The Fourier Transform

Begin with the exponential Fourier series of Section 17.6, written over a symmetric period:

Now let \(T\) grow. Three things happen together:

• The spacing between adjacent harmonics, \(\Delta\omega = \omega_{0} = 2\pi/T\), becomes vanishingly small. Write it as \(d\omega\).

• The discrete variable \(n\omega_{0}\) becomes a continuous variable \(\omega\).

• Each \(c_{n}\) tends to zero, because of the \(1/T\) factor. So we work instead with the product \(Tc_{n}\), which does not vanish.

Define that surviving quantity as the transform:

Reversing the argument recovers \(f(t)\). Substituting \(c_{n} = F(\omega)/T = F(\omega)\,d\omega/2\pi\) into the series turns the sum into an integral:

Note the asymmetry. The factor \(1/2\pi\) appears in the inverse transform but not the forward one — a consequence of using \(\omega\) in rad/s rather than \(f\) in hertz. Some texts split it as \(1/\sqrt{2\pi}\) on each side, and engineers working in hertz avoid it altogether. The convention above is the standard one in circuit analysis; only consistency matters.

Existence. A sufficient condition is that \(f(t)\) be absolutely integrable:

This excludes a great deal — a constant, a step, an unending sinusoid all fail it, because none decays. Yet all three have transforms in practice, obtained as limits and expressed using impulses. The condition is sufficient but not necessary, and the impulse function rescues the important exceptions.

\(F(\omega)\) is complex, so it is described by two plots:

the continuous amplitude spectrum and the phase spectrum. The contrast with Chapter 17 is the whole point: a periodic signal has energy at isolated frequencies, a non-periodic one across a continuum.

The impulse. The sifting property does all the work, as it did in Section 15.3:

An impulse contains every frequency equally. This is why a sharp tap excites all the resonances of a structure at once, and why the impulse response characterises a system completely.

The causal exponential.

Compare with the Laplace transform \(1/(s+a)\) — identical with \(s = j\omega\). And compare with the RC low-pass response of Section 14.2: the same expression again, because the two problems are the same problem.

The rectangular pulse, of height \(A\) and width \(\tau\) centred on the origin:

This is the sinc function, the single most important shape in signal processing. Its features are worth memorising: it peaks at \(F(0) = A\tau\), which is the pulse's area; it crosses zero at \(\omega = 2\pi n/\tau\); and its side lobes decay as \(1/\omega\).

The reciprocal-spreading principle. A narrow pulse has widely spaced zeros and hence a broad spectrum; a wide pulse has a narrow one. Compressing a signal in time stretches it in frequency, and the product of the two widths is fixed. This is the engineering version of the uncertainty principle, and it explains why a fast digital link needs bandwidth: short bits demand a wide spectrum, with no way around it.

The time-shift property is worth dwelling on. Delaying a signal multiplies its transform by \(e^{-j\omega t_{0}}\), whose magnitude is exactly 1. So a delay changes the phase spectrum but leaves the amplitude spectrum untouched. This makes physical sense — moving an event in time does not alter which frequencies it contains — and it explains why amplitude spectra alone cannot reveal timing.

The differentiation property shows why differentiators are noisy: multiplying by \(j\omega\) amplifies high frequencies without limit, and noise lives at high frequencies. Integration divides by \(j\omega\) and therefore smooths — the same conclusion reached about the RL circuit in Example 17.5, now stated generally.

Symmetry for real signals. If \(f(t)\) is real then \(F(-\omega) = F^{*}(\omega)\), so the amplitude spectrum is even and the phase spectrum odd. Half the spectrum is therefore redundant, which is why single-sided spectra are commonly plotted.

The convolution property makes circuit analysis with non-periodic inputs straightforward. Since convolution in time is multiplication in frequency,

where \(\mathbf{H}(\omega)\) is the transfer function of Chapter 14 — the same object, unchanged. The procedure mirrors the Laplace procedure of Chapter 16:

1. Transform the input to obtain \(V_{i}(\omega)\).

2. Find \(\mathbf{H}(\omega)\) using impedances \(R\), \(j\omega L\) and \(1/j\omega C\).

3. Multiply, then invert by partial fractions.

The one restriction to remember: this assumes zero initial conditions. The Fourier transform has no mechanism for injecting them, which is a genuine limitation and the reason Laplace remains the tool of choice for switching problems.

