Impulse Trains

The impulse signal $\delta(t)$ (defined in §B.10) has a constant Fourier transform:

$$\hbox{\sc FT}_f(\delta) \isdef \int_{-\infty}^\infty \delta(t) e^{-j2\pi f t}\,dt = 1, \quad \forall f\in\mathbb{R}$$ (B.43)

An impulse train can be defined as a sum of shifted impulses:

$$\psi_P(t) \isdef \sum_{m=-\infty}^\infty \delta(t-mP)$$ (B.44)

Here, $P$ is the period of the impulse train, in seconds--i.e., the spacing between successive impulses. The $P$ -periodic impulse train can also be defined as

$$\psi_P(t)\eqsp \frac{1}{P}\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}\left(\frac{t}{P}\right), \protect$$ (B.45)

where $\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(t)$ is the so-called shah symbol [23]:

$${\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(t) \, \isdef \sum_{m=-\infty}^\infty \delta(t-m)}$$ (B.46)

Note that the scaling by $1/P$ in (B.46) is necessary to maintain unit area under each impulse.

We will now show that

$$\zbox {\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(t)\;\longleftrightarrow\;\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(f).}$$ (B.47)

That is, the Fourier transform of the normalized impulse train $\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(t)$ is exactly the same impulse train $\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(f)$ in the frequency domain, where $t$ denotes time in seconds and $f$ denotes frequency in Hz. By the scaling theorem (§B.4),

$${\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}\left(\frac{t}{P}\right) \;\longleftrightarrow\;P\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(Pf),}$$ (B.48)

so that the $P$ -periodic impulse-train defined in (B.46) transforms to

$$\begin{eqnarray*} \psi_P(t) &=& \frac{1}{P}\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}\left(\frac{t}{P}\right) \;\longleftrightarrow\;\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(Pf) \eqsp \sum_{m=-\infty}^\infty \delta(Pf-m)\\ &=& \frac{1}{P}\sum_{m=-\infty}^\infty \delta\left(f-\frac{m}{P}\right) \eqsp \frac{1}{P}\psi_{\frac{1}{P}}(f) \eqsp \Psi_P(f). \end{eqnarray*}$$

Thus, the $P$ -periodic impulse train transforms to a $(1/P)$ -periodic impulse train, in which each impulse contains area $1/P$ :

$${\Psi_P(f) \isdefs \hbox{\sc FT}_f(\psi_P) \eqsp \frac{1}{P}\psi_{\frac{1}{P}}(f)}$$ (B.49)

Proof: Let's set up a limiting construction by defining

$$\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}_M(t) \isdefs \sum_{m=-M}^M \delta(t-m),$$ (B.50)

so that $\lim_{M\to\infty}\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}_M(t)=\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(t)$ . We may interpret $\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}_M(t)$ as a sampled rectangular pulse of width $2M$ seconds (yielding $2M+1$ samples).By linearity of the Fourier transform and the shift theorem (§B.5), we readily obtain the transform of $\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}_M(t)$ to be

$$\begin{eqnarray*} \hbox{\sc FT}_f(\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}_M) &\isdef & \hbox{\sc FT}_f\left[\sum_{m=-M}^M \hbox{\sc Shift}_{m}(\delta)\right]\\ &=& \sum_{m=-M}^M \hbox{\sc FT}_f[\hbox{\sc Shift}_{m}(\delta)] \eqsp \sum_{m=-M}^M e^{-j2\pi f m}. \end{eqnarray*}$$

