3.1. Poisson Sum Formula

3.1.1. Time-domain Poisson sum formula

Let us re-develop a slightly more general version of the Poisson sum formula:

• Consider \(x(t) \in L^1 \cap L^2\). Fix \(T>0\). Then

  • \(\sum_{n=-\infty}^{\infty} |x(t+nT)| < \infty\) for almost everywhere on \(\mathbb{R}\) and

  • \(x(t)\) has its FT \(X(\omega) \in L^2\).

  Thus, \(x_T (t) = T \sum_{n=-\infty}^{\infty} x(t+nT)\) is well-defined and finite. Clearly, \(x_T(t)\) is a periodic signal with period \(T\) and it is of finite power, i.e., \(x_T(t) \in L^2 \left[-\frac{T}{2}, \frac{T}{2}\right]\). As a consequence, \(x_T (t)\) has a FS expansion.

• Now, according to the forward FT formula (2.12)

  \[\begin{align*} X(\omega) &= \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \, dt \notag \\ &= \sum_{n=-\infty}^{\infty} \int_{nT-\frac{T}{2}}^{nT+\frac{T}{2}} x(t) e^{-j\omega t} \, dt \notag \\ &= \sum_{n=-\infty}^{\infty} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t+nT) e^{-j\omega (t+nT)} \, dt \notag \\ &= \int_{-\frac{T}{2}}^{\frac{T}{2}} \left( \sum_{n=-\infty}^{\infty} x(t+nT) e^{-j\omega (t+nT)} \right) \, dt & \text{(dominated convergence)} \end{align*}\]

  In particular, at \(\omega = \frac{2\pi k}{T}\), for \(k \in \mathbb{Z}\), we get

  \[\begin{equation*} X\left(\frac{2\pi k}{T}\right) = \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x_T(t) e^{-j\frac{2\pi k t}{T}} \, dt. \end{equation*}\]

  That is, \(X\left(\frac{2\pi k}{T}\right)\) is simply the \(k\)th FS coefficient of \(x_T(t)\). Hence, \(\left\{X\left(\frac{2\pi k}{T}\right) \right\}_{k \in \mathbb{Z}} \in l^2\). The FS synthesis formula (2.9) then gives the following version of the Poisson sum formula:

  (Time-domain) Poisson Sum Formula

  For any \(x(t) \in L^1 \cap L^2\) and \(T>0\),

  (3.1)\[\begin{equation} \sum_{n=-\infty}^{\infty} x(t+nT) = \frac{1}{T} \sum_{k=-\infty}^{\infty} X\left(\frac{2\pi k}{T}\right) e^{j\frac{2\pi k t}{T}} \end{equation}\]

  where \(x(t) \stackrel{\text{FT}}{\longleftrightarrow} X(\omega)\).

3.1.2. Periodic vs aperiodic Dirac delta

• Consider the time-domain versions of the periodic and aperiodic Dirac delta function:

  • Periodic: \(\frac{2\pi}{T} \delta (e^{j\frac{2\pi t}{T}}) = \lim_{\lambda \rightarrow 1} \frac{W_{\lambda} (e^{j\frac{2\pi t}{T}})}{T}\)

  • Aperiodic: \(\delta(t) = \lim_{\mu \rightarrow 0} \tilde{w}_{\mu}(t)\)

  These two Dirac deltas are related to one another and the relationship can be established using the Poisson sum formula in (3.1).

• Let \(x(t) = \tilde{w}_{\mu}(t) \stackrel{\text{FT}}{\longleftrightarrow} X(\omega) = \tilde{W}_{\mu}(\omega)\) in (3.1). We get

  (3.2)\[\begin{split}\begin{align} \sum_{n=-\infty}^{\infty} \tilde{w}_{\mu}(t+nT) &= \frac{1}{T} \sum_{k=-\infty}^{\infty} \tilde{W}_{\mu} \left( \frac{2\pi k}{T}\right) e^{j\frac{2\pi k t}{T}} \notag \\ &= \frac{1}{T} \sum_{k=-\infty}^{\infty} e^{-\mu\frac{2\pi |k|}{T}} \cdot e^{j\frac{2\pi k t}{T}} \notag \\ &= \frac{1}{T} \cdot \frac{1 - e^{-\frac{4\pi\mu}{T}}}{1 - 2e^{-\frac{2\pi\mu}{T}} \cos\left(\frac{2\pi t}{T}\right) +e^{-\frac{4\pi \mu}{T}}} \notag \\ &= \frac{1}{T} \cdot W_{e^{-\frac{2\pi\mu}{T}}} \left(e^{j\frac{2\pi t}{T}} \right). \end{align}\end{split}\]

• Taking “limit” as \(\mu \rightarrow 0\) on both sides of (3.2), we obtain

  (3.3)\[\begin{equation} \sum_{n=-\infty}^{\infty} \delta(t+nT) = \frac{2\pi}{T} \delta \left(e^{j\frac{2\pi t}{T}} \right) \end{equation}\]

  which confirms the pictorial intuition that the periodic Dirac delta \(\frac{2\pi}{T} \delta (e^{j\frac{2\pi t}{T}})\) is simply a train of aperiodic Dirac deltas separated by \(T\).