The total energy delivered by a signal \(f(t)\) into a 1 \(\Omega\) resistor is \(\int f^{2}(t)\,dt\). Parseval's theorem says this may be computed equally well in the frequency domain:

This is the non-periodic counterpart of the RMS result in Section 17.8, and it carries the same message: energy is conserved by the transform. Neither domain is privileged.

The quantity \(\left|F(\omega)\right|^{2}\) is the energy spectral density, in joules per hertz of bandwidth. Integrating it over a band gives the energy in that band, so the theorem answers a practical question directly: how much of a signal's energy would survive a given filter?

Using the even symmetry of \(\left|F\right|^{2}\) for real signals, the integral may be taken over positive frequencies only:

Two transforms now serve overlapping purposes, and knowing which to reach for matters.

The formal relationship is simple. Compare the definitions:

For a causal signal the lower limits agree, and setting \(s = j\omega\) makes the two identical:

So the Fourier transform is the Laplace transform evaluated on the \(j\omega\) axis — the boundary of the region of convergence discussed in Section 15.2. The condition on the poles matters: if any lies on or right of the axis, the Laplace transform exists but the Fourier transform does not converge, which is exactly why an unstable system has no meaningful frequency response.

The practical rule. Use Laplace when the question involves switching, initial energy, or stability — anything where the transient matters. Use Fourier when the question is about frequency content: bandwidth, filtering, modulation, noise. The Laplace transform is the better calculating tool; the Fourier transform is the better describing tool, because \(\left|F(\omega)\right|\) is something an instrument can display.

Radio transmission poses a problem the transform solves neatly. A speech signal occupies roughly 0 to 5 kHz, and an antenna efficient at those frequencies would need to be tens of kilometres long. Worse, every transmitter would occupy the same band and interfere with every other.

The solution is to shift the signal to a much higher frequency, and the frequency-shift property does exactly that. Multiplying by a carrier \(\cos\omega_{c}t\) and using the transform pair for the cosine:

Multiplication in time is frequency shifting. The message spectrum is copied to \(\pm\omega_{c}\), at half amplitude, forming the sidebands. The shape is unchanged; only its position on the frequency axis has moved.

Two consequences follow immediately from the picture. The transmitted bandwidth is twice the message bandwidth, because both an upper and a lower sideband are produced. And since each station occupies a distinct band around its own carrier, many may share the airwaves without interference — the tuned circuit of Section 14.11 selects the wanted one.

Demodulation reverses the shift: multiplying the received signal by \(\cos\omega_{c}t\) again produces copies at 0 and at \(\pm2\omega_{c}\), and a low-pass filter discards the latter, recovering the message.

Every digital system begins by sampling a continuous signal at regular intervals. A question of some importance is whether anything is lost by doing so.

Sampling multiplies the signal by a train of impulses spaced \(T_{s}\) apart. By the convolution-in-frequency property, multiplication in time is convolution in frequency — and convolving with an impulse train produces copies of the spectrum, repeated at intervals of the sampling frequency \(\omega_{s} = 2\pi/T_{s}\).

Everything now depends on whether those copies overlap. If the signal is band-limited to \(\omega_{m}\) and the copies are spaced \(\omega_{s}\) apart, they remain separate provided \(\omega_{s} > 2\omega_{m}\). This is the sampling theorem:

The minimum rate \(2f_{m}\) is the Nyquist rate. Sample faster than this and the original may be recovered exactly by low-pass filtering — no information is lost whatever, despite the signal having been reduced to a list of numbers. This result underpins all digital audio, video and telecommunications.

Aliasing. Sample too slowly and the copies overlap. High-frequency content then folds back and appears as a spurious low frequency, indistinguishable from genuine signal. A component at \(f\) appears at

for whichever integer \(n\) brings it closest to zero. Aliasing is irreversible — once the frequencies have merged, no processing can separate them. The only defence is to remove the offending content before sampling, using an anti-aliasing filter: a low-pass filter, of the kind designed in Chapter 14, placed ahead of the converter.

