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Using ROM decomposition method of DA architecture we can reduce ROM size, however complexity of system design still exists. Hence to study DA architecture in FIR filter design a lower order filter is designed. Decrease in the order of the filter affects the frequency response of the filter, but it is time being neglected. | Chapter 11 Fourier Series Method of FIR Filter Design 11.1 Basis of the Fourier Series Method This Fourier series method of FIR filter design is based on the fact that the frequency response of a digital filter is periodic and is therefore representable as a Fourier series. A desired target frequency response is selected and expanded as a Fourier series. This expansion is truncated to a finite number of terms that are then used as the filter coefficients or tap weights. The resulting filter has a frequency response that approximates the original desired target response. Algorithm 11.1 Designing FIR filters via the Fourier series method Step 1. Specify a desired frequency response Hd . . Step 2. Specify the desired number of filter taps N. Step 3. Compute the filter coefficients fi n for n 0 1 2 . . N 1 using 7i n frf . cos mz j sin mz dz 11-1 where m n N l 2. Simplifications of 11.1 are presented below for the cases in which Hd is the magnitude response of ideal lowpass highpass bandpass or bandstop filters. Step 4. Using the techniques presented in Secs. 10.2 and 10.3 compute the actual frequency response of the resulting filter. If the performance is not adequate change N or Hd 2 and go back to step 3. 171 172 Chapter Eleven Hd ejX X -2 r 5 Figure 11.1 Desired frequency response for Example 11.1. Example 11.1 Use the Fourier series method to design a 21-tap FIR filter that approximates the amplitude response of an ideal lowpass filter with a cutoff frequency of 2 kHz assuming a sampling frequency of 5 kHz. solution The normalized cutoff is I 2- 5. The desired frequency response is depicted in Fig. 11.1. Using Eq. 11.1 we can immediately write J i 27t 5 i i 2n 5 i n cos znA d . j sin mz dz 2 J-2IT 5 J_ 2n 5 Since the second integrand is an odd function and the limits of integration are symmetric about zero the second integral equals zero. Therefore sin mz 2nl 2m7t 2_ 2n 5 sin 2m7t 5 mn 11-2 where m n 10. L Hospital s rule can be used to evaluate 11.2 for the case
