Efficient Calculation: Fourier Transform via Convolution

Given the importance of the Fourier transforms of periodic functions, is there not a computationally more efficient way of finding these transforms? The answer is ``yes'', and it hinges on the remarkable properties of the convolution integral²³

$$f\ast g(x) = \int_{-\infty}^\infty f(x-\xi)g(\xi)\, d\xi$$ (256)

$$= \int_{-\infty}^\infty g(x-\xi)f(\xi)\, d\xi$$

of the two functions $f$ and $g$ . Before identifying these properties we first describe the mental process which leads to the graph of this integral:

(i)

Take the graph of the function $f(\xi)$ and flip it around the vertical axis $\xi=0$ . This yields the graph of the new function $f(-\xi)$ .

(ii)

Slide that flipped graph to the right by an amount $x$ by letting $\xi \rightarrow \xi-x$ , and thus obtain the graph of $f(x-\xi)$ .

(iii)

Multiply this graph by the graph of $g(\xi)$ to obtain the graph of the product function $f(x-\xi)g(\xi)$ .

(iv)

Find the area under this product function.

As one slides the flipped graph to the right, this area generates the graph of $f\ast g(x)$ .

Example 3 (Periodic train of Gaussians via convolution)

Consider the graph of the Gaussian

$$f(\xi)=e^{-(\xi-c)^2/2b^2}$$ (257)

having full width $2b$ centered around $\xi=c$ , and let

$$g(\xi)=\sum_{n=-\infty}^\infty \delta (\xi -n)$$ (258)

be a periodic train of Dirac delta functions. To form the convolution $f\ast g(x)$ , flip the function $f$ to obtain

$$f(-\xi)=e^{-(-\xi-c)^2/2b^2}~,$$

which is centered around $\xi=-c$ , shift it to the right by an amount $x$ to obtain

$$f(x-\xi)=e^{-(x-\xi-c)^2/2b^2}~,$$

and finally do the integral

$$\int_{-\infty}^\infty f(x-\xi)\sum_{n=-\infty}^\infty \delta (\xi -n) \, d\xi = \sum_{n=-\infty}^\infty f(x-n)$$

$$= \sum_{n=-\infty}^\infty e^{-(x-n-c)^2/2b^2}~.$$

This is a periodic train of Gaussians, and the period is $\Delta x=1$ . This result also illustrates how a periodic function, in our case

$$h(x)=\sum_{n=-\infty}^\infty f(\xi-n) \equiv \sum_{n=-\infty}^\infty e^{-(x-n-c)^2/2b^2}~,$$

can be represented as the convolution

$$\boxed{\sum_{n=-\infty}^\infty f(x-n)=f*g(x)}$$

where $f$ and $g$ are given by Eqs.(2.57) and (2.58). The graph of this convolution is the Gaussian train in Figure 2.7. Its Fourier transform, Figure 2.8 is calculated below using a fundamental property of the convolution integral.

Figure 2.7: Periodic train of Gaussian pulses obtained by convolving a single Gaussian, on the very left, with a periodic train of Dirac delta functions, whose infinite amplitudes are represented in this figure by the heavy dots.The Fourier transform of the train of Gaussians is shown in Figure 2.8

Exercise 23.8 (PERIODIC FUNCTION AS A CONVOLUTION)
Show that any periodic function $f(\xi)=f(\xi+a)$ is the convolution of a nonperiodic function with a train of Dirac delta functions.

The convolution of two functions has several fundamental properties (commutativity, associativity, distributivity), but its most appealing property is that its Fourier transform is simply the product of the Fourier transforms of the respective functions,

$$\int_{-\infty}^\infty f\ast g(x) \frac{e^{-ikx}}{\sqrt{2\pi}}\, dx = \int_{-\infty}^\infty \int_{-\infty}^\infty f(x-\xi) g(\xi)\, d\xi \frac{e^{-ikx}}{\sqrt{2\pi}}\, dx$$

$$= \sqrt{2\pi}\hat{f}(k) \hat g (k)$$ (259)

