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46  Fourier transforms

From periodic decomposition to continuous frequency analysis

The Fourier series produces a discrete spectrum: a list of amplitudes at frequencies \(\pi/L, 2\pi/L, 3\pi/L, \ldots\) spaced \(\pi/L\) apart. As \(L \to \infty\), the spacing between frequencies shrinks to zero, the list of amplitudes becomes a continuous function of frequency, and the sum becomes an integral. This is the Fourier transform.

The transition from series to transform is not just a mathematical technicality. It extends the analysis to any signal that decays to zero as \(|t| \to \infty\) — a one-shot pulse, a finite-duration transient, or any square-integrable function. The series was restricted to functions that repeat forever. The transform has no such restriction.

46.1 From Fourier series to the Fourier integral

Consider a function \(f\) of period \(2L\) with Fourier series:

\[f(x) = \sum_{n=-\infty}^{\infty} c_n\,e^{in\pi x/L}, \qquad c_n = \frac{1}{2L}\int_{-L}^{L} f(x)\,e^{-in\pi x/L}\,dx\]

Define \(\omega_n = n\pi/L\) and \(\Delta\omega = \pi/L\) (the spacing between successive frequencies). Then:

\[f(x) = \sum_{n=-\infty}^{\infty} \frac{\Delta\omega}{2\pi}\left[\int_{-L}^{L} f(t)\,e^{-i\omega_n t}\,dt\right]e^{i\omega_n x}\]

The rewrite uses a single substitution: since \(\Delta\omega = \pi/L\), the prefactor \(1/(2L)\) in \(c_n\) equals \(\Delta\omega/(2\pi)\). Replacing it turns each coefficient into \((\Delta\omega/2\pi)\times[\text{integral}]\), and the whole sum becomes a Riemann sum with step size \(\Delta\omega\) — ready to become an integral.

Let \(L \to \infty\): \(\Delta\omega \to d\omega\), the sum becomes an integral, and we obtain the Fourier integral representation:

\[f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty} f(t)\,e^{-i\omega t}\,dt\right]e^{i\omega x}\,d\omega\]

The inner bracket is the Fourier transform of \(f\) evaluated at \(\omega\).

46.2 The Fourier transform

The Fourier transform of \(f(t)\) is:

\[F(\omega) = \mathcal{F}\{f\}(\omega) = \int_{-\infty}^{\infty} f(t)\,e^{-i\omega t}\,dt\]

The inverse Fourier transform recovers \(f\) from \(F\):

\[f(t) = \mathcal{F}^{-1}\{F\}(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega)\,e^{i\omega t}\,d\omega\]

Together these are a transform pair: \(f(t) \leftrightarrow F(\omega)\). Knowing either one determines the other completely (for well-behaved functions).

Conventions. There are three common conventions for where the \(2\pi\) factor sits and whether the sign in the exponent is \(-i\omega t\) or \(+i\omega t\). The one above is the engineering convention (also used by Kreyszig and Stroud). The physics convention places \(1/\sqrt{2\pi}\) on both the forward and inverse transform. When using a reference or software tool, check which convention it follows.

Existence. The Fourier transform is guaranteed to exist (as an ordinary integral) when \(f\) is absolutely integrable: \(\int_{-\infty}^{\infty}|f(t)|\,dt < \infty\). For the larger class of square-integrable functions (\(\int|f|^2 < \infty\)), it exists in the \(L^2\) sense. Important special cases — the unit step, the delta function, the constant function — require the theory of distributions but yield well-defined transforms.

46.3 Key transform pairs

46.3.1 Rectangular pulse

The rectangular pulse (or gate function) of width \(2a\):

\[f(t) = \begin{cases} 1 & |t| < a \\ 0 & |t| > a \end{cases}\]

Its transform:

\[F(\omega) = \int_{-a}^{a} e^{-i\omega t}\,dt = \left[\frac{e^{-i\omega t}}{-i\omega}\right]_{-a}^{a} = \frac{e^{-i\omega a} - e^{i\omega a}}{-i\omega} = \frac{2\sin(\omega a)}{\omega} = 2a\,\operatorname{sinc}(\omega a)\]

where \(\operatorname{sinc}(x) = \sin(x)/x\).

The time–bandwidth tradeoff. The width of the main lobe of \(|F(\omega)|\) is approximately \(2\pi/a\). A narrow pulse (\(a\) small) has a wide spectrum. A wide pulse (\(a\) large) has a narrow spectrum. This is the mathematical statement of the tradeoff the bat exploits.