The familiar visual example is a wagon wheel in a film appearing to rotate backwards. The frame rate is the sampling rate, the wheel's rotation exceeds half of it, and the alias is a slow reverse rotation.

Computers cannot evaluate an integral over infinite time. The discrete Fourier transform replaces the integral with a finite sum over \(N\) samples, and the fast Fourier transform is an algorithm that computes it in \(N\log_{2}N\) operations instead of the obvious \(N^{2}\).

The saving is not marginal. For \(N = 1024\) it is a factor of about 100; for \(N = 10^{6}\), a factor of 50 000. Cooley and Tukey's 1965 publication of the algorithm made real-time spectral analysis practical, and essentially every digital instrument that displays a spectrum — oscilloscope, audio analyser, vibration monitor, medical scanner — computes an FFT to do it.

• The Fourier transform is the limit of the Fourier series as \(T \to \infty\): the sum becomes an integral and the line spectrum becomes continuous.

• An impulse contains all frequencies equally; a constant is a single impulse at \(\omega = 0\).

• A rectangular pulse transforms to a sinc function whose peak is the pulse area and whose zeros are at \(2\pi n/\tau\).

• Reciprocal spreading: narrow in time means broad in frequency, and the product of the widths is fixed.

• Duality turns every transform pair into a second one.

• A time delay changes only the phase spectrum, never the amplitude spectrum.

• Circuit analysis uses \(V_{o}(\omega) = \mathbf{H}(\omega)V_{i}(\omega)\), valid only for zero initial conditions.

• Parseval's theorem computes energy in either domain; \(\left|F(\omega)\right|^{2}\) is the energy spectral density.

• The Fourier transform is the Laplace transform on the \(j\omega\) axis, for causal signals with left-half-plane poles.

• Modulation shifts a spectrum to a carrier, doubling the bandwidth; sampling replicates it, requiring \(f_{s} > 2f_{m}\) to avoid irreversible aliasing.

• Putting the \(1/2\pi\) in the forward transform. It belongs in the inverse. Mixing conventions between two sources is a frequent source of factor-of-\(2\pi\) errors.

• Using the Fourier transform when initial conditions are non-zero. It has no way to represent them — use Laplace.

• Integrating from 0 instead of \(-\infty\). The Fourier transform is two-sided. The lower limit is only 0 in effect when the signal is causal.

• Expecting \(F(\omega)\) to be positive. It is complex in general, and even when real it may be negative — a negative real value means a phase of 180°.

• Forgetting the impulse in \(\mathcal{F}\left[u(t)\right]\). It is \(\pi\delta(\omega)+1/j\omega\), not just \(1/j\omega\). The impulse carries the step's DC content.

• Assuming a wider pulse gives a wider spectrum. The relation is reciprocal — wider in time is narrower in frequency.

• Taking the AM bandwidth as equal to the message bandwidth. It is twice, because both sidebands are transmitted.

• Sampling at exactly the Nyquist rate. The theorem requires \(f_{s}\) strictly greater than \(2f_{m}\), and practical filters demand a margin beyond that.

• Believing aliasing can be filtered out afterwards. Once frequencies have folded together they are indistinguishable. The filter must precede the sampler.

• Applying \(s = j\omega\) to an unstable system. If any pole is on or right of the axis, the Fourier transform does not converge and the substitution is meaningless.

Work these before opening the answers. Note whether each signal is causal.

1. P18.1 Find the Fourier transform of \(f(t) = 5e^{-3t}u(t)\), and its magnitude at \(\omega = 3\) and \(\omega = 4\).

   Show answer

   \[F(\omega) = \frac{5}{3+j\omega}, \qquad \left|F\right| = \frac{5}{\sqrt{9+\omega^{2}}}\]

   At \(\omega = 3\): \(5/\sqrt{18} = 1.179\), with phase \(-45^{\circ}\).
   At \(\omega = 4\): \(5/5 = 1.000\), with phase \(-53.13^{\circ}\).

2. P18.2 A rectangular pulse of height 4 lasts from \(t = -0.5\) to \(+0.5~\mathrm{s}\). Find \(F(0)\) and the first zero of its spectrum.

