Chapter 7

Finite Impulse Response(FIR)

Filter Design
 Filter design steps
- The design of a digital filter involves five steps:
(1) Specifications of the filter requirements
(2) calculation of suitable filter coefficients
(3) Representation of the filter by a suitable structure (realization)
(4) Analysis of the effects of finite wordlength on the filter performance
(5) Implementation of filter in software and/or hardware

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 Filter design
– Need to decide :
• Type of filter

• Amplitude and/or phase responses

• Tolerances

• Sampling frequency

• Wordlength of the input data

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 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

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 ILPF

Fig. 7-3.

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 FIR coefficient calculation
N
y (n)  h(k ) x(n  k )
k 0

N
H ( z )  h(k ) z  k
k 0

– Most common methods used for calculating h(k )

• Window method
• Optimal method

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 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 h( n) hI (n) w(n)
of the actual FIR coefficient, , by multiplying by
h(n) hI (n) w(n) (7-
26)

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 Window method
 Design of FIR filter using window method
– Frequency response of filter, H I ( )
– Impulse response, hI (n)
1 
hI (n) 
2 

H I ( )e j n d  (7-
19)

– Ideal lowpass response

1  1 c
 
j n
hI (n)  1 e d  e j n d 
2  2  c

 2 f c sin(nc )
 , n 0, - n 
=  n  c
(7-
 2 fc , n 0 20)


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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

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 Fig. 7-4.
(The frequency scale is normalized by T = 1)

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 Truncation for 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.

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Fig. 7-6.

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Fig. 7-7.

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 Some common window functions
– Hamming window

0.54  0.46 cos 2 n / N  , 0 n  N

w n   (7-
 0 , elsewhere 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

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– Properties of common window functions

Fig. 7-8. Window functions: (a) Rectangular, (b) Hamming, (c) Blackman
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Table 7.3 Summary of important features of common window functions

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– Kaiser window
• Trade-off transition width against ripple
– using a ripple control parameter, 

  2 n  N
2 
 I0   1    

   N  
w n    , 0 n  N (7-
 I 0   23)

 0 , elsewhere

where I 0 ( x) is zero-order modified Bessel function of the first kind

2
L
 ( x / 2) k 
I 0 ( x) 1   
k 1  k ! 

where typically L 25

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• Determination of parameter 
– by using the stopband attenuation requirement through empirical
relationships below
0, if A 21dB

 0.5842( A  21)0.4  0.07886( A  21), if 21dB  A  50dB
0.1102( A  8.7) if A  50dB

where A  20 log10 ( ) is the stopband attenuation value and
 min( p ,  s ) since the passband and stopband ripples are nearly equal

• The number of filter coefficients N

A  7.95
N  (7-
14.36f
25)

where f is the normalized transition width

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 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
 c f , n 0
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
h(n) hI (n) w(n) 0 n 54

w(n) 0.54  0.46 cos(2 n / 54) 0 n 54

– Considering the smearing effect of the window function

f c'  f p  f / 2 (1.5  0.25) 1.75[kHz ] (centered on the transition band)

 f c  f c' / f s 1.75 / 8 0.21875 (normalized)

  N 
 2 f c sin  ( n  )c 
2 N
 , n  , 0 n N
 N 2
hI (n)  (n  )c
 2
 N
 2 fc , n
2
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– Symmetrical function h(n)
» Calculation of h(0), h(1), , h(27)
» Using the symmetry property to obtain the other coefficients

2 0.21875
n 0 : hI (0)  sin( 27 2 0.21875)
 27 2 0.21875
 0.00655
w(0) 0.54  0.46 cos(0) 0.08
h(0) hI (0) w(0)  0.00052398

2 0.21875
n 1 : hI (1)  sin( 26 2 0.21875)
 26 2 0.21875
 0.011311
w(1) 0.54  0.46 cos(2 / 54)
0.08311
h(1) hI (1) w(1)  0.000940057
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 2 0.21875
n 2 : hI (2)  sin(  25 2 0.21875)  0.00248397
 25 2 0.21875
w(2) 0.54  0.46 cos(2 2 / 54)
0.092399
h(2) hI (2) w(2) 0.000229516
  

2 0.21875
n 26 : hI (26)  sin(  12 0.21875)
 12 0.21875
0.312936
w(26) 0.54  0.46 cos(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.

