Summary of Widowed Fourier Series Method for Calculating FIR Filter Coefficients

Step 1: Specify ‘ideal’ or desired frequency response of filterStep 2: Obtain impulse response, hD[n] of the desired filter by

evaluating the inverse Fourier Transform as summarized in Table in next slide.

Step 3: Select a window function that satisfies the passband or attenuation specifications and then

determine the number of coefficients using the appropriate relationship between the filter length and the transition width

Step 4: Obtain values of w[n] for the chosen window function and the values of the actual FIR coefficients, h[n] by multiplying hD[n] by w[n]

( )DH

Summary of ideal impulse response of standard frequency selective filters

Filter type Ideal impulse response hD[n]

hD [0]

Low Pass

High Pass 1

Band Pass 2f2sinc(n2)- 2f1sinc(n1) 2(f2-f1)

Band Stop 1-[2f2sinc(n2)- 2f1 inc(n1)] 1-2(f2-f1)

' sin( )2 cc

c

nf

n

'2 cf

' sin( )2 cc

cf

n

'2 cf

Summary of Important Features of Common Window Functions

Window Representation ExpressionRectangular wR[n] 1Hanning whn [n] 0.50 + 0.50 cos{2n/(N)}Hamming whm [n] 0.54 + 0.46 cos{2n/(N)}Blackman wb [n] 0.42 +0.50 cos{2n/(N-1)}

+0.08 cos{4n/(N-1)}

' _, where 2 _

cc

s

f f Transition widthf fF Sampling Frequency

Order of filter = N = where c the coefficients depend on type of windows being used as in table next slide

cf

Summary of Important Features of Common Window Functions….Cont

Window Transition Passband Stopband Width Ripple Attenuation (dB)

(normalized) (dB) (maximum allowed)Rectangular 0.9/N 0.7416 21 Hanning 3.1/N 0.0546 44

Hamming 3.3/N 0.0194 53

Blackman 5.5/N 0.0017 75

f

Example:Design a low-pass FIR filter to meet the following specs:Pass band edge frequency: 1500 HzTransition width: 500 Hz.Stop-band attenuation AWS= > 50 dBSampling frequency fs = 8000 Hz.

Problem Statement:1. Meaning of given specifications are:

Sampling frequency fs = 8000 Hz.

Pass band edge frequency: fc =1500/8000

Transition width f = 500/8000.Stop-band attenuation AWS= > 50 dB

Design considerations contd…2. The filter function is

3. Because of stop-band attenuation characteristics, either of the Hamming,

Blackman or,Kaiser windows

can be used. We use Hamming window:

whm[n] =0.54 + 0.46 cos{2n/(N-1)}

' sin( )2 cc

c

nf

n

Design considerations contd…4. f = transition band width/sampling frequency

= 0.5/8 =0.0625 = 3.3/N.Thus N = 52.8 53 i.e. for symmetrical window

–26 n 26.

fc’ = fc + f/2 = (1500+ 250)/8000 = 0.21875.5. Calculate values of hD [n] and whm[n] for

–26 n 26 Add 26 to each index so that the indices range

from 0 to 52.6. Plot the response of the design and verify the

specifications.

Calculations:• c = 2fc

’ = 1.3745

2 fc’ =1.3745/ = 0.4375

• hD(n) = fc’ [sin(nc)/ nc]

wn = [0.54 + 0.46cos(2n/N) The input signal to the filter function is a series of

pulses of known width but of different heights manipulated as per the window function.

• The overall is the multiplication of two.

Calculations…

h(n) = hD [n] w D[n] = 0.4375 {[sin(nc)/ nc]}

x {[0.54 + 0.46cos(2n/N)}at n=0, since sin(nc)/nc = 1, and cos(0) = 1; h(0) = 0.4375 x[0.54 + 0.46] = 0.4375.Again since 2 fc

’ / c = 1/h(n)= [sin(1.3745n)/n] [0.54 +0.46cos(2n/53)]

Coefficient Calculationsn 1 26.. hn

sin 1.3745n( )

n wn 0.54 0.46( )cos 2

n53

hn

0.3120.061-0.088-0.0560.0350.049

-3-8.888·10-0.04

-3-6.883·100.0290.016-0.019-0.02

-38.716·100.021

-5-1.694·10-0.018

-3-6.752·100.0140.011

-3-8.435·10-0.013

-32.717·100.013

-32.467·10-0.011

wn

0.9930.9720.9370.89

0.8290.7580.6750.5830.4830.3760.2640.1480.03

-0.089-0.206-0.32-0.43

-0.534-0.63

-0.718-0.795-0.861-0.915-0.956-0.984-0.998

hn wn

0.310.059-0.083-0.050.0290.037

-3-5.999·10-0.023

-3-3.323·100.011

-34.234·10-3-2.771·10-4-6.03·10-4-7.74·10-3-4.286·10-65.425·10-37.899·10-33.604·10-3-8.783·10-3-8.066·10-36.705·10

0.012-3-2.486·10

-0.013-3-2.428·10

0.011

n

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