Suppose we want to average daily stock prices taken over last year to yield a running weekly average (average over five trading sessions). The filter we want is a length-5 averager (as shown in the unit-sample response), and the input's duration is 253 (365 calendar days minus weekend days and holidays). The output duration will be 253+5-1=257 253 5 1 257 , and this determines the transform length we need to use. Because we want to use the FFT, we are restricted to power-of-two transform lengths. We need to choose any FFT length that exceeds the required DFT length. As it turns out, 256 is a power of two ( 28=256 2 8 256 ), and this length just undershoots our required length. To use frequency domain techniques, we must use length-512 fast Fourier transforms.

Figure 1 shows the input and the filtered output. The MATLAB programs that compute the filtered output in the time and frequency domains are

	Time Domain 
	h = [1 1 1 1 1]/5; 
	y = filter(h,1,[djia zeros(1,4)]);

	Frequency Domain
	h = [1 1 1 1 1]/5; 
	DJIA = fft(djia, 512);
	H = fft(h, 512);
	Y = H.*X;
	y = ifft(Y);
      

MATLAB's fft function automatically zero-pads its input if the specified transform length (its second argument) exceeds the signal's length. The frequency domain result will have a small imaginary component — largest value is 2.2×10-11 2.2-11 — because of the inherent finite precision nature of computer arithmetic. Because of the unfortunate misfit between signal lengths and favored FFT lengths, the number of arithmetic operations in the time-domain implementation is far less than those required by the frequency domain version: 514 versus 62,271. If the input signal had been one sample shorter, the frequency-domain computations would have been more than a factor of two less (28,696), but far more than in the time-domain implementation.

An interesting signal processing aspect of this example is demonstrated at the beginning and end of the output. The ramping up and down that occurs can be traced to assuming the input is zero before it begins and after it ends. The filter "sees" these initial and final values as the difference equation passes over the input. These artifacts can be handled in two ways: we can just ignore the edge effects or the data from previous and succeeding years' last and first week, respectively, can be placed at the ends.