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 Example 7-3
– Obtain coefficients using Kaiser 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,  p 0.00115

 20 log( r ) 40dB,  r 0.01

  r  p 0.00115
 20 log(0.00115) 58.8dB
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• Using Kaiser window
– The number of filter order N
A  7.95 58.8  7.95
N  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

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– 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)

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– Symmetrical function h(n)
» Calculation of h(0), h(1), , h(35)
» Using the symmetry property to obtain the other coefficients

2 0.145
n 0 : hI (0)  sin(  35.5 2 0.145)
 35.5 2 0.145
0.00717
  2n  N  
2

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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 2 0.145
n 1: hI (1)  sin(  34.5 2 0.145)
 34.5 2 0.145

0.0001449

   69  
2

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

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 2 0.145
n 2 : hI (2)  sin(  34.5 2 0.145)
 33.5 2 0.145

0.007415484

   69  
2

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

  

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 2 0.145
n 35 : hI (35)  sin(  0.5 2 0.145)
 0.5 2 0.145

0.280073974

   1 
2

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

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Fig. 7-10.

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 The optimal method

 Basic concepts
– Equiripple passband and stopband
E ( ) W ( )[ H I ( )  H ( )]

Weighted Weighting Ideal desired Practical

Approx. error function response response

– Mathematically expressed as
min[max E ( ) ]
over the passbands and stopbands

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 Practical response
Ideal response

Fig. 7-11.

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 min[max E ( ) ]
 equiripple passband and stopband of the resulting filter response,
with the ripple alternating in sign between equal amplitude levels
 The minima and maxima are known as extrema.
 For linear lowpass filters, for example, there are either m+1 or m+2
extrema,
m=(N+1)/2, type 1 filter
m= N/2, type 2 filter

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Fig. 7-12. Extremal frequencies

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– The procedure of optimal method
• Use the alternation theorem (Parks and McClellan) 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
– Transfer function of lowpass filter
N
H ( z )  h(k ) z  k (7-
k 0 28)
where h(n) h(  n)

• Symmetric property gives

N /2 N /2
H ( ) h(0)   2h(k ) cos kT  a(k ) cos kT
k 1 k 0

where a (0) h(0) and a (k ) 2h(k ) , k 1, 2, , N / 2

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• Let f  / 2 f s T / 2 be defined by normalized frequency, then
– Normalize as f p / f s  f p , f r / f s  f r
-> Normalized passband :  0, f p 
-> Normalized stopband :  f r , 0.5

• Desired magnitude response

1, 0  f  fp (7-
H I ( f ) 
0, f r  f 0.5 30)

• Weighting function
1
 , 0  f  fp
W  f   k (7-
 1 , f r  f 0.5 31)

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 a(k ) k 0,1, 2, , N / 2

min[max E ( f ) ]

N /2
H ( f )  H (e j 2 f
)  a (k ) cos k 2 f (7-
k 0 32)

E ( f ) W ( f )[ H ( f )  H I ( f )] (7-
33)

for f in  0, f p  and  f r , 0.5

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– Alternation theorem
• Let F  0, f p    f r , 0.5
• H ( f ) H I ( f ) is the unique best approximation if and only if E ( f ) is
equiripple at F and has at least m+2 extremal points in , that is, there
exists
f 0  f1   f l (l m  1)

such that
E ( f i )  E ( f i 1 )  e, i 0,1, , l  1 (7-
34)

where e max E( f )
f  0, f p  &  f r ,0.5 

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• From equations (7-33) and (7-34)

W ( fi )  H ( fi )  H I ( fi )   ( 1)i e, i 0,1, 2, , m  1 (7-

35)