This result can be an enormous time saver. Let us apply it to the problem of finding the Fourier transform of $h(x)$ , the periodic train of Gaussians considered in Example 3, but with $c=0$ , i.e. centered around the origin. The calculation yields

$$\hat{f}(k) = \int_{-\infty}^\infty e^{-\xi^2/2b^2}\frac{e^{-ik\xi}}{\sqrt{2\pi}}d\xi$$

$$= b e^{-b^2k^2/2}$$ (260)

and

$$\hat g (k) = \int^\infty_{-\infty}\sum_{n=-\infty}^\infty \delta (\xi -n)\frac{e^{-ik\xi}}{\sqrt{2\pi}}d\xi$$

$$= \sum_{m=-\infty}^\infty \sqrt{2\pi}\delta (k-2\pi m)~.$$ (261)

It follows that the Fourier transform of that train yields

$$\int_{-\infty}^\infty \sum_{n=-\infty}^\infty e^{-(x-n)^2/2b^2}\... ...{2\pi}dx= be^{-b^2k^2/2}\sum_{m=-\infty}^\infty \sqrt{2\pi}\delta (k-2\pi m)~.$$

Figure 2.8 shows the real part of this transform. Study the relationship between this figure and Figure 2.7 carefully. They highlight the archetypical properties of the Fourier transform. To name a few:

• Local properties in the given domain get transformed into global properties in the Fourier domain.
• Jaggedness in the given domain gets transformed into broad spectral behaviour in the Fourier domain.
• Narrow pulses get transformed into wide envelopes, and vice versa.
• Periodicity in the given domain gets transformed into equally spaced (but in general nonperiodic) spectral lines in the Fourier domain.

And there are others.

Figure 2.8: Set of equally spaced (here with $\Delta k=2\pi$ ) spectral lines resulting from the Fourier transform of the periodic train of Gaussians in Fig. 2.7. The spectral envelope, here again a Gaussian, is the Fourier transform of one of the identical pulses which make up the train.

The pulses that make up the periodic train of Gaussians, Fig. 2.7, have no internal structure. Thus the natural question is: What is the Fourier transform of a periodic train of pulses, each one made up of a finite number of oscillations as in Fig. 2.9? The next example addresses this question.

Figure 2.9: Periodic train of optical pulses emitted by a ``mode-locked'' laser. In this figure the pulse separation is highly understated. In an actual train the pulse separation is typically more than a million times the full width of each pulse. In spite of this, the optical phase (relative to the Gaussian envelope) shifts by a fixed and controllable amount from one pulse to the next. In this figure that phase shift is zero: the optical oscillation amplitude profile in each pulse is congruent to that in all the others.

Example 4 (Fourier transform of light from a mode-locked laser)

A mode-locked laser generates light in the form of a periodic train of pulses of light. This periodicity is expressed in terms of the separation between successive pulses, and each pulse is characterized by three properties:

1. pulse envelope,
2. optical (``carrier'') frequency and the
3. phase of the optical carrier vibrations relative to the pulse envelope.

The temporal amplitude profile of the the $n$ th pulse is

$$f_n(t)=e^{-(t-nT)^2 /2b^2} e^{i\omega_0 (t-nT)} e^{i\delta_n}$$

The constant $T$ is the separation between successive pulses. The first factor is the pulse envelope, which we take to be a Gaussian of full width $2b$ centered around time $t=nT$ . The second factor expresses the oscillations of the optical carrier whose frequency is $\omega_0$ . The last factor expresses the phase shift of the optical carrier relative to the pulse envelope. The optical pulse train is the sum

$$f(t)=\sum_{n=-\infty}^\infty e^{-(t-nT)^2 /2b^2} e^{i\omega_0 (t-nT)} e^{i\delta_n}~.$$

The width of the pulse envelope in lasers nowadays (2002) is less than 10 femtoseconds (=10$^{-14}$ sec.). This corresponds to light travelling a distance of less than three microns. Such a pulse width is achieved by the constructive interference of more than a million longitudinal laser modes phase-locked to oscillate coherently.

The pulse repetition rate for a phase-locked laser is determined by the round trip travelling time inside the laser cavity with a partially silvered mirror at one end. For a laser 1.5 meters long the pulses emerge therefore at one end at a rate of $1/T$ =100 megaHertz, corresponding to a pulse separation of 3 meters of light travelling time between two pulses. In between two such pulses there is no light, no electromagnetic energy whatsoever. The destructive interference of the above-mentioned million laser modes guarantees it.