Using the closed form of a geometric series,

$$\sum_{m=L}^U r^m \eqsp \frac{ r^L - r^{U+1}}{1-r},$$ (B.51)

with $r=e^{-j\pi f}$ , we can write this as

$$\begin{eqnarray*} \hbox{\sc FT}_f(\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}_M) &=& \frac{e^{j2\pi f M } - e^{-j2\pi f M } e^{-j2\pi f }}{1-e^{-j2\pi f }}\\ [10pt] &=& \frac{e^{-j\pi f}}{e^{-j\pi f}} \cdot \frac{e^{j\pi f (2M+1) } - e^{-j\pi f (2M+1) }}{e^{j\pi f}-e^{-j\pi f}}\\ [10pt] &=& \frac{\sin[\pi f (2M+1) ]}{\sin(\pi f)}\\ [5pt] &\isdef & (2M+1)\,\hbox{asinc}_{2M+1}(2\pi f ) \end{eqnarray*}$$

where we have used the definition of $\hbox{asinc}$ given in Eq.(3.5) of §3.1. As we would expect from basic sampling theory, the Fourier transform of the sampled rectangular pulse is an aliased sinc function. Figure 3.2 illustrates one period $M\cdot\hbox{asinc}_M(\omega)$ for $M=11$ .

The proof can be completed by expressing the aliased sinc function as a sum of regular sinc functions, and using linearity of the Fourier transform to distribute $\hbox{\sc FT}_f$ over the sum, converting each sinc function into an impulse, in the limit, by §B.13:

$$\begin{eqnarray*} (2M+1)\,\hbox{asinc}_{2M+1}(2\pi f) &\isdef & \frac{\sin[\pi f (2M+1) ]}{\sin(\pi f)}\\ [5pt] &=& \sum_{k=-\infty}^{\infty} \mbox{sinc}(2Mf-k)\\ [5pt] &\to& \sum_{k=-\infty}^{\infty} \delta(f-k) \end{eqnarray*}$$

by §B.13. Note that near $f=0,2,4,\ldots$ , we have

$$\begin{eqnarray*} \hbox{\sc FT}_f(\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}_M) &=& \frac{\sin[\pi f (2M+1) ]}{\sin(\pi f)} \;\;\approx\;\; \frac{\sin[\pi f (2M+1) ]}{\pi f}\\ [5pt] &=&(2M+1)\mbox{sinc}[(2M+1)f] \;\;\to\;\;\delta(f) \end{eqnarray*}$$

as $M\to\infty$ , as shown in §B.13. Similarly, near $f=1,3,5,\ldots$ , we have

$$\hbox{\sc FT}_f(\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}_M) \;\;\approx\;\; \frac{\sin[\pi f (2M+1) ]}{-\pi f} \;\;\to\;\;\delta(f)$$ (B.52)

as $M\to\infty$ . Finally, we expect that the limit for non-integer $f$ can be neglected since

$$\lim_{M\to\infty}\int_a^b \frac{\sin(M\pi f)}{\pi f} df \eqsp 0,$$ (B.53)

whenever $n<a\leq b<n+1$ and $n$ is some integer, as implied by §B.13.

See, e.g., [23,79] for more about impulses and their application in Fourier analysis and linear systems theory.

Exercise: Using a similar limiting construction as before,

$$\Psi_P(f) = \lim_{L\to\infty} \Psi_{P,L}(f) \isdefs \lim_{L\to\infty} \frac{2\pi}{P}\sum_{l=-L}^L \delta\left(2\pi f-l\frac{2\pi}{P}\right),$$ (B.54)

show that a direct inverse-Fourier transform calculation gives

$$\psi_{P,L}(t) = \frac{\sin\left[\pi(2L+1)\frac{t}{P}\right]}{\sin\left( \pi \frac{t}{P}\right)},$$ (B.55)

and verify that the peaks occur every $P$ seconds and reach height $(2L+1)/P$ . Also show that the peak widths, measured between zero crossings, are $P/(2L+1)$ , so that the area under each peak is of order 1 in the limit as $L\to\infty$ . [Hint: The shift theorem for inverse Fourier transforms is $e^{j\nu t}x(t) \;\leftrightarrow\; X(f-\nu)$ , and $\hbox{\sc IFT}_t(\delta)=1/(2\pi)$ .]