SUMMARY: AMPLITUDE FORMULAS

where M=N−12 M N 1 2

AMPLITUDE RESPONSE CHARACTERISTICS

To analyze or design linear-phase FIR filters, we need to know the characteristics of the amplitude response Aω A ω .

Figure 1

EVALUATING THE AMPLITUDE RESPONSE

The frequency response H f ω H f ω of an FIR filter can be evaluated at LL equally spaced frequencies between 0 and π using the DFT. Consider a causal FIR filter with an impulse response hn h n of length-NN, with N≤L N L . Samples of the frequency response of the filter can be written as H2πLk=∑n=0N−1hnⅇ-ⅈ2πLnk H 2 L k n 0 N 1 h n 2 L n k Define the LL-point signal {gn|0≤n≤L−1} g n 0 n L 1 as gn=hnif0≤n≤N−10ifN≤n≤L−1 g n h n 0 n N 1 0 N n L 1 Then H2πLk=Gk= DFT L gn H 2 L k G k DFT L g n where Gk G k is the LL-point DFT of gn g n .

Example 1: EVALUATING THE AMP RESP (TYPE I)

In this example, the filter is a Type I FIR filter of length 7. An accurate plot of Aω A ω can be obtained with zero padding.

Figure 2

The following Matlab code fragment for the plot of Aω 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 Aω A ω due to hn h n being real-valued. Because of this symmetry, Aω A ω is usually plotted for 0≤ω≤π 0 ω only.

Example 2: EVALUATING THE AMP RESP (TYPE II)

Figure 3

The following Matlab code fragment produces a plot of Aω 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 Aπ=0 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 Aω A ω for Type III and Type IV FIR filters. Modify the Matlab code above for these types. Do you notice that Aω=0 A ω 0 always for special values of ωω?

Modules for Further Study