One approach to the design of FIR filters is to ask that A⁢ω A ω pass through a specified set of values. If the number of specified interpolation points is the same as the number of filter parameters, then the filter is totally determined by the interpolation conditions, and the filter can be found by solving a system of linear equations. When the interpolation points are equally spaced between 0 and 2⁢π 2 , then this interpolation problem can be solved very efficiently using the DFT.

To derive the DFT solution to the interpolation problem, recall the formula relating the samples of the frequency response to the DFT. In the case we are interested here, the number of samples is to be the same as the length of the filter ( L=N L N ).

The following Matlab code fragment illustrates how to use this approach to design a length 11 Type I FIR filter for which ∀k,(0≤k≤N−1)∧(N=11):A⁢2⁢πN⁢k=11100000011T k 0 k N 1 N 11 A 2 N k 1 1 1 0 0 0 0 0 0 1 1 .

	>> N = 11;
	>> M = (N-1)/2;
	>> Ak = [1 1 1 0 0 0 0 0 0 1 1};   % samples of A(w)
	>> k = 0:N-1;
	>> W = exp(j*2*pi/N);
	>> h = ifft(Ak.*W.^(-M*k));
	>> h'

	ans = 

	   0.0694 - 0.0000i
	  -0.0540 - 0.0000i
	  -0.1094 + 0.0000i
	   0.0474 + 0.0000i
	   0.3194 + 0.0000i
	   0.4545 + 0.0000i
	   0.3194 + 0.0000i
	   0.0474 + 0.0000i
	  -0.1094 + 0.0000i
	  -0.0540 - 0.0000i
	   0.0694 - 0.0000i

      

Observe that the filter coefficients h are real and symmetric; that a Type I filter is obtained as desired. The plot of A⁢ω A ω for this filter illustrates the interpolation points.

	L = 512;
	H = fft([h zeros(1,L-N)]);
	W = exp(j*2*pi/L);
	k = 0:L-1;
	A = H .* W.^(M*k);
	A = real(A);
	w = k*2*pi/L;
	plot(w/pi,A,2*[0:N-1]/N,Ak,'o')
	xlabel('\omega/\pi')
	title('A(\omega)')

      

Figure 1

An exercise for the student: develop this DFT-based interpolation approach for Type II, III, and IV FIR filters. Modify the Matlab code above for each case.

For an NN-point linear-phase FIR filter h⁢n h n , we summarize: