Impulse Trains

The impulse signal $\delta(t)$ (defined in §2.4.9) has a constant Fourier transform:

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

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

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

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

$$\psi_P(t)=\frac{1}{P}\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}\left(\frac{t}{P}\right), \protect$$ (3.9)

where $\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(t)$ is the so-called shah symbol [22]:

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

Note that the scaling by $1/P$ in (2.9) is necessary to maintain unit area under each impulse.

We will now show that

$$\zbox {\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\e... ...sebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(f).}$$

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 (§2.4.4),

$${\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensurem... ...ebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(Pf),}$$

so that the $P$-periodic impulse-train defined in (2.9) transforms to

$$\begin{eqnarray*} \psi_P(t) &=& \frac{1}{P}\,\raisebox{0.8em}{\rotatebox{-90}{\r... ...ac{m}{P}\right) = \frac{1}{P}\psi_{\frac{1}{P}}(f) = \Psi_P(f). \end{eqnarray*}$$

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

$${\Psi_P(f) \isdef \hbox{\sc FT}_f(\psi_P) = \frac{1}{P}\psi_{\frac{1}{P}}(f)}$$

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

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

so that $\lim_{M\to\infty}\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensu... ...raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}(t)$. 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 (§2.4.5), we readily obtain the transform of $\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{1em}{1em}{\ensuremath{\exists}}}}_M(t)$ to be

$$\begin{eqnarray*} \hbox{\sc FT}_f(\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{... ...[\hbox{\sc Shift}_{m}(\delta)] \;= \sum_{m=-M}^M e^{-j2\pi f m}. \end{eqnarray*}$$

Using the closed form of a geometric series,

$$\sum_{m=L}^U r^m = \frac{ r^L - r^{U+1}}{1-r},$$

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

$$\begin{eqnarray*} \hbox{\sc FT}_f(\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{... ...n(\pi f)}\\ [5pt] &\isdef & (2M+1)\,\hbox{asinc}_{2M+1}(2\pi f ) \end{eqnarray*}$$

where we have used the definition of $\hbox{asinc}$ given in Eq.$\,$(1.4) of §1.5. As we would expect from basic sampling theory, the Fourier transform of the sampled rectangular pulse is an aliased sinc function. Figure 1.8 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 §2.4.12:

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

by §2.4.12.

Note that near $f=0,2,4,\ldots$, we have

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

as $M\to\infty$, as shown in §2.4.12. Similarly, near $f=1,3,5,\ldots$, we have

$$\begin{eqnarray*} \hbox{\sc FT}_f(\,\raisebox{0.8em}{\rotatebox{-90}{\resizebox{... ...pprox& \frac{\sin[\pi f (2M+1) ]}{-\pi f}\\ [5pt] &\to&\delta(f) \end{eqnarray*}$$

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

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

whenever $n<a\leq b<n+1$ and $n$ is some integer, as implied by §2.4.12.

See, e.g., [22,69] for more about impulses and their application in Fourier analysis and linear systems theory.

Exercise: Using a similar limiting construction as before,

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

show that a direct inverse-Fourier transform calculation gives

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

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 $\hbox{\sc IFT}_t(\delta)=1/(2\pi)$.]