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With the window design method, the amplitudes of the ripples in each band of a multiband filter will be approximately equal, and the transition bands will have approximately the same width. Because the requirements on the peak ripple in the three bands of this bandpass filter are not the same, it is necessary to design the filter so that it has the smallest ripple in all three bands, which, in this case, requires that we set 6, = 0.0025. In addition, because the transition bands do not have the same width, it is necessary to set the desired transition width, Aw, equal to the smaller of the two (Aw = 0 . 1 5 ~ ) . With a, = -20 log 6, = 52 dB. it follows that we may use a Hamming window, and with NAf = 3.3

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For a type I1 filter, however, N must be odd, so we set N = 45. Now we must find the unit sample response of the ideal bandpass filter that is to be windowed. Because the , width of both the upper and lower transition bands will be approximately Aw = 0 . 1 5 ~for the ideal filter we set the lower cutoff frequency equal to

AB )

lim f (x) = 1

1 H,,(ul"')( =

1 2 (80

Repeating the steps in the derivaticn of the unit sample response of an ideal bandpass fi lter given in Prob. 9.8, using the given cutoff frequencies and a delay of N j 3 = 22.5. we have

1 2 (70

Use the window design method to design a multiband filter that meets the following specifications:

120 )

lim f (x) = 2

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To design a multiband filter that meets these specifications usmg the window design method, we begin by finding the ideal unit sample response. For the frequency response of the ideal filter. we set the cutoff frequencies equal to the midpoint of the transition bands. Therefore, we have

1 2(

The unit sample response of this ideal lilter may be found easily by noting that Hd(ei")may be written as an allpass plus a low-pass filter with a gain of 0.5 minus a low-pass ti lter with a gain of 0.5 and a cutoff frequency w, = 0 . 5 7 5 ~ filter with a gain of I and a cutoff frequency of w2 = 0.325n. Therefore. if we assume that H,,(eJW) linear phase has with a delay of n d ,

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Having found the ideal unit sample response. the next step is to choose an appropriate window. When h d ( n ) is with multiplied by a window w(rr),the frequency response is the convolution of the transform of the window W ( e J W ) Hd(el'").Assuming that the length of the filter is long compared to the inverse of the transition width, so that the discontinuities between the bands may be treated independently. the ripples in the three bands will be approximately the same as they would be for a low-pass filter, except that they will be scaled by the amplitude of the discontinuities at the band edge. Therefore. if the ripple in the lower passband and the stopband are 4, the ripple in the upper passband will be 6 , , / 2 . Consequently. we must use a window that would produce a low-pass filter with a ripple no larger than 0.01. Thus. we may use a Hanning window. Finally, to determine the filter order, note that because the widths of both transition bands are the same. Ao) = 0 . 0 5 ~an. estimate of the filter order is ~

Fig. 6-41

Note that another way to design this tilter would have been to design a network of three filters in parallel: a low-pass filter, a bandpass filter, and a high-pass filter. This approach would give greater control over the ripple amplitudes and the transition widths but would require a trial and error approach to establish the specifications for the three filters.

Fig. 9-1

1 2 (200 1 2

Shown in the following figure is the magnitude of the frequency response of a type I high-pass filter that was designed using the Parks-McClellan algorithm.

120 . 66 .

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