The pulses can therefore be pictured as micron-sized ``light bullets'' shot out by the laser. Because of their small size these bullets have an enormous amount of energy per unit volume, even for modestly powered lasers.

Ordinarily the phase $\delta _n$ varies randomly from one pulse to the next. In that case $f$ is merely a train of pulses with incoherent phases. The Fourier transform of such a train would be a correspondingly irregular superposition of Fourier transforms. This superposition is exhibited in Figure 2.10

Figure 2.10: Average of the Fourier spectra of 81 pulses of a train like that in Fig. 2.9, but each pulser having random phase $\delta _n$ from one to the next. Compare the Fourier spectrum in this figure with the one in Fig. 2.12 whose pulse train is coherently phased (i.e. $\delta _n =0$ ) and of infinite length.

However, a recent discovery shows that light generated by a laser operating in a ``locked-mode'' way can be made to produce pulses which are phase coherent, even though they are separated by as much as three meters. Indeed, experiments show that the phase $\delta _n$ increases by a constant amount from one pulse to the next. Evidently the amplifying medium in the laser must somehow ``remember'' the phase of the carrier oscillations from one emitted pulse to the next. Thus

$$\delta_n=n\Delta \phi ~.$$

where $\Delta \phi$ is a constant as in Figure 2.11.

Figure 2.11: Overlay of two successive pulses with phase difference $\Delta \phi =\pi /2$ .

In that case $f$ is a periodic function,

$$f(t) = \sum_{n=-\infty}^\infty e^{-(t-nT)^2 /2b^2} e^{i\omega_0 (t-nT)} e^{in\Delta \phi}$$ (262)

$$= \sum_{n=-\infty}^\infty e^{-(t-nT)^2 /2b^2} e^{i\omega_0 t} e^{in(\Delta \phi-\omega_0 T)}~.$$

Here

$$\omega_0 T=\left( \textrm{\char93 of optical carrier cycles between adjacent pulses} \right)\times2\pi$$

What is the Fourier spectrum of such a periodic train? The result is depicted in Figure 2.12.

The line of reasoning leading to this result is as follows: Observe that the periodic train can be written as the convolution integral

$$f(t) = \int\limits_{-\infty}^\infty e^{-(t-\xi)^2 /2b^2} e^{i\omega_0 (t-\xi)} \sum_{n=-\infty}^\infty e^{in\Delta \phi}\delta(\xi-nT)\,d\xi$$

$$\equiv \textrm{pulse}\ast \textrm{comb}_T (t;\Delta\phi)$$

where

$$\textrm{pulse}(t)\equiv e^{-t^2/2b^2}e^{i\omega_0t}$$

is a carrier amplitude modulated by a Gaussian, and

$$\textrm{comb}_T(t;\Delta\phi)\equiv \sum_{n=-\infty}^\infty e^{in\Delta \phi }\delta(t-nT)$$

is a periodic train of linearly phased Dirac delta functions with fixed phase difference $\Delta \phi$ from one delta function to the next.

Figure 2.12: Set of equally spaced spectral lines resulting from the Fourier transform of the optical train of pulses in Fig. 2.9. The line spacings in the figure have the common value $\Delta \omega /2\pi =1/T$ Hertz. For an actual laser generated train the typical value is $\Delta \omega /2\pi \sim$ 10$^8$ Hertz, which is precisely the rate at which energy pulses back and forth in a laser cavity of length $\sim$ 1.5 meters. For a laser which generates 10 femtosecond pulses, the Gaussian spectral envelope encompasses $\sim$ 10$^6$ spectral lines instead of only the 20 depicted in this figure.

The respective Fourier transforms are

$${\mathcal F}[\textrm{pulse}](\omega) = \int^\infty_{-\infty} e^{-t^2/2b^2}e^{i\omega_0t} \frac{e^{-i\omega t}}{\sqrt{2\pi}}dt$$

$$= be^{-(\omega-\omega_0)^2b^2/2}~,$$ (263)

a Gaussian in frequency space centered around $\omega_0$ , and, with the help of Poisson's sum formula, Eq.(2.27),

$${\mathcal F}[\textrm{comb}](\omega) = \int^\infty_{-\infty}\textrm{comb}_T(t;\Delta\phi) \frac{e^{-i\omega t}}{\sqrt{2\pi}}dt$$

$$= \sum_{n=-\infty}^\infty \frac{e^{i(\Delta\phi- \omega T)n}}{\sqrt{2\pi}}$$

$$= \sum_{m=-\infty}^\infty \sqrt{2\pi}\delta(\omega T -2\pi m -\Delta\phi)~.$$ (264)

