Part 1 · Oscillators and waves

1.7 The Fourier transform and the Dirac delta

Source: Postorino, week 2 slides, pp. 6–10, and the readings posted with them (Zucchelli on Fourier; the Heaviside step)

Objectives

Picture. A non-periodic function is a periodic one with infinite period: the harmonics merge into a continuous spectrum.
Recipe. Compute the transform of a pulse, using parity to kill $b (ω)$ .
Theorem. The boxcar’s transform is a sinc, with $Δ ν \cdot τ = 2$ : the shorter the pulse, the wider its spectrum.
Vocabulary. Fourier transform, boxcar, sinc, Dirac delta, sifting property.
Key. An impulse contains every frequency equally; a system with narrow resonances answers it at those only. That is why we see sinusoids.

From the series to the transform

The periodicity condition can be relaxed. A non-periodic function can be seen as a periodic function whose period is infinite. Send $T \to \infty$ : the spacing between harmonics $ω = 2 π / T \to 0$ , the quantity $nω$ , which jumped in steps of $ω$ , becomes a continuous variable, and sums become integrals: $n = 0 \sum \infty const_{n} f (nω t) \to \int_{0}^{\infty} const (ω) f (ω t) d ω .$ This is not rigorous, but it captures the process; the careful version is Zucchelli’s.

Definition 1.7.1Fourier transformEssentialslides w2, p. 6

Real form. A function $f (t)$ is a continuous sum of cosines and sines, $f (t) = \int_{0}^{\infty} [a (ω) cos (ω t) + b (ω) sin (ω t)] d ω,$ $a (ω) = \frac{1}{π} \int_{- \infty}^{+ \infty} f (t) cos (ω t) d t, b (ω) = \frac{1}{π} \int_{- \infty}^{+ \infty} f (t) sin (ω t) d t .$ Complex form: $f (t) = \int_{- \infty}^{+ \infty} c (ω) e^{iω t} d ω, c (ω) = \frac{1}{2 π} \int_{- \infty}^{+ \infty} f (t) e^{- iω t} d t, c (ω) \in C .$

$a (ω) d ω$ and $b (ω) d ω$ say how much cosine and sine of frequency between $ω$ and $ω + d ω$ the function contains: they play the role of $A_{n}$ and $B_{n}$ . By the same Euler step as for the series, for $ω > 0$ beyond the lectures, $c (ω) = \frac{1}{2 π} \int f (t) [cos ω t - i sin ω t] d t = \frac{a ( ω ) - ib ( ω )}{2},$ the analogue of $C_{n} = (A_{n} - i B_{n}) /2$ (Proposition 1.6.7).

RemarkOther conventionsbeyond the lectures

Where the $2 π$ goes is a convention. Zucchelli’s reading, like most signal-processing texts, uses $F (ω) = \int f (t) e^{- iω t} d t$ and $f (t) = \frac{1}{2 π} \int F (ω) e^{iω t} d ω$ , so $F (ω) = 2 π c (ω)$ . The shapes of all transforms are the same; only the constant in front changes.

Key observation 1.7.2The power spectrumbeyond the lectures

$∣ c (ω) ∣^{2}$ is the power spectrum: how the energy (squared amplitude, Key observation 1.3.8) of the signal is shared out over frequencies. It is what a spectrometer measures (Key observation 2.5.5).

The boxcar function

A key example: a rectangular pulse of height $C$ and duration $τ$ , the boxcar (on Wikipedia, “rectangular function”): $f (t) = {C 0 - \frac{τ}{2} \leq t \leq + \frac{τ}{2}, otherwise .$ Its area is kept fixed, $C τ = const$ . It is aperiodic, so it needs the transform.

Theorem 1.7.3Transform of the boxcarslides w2, p. 7–8

$b (ω) = 0, a (ω) = \frac{C τ}{π} \frac{sin ( ω τ /2 )}{ω τ /2} = \frac{C τ}{π} sinc (ω τ /2), x \to 0 lim \frac{sin x}{x} = 1.$

Proof

Parity. $sin (ω t)$ is odd, $sin (- ω t) = - sin (ω t)$ ; $cos (ω t)$ is even; $f$ is even. So $f (t) sin (ω t)$ is odd, and its integral over a symmetric interval vanishes: $b (ω) = 0$ . What is left: $a (ω) = \frac{1}{π} \int_{- τ /2}^{+ τ /2} C cos (ω t) d t = \frac{C}{π} \frac{sin ( ω t )}{ω}_{- τ /2}^{+ τ /2} = \frac{2 C sin ( ω τ /2 )}{πω} = \frac{C τ}{π} \frac{sin ( ω τ /2 )}{ω τ /2} .$ The evaluation uses $sin (ω τ /2) - sin (- ω τ /2) = 2 sin (ω τ /2)$ ; the last step multiplies top and bottom by $τ /2$ .

WarningSlide slip

The slide justifies $b = 0$ with “ $sin (- ω t) = sin (ω t)$ ”. The sine is odd: $sin (- ω t) = - sin (ω t)$ .

The time domain shows a pulse of width $τ$ ; the frequency domain a sinc: a large central lobe flanked by side lobes that alternate in sign and die away. At $ω = 0$ , $a = C τ / π$ : the area of the pulse, over $π$ .

Proposition 1.7.4Width of the central lobeslides w2, p. 8

The first zeros are at $ω_{\pm} = \pm 2 π / τ$ , so $Δ ω = \frac{4 π}{τ}, Δ ν = \frac{Δ ω}{2 π} = \frac{2}{τ}, Δ ν \cdot τ = 2.$

Proof

$\frac{s i n ( ω τ /2 )}{ω τ /2}$ first vanishes away from zero where the sine does: $ω τ /2 = \pm π$ .

Key observation 1.7.5Short in time, wide in frequencyslides w2, p. 8

The shorter the pulse, the wider its spectrum: $Δ ν \cdot τ = 2$ , a constant. To make something short in time you need a wide range of frequencies. This trade-off comes back as the uncertainty principle beyond the lectures.

Loading the demo…

Figure 1.7.1 A boxcar of width

τ

and fixed area 1, and its transform

a (ω)

. Drag the edge of the pulse, or the slider. The central lobe’s width

Δ ω = 4 π / τ

grows as the pulse shrinks; at

τ \to 0

the transform flattens to

1/ π

at every frequency. Switch to a Gaussian: its transform is a Gaussian too, with the same trade-off.

The limit $τ \to 0$ . Keeping $C τ$ fixed, $C$ grows as $τ$ shrinks; at $τ \to 0$ , an impulse, $C \to \infty$ .

The width of $sinc (ω τ /2)$ goes to $\infty$ : its first zero $2 π / τ$ runs away.
So $a (ω) \to C τ / π = const$ , the same at every frequency.
Empirically, we have built an impulse function and its Fourier transform.

The Dirac delta

An impulse function is non-zero only in a small interval $τ$ . In the limit $τ \to 0$ it keeps a non-zero integral, so at that point it must go to $\infty$ .

Definition 1.7.6Dirac deltaslides w2, p. 9

By analogy with the boxcar, the Dirac delta is $δ (t) = \frac{1}{2 π} \int_{- \infty}^{+ \infty} e^{iω t} d ω, δ (t - t_{0}) = \frac{1}{2 π} \int_{- \infty}^{+ \infty} e^{iω (t - t_{0})} d ω .$

Reading it beyond the lectures: this is the complex Fourier integral with $c (ω) = \frac{1}{2 π}$ at every frequency, the flat spectrum the boxcar tended to. All frequencies add up in phase at $t = 0$ and cancel everywhere else.

