Laboratory Exercise 6.7b (continued)
Starting from Laboratory Exercise 6.7a, the same random signal of duration -- seconds and the same filter are used but this time the technique used to produce g[n] is through the frequency domain. First the Discrete Fourier Transform (DFT) of the filter is computed yielding H(Ωk). The FIR filter consists of 43 values so the DFT also has 43 complex values, that is, k = 1,2,3,...,43.

Our intention is to multiply the discrete spectrum of the filter H(Ωk) by the discrete spectrum of the signal G(Ωk). To do so, however, requires that the two spectra have the same number of discrete frequencies.
  1. Why must the two spectra have the same number of frequencies?
The signal spectrum has over 40,000 frequencies so, to make the two equal in length, we use “zero-padding” on the filter in the time domain.
  1. Why will the signal spectrum have over 40,000 frequencies?

  2. If there are 40,000 frequencies and a sampling frequency of 44.1 kHz, what is the difference between two adjacent frequencies Δω = ωk - ωk-1?
We append sufficient zeroes to the FIR filter so that h[n] and g[n] have the same length. Next, we compute the DFT of the padded filter and use this result as H(Ωk). See Section 13-6 of McClellan1.

We multiply the two spectra, GF(Ωk) = H(Ωk)G(Ωk), and then compute the Inverse Discrete Fourier Transform (IDFT) to produce gF [n]. In formula:
(6.44) g[n] = F  –1 {{ h[n] } • { g[n] }}
where {•} is the Fourier transform operator. Examine and listen to g[n], the original signal, and gF [n], the frequency-domain filtered result.
  1. Do the frequencies where the passband changes to stopband (and vice versa) still correspond to the frequencies you expected?

  2. Is the width of the probability distribution associated with the amplitudes of gF [n] still smaller than the width associated with g[n]?

  3. Does the autocorrelation function φ FF (τ=kT), shown below, resemble the autocorrelation function φFF (τ=kT) shown in section 6.7a? Use the “Zoom” function to examine the results.

  4. Does the ratio SFF (Ω)/Sgg (Ω) computed through the Fourier domain and shown below resemble the ratio SFF (Ω)/Sgg (Ω) computed through convolution and shown on the previous page? Use the “Zoom” function to examine the results.
Proceeed to another part of this exercise by choosing the variant below.

1McClellan, J. H., R. W. Schafer and M. A. Yoder (2003). Signal Processing First. Upper Saddle River, NJ, USA, Pearson Education, Inc.   


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Zoom: N = ----- samples
= ----- ms

  Play signal g(t)      

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