46.3.2 Gaussian

The Gaussian \(f(t) = e^{-\alpha t^2}\) (\(\alpha > 0\)) transforms to:

\[F(\omega) = \sqrt{\frac{\pi}{\alpha}}\,e^{-\omega^2/(4\alpha)}\]

also a Gaussian. The Gaussian is its own Fourier transform (up to scaling). It minimises the time–bandwidth product, which is why it is used in pulse shaping and wavelet theory.

46.3.3 Dirac delta

The Dirac delta is not an ordinary function — it has no well-defined value at any point, and you cannot draw its graph. It is defined entirely by what it does inside an integral: \(\int_{-\infty}^{\infty}\delta(t)\phi(t)\,dt = \phi(0)\) for any smooth \(\phi\). Its transform:

\[\mathcal{F}\{\delta(t)\} = \int_{-\infty}^{\infty}\delta(t)\,e^{-i\omega t}\,dt = e^{0} = 1\]

The delta function has a flat (white) spectrum — it contains all frequencies equally. This is why a sharp impulse (the tap of a hammer on a structure) excites all resonant frequencies simultaneously.

The inverse: \(\mathcal{F}\{1\} = 2\pi\delta(\omega)\) — a constant signal contains only zero frequency.

46.4 Properties of the Fourier transform

These properties make the Fourier transform useful in practice.

Linearity. \(\mathcal{F}\{\alpha f + \beta g\} = \alpha F + \beta G\).

Time shifting. \(\mathcal{F}\{f(t - t_0)\} = e^{-i\omega t_0}F(\omega)\). Shifting a signal in time multiplies its spectrum by a complex exponential — it changes the phase but not the magnitude.

Frequency shifting. \(\mathcal{F}\{e^{i\omega_0 t}f(t)\} = F(\omega - \omega_0)\). Multiplying by a complex exponential shifts the spectrum. This is the mathematical basis of modulation: shifting a baseband signal to a carrier frequency. In engineering practice, modulation uses a real cosine \(\cos(\omega_0 t) = \frac{1}{2}(e^{i\omega_0 t} + e^{-i\omega_0 t})\), which shifts the spectrum to both \(+\omega_0\) and \(-\omega_0\) (creating upper and lower sidebands); the complex exponential form isolates the single-sideband case.

Scaling. \(\mathcal{F}\{f(at)\} = \frac{1}{|a|}F\!\left(\frac{\omega}{a}\right)\). Compressing time by factor \(a\) expands the spectrum by the same factor — another form of the time–bandwidth tradeoff.

Differentiation. \(\mathcal{F}\{f'(t)\} = i\omega\,F(\omega)\). Differentiation in time corresponds to multiplication by \(i\omega\) in frequency. This turns differential equations into algebraic equations.

Convolution theorem.

\[\mathcal{F}\{f * g\}(\omega) = F(\omega)\cdot G(\omega)\]

where \((f*g)(t) = \int_{-\infty}^{\infty}f(\tau)g(t-\tau)\,d\tau\) is the convolution of \(f\) and \(g\). Equivalently, \(\mathcal{F}\{fg\} = \frac{1}{2\pi}F * G\).

Parseval’s theorem. \(\int_{-\infty}^{\infty}|f(t)|^2\,dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}|F(\omega)|^2\,d\omega\).

46.5 The convolution theorem in engineering

The convolution theorem is the most practically important property. A linear time-invariant (LTI) system is completely characterised by its impulse response \(h(t)\): if you drive the system with input \(f(t)\), the output is the convolution \(y(t) = (f * h)(t)\).

Computing convolution directly requires integrating over all past input values. In the frequency domain, this becomes multiplication:

\[Y(\omega) = F(\omega)\cdot H(\omega)\]

where \(H(\omega) = \mathcal{F}\{h\}\) is the transfer function of the system.

This is why filters are designed in the frequency domain. A low-pass filter that passes frequencies \(|\omega| < \omega_c\) and blocks higher frequencies has \(H(\omega) = 1\) for \(|\omega| < \omega_c\) and \(H(\omega) = 0\) otherwise. Computing \(H(\omega) \cdot F(\omega)\) is multiplication of two functions. Computing the equivalent convolution would mean implementing an integral.

In discrete time, the same structure holds: the discrete Fourier transform converts discrete convolution to pointwise multiplication. The FFT computes this in \(O(N\log N)\) time, which is why all modern digital signal processing systems work this way.

