m+2 f i point in F
Step 3. Calculate a(k ) and e using equation (7)
Step 4. Calculate E ( f ) using equation (5). If
E( f )  e
in some
of f , go to step 5, otherwise go to step 6
Step 5. Determine m local minma or maxma points
Step 6. Calculate h(0)  a(0), h(k )  a(k ) / 2 when k  1, 2, , m
where H ( z ) 
N
 h( k ) z
k
k 0
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– Example 7-4
• Specification of desired filter
– Ideal low pass filter
– Filter length : 3
– pT  1[rad], rT  1.2[rad]
– k2
• Normalized frequency
f0  0.5 / 2 , f1  1/ 2 , f2  1.2 / 2
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• From H I ( f0 )  H I ( f1 )  1, H I ( f2 )  0, W ( f0 )  W ( f1 )  1/ 2 and W ( f2 )  1
1 0.8776 2   a(0)  1 
1 0.5403 2   a(1)   1 


  
1 0.3624 1   e   0 
a(0)  0.645, a(1)  2.32, e  0.196
H ( f )  0.645  2.32cos 2 f : not the optimal filter
• Cutoff frequency
f 0 
1
2
f1 
1.2
2
f 2  0.5
H I ( f0)  1, H I ( f1)  H I ( f 2)  0, W ( f0)  0.5, and W ( f1)  W ( f 2)  1
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1 0.5403 2   a(0)  1 
1 0.3624 1  a(1)   0 


  
1
1
1   e  0 
a(0)  0.144, a(1)  0.45, e  0.306
H ( f )  0.144  0.45cos 2 f
: has the minimum max E( f )
(N=3)
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Fig. 7-13.
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 Optimization
using MATLAB
– Park-McClellan
– Remez
b  remez(N,F, M )
b  remez(N,F, M, WT )
where N is the filter order (N+1 is the filter length)
F is the normalized frequency of border of pass band
M is the magnitude of frequency response
WT is the weight between ripples
60/74
– Example 7-5
• Specification of desired filter
– Band pass region : 0 – 1000Hz
– Transition region : 500Hz
– Filter length : 45
– Sampling frequency : 10,000Hz
• Normalized frequency of border of passband
F  [0, 0.2, 0.3, 1]
• Magnitude of frequency response
M  [11 0 0]
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Table 7-4.
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Fig.7-14.
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– Example 7-6
• Specification of desired filter
– Band pass region : 3kHz – 4kHz
– Transition region : 500Hz
– Pass band ripple : 1dB
– Rejection region : 25dB
– Sampling frequency : 20kHz
• Frequency of border of passband
F  [2500, 3000, 4000, 4500]
M  [0 1 0]
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• Transform dB to normal value
Ap
p 
10 20  1
Ap
10
where
band
20
1
 r  10
 Ar
20
A p and A r ripple value(dB) of pass band and rejection
• Filter length
– Remezord (MATLAB command)
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Table 7-5.
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Fig. 7-15.
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7. Frequency sampling method
 Frequency
sampling filters
– Taking N samples of the frequency response at intervals
of kFs / N k  0,1, , N  1
– Filter coefficients h(n)
1 N 1
h(n)   H (k )e j (2 / N ) nk
N k 0
where H (k ), k  0,1,
, N  1 are samples of the ideal or target frequency response
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– For linear phase filters (for N even)
 N2 1

1
h(n) 
2 H (k ) cos[2 k (n   ) / N ]  H (0) 


N  k 1


where
  ( N  1) / 2
– For N odd
• Upper limit in summation is ( N  1) / 2
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Fig. 7-16.
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– Example 7-7
1  N /21

(1) Show the h(n)    2 H (k ) cos[2 k (n   ) / N ]  H (0) 
N  k 1

• Expanding the equation
1 N 1
h(n)   H (k ) e j 2 nk / N
N k 0
1 N 1
  H (k ) e  j 2 k / N e j 2 kn / N
N k 0
1 N 1
  H (k ) e j 2 ( n  ) k / N
N k 0
h(n) is real value
N 1
1

N
 H (k ) cos[2 k (n   ) / N ]  j sin[2 k (n   ) / N ]
1

N
N 1
k 0
 H (k ) cos[2 k (n   ) / N ]
k 0
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(2) Design of FIR filter
– Band pass region : 0 – 5kHz
– Sampling frequency : 18kHz
– Filter length : 9
Fig. 7-17.
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• Samples of magnitude in frequency
 1,
H (k )  
 0,
k  0,1, 2
k  3, 4
Table 7-6.
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8. Comparison of most commonly
method
– Window method
• The easiest, but lacks flexibility especially when passband and
stopband ripples are different
– Frequency sampling method
• Well suited to recursive implementation of FIR filters
– Optimal method
• Most powerful and flexible
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