Chapter 7
Finite Impulse Response(FIR)
Filter Design
Properties of FIR filter

FIR filter has characteristics:
N
y ( n)   h( k ) x ( n  k )
k 0
(7-1)
N
H ( z )   h( k ) z  k
(7-2)
k 0
– Always stable
– Exactly linear phase response
– Simple to implement
• The effect of finite wordlength in FIR is less than in IIR filters
2/77
Linear phase response

Phase response of FIR filter
H (e
jT
N
)   h(k )e  j kT
k 0
= H (e jT ) e j ( )
(7-3)
where  ( )  arg  H (e jT ) 
(7-4)
– Phase delay  p and group delay  g
p  
 ( )

g  
d ( )
d
(7-5)
(7-6)
3/77
– Conditions to be linear phase response
 ( )  
(7-7)
 ( )    
(7-8)
where  and  are constants
 Constant group and constant phase delay responses
4/77
• Impulse response of the filter should have positive symmetry
to satisfy Eq. (7-7), thus
 ( )  
N
= tan 1
 h(n) sin  nT
n 0
N
 h(n) cos nT
(7-9)
n 0
N
tan =
 h(n) sin nT
n 0
N
 h(n) cos nT
(7-10)
n 0
5/77
N
 h(n) cos nT sin   sin nT cos    0
(7-11)
n 0
N
 h(n) sin(  nT )  0
(7-12)
n 0
– For symmetry condition
h(n)  h( N  n), 0  n  N
  NT / 2
(7-13)
(7-14)
6/77
– For the condition given in Eq. (7-8)
» Constant group delay only
» Negative symmetric impulse response
h(n)  h( N  n), 0  n  N
  NT / 2
(7-15)
(7-16)
  /2
7/77
Fig. 7-1.
8/77

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) Using the symmetry condition
• For N  10
h(0)  h(10)
h(1)  h(9)
h(2)  h(8)
h(3)  h(7)
h(4)  h(6)
9/77
• 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 ]
– Using the symmetry condition
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)]
– Compact form
5
H ( )=  a(k ) cos(kT )e  j 5T  H ( ) e j ( )
k 0
where
5
H ( ) =  a(k ) cos(kT )
k 0
 ( )  5T
10/77
3) N  9
h(0)  h(9)
h(1)  h(8)
h(2)  h(7)
h(3)  h(6)
h(4)  h(5)
• Using the symmetry condition
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 )  2h(
N 1
 k ), k  1, 2,
2
,
N 1
2
11/77
Table 7.1 A summary of the key point about the four types of linear phase FIR filters
12/77
Zeros of FIR filters

Transfer function for FIR filter
N
H ( z )   h( k ) z  k
k 0
– Positive symmetry (types 1 and 2)
N
H ( z )   h( N  k ) z  k
k 0
0
=  h( k ) z k z  N
k  N
=z  N H ( z 1 )
(7-17)
13/77
1) H ( z ) has zero at
z  z0 (i.e., z0  re j ) , then should have zero at
z0 1  r 1e j
If h(n) is real and z0 is complex number, should have conjugate zero at
z0*  re j
This means that ( z0* ) 1  r 1e j should be zero.
If each complex zero does not exist on the unit circle,
then have 4 conjugate reciprocal zeros :
(1  re j z 1 )(1  re j z 1 )(1  r 1e j z 1 )(1  r 1e j z 1 )
14/77
2) If zero exists on unit circle, then r  1, 1/ r  1 , i.e.,
z0  e j
becomes
z0 1  e j  z0*
(1  e j z 1 )(1  e j z 1 )
3) If zeros are real and do not exist on the unit circle, then
(1  rz 1 )(1  r 1 z 1 )
4) If zeros exist on z  1 , then
(1  z 1 )
15/77
• If zeros exist on z  1 , then
H (1)  (1) N H (1)
– For symmetry and odd N,
Should have mandatory zero at z  1
– Impulse response of type 3 (even N) and type 4 (odd N)
H ( z )  ( z ) N H ( z 1 )
(7-18)
• For z = 1
Mandatory zero at z  1
• For z  1 and even N
Mandatory zero at z  1
16/77
Mandatory
zero
Mandatory
zero
Mandatory
zero
Mandatory
zero
Fig. 7-2.
17/77
FIR filter design