Theorem 1.7.7Properties of the Dirac deltaslides w2, p. 9

$\int_{- \infty}^{+ \infty} f (t) δ (t) d t = f (0), \int_{- \infty}^{+ \infty} f (t) δ (t - t_{0}) d t = f (t_{0}) (sifting), \int_{- \infty}^{+ \infty} δ (t) d t = 1.$

Why sifting worksbeyond the lectures

Picture $δ$ as the boxcar with $C = 1/ τ$ (unit area) and $τ \to 0$ . Then $\int f δ = \frac{1}{τ} \int_{- τ /2}^{τ /2} f (t) d t$ , the mean of $f$ over a shrinking interval around 0, which tends to $f (0)$ if $f$ is continuous there. Unit area is the case $f = 1$ .

The transform of the delta

Example 1.7.8Transform of the Dirac deltaslides w2, p. 10

With the sifting property, $a (ω) = \frac{1}{π} \int_{- \infty}^{+ \infty} δ (t) cos (ω t) d t = \frac{1}{π} cos 0 = \frac{1}{π}, b (ω) = \frac{1}{π} \int_{- \infty}^{+ \infty} δ (t) sin (ω t) d t = \frac{1}{π} sin 0 = 0.$ It matches the boxcar’s limit $C τ / π$ with unit area.

Key observation 1.7.9An impulse drives every frequencyslides w2, p. 10

An impulsive kick is, by Fourier, a combination of harmonics at all frequencies with the same amplitude. The squared modulus is the intensity, so all frequencies also carry the same intensity.
Kicking a system with an impulse means driving it at every possible frequency at once, all with the same amplitude.
If the system has natural frequencies (resonance frequencies), its response is very strong near them and very weak far from them (Theorem 1.4.3). If the resonances are narrow, it oscillates with sizeable amplitude only at its resonance frequencies.
This is why we so often see sinusoidal waves: the Fourier components picked out by resonances. Strike a bell and it rings at its own notes, whatever the shape of the blow.

Readings

The two readings posted in week 2. Lectures 4–6 fall in that week, which is probably what the “4-5-6” in the file name means.

Zucchelli: from the series to the transform

La Trasformata di Fourier, M. Zucchelli, Verona 2017, pp. 1–3. It writes $w$ for the angular frequency and $F (w) = \int f (t) e^{- i wt} d t$ for the transform. This first part is the careful version of the slides’ $T \to \infty$ argument.

Series (p. 1). $f (t) = a_{0} + \sum_{n = 1}^{\infty} [a_{n} cos (n w_{0} t) + b_{n} sin (n w_{0} t)]$ , with $a_{0} = \frac{1}{T} \int_{T} f$ (the mean, no $\frac{1}{2}$ ), $a_{n}, b_{n} = \frac{2}{T} \int_{T} f cos, sin$ , and $w_{0} = 2 π / T$ . Complex form $f = \sum c_{n} e^{in w_{0} t}$ , $c_{n} = \frac{1}{T} \int_{T} f e^{- in w_{0} t} d t$ .
Limit (p. 2). Repeat the aperiodic $f$ every $T$ to get a periodic $f_{T}$ , expand $f_{T}$ in a series, and let $T \to \infty$ , where $f_{T} \to f$ .
1. Define $F (n w_{0}) = lim T c_{n}$ , which gives $F (w) = \int_{- \infty}^{\infty} f (t) e^{- i wt} d t$ .
2. Then $f = \sum \frac{F ( n w _{0} )}{T} e^{in w_{0} t} = \sum \frac{F ( n Δ w ) Δ w}{2 π} e^{in Δ wt}$ , with $Δ w = w_{0} = 2 π / T \to 0$ .
3. That is a rectangle-rule sum for an integral, so $f (t) = \frac{1}{2 π} \int F (w) e^{i wt} d w$ . (The PDF prints $d t$ in that last integral, a typo.)
Ordinary frequency (p. 3). With $w = 2 π f$ : $X (f) = \int x (t) e^{- 2 πi f t} d t$ and $x (t) = \int X (f) e^{2 πi f t} df$ . The PDF writes this with a stray $\frac{1}{2 π}$ and $d t$ ; with $f$ in place of $w$ the $2 π$ goes in the exponent and the prefactor disappears.