This is a periodic set of Dirac delta functions in the frequency domain, but collectively shifted by the common amount $\Delta \phi$ . The convolution theorem, Eq.(2.59), implies that the Fourier transform of the train of laser pulses, Eq.(2.62) is simply the product of Eqs.(2.63) and (2.64):

$${\mathcal F}[f](\omega) = \int^\infty_{-\infty} \frac{e^{-i\omega t}}{\sqrt{2\pi}} f(t)dt$$

$$= \underbrace{\sqrt{2\pi}be^{-(\omega-\omega_0)^2b^2/2}}_{\textrm{s... ...\infty}^\infty \delta(\omega T -\Delta\phi -2\pi m )}_{\textrm{spectral lines}}$$ (265)

$$= \sqrt{2\pi}be^{-(\omega-\omega_0)^2b^2/2} \frac{1}{2\pi T} \sum_{... ...ft(\frac{\omega}{2\pi}-m\frac{1}{T} -\frac{\Delta\phi}{2\pi}\frac{1}{T} \right)$$ (266)

This is a discrete spectrum of equally spaced sharp spectral lines. The separation between them is

$$\frac{\Delta\omega}{2\pi}=\frac{1}{T}~,$$

which is the pulse repetition rate.

From one pulse to the next there is a change in the optical phase relative to the envelope. This phase change, $\Delta \phi$ (exhibited in Figure 2.11) results in all frequencies of the spectral lines under a pulse envelope being shifted by the common amount

$$\underbrace{\frac{(\textrm{fraction of a cycle})}{(\textrm{pulse}... ...textrm{(pulses)}}{\text{(time)}}}_{\displaystyle\textrm{pulse repetition rate}} = \underbrace{\frac{\Delta\phi}{2\pi}\times\frac{1}{T}}_{\displaystyle\textrm{\lq\lq frequency offset''}}$$

$$= \frac{\textrm{(cycles)}}{\textrm{(time)}}$$

Figure 2.13: Two sets of equally spaced spectral lines resulting from the Fourier transform of two optical trains of pulses. The first (dotted) spectrum (which is identical to that of Fig. 2.12) is due to the train whose pulses have zero ( $\Delta \phi =0$ ) interpulse carrier phase shift of the optical carrier relative to the envelope. The second (solid) spectrum is due to pulses with nonzero ( $\Delta \phi \not =0$ ) interpulse carrier phase shift.

This frequency offset does not apply to the spectral envelope, which remains fixed as one changes $\Delta \phi$ . Instead, it applies only to the position of the spectral lines, which get shifted by this frequency offset. This is illustrated in Figure 2.13.

Finally note that, with light oscillating at its carrier frequency $\omega_0$ , the Gaussian envelope in Figure 2.12 is centered around the carrier frequency $\omega=\omega_0$ in the frequency domain. When $\omega_0=0$ , Figs. 2.9 and 2.12 reduce to Figs. 2.7 and 2.8 of Example 3.

Exercise 23.9 (FINITE TRAIN OF PULSES)
Find the Fourier spectrum of a finite train of identical coherent ( $\delta _n =0$ for $n=0,\pm 1,\cdots ,\pm N$ ) pulses of the kind shown in Fig. 2.9. Describe the result in terms of a picture and a mathematical formula. Point out how the result differs from Figs. 2.10 and 2.12.

Exercise 23.10 (FOURIER SERIES OF A TRAIN OF GAUSSIANS)
Verify that

$$f(t)=\sum_{n=-\infty}^\infty e^{-(t-nT)^2 /2b^2} e^{i\omega_0 (t-nT)}$$

is a function periodic in $t$ : $f(t+T)=f(t)$ .

Find the Fourier series representation

$$f(t)=\sum_{m=-\infty}^\infty c_m e^{i\omega_m t}$$

of $f(t)$ by determining $\omega_m$ and $c_m$ .

Lecture 15

Footnotes

... integral²³

Not to be confused with the auto-correlation integral, Eq.(2.47) on page .