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
2/74
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
5/74
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)
6/74
• 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
9/74
 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)
10/74
• 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
N 1
b ( k )  2h(
 k ), k  1, 2,
2
,
N 1
2
12/74
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 )
13/74
 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 )
14/74
– 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 )
15/74
Necessary
zero
Necessary
zero
Necessary
zero
Necessary
zero
Fig. 7-2.
16/74
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
17/74
ILPF
Satisfies spec’s
Fig. 7-3.
18/74
 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)
H(e j ) 
1
2

1
2
"windowing"
H I (e j ) W(e j )



H I (e j )W(e j (  ) )d
– Rectangular Window
1 , n  0,1,..., N
w(n)  
0 , elsewhere
22/74
Fig. 7-5.
23/74
Fig. 7-6.
24/74
Fig. 7-7.
25/74
Table 7.2 summary of ideal impulse responses for standard frequency selective filters
hI (n)
hI (n)
hI (n)
f c , f1 and f 2 are the normalized passband or stopband edge frequencies; N is the length of filter
26/74
 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
30/74
• 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

N
A  7.95
14.36F
31/74
 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 

2
f
, n0
 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
32/74
• Using Hamming window
0  n  54
h(n)  hI (n) w(n)
w(n)  0.54  0.46 cos(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
33/74
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
34/74
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
36/74
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 : 1200 Hz
20 log(1   p )  0.01dB,
20log( r )  40dB,
 p  0.00115
 r  0.01
   r   p  0.00115
38/74
• 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
f c  f c' / 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

39/74
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 

2
n

N



I0   1  


 N  
I (0)

w(0) 
 0
 0.023
I0 ( )
I 0 (5.52)
h(0)  hI (0) w(0)  0.000164935
40/74
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
41/74
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
42/74
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
43/74
Fig. 7-10.
44/74
 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)
45/74
 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
46/74
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)
47/74
Practical response
Ideal response
Fig. 7-11.
48/74
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
50/74
 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
51/74
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  f r ,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 )  H I ( 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,
where H ( z ) 
N
 h( k ) z
,m
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 , f 2  1.2 / 2
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• From H I ( f0 )  H I ( f1 )  1, H I ( f 2 )  0, W ( f0 )  W ( f1 )  1/ 2 and W ( f 2 )  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  2.32 cos 2 f : not the optimal filter
• Cutoff frequency
f 0 
1
2
f1 
1.2
2
f 2  0.5
H I ( f 0)  1, H I ( f1)  H I ( f 2)  0, W ( f 0)  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.
59/74
 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  [1 1 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 ripple value(dB) of pass band and rejection
r
• 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
h( n) 
N
N 1
j (2 / N ) nk
H
(
k
)
e

k 0
where H (k ), k  0,1,
, N  1 are samples of the ideal or target frequency response
68/74
– 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
h( n) 
N
h(n) is real value
N 1
 H (k ) e
j 2 nk / N
k 0
N 1
1

N
 H (k ) e
1

N
N 1
 H (k ) e
1

N
N 1
 H (k ) cos[2 k (n   ) / N ]  j sin[2 k (n   ) / N ]
1

N
N 1
 j 2 k / N
e j 2 kn / N
k 0
j 2 ( n  ) k / N
k 0
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.
73/74
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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