Zucchelli, sections 3 to 6: more transforms, properties, sampling

Pp. 3–18. Not examined; read for understanding.

rect (p. 3–4). $rect (t) = 1$ for $∣ t ∣ < \frac{1}{2}$ , $\frac{1}{2}$ at $∣ t ∣ = \frac{1}{2}$ , 0 outside. $F (w) = \frac{2 s i n ( w /2 )}{w} = sinc (w /2)$ : the boxcar with $C = τ = 1$ .
Gaussian (p. 3–5). $f = e^{- t^{2} /2 σ^{2}}$ gives $F (w) = 2 π σ^{2} e^{- σ^{2} w^{2} /2}$ , again a Gaussian, of width $1/ σ$ : the same trade-off as $Δ ν τ = 2$ . The proof completes the square.
Delta (p. 5–7). $δ (t) = lim_{n \to \infty} \frac{n}{π} e^{- n^{2} t^{2}}$ , a unit-area Gaussian getting narrower; its transform $lim e^{- w^{2} /4 n^{2}} = 1$ is flat. A shifted delta $δ (t - τ)$ has $F = e^{- i w τ}$ : flat modulus, phase growing with $w$ . (The PDF’s eq. 33 has $- w^{2} / n^{2}$ for $- w^{2} /4 n^{2}$ , irrelevant in the limit.)
Properties (p. 7), with $f \leftrightarrow F$ :

operation	$f (t)$	$F (w)$
addition	$f_{1} + f_{2}$	$F_{1} + F_{2}$
scalar	$k f$	$k F$
symmetry	$F (t)$	$2 π f (- w)$
dilation, $a$ real	$f (a t)$	$\frac{1}{∣ a ∣} F (w / a)$
time shift	$f (t - τ)$	$F (w) e^{- i w τ}$
convolution	$f_{1} * f_{2}$	$F_{1} F_{2}$
product	$f_{1} f_{2}$	$\frac{1}{2 π} F_{1} * F_{2}$

Dilation is the width trade-off in general: squeezing $f$ by $a$ stretches $F$ by $a$ .

Exponential and cosine (p. 8). $e^{i w_{0} t} \leftrightarrow 2 π δ (w - w_{0})$ : a pure frequency is a single spike. $cos (w_{0} t) \leftrightarrow π [δ (w - w_{0}) + δ (w + w_{0})]$ .
Delta and convolution (p. 8–9). Convolving with $δ (t - τ)$ shifts: $f * δ (t - τ) = f (t - τ)$ .
Dirac comb (p. 9–11). $Δ_{T} (t) = \sum_{n} δ (t - n T)$ has all Fourier coefficients $1/ T$ ; its transform is another comb, $w_{0} Δ_{w_{0}} (w)$ with $w_{0} = 2 π / T$ : fine spacing in time, coarse in frequency.
Periodic functions (p. 11–13). Repeating $f$ is convolving it with a comb, so its transform is $F (w)$ sampled at multiples of $2 π / N$ . A periodic function has a line spectrum, the Fourier series, and the lines get denser as the period grows: the slides’ $T \to \infty$ argument from the other side.
DTFT (p. 13–16). Sampling $f$ every $T$ makes the spectrum periodic; if $T$ is too large the copies overlap (aliasing). Avoiding it needs a sampling rate at least twice the highest frequency (Nyquist–Shannon). From section 5 the author writes $j$ for $i$ .
DFT (p. 17–18). $X [k] = \sum_{n = 0}^{N - 1} f [n] e^{- jkn 2 π / N}$ , inverse $x [n] = \frac{1}{N} \sum_{k} X [k] e^{jkn 2 π / N}$ . Resolution $2 π / NT$ ; refine it by zero-padding a signal that already decays to zero.

The Heaviside step

An Italian Wikipedia page, printed. Not examined.