• Using (3.3), the earlier version of the Poisson sum formula given in (2.11) can now be re-written as

  (3.4)\[\begin{equation} \sum_{n=-\infty}^{\infty} \delta(t+nT) = \frac{1}{T} \sum_{n=-\infty}^{\infty} e^{j\frac{2\pi n t}{T}} \end{equation}\]

  which can be thought of as the extension of (3.1) to the Dirac delta signal \(\delta(t)\) because \(\delta(t) \stackrel{\text{FT}}{\longleftrightarrow} 1\) from Example 6. in Section 2.7. Of course, one should interpret (3.4) as a mnemonic for the limiting process of (3.1) with \(x(t) = \tilde{w}_{\mu}(t)\) and \(X(\omega) = \tilde{W}_{\mu}(\omega)\) as \(\mu \rightarrow 0\).

3.1.3. FT of a Dirac delta train

• Example:

  7. Taking FT on both sides of the Poisson sum formula (3.4) and using the linearity property of FT and Example 1. in Section 2.7, we get

     \[\begin{equation*} \sum_{n=-\infty}^{\infty} \delta(t+nT) = \frac{1}{T} \sum_{n=-\infty}^{\infty} e^{j\frac{2\pi n t}{T}} ~~ \stackrel{\text{FT}}{\longleftrightarrow} ~~ \frac{2\pi}{T} \sum_{n=-\infty}^{\infty} \delta \left(\omega + \frac{2\pi n}{T} \right) \end{equation*}\]

3.1.4. Frequency-domain Poisson sum formula

Let us re-develop the Poisson sum formula from a frequency-domain perspective:

• Consider \(X(\omega) \in L^1 \cap L^2\). Fix \(f_s>0\) and let \(\hat\omega = \frac{\omega}{fs}\). Then

  • \(\sum_{k=-\infty}^{\infty} |X((\hat\omega+2\pi k)f_s)| < \infty\) for almost everywhere on \(\mathbb{R}\) and

  • \(X(\omega)\) has its inverse FT \(x(t) \in L^2\).

  Thus, \(X(e^{j\hat\omega}) = f_s \sum_{k=-\infty}^{\infty} X((\hat\omega+2\pi k)f_s) \) is well-defined and finite. Clearly, \(X(e^{j\hat\omega})\) is a periodic function in \(\hat\omega\) with period \(2\pi\) and \(X(e^{j\hat\omega}) \in L^2[-\pi,\pi]\).

• Now, consider sampling \(x(t)\) with the sampling rate \(f_s\) samples per second to obtain the discrete-time signal \(x[n]\). Then

  \[\begin{align*} x[n] &= x\left(\frac{n}{f_s}\right) \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\frac{\omega n}{f_s}} \, d\omega & \text{(inverse FT)} \notag \\ &= \sum_{k=-\infty}^{\infty} \frac{1}{2\pi} \int_{2\pi k f_s - \pi f_s}^{2\pi k f_s + \pi f_s} X(\omega) e^{j\frac{\omega n}{f_s}} \, d\omega \notag \\ &= \sum_{k=-\infty}^{\infty} \frac{f_s}{2\pi} \int_{-\pi}^{\pi} X((\hat\omega+2\pi k)f_s) e^{j\hat\omega n} \, d\hat\omega \notag \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left( f_s \sum_{k=-\infty}^{\infty} X((\hat\omega+2\pi k)f_s) \right) e^{j\hat\omega n} \, d\hat\omega & \text{(dominated convergence)} \notag \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\hat\omega}) e^{j\hat\omega n} \, d\hat\omega. \end{align*}\]

  That is, \(X(e^{j\hat\omega})\) is simply the DTFT of \(x[n]\). The forward DTFT formula (2.1) then gives the following frequency-domain version of the Poisson sum formula:

  (Frequency-domain) Poisson Sum Formula

  For any \(X(\omega) \in L^1 \cap L^2\) and \(f_s>0\),

  (3.5)\[\begin{equation} f_s \sum_{k=-\infty}^{\infty} X((\hat\omega+2\pi k)f_s) = \sum_{n=-\infty}^{\infty} x\left(\frac{n}{f_s}\right) e^{-j\hat\omega n} \end{equation}\]

  where \(x(t) \stackrel{\text{FT}}{\longleftrightarrow} X(\omega)\).