EVALUATING THE AMPLITUDE RESPONSE The frequency response H f ω of an FIR filter can be evaluated at L equally spaced frequencies between 0 and using the DFT. Consider a causal FIR filter with an impulse response h n of length-N, with N L . Samples of the frequency response of the filter can be written as H 2 L k n 0 N 1 h n 2 L n k Define the L-point signal g n 0 n L 1 as g n h n 0 n N 1 0 N n L 1 Then H 2 L k G k DFT L g n where G k is the L-point DFT of g n .

Types I and II Suppose the FIR filter h n is either a Type I or a Type II FIR filter. Then we have from above H f ω A ω M ω or A ω H f ω M ω Samples of the real-valued amplitude A ω can be obtained from samples of the function H f ω as: A 2 L k H 2 L k M 2 L k G k W L M k Therefore, the samples of the real-valued amplitude function can be obtained by zero-padding h n , taking the DFT, and multiplying by the complex exponential. This can be written as: A 2 L k DFT L h n 0 L - N W L M k

Types III and IV For Type III and Type IV FIR filters, we have H f ω M ω A ω or A ω H f ω M ω Therefore, samples of the real-valued amplitude A ω can be obtained from samples of the function H f ω as: A 2 L k H 2 L k M 2 L k G k W L M k Therefore, the samples of the real-valued amplitude function can be obtained by zero-padding h n , taking the DFT, and multiplying by the complex exponential. A 2 L k DFT L h n 0 L - N W L M k

EVALUATING THE AMP RESP (TYPE I) In this example, the filter is a Type I FIR filter of length 7. An accurate plot of A ω can be obtained with zero padding.

The following Matlab code fragment for the plot of A ω for a Type I FIR filter.


	  h = [3 4 5 6 5 4 3]/30;
	  N = 7;
	  M = (N-1)/2;
	  L = 512;
	  H = fft([h zeros(1,L-N)]);
	  k = 0:L-1;
	  W = exp(j*2*pi/L);
	  A = H .* W.^(M*k);
	  A = real(A);

	  figure(1)
	  w = [0:L-1]*2*pi/(L-1);
	  subplot(2,1,1)
	  plot(w/pi,abs(H))
	  ylabel('|H(\omega)| = |A(\omega)|')
	  xlabel('\omega/\pi')
	  subplot(2,1,2)
	  plot(w/pi,A)
	  ylabel('A(\omega)')
	  xlabel('\omega/\pi')
	  print -deps type1

The command A = real(A) removes the imaginary part which is equal to zero to within computer precision. Without this command, Matlab takes A to be a complex vector and the following plot command will not be right. Observe the symmetry of A ω due to h n being real-valued. Because of this symmetry, A ω is usually plotted for 0 ω only. EVALUATING THE AMP RESP (TYPE II)

The following Matlab code fragment produces a plot of A ω for a Type II FIR filter.


	  h = [3 5 6 7 7 6 5 3]/42;
	  N = 8;
	  M = (N-1)/2;
	  L = 512;
	  H = fft([h zeros(1,L-N)]);
	  k = 0:L-1;
	  W = exp(j*2*pi/L);
	  A = H .* W.^(M*k);
	  A = real(A);

	  figure(1)
	  w = [0:L-1]*2*pi/(L-1);
	  subplot(2,1,1)
	  plot(w/pi,abs(H))
	  ylabel('|H(\omega)| = |A(\omega)|')
	  xlabel('\omega/\pi')
	  subplot(2,1,2)
	  plot(w/pi,A)
	  ylabel('A(\omega)')
	  xlabel('\omega/\pi')
	  print -deps type2

The imaginary part of the amplitude is zero. Notice that A 0 . In fact this will always be the case for a Type II FIR filter. An exercise for the student: Describe how to obtain samples of A ω for Type III and Type IV FIR filters. Modify the Matlab code above for these types. Do you notice that A ω 0 always for special values of ω?