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Mathematical Methods

Fourier transform in depth: Gaussian, Parseval, spectral decay, uncertainty and sampling

Transform of the Gaussian, properties and Parseval's theorem proved, signal smoothness and spectral decay, uncertainty principle, sampling theorem with the reconstruction formula, tables of pairs.

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This page goes deeper into the lesson on the Fourier transform. It computes step by step the transform of the Gaussian, proves the properties and Parseval’s theorem, explains why a spectrum decays more or less quickly depending on the smoothness of the signal, derives the uncertainty principle and proves the sampling theorem with the reconstruction formula. At the end are the tables to remember of transform pairs and properties.

Convention: F(\omega) = \int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt, f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}d\omega.

The Gaussian: a function that transforms into itself

Theorem. \displaystyle\mathcal F\left\{e^{-t^2/(2\sigma^2)}\right\} = \sigma\sqrt{2\pi}\,e^{-\sigma^2\omega^2/2}.

Proof. Let f(t) = e^{-at^2} with a = \frac{1}{2\sigma^2}. The function satisfies the differential equation f'(t) = -2at\,f(t). Transform both sides: on the left the derivative becomes i\omega F(\omega); on the right, since \mathcal F\{t\,f(t)\} = i\,F'(\omega) (differentiating the definition with respect to \omega under the integral sign), we get -2a\,i\,F'(\omega). So

    \[ i\omega F = -2ai\,F' \quad\Rightarrow\quad F'(\omega) = -\frac{\omega}{2a}F(\omega) \]

It is the same equation, with \frac{1}{4a} in place of a: F(\omega) = F(0)\,e^{-\omega^2/(4a)} = F(0)\,e^{-\sigma^2\omega^2/2}. Finally F(0) = \int e^{-at^2}dt = \sqrt{\pi/a} = \sigma\sqrt{2\pi} (Gaussian integral). ∎

A Gaussian that is narrow in time (small σ) is wide in frequency (width 1/\sigma), and vice versa: the product of the two widths is constant.

Gaussians in time and their transforms
On the left two Gaussians, a wide one in blue and a narrow one in orange; on the right the transforms: the Gaussian that is wide in time has a narrow spectrum, the narrow one a wide and lower spectrum.

The properties, proved

They all follow from the definition with a change of variable.

  • Delay. \int f(t - t_0)e^{-i\omega t}dt, with u = t - t_0, becomes e^{-i\omega t_0}\int f(u)e^{-i\omega u}du = e^{-i\omega t_0}F(\omega).
  • Modulation. \int e^{i\omega_0 t}f(t)e^{-i\omega t}dt = \int f(t)e^{-i(\omega - \omega_0)t}dt = F(\omega - \omega_0). With cosine, \cos\omega_0t = \frac{e^{i\omega_0t} + e^{-i\omega_0t}}{2}, the spectrum splits into two halved copies centred at \pm\omega_0.
  • Scaling. For a \gt 0, with u = at: \int f(at)e^{-i\omega t}dt = \frac{1}{a}\int f(u)e^{-i(\omega/a)u}du = \frac{1}{a}F(\omega/a).
  • Derivative. By parts, if f \to 0 at infinity: \int f'(t)e^{-i\omega t}dt = \left[f e^{-i\omega t}\right] + i\omega\int f e^{-i\omega t}dt = i\omega F(\omega).
  • Real signal. If f is real, F(-\omega) = \overline{F(\omega)}: the magnitude is even and the phase odd. This is why instruments show only positive frequencies.
Effect of modulation on the spectrum
On the left the spectrum of a baseband signal; on the right the spectrum after multiplication by a cosine of angular frequency omega zero: two copies of half height, centred at plus and minus omega zero.

Theorem (Parseval). \displaystyle\int_{-\infty}^{\infty}|f(t)|^2dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}|F(\omega)|^2d\omega.

Proof. Write |f|^2 = f\,\bar f and replace \bar f with the conjugate inversion formula:

    \[ \int f(t)\overline{f(t)}dt = \int f(t)\,\frac{1}{2\pi}\int\overline{F(\omega)}e^{-i\omega t}d\omega\,dt = \frac{1}{2\pi}\int\overline{F(\omega)}\left(\int f(t)e^{-i\omega t}dt\right)d\omega = \frac{1}{2\pi}\int|F|^2d\omega \]

swapping the order of the integrals. ∎

Smoothness and spectral decay

How fast |F(\omega)| goes to zero as \omega \to \infty depends on how “smooth” the signal is:

  • if f has a jump (rectangular pulse), |F| \sim 1/\omega;
  • if f is continuous but its derivative has a jump (triangular pulse), |F| \sim 1/\omega^2;
  • in general, if the first k - 1 derivatives are continuous and the k-th has a jump, |F| \sim 1/\omega^{k+1};
  • if f has all derivatives (Gaussian), F decays faster than any power.

Why. Applying the derivative rule k times, \mathcal F\{f^{(k)}\} = (i\omega)^kF. If f^{(k)} has a jump of size J at t_0, its derivative contains J\,\delta(t - t_0), whose transform does not decay: so \omega^{k+1}|F| stays bounded and non-zero. ∎

This is of great practical importance: the steep edges of digital signals generate harmonics up to very high frequencies and hence interference (electromagnetic compatibility). This is why fast signals are “rounded” with edges of controlled duration.

