DFT of Aperiodic Discrete-Time Signals

10.4.2 DFT of Aperiodic Discrete-Time Signals

We obtain the DFT of an aperiodic signal y[n] by sampling its DTFT, Y(e jω ) , in frequency. Suppose we choose {ω k = 2πk/L, k = 0, . . . , L − 1} as the sampling frequencies, where an appropriate value for the integer L > 0 needs to be determined. Analogous to the sampling-in-time we did before, sampling-in-frequency generates a periodic signal in time:

˜y[n] =

Now, if y[n] is of finite length N, then when L ≥ N the periodic expansion ˜y[n] clearly displays a first period equal to the given signal y[n] (with some zeros attached at the end when L > N). On the other hand, if the length L < N the first period of ˜y[n] does not coincide with y[n] because of superposition of shifted versions of it (this corresponds to time aliasing, the dual of frequency aliasing, which occurs in time sampling).

Assuming y[n] is of finite length N and that L ≥ N, as the dual of sampling in time we then have that

X −1

˜y[n] =

y[n + rL] ⇔ Y[k] = Y(e

j2π k/L

y[n]e −j2πnk/L k = 0, . . . , L − 1 (10.48)

r =−∞

n =0

The equation on the right is the DFT of y[n]. The inverse DFT is the Fourier series representation of ˜y[n] (normalized with respect to L) or its first period

X −1 y[n] j2π nk/L = Y[k]e

0 ≤n≤L−1

where Y[k] = Y(e j2π k/L ) . Thus, instead of the frequency aliasing that sampling-in-time causes, we have time-aliasing whenever

the length N of y[n] is greater than the chosen L in the sampling-in-frequency. In practice, the gen- eration of the periodic extension ˜y[n] is not needed—we just need to generate a period that either coincides with y[n] when L = N, or when L > N that coincides with y[n] with a sequence of L − N zeros attached to it (i.e., y[n] is padded with zeros). To avoid time aliasing we do not consider choosing L < N.

If the signal y[n] is a very long signal, in particular if N → ∞, it does not make sense to compute its DFT, even if we could. Such a DFT would give the frequency content of the whole signal and since an infinite-length signal could have all types of frequencies its DFT would just give no valuable information. A possible approach to obtain, over time, the frequency content of a signal with a large time support is to window it and compute the DFT of each of these segments. Thus, when y[n] is of infinite length, or its length is much larger than the desired or feasible length L, we use a window W L [n] of length L, and represent y[n] as the superposition

y[n] X = y

m [n] where y m [n] = y[n]W L [n − mL]

10.4 Discrete Fourier Transform 617

Therefore, by the linearity of the DFT, we have the DFT of y[n] is

X Y[k] X = DFT(y

m [n]) =

Y m [k]

where each Y m [k] provides a frequency characterization of the windowed signal or the local frequency content of the signal. Practically, this would be more meaningful than finding the DFT of the whole signal. Now we have frequency information corresponding to segments of the signal and possibly evolving over time.

The DFT of an aperiodic signal x[n] of finite length N is found as follows:

Choose an integer L ≥ N that is the length of the DFT to be the period of a periodic extension ˜x[n] having x[n] as a period with padded zeros if necessary.

Find the normalized Fourier series representation of ˜x[n],

1 L X −1 ˜x[n] = j2π nk/L ˜X[k]e

≤n≤L−1

˜X[k] =

˜x[n]e −j2πnk/L

0 ≤k≤L−1

X[k] = ˜X[k] for 0 ≤ k ≤ L − 1 is the DFT of x[n].

x[n] = ˜x[n]W[n] where W[n] = u[n] − u[n − L] is a rectangular window of length N, is the IDFT of X[k]. The IDFT x[n] is defined for 0 ≤ n ≤ L − 1.