5.1. Frequency-domain Sampling#

  • The sampling theorem (see Section 3.2) tells us that we can reconstruct a bandlimited continuous-time signal \(x(t)\) with bandwidth \(B\) Hz from its samples \(x[n]=x\left(\frac{n}{f_s}\right)\) as long as we oversample, i.e., choose the sampling frequency \(f_s > 2B\).

  • One may ask whether a similar result applies to sampling the FT \(X(\omega)\) in the frequency domain. That is, can we get back \(X(\omega)\) (and hence \(x(t)\)) from its samples? If so, what is the condition under which we can do that?

  • We will answer the above questions in the context of a discrete-time signal \(x[n]\) and its DTFT \(X(e^{j\hat\omega})\), leaving the continuous-time counterpart as an exercise. More specifically, consider \(x[n] \stackrel{\text{DTFT}}{\longleftrightarrow} X(e^{j\hat\omega})\). Suppose we take \(M\) samples in one period of \(X(e^{j\hat\omega})\) at the normalized radian frequencies \(\hat\omega = \frac{2\pi k}{M}\) for \(k=0,1,\ldots,M-1\).

    Tip

    Recall that \(X(e^{j\hat\omega})\) is periodic in \(\hat\omega\) with period \(2\pi\). In particular, \(\displaystyle X(e^{j\hat\omega})\Big|_{\hat\omega = \frac{2\pi k}{M}} = X(e^{j\hat\omega})\Big|_{\hat\omega = \frac{2\pi (M+k)}{M}}\). Thus there is no difference between taking samples over the periods \([0,2\pi)\) and \([-\pi,\pi)\). The former is usually considered in the context of frequency-domain sampling.

    We want to determine whether, and if so the condition under which, the DTFT \(X(e^{j\hat\omega})\) can be recovered from the \(M\) frequency-domain samples \(\left\{X(e^{j\frac{2\pi k}{M}}) \right\}_{k=0}^{M-1}\). This frequency-domain sampling question can again be answered by way of the Poisson sum formula(s).

  • Let \(x[n]\) be obtained from sampling a continuous-time signal \(x(t)\) at the sampling rate \(f_s = \frac{1}{T_s}\) samples per second, i.e., \(x[n] = x(nT_s)\) and set \(T=MT_s\). Then the time-domain Poisson sum formula (3.1) gives

    \[\begin{equation*} \sum_{m=-\infty}^{\infty} x(t+mMT_s) = \frac{1}{MT_s} \sum_{l=-\infty}^{\infty} X\left(\frac{2\pi l}{MT_s} \right) e^{j\frac{2\pi l t}{MT_s}} \end{equation*}\]

    where \(x(t) \stackrel{\text{FT}}{\longleftrightarrow} X(\omega)\). At \(t=nT_s\), we get

    (5.1)#\[\begin{split}\begin{align*} \sum_{m=-\infty}^{\infty} x[n+mM] &= \frac{1}{MT_s} \sum_{l=-\infty}^{\infty} X\left(\frac{2\pi l}{MT_s} \right) e^{j\frac{2\pi l n}{M}} \\ &= \frac{1}{M} \sum_{k=0}^{M-1} f_s \sum_{m=-\infty}^{\infty} X\left( \frac{2\pi (k+mM)}{MT_s} \right) e^{j\frac{2\pi (k+mM) n}{M}} \\ &= \frac{1}{M} \sum_{k=0}^{M-1} X(e^{j\frac{2\pi k}{M}}) e^{j\frac{2\pi k n}{M}} \end{align*}\end{split}\]

    where the last equality is due to the frequency-domain Poisson sum formula (3.6). Let \(x_M[n] = \sum_{m=-\infty}^{\infty} x[n+mM]\) be the periodic signal (with period \(M\)) on the LHS of (5.1). We often call \(x_M[n]\) the periodic extension of \(x[n]\). Then, (5.1) is simply the synthesis formula of the DFS of \(x_M[n]\) (see Section 2.4.4). That is, \(x_M[n]\) can be constructed as the sum of the \(M\) weighted harmonics at frequencies \(\{\frac{2\pi k}{M}\}_{k=0}^{M-1}\). The frequency-domain samples \(\left\{\frac{1}{M} X(e^{j\frac{2\pi k}{M}}) \right\}_{k=0}^{M-1}\) serve as the DFS coefficients to weigh the harmonics.

  • In particular, if \(x[n]=0\) except for \(n=0,1,\ldots, M-1\), then \(x_M[n] = x[n]\) for \(n=0,1,\ldots, M-1\). We often say that the length of the signal \(x[n]\) is at most \(M\). Thus, as long as the length of \(x[n]\) is at most \(M\), we can easily obtain \(x[n]\), and hence \(X(e^{j\hat\omega})\), back from the frequency-domain samples \(\left\{X(e^{j\frac{2\pi k}{M}}) \right\}_{k=0}^{M-1}\) using (5.1).

  • On the other hand, if the length of \(x[n]\) is larger than \(M\), then \(x_M[n] \neq x[n]\) over any period. The DFS synthesis method in (5.1) can only give us \(x_M[m]\) but not \(x[n]\). We often call this reconstruction suffers from time-aliasing.