Design of FIR Filter
– Need to decide :
• Type of filter
• Amplitude and/or phase responses
• Tolerances
• Sampling frequency
• Wordlength of the input data
– To calculate filter coefficients, select the method from:
• Window method, Optimal method, Frequency sampling method
18/77

Filter specifications
– Important parameters
 p peak passband deviation (or ripples)
 r stopband deviation
 p passband edge frequency (  2 f )
p
p
r stopband edge frequency (r  2 f r )
s sampling frequency (s  2 f s )
– Another important parameter
Filter order N
19/77
ILPF
Fig. 7-3.
20/77

FIR coefficient calculation
N
y ( n)   h( k ) x ( n  k )
k 0
N
H ( z )   h( k ) z  k
k 0
– Most commonly used methods for obtaining h(k )
• Window, optimal, and frequency sampling methods
21/77
Window method

Design of FIR filter using window methods
– Frequency response of filter, H I ( )
– Impulse response, hI (n)
hI (n) 

1
2

1
2


H I ( )e j n d 
(7-19)
– Ideal lowpass response
hI (n) 


1 e j n d  
 2 f c sin(nc )
,

n

= 
c

2 fc ,

1
2
c


e j n d 
c
n  0, -  n  
n0
(7-20)
22/77
Fig. 7-4.
23/77

Truncation for 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
24/77
Fig. 7-5.
25/77
Fig. 7-6.
26/77
Fig. 7-7.
27/77
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 cutoff frequencies
28/77

Some common window functions
– Hamming window
0.54  0.46cos  2 n / N  , 0  n  N
w n  
0
, elsewhere

(7-21)
• Appropriate relationship between transition width and filter length
F  3.32 / N
(7-22)
where N is filter order and
F is normalized transition width
29/77
– Properties of common window functions
Fig. 7-8.
30/77
Table 7.3 summary of important features of common window functions
31/77
– Kaiser window
• Trade-off transition width against ripple
– Using a ripple control parameter 
2 
 
2
n

N


 I0   1  


 
 N  
wn   
, 0n N
I0   


0
, elsewhere

(7-23)
where I 0 ( x) is 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
32/77
• Determination of parameter 
– Using the stopband attenuation requirements
– Using empirical relationships
0,

  0.5842( A  21)0.4  0.07886( A  21),
0.1102( A  8.7)

if A  21dB
if 21dB  A  50dB
if A  50dB
where A  20log10 ( ) is the stopband attenuation
  min( p ,  s ) , since the passband and stopband ripples are nearly equal
• The number of filter coefficients N
N 
A  7.95
14.36f
(7-25)
where  f is the normalized transition width
33/77

The window method of calculating FIR filter coefficients
– Step 1 : specify the desired frequency response of filter, H I ( )
– Step 2 : obtain the impulse response, hI (n) , of desired filter by evaluating
the inverse Fourier transform
– Step 3 : select a window function and then determine the number of
coefficients using the appropriate relationship between the filter
length and the transition width, f
– Step 4 : obtain values of w( n ) for chosen window function and the values
of the actual FIR coefficient, h(n) , by multiplying hI (n) by w( n )
h(n)  hI (n) w(n)
(7-26)
34/77
 Example
7-2
– Obtain coefficients of FIR lowpass filter 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
35/77
• 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
– Considering the smearing effect of the window function
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
36/77
– Symmetrical function h(n)
» Calculation of h(0), h(1), , h(27)
» Using the symmetry property to obtain the other coefficients
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)   0.000940057
37/77
2  0.21875
sin( 25  2  0.21875)  0.00248397
25  2  0.21875
w(2)  0.54  0.46 cos(2  2 / 54)
n  2 : hI (2) 
 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) 
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
38/77
Fig. 7-9.
39/77
 Example
7-3
– Obtain coefficients using Kaiser or Blackman window
Stopband attenuation
: 40dB
passband attenuation
: 0.01dB
Transition region
f : 500 Hz
Sampling frequency
f s : 10kHz
Passband cutoff frequency
f p : 1200 Hz
• From filter specifications
20 log(1   p )  0.01dB,
20log( r )  40dB,
 p  0.00115
 r  0.01
   r   p  0.00115
20 log(0.00115)  58.8dB
40/77
• Using Kaiser window
– The number of filter order N
N
A  7.95
58.8  7.95