Definition. $Θ (x)$ (also $H$ , $u$ ) is $0$ for $x < 0$ and $1$ for $x > 0$ ; at $0$ it is a convention, most often $\frac{1}{2}$ , giving $Θ = \frac{1}{2} (1 + sgn x)$ .
Link to the delta. $\frac{d}{d x} Θ (x) = δ (x)$ as a distribution, and $Θ (x) = \int_{- \infty}^{x} δ (t) d t$ : the unit jump is all of the derivative. As a distribution, $\int Θ f^{'} = - f (0)$ for smooth $f$ decaying at infinity (integration by parts).
Ramp. $R (x) = \int_{- \infty}^{x} Θ = x Θ (x)$ .
Discrete form. $Θ [n] = \sum_{k \leq n} δ [k]$ , with the Kronecker $δ [k] = δ_{k, 0}$ (the page says “Dirac” there, a slip).
Fourier transform. $\frac{1}{iω} + π δ (ω)$ : a $1/ ω$ spectrum plus a spike at $0$ carrying the mean $\frac{1}{2}$ .
It is the cumulative distribution of a variable that is almost surely 0 (§4.1 Probability and random variables); in signal processing, a signal switched on and left on.

Exercises

Exercise 1.7.1beyond the lectures

A laser pulse lasts $τ = 10$ fs. Estimate the spread of frequencies it contains.

Solution

For a boxcar, $Δ ν = 2/ τ = 2 \times 1 0^{14}$ Hz. Visible light is around $5 \times 1 0^{14}$ Hz, so the pulse spans a big fraction of the visible: very short pulses can’t be monochromatic.

Exercise 1.7.2beyond the lectures

Compute $\int_{- \infty}^{\infty} cos (3 t) δ (t - π) d t$ and $\int_{- \infty}^{\infty} e^{- t^{2}} δ (t) d t$ .

Solution

By sifting: $cos (3 π) = - 1$ and $e^{0} = 1$ .

Exercise 1.7.3beyond the lectures

Show that for real $f$ , $c (- ω) = c (ω)^{*}$ , and so $∣ c (ω) ∣^{2}$ is even.

Solution

$c (- ω) = \frac{1}{2 π} \int f (t) e^{+ iω t} d t = [\frac{1}{2 π} \int f (t) e^{- iω t} d t]^{*} = c (ω)^{*}$ , since $f$ is real. Then $∣ c (- ω) ∣^{2} = ∣ c (ω) ∣^{2}$ .

Exercise 1.7.4beyond the lectures

Find $a (ω)$ and $b (ω)$ for the odd pulse $f (t) = + 1$ for $0 < t < τ$ , $- 1$ for $- τ < t < 0$ , 0 elsewhere.

Solution

$f$ is odd, so $a = 0$ . $b (ω) = \frac{1}{π} \cdot 2 \int_{0}^{τ} sin ω t d t = \frac{2}{πω} (1 - cos ω τ) = \frac{4}{πω} sin^{2} (ω τ /2)$ . It vanishes at $ω = 0$ : the pulse has zero mean.

Exercise 1.7.5beyond the lectures

A piano hammer gives a string a short blow. Why do we hear the string’s notes, not a click?

Solution

The blow is nearly an impulse, so it drives all frequencies about equally (Key observation 1.7.9). The string’s resonances are narrow, so it responds strongly only at its own frequencies, which keep ringing after the blow; the other frequencies die at once.

Exercise 1.7.6beyond the lectures

Shift the boxcar to start at $t = 0$ : $f = C$ for $0 \leq t \leq τ$ . Find $∣ c (ω) ∣$ . Did the spectrum’s width change?

Solution

$c (ω) = \frac{C}{2 π} \int_{0}^{τ} e^{- iω t} d t = \frac{C}{2 π} \frac{1 - e ^{- iω τ}}{iω} = \frac{C τ}{2 π} e^{- iω τ /2} sinc (ω τ /2)$ . So $∣ c ∣ = \frac{C τ}{2 π} ∣ sinc (ω τ /2) ∣$ : the same as the centred boxcar. A shift in time only adds a phase (the time-shift rule).