• Equation (7-32) into equation (7-35) yields

m
e

k 0
a(k ) cos 2 kf i  H I ( f i )  ( 1)i
W ( f i )
, i 0,1, 2, , m 1

• Matrix form

 1 cos 2 f 0 cos 4 f 0  cos 2 mf 0 1/ W ( f 0 )   a(0)   H I ( f 0 ) 

 1 cos 2 f cos 4 f1  cos 2 mf1  1/ W ( f1 )   a(1)   H I ( f1 ) 
 1  
            
    
 1 cos 2 f m cos 4 f m  cos 2 mf m (  1) m / W ( f m )   a(m)   H I ( f m ) 
 1 cos 2 f m 1 cos 4 f m 1  cos 2 mf m 1 (  1) m ! / W ( f m 1 )   e   H I ( f m 1 ) 

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– Summary
Step 1. Choose filter length as 2m+1
Step 2. Choose m+2 points of fi 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
h(0)mlocal
a(0), minima
h(k ) a(or
k )maxima
/ 2 k points
1, 2, , m
Step 6. Obtain , ,
then the optimal filter Nis given by
H ( z )  h(k ) z  k
k 0

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 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  m  2 3

• Choose three frequencies and normalize them,

two of them are cutoff frequencies, the third one arbitrarily
f 0 0.5 / 2 , f1 1/ 2 , f 2 1.2 / 2

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• From H I ( f 0 ) H I ( f1 ) 1, H I ( f 2 ) 0, W ( f 0 ) 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)
• The error at f 0 and f1 is 2x0.196= 0.392, and the error at f 2 is 0.196.
=> not have E ( f )  e
=> Not the characteristic of the optimal filter

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• Choose a new set of fi
1 1.2
f 0  , f1 , f 2 0.5
2 2

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,

For this filter for all f in
e 0.306

( f ) 0.306
•Transfer of the optimalEfunction  0,1/ 2  and 1.2 / 2 , 0.5
H ( f ) 0.144  0.45cos 2 f (7-
38)

H ( z ) 0.225  0.144 z  1  0.225 z  2 (7-39)

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Fig. 7-13. Characteristics of filter H(f)

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 Optimal method using MATLAB
– Based on Parks-McClellan and Remez algorithm
– Calculation of coefficients 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

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 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]

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Table 7-4. Impulse response coefficients

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Fig.7-14. Frequency response of filter

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 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]

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• Estimation of filter length
– Pass and rejection band ripples
Ap

10 20  1  Ar
 p  Ap  r 10 20

10 20  1
where A p and Ar are ripples of dB scale in pass and rejection band

– Using Remezord in MATLAB

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Table 7-5. Impulse response coefficients of
filter

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Fig. 7-15. Frequency response of filter

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 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)

N1
1
h( n) 
N
 H
k 0
( k ) e j (2 / N ) nk
(7-
40)

where H (k ) , k 0,1,  , N  1 are samples of desired frequency response

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– 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)  (7-
N  k 1  41)
 

where  ( N  1) / 2

• For N odd
– Upper limit in the 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 
• From equation (7-40)

N1
1
h( n) 
N
 H (k ) e
k 0
j 2 nk / N

N1
1

N
 H (k ) e
k 0
 j 2 k / N
e j 2 kn / N
N1
1

N
 H (k ) e
k 0
j 2 ( n   ) k / N

N1
1

N
 H (k ) cos[2 k (n   ) / N ]  j sin[2 k (n   ) / N ]
k 0
h(n) is real value N1
1
•

is symmetryN
 H (k ) cos[2 k (n   ) / N ]
k 0

h( n)

1  N /2 1 
h(n)    2 H (k ) cos[2 k ( n   ) / N ]  H (0)  (7-
N  k 1  42)
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(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, k 0,1, 2
H (k ) 
 0, k 3, 4

Fig. 7-17. 61/77

• Coefficient of FIR filter using equation (7-42)

Table 7-6.

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 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

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