46.6 Solving differential equations with Fourier transforms

The differentiation property converts a linear ODE with constant coefficients into an algebraic equation in the frequency domain.

Example. Solve \(y' + 2y = e^{-|t|}\) for \(y(t)\).

Take the Fourier transform of both sides:

\[i\omega\,Y(\omega) + 2Y(\omega) = F(\omega)\]

\[(i\omega + 2)Y(\omega) = F(\omega)\]

\[Y(\omega) = \frac{F(\omega)}{i\omega + 2}\]

Now we need \(F(\omega) = \mathcal{F}\{e^{-|t|}\} = \int_{-\infty}^{\infty}e^{-|t|}e^{-i\omega t}\,dt = \frac{2}{1+\omega^2}\).

So \(Y(\omega) = \frac{2}{(1+\omega^2)(i\omega+2)}\).

Write this in partial fractions:

\[\frac{2}{(1+\omega^2)(i\omega+2)} = \frac{A}{i\omega+2} + \frac{B\omega + C}{1+\omega^2}\]

Matching numerators gives \(A = 2/5\), \(B = -2/5\), \(C = 4/5\), so:

\[Y(\omega) = \frac{2/5}{i\omega+2} + \frac{(-2\omega + 4)/5}{1+\omega^2}\]

Each term matches a known transform pair: \(\mathcal{F}^{-1}\{1/(i\omega+2)\} = e^{-2t}\) for \(t > 0\) (causal exponential); \(\mathcal{F}^{-1}\{1/(1+\omega^2)\} = \frac{1}{2}e^{-|t|}\); and \(\mathcal{F}^{-1}\{\omega/(1+\omega^2)\}\) gives a derivative term. Inverting term by term:

\[y(t) = \frac{1}{5}\left(2e^{-2t} - e^{-|t|} + \operatorname{sgn}(t)e^{-|t|}\right)\]

The Fourier transform has reduced the differential equation to algebra.

46.7 Connection to the Laplace transform

If you have met the Laplace transform in an ODE course, this section shows how it connects to the Fourier transform. If not, you can skip it — nothing later in this chapter depends on it.

The Laplace transform \(\mathcal{L}\{f\}(s) = \int_0^{\infty}f(t)e^{-st}\,dt\) is a one-sided transform: it only sees \(t \geq 0\) (causal signals). Replacing \(s = i\omega\) recovers the Fourier transform of the causal part of \(f\).

The Laplace transform is preferred for initial-value problems (ODEs with \(t \geq 0\) and given initial conditions) because it handles them directly via \(\mathcal{L}\{f'\} = sF(s) - f(0)\). The Fourier transform is preferred for steady-state and bilateral analysis (signals for all \(t\)).

In circuit analysis: the transfer function \(H(s)\) in Laplace domain becomes \(H(i\omega)\), the Fourier domain transfer function, when you restrict to purely sinusoidal steady-state inputs.

46.8 Exercises

46.8.1 Exercise 1: Transform of a rectangular pulse

Compute the Fourier transform of \(f(t) = \begin{cases}1 & |t| \leq 1 \\ 0 & |t| > 1\end{cases}\). Identify the first zero of \(|F(\omega)|\).

46.8.2 Exercise 2: Time-shifting theorem

Given that \(\mathcal{F}\{e^{-|t|}\} = \frac{2}{1+\omega^2}\), find \(\mathcal{F}\{e^{-|t-3|}\}\) and \(\mathcal{F}\{e^{-|t|}e^{2it}\}\).

46.8.3 Exercise 3: Differentiation property — solving an ODE

Use the Fourier transform to solve \(y' + 3y = e^{-2|t|}\).

(The inversion uses only partial fractions and the known pairs from this chapter — no complex analysis required.)

46.8.4 Exercise 4: Convolution theorem — LTI system response

A low-pass filter has impulse response \(h(t) = e^{-2t}u(t)\) (where \(u(t)\) is the unit step: \(u(t) = 1\) for \(t \geq 0\), \(0\) otherwise). The input is \(f(t) = e^{-t}u(t)\). Find the output \(y(t) = (f * h)(t)\) using the Fourier transform.

46.8.5 Exercise 5: Parseval’s theorem — energy in the frequency domain

The voltage pulse across a 1 Ω resistor is \(v(t) = e^{-|t|}\). Use Parseval’s theorem to find the total energy dissipated.