 70.82
14.36F 14.36(500 /10000)
N  71
– The ripple parameter
  0.1102(58.8  8.7)  5.52
– Normalized cutoff frequency
f c  1200  f / 2  1450 Hz
f c  f c' / f s  1, 450 /10, 000  0.145
41/77
– Calculation of FIR coefficients

N 
2 f c sin  n   c 
2 

hI (n) 
, 0n N
N

 n   c
2

h(n)  hI (n) w(n)
42/77
– Symmetrical function h(n)
» Calculation of h(0), h(1), , h(35)
» Using the symmetry property to obtain the other coefficients
2  0.145
sin(35.5  2  0.145)
35.5  2  0.145
 0.00717
n  0 : hI (0) 
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
43/77
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)  hI (1) w(1)  0.000004897
44/77
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)  hI (2) w(2)  0.000345413
45/77
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)  hI (35)w(35)  0.279934818
46/77
Fig. 7-10.
47/77

Advantages and disadvantages
– Simplicity
– Lack of flexibility
– The passband and stopband edge frequencies cannot be precisely
specified
– For a given attenuation specification, filter designer must find a suitable
window
48/77
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.
50/77
Fig. 7-12.
51/77
– The procedure of optimal method
• 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
52/77

Optimal FIR filer design
– Transfer function of lowpass filter
N
H ( z )   h( k ) z  k
(7-28)
k 0
where h(n)  h(n)
• Symmetric property gives
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
53/77
• Let f   / 2 f s  T / 2
– f p / fs  f p , fr / fs  fr
– Normalized passband : 0, f p 
– Normalized stopband :  f r ,0.5
• Desired magnitude response
0  f  fp
1,
HI ( f )  
0,
f r  f  0.5
(7-30)
• Weighting function
1

W  f   k
 1
,
,
0  f  fp
f r  f  0.5
(7-31)
54/77
• Find a(k ) with min[max E( f ) ] , k  0,1, 2, , N / 2
H ( f )  H (e
j 2 f
N /2
)   a (k ) cos k 2 f
(7-32)
k 0
E ( f )  W ( f )[ H ( f )  H I ( f )]
(7-33)
where f are 0, f p  and  f r ,0.5
55/77
– Alternation theorem
• Let F  0, f p    f r , 0.5
• If E ( f ) has equiripple inside F and exhibit at least m+2 alternations, then
H ( f )  HI ( f )
E ( fi )   E ( fi 1 )  e,
f0  f1 
where
 fl
e
i  0,1,
, l 1
(7-34)
(l  m  1)
max
f 0, f p  &  f r ,0.5 
E( f )
56/77
• From equation (7-33) and (7-34)
W ( fi )  H ( fi )  H I ( fi )  (1)i e,
i  0,1, 2,
, m 1
(7-35)
• From 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 )
57/77
– Summary
Step 1. Select filter length as 2m+1
Step 2. Select m+2 points of f i in F
Step 3. Calculate a(k ) and e using equation (7-36)
Step 4. Calculate E ( f ) using equation (7-29). If E( f )  e , go to step 5,
otherwise go to step 6
Step 5. Determine m local minima or maxima points
Step 6. Calculate h(0)  a(0), h(k )  a(k ) / 2 , k  1, 2,
,m
N
H ( z )   h( k ) z  k
k 0
58/77