Spectral decay in decibels
The spectrum of the rectangular pulse, in blue, falls by 20 dB per decade; that of the triangular pulse, in orange, by 40 dB per decade; that of the Gaussian, in green, collapses much faster.
Triangular pulse and its spectrum
The triangular pulse is the convolution of two rectangular pulses, so its spectrum is the square of a sinc: positive and with side lobes much lower than those of the plain sinc, dashed.

The uncertainty principle

Define the duration and the bandwidth of a signal as standard deviations, weighted by energy:

    \[ \Delta t^2 = \frac{\int t^2|f|^2dt}{\int|f|^2dt},\qquad \Delta\omega^2 = \frac{\int\omega^2|F|^2d\omega}{\int|F|^2d\omega} \]

Theorem. For every signal (decaying fast enough at infinity) \Delta t\,\Delta\omega \ge \frac{1}{2}, with equality only for the Gaussian.

Idea of the proof. By Parseval and the derivative rule, \int\omega^2|F|^2d\omega = 2\pi\int|f'|^2dt. By the Cauchy-Schwarz inequality, \left|\int t f\,\bar f'\,dt\right|^2 \le \int t^2|f|^2dt\cdot\int|f'|^2dt, and integrating by parts the left-hand side is at least \frac{1}{4}\left(\int|f|^2\right)^2. Putting things together gives the inequality; equality in Cauchy-Schwarz requires f' \propto t f, i.e. the Gaussian. ∎

It is the same as Heisenberg’s principle in quantum mechanics, where time and frequency become position and momentum.

The sampling theorem, proved

Theorem (Nyquist-Shannon). If F(\omega) = 0 for |\omega| \ge \omega_M and one samples with period T such that \omega_c = \frac{2\pi}{T} \gt 2\omega_M, then

    \[ f(t) = \sum_{n=-\infty}^{\infty}f(nT)\,\text{sinc}\left(\frac{\pi(t - nT)}{T}\right),\qquad \text{sinc}(x) = \frac{\sin x}{x} \]

Proof. The sampled signal is f_c(t) = f(t)\sum_n\delta(t - nT). The Dirac comb has transform \omega_c\sum_k\delta(\omega - k\omega_c) (lesson on the delta), and the product in time becomes a convolution in frequency divided by 2\pi: F_c(\omega) = \frac{1}{T}\sum_k F(\omega - k\omega_c). The spectrum repeats with period \omega_c. If \omega_c \gt 2\omega_M the copies do not overlap, and an ideal low-pass filter H(\omega) = T for |\omega| \lt \omega_c/2 gives back F(\omega). The impulse response of that filter is h(t) = \text{sinc}(\pi t/T), and the output f_c * h is the sum in the formula. ∎

Each sample contributes a sinc centred at its instant, which is 1 there and 0 at all other sampling instants: the formula interpolates exactly. If instead \omega_c \lt 2\omega_M, the copies overlap and high frequencies “fold” onto low ones (aliasing), irreparably: this is why every A/D converter is preceded by an anti-aliasing filter.

Spectrum of a sampled signal
On the left the sampling frequency is high enough and the copies of the spectrum are separate. On the right it is too low: the copies, dashed, overlap and their sum, in red, no longer allows the original spectrum to be recovered.

Tables to remember

Transform pairs (a \gt 0; u is the step, \text{rect}_T is 1 for |t| \lt T/2, \text{tri}_T is the triangle of base 2T and height 1)

f(t)F(\omega)
\delta(t)1
12\pi\delta(\omega)
\text{rect}_T(t)T\,\text{sinc}(\omega T/2)
\text{tri}_T(t)T\,\text{sinc}^2(\omega T/2)
e^{-at}u(t)\dfrac{1}{a + i\omega}
e^{-a|t|}\dfrac{2a}{a^2 + \omega^2}
\dfrac{1}{a^2 + t^2}\dfrac{\pi}{a}e^{-a|\omega|}
e^{-t^2/(2\sigma^2)}\sigma\sqrt{2\pi}\,e^{-\sigma^2\omega^2/2}
\cos\omega_0 t\pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]
\sin\omega_0 t-i\pi[\delta(\omega - \omega_0) - \delta(\omega + \omega_0)]
u(t)\pi\delta(\omega) + \dfrac{1}{i\omega}
\text{sgn}(t)\dfrac{2}{i\omega}
\sum_n\delta(t - nT)\omega_c\sum_k\delta(\omega - k\omega_c), \omega_c = \frac{2\pi}{T}

Properties (besides those in the lesson): \mathcal F\{t f(t)\} = iF'(\omega); \mathcal F\{F(t)\} = 2\pi f(-\omega) (duality); \mathcal F\{f(-t)\} = F(-\omega); F(0) = \int f\,dt and f(0) = \frac{1}{2\pi}\int F\,d\omega (areas); \int f\bar g\,dt = \frac{1}{2\pi}\int F\bar G\,d\omega (generalised Parseval).

Exercises from basic to exam level are on the exercise page linked at the top.

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