Example 7-4
– Specification of desired filter
• Filter length : 3
•  pT  1[rad], rT  1.2[rad] , k  2
N  2m  1  3
m  1
– Normalized frequency
f0  0.5 / 2 , f1  1/ 2 , f 2  1.2 / 2
59/77
• 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
(7-37)
60/77
• Selection of new f i
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
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
(7-38)
– Transfer function of optimal filter
H ( z )  0.225  0.144 z 1  0.225z 2
(7-39)
61/77
Fig. 7-13.
62/77

Optimal method using MATLAB
– Based on Park-McClellan and Remez algorithm
– Calculation of coefficient for FIR filter using Remez
b  remez(N,F, M )
b  remez(N,F, M, WT )
where N is the filter length
F is the normalized frequency band edges
M is the magnitude response
WT is the relative weight between ripples
63/77

Example 7-5
– Specification of desired filter
• Pass band
: 0 – 1000Hz
• Transition band
: 500Hz
• Filter length
: 45
• Sampling frequency : 10,000Hz
– Normalized frequency band edges
F  [0, 0.2, 0.3, 1]
– Magnitude response
M  [1 1 0 0]
64/77
Table 7-4.
65/77
Fig.7-14.
66/77

Example 7-6
– Specification of desired filter
• Pass band
: 3kHz – 4kHz
• Transition band
: 500Hz
• Pass band ripple
: 1dB
• Rejection band attenuation
: 25dB
• Sampling frequency
: 20kHz
– Frequency band edges and magnitude response
F  [2500, 3000, 4000, 4500]
M  [0 1 0]
67/77
• Estimation of filter length
– Pass and rejection band ripples
Ap
p 
10 20  1
Ap
10 20  1
where A p and
 r  10
 Ar
20
Ar are ripples of dB scale in pass and rejection band
– Using Remezord in MATLAB
68/77
Table 7-5.
69/77
Fig. 7-15.
70/77
Frequency sampling method

Design of FIR filter
– 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
where H ( k ), k  0,1,
N 1
 H ( k )e
j (2 / N ) nk
(7-40)
k 0
, N  1 are samples of desired frequency response
71/77
– Linear phase filters with positive symmetrical impulse response
• For N even
 N2 1

1
h( n) 
2 H (k ) cos[2 k (n   ) / N ]  H (0) 


N  k 1


(7-41)
where   ( N  1) / 2
• For N odd
– Upper limit in the summation is ( N  1) / 2
72/77
Fig. 7-16.
73/77

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

• From equation (7-40)
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
• h(n) is symmetry
1  N /21

h(n)    2 H (k ) cos[2 k (n   ) / N ]  H (0) 
N  k 1

(7-42)
74/77
(2) Design of FIR filter
• Specification of desired filter
– Pass band
: 0 – 5kHz
– Sampling frequency : 18kHz
– Filter length
:9
• Selection of frequency samples at intervals of kf s / N
 1,
H (k )  
 0,
k  0,1, 2
k  3, 4
Fig. 7-17.
75/77
• Coefficient of FIR filter using equation (7-42)
Table 7-6.
76/77
Comparison of the window, optimum
and frequency sampling methods

Optimal method
– Easy and efficient way of computing FIR filter coefficients
– Making filter with good amplitude response characteristics for reasonable
values of N

Window method
– In the absence of the optimal software or when the passband and stopband
ripples are equal, the window method represents a good choice
– Particularly simple method to apply and conceptually easy to understand

Frequency sampling method
– Filters with arbitrary amplitude-phase response can be easily designed
– Lack of precise control for the location of the bandedge frequencies or the
passband ripples
77/77