   Show answer

   \[F(0) = A\tau = \left(4\right)\left(1\right) = 4 \quad\text{(the area)}\]

   \[\text{First zero at } \omega = \frac{2\pi}{\tau} = 2\pi = 6.283~\mathrm{rad/s}\]

   Compare Example 18.1: halving the pulse width has doubled the spectral width, exactly as reciprocal spreading requires.

3. P18.3 Use duality to find the transform of \(f(t) = \dfrac{1}{2+jt}\).

   Show answer

   We know \(e^{-2t}u(t) \leftrightarrow 1/(2+j\omega)\). Duality gives \(F(t) \leftrightarrow 2\pi f(-\omega)\):

   \[\frac{1}{2+jt} \leftrightarrow 2\pi e^{2\omega}u(-\omega)\]

4. P18.4 An RL circuit with \(R = 2~\Omega\) and \(L = 1~\mathrm{H}\) has \(v_{i}(t) = 6e^{-t}u(t)~\mathrm{V}\) applied. Find the current, taking \(\mathbf{H}(\omega) = 1/\left(R+j\omega L\right)\).

   Show answer

   \[I(\omega) = \frac{6}{\left(1+j\omega\right)\left(2+j\omega\right)} = \frac{6}{1+j\omega}-\frac{6}{2+j\omega}\]

   \[i(t) = 6\left(e^{-t}-e^{-2t}\right)u(t)~\mathrm{A}\]

   Peak at \(t = \ln 2\), of value \(6\left(0.5-0.25\right) = 1.500~\mathrm{A}\).

5. P18.5 Find the total energy in \(f(t) = 3e^{-t}u(t)\) by both domains.

   Show answer

   Time domain: \(W = \int_{0}^{\infty}9e^{-2t}dt = 9/2 = 4.500~\mathrm{J}\)
   Frequency domain: \(\left|F\right|^{2} = 9/(1+\omega^{2})\), so

   \[W = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{9\,d\omega}{1+\omega^{2}} = \frac{9}{2\pi}\left(\pi\right) = 4.500~\mathrm{J} \;\checkmark\]

6. P18.6 What fraction of the energy in P18.5 lies below \(\omega = 1~\mathrm{rad/s}\)?

   Show answer

   \[\frac{\int_{-1}^{1}d\omega/\left(1+\omega^{2}\right)}{\int_{-\infty}^{\infty}d\omega/\left(1+\omega^{2}\right)} = \frac{2\tan^{-1}(1)}{\pi} = \frac{\pi/2}{\pi} = 0.5000\]

   Fifty per cent — the same result as Example 18.3, and for the same reason: the pole frequency is the energy median.

7. P18.7 A 3 kHz signal modulates a 600 kHz carrier. Find the sidebands and the bandwidth.

   Show answer

   Sidebands from 597 kHz to 600 kHz and from 600 kHz to 603 kHz.

   \[\text{Bandwidth} = 2\left(3\right) = 6~\mathrm{kHz}\]

8. P18.8 Find the Nyquist rate for a signal band-limited to 12 kHz. If it is sampled at 20 kHz instead, where does a 12 kHz component appear?

   Show answer

   \[f_{s,\min} = 2\left(12\right) = 24~\mathrm{kHz}\]

   At 20 kHz the rate is too low, so 12 kHz aliases to

   \[f_{\text{alias}} = \left|12-20\right| = 8~\mathrm{kHz}\]

   and is then indistinguishable from a genuine 8 kHz component.

9. P18.9 A delayed pulse \(f(t-3)\) replaces \(f(t)\). How do the amplitude and phase spectra change?

   Show answer

   The transform becomes \(e^{-j3\omega}F(\omega)\). Since \(\left|e^{-j3\omega}\right| = 1\), the amplitude spectrum is unchanged. The phase spectrum acquires an extra term \(-3\omega\) — a linear phase shift, growing with frequency.

10. P18.10 Explain why an unstable system has no Fourier transform of its impulse response, although it has a Laplace transform.

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    An unstable system has a pole with \(\mathrm{Re}(p) \ge 0\), so \(h(t)\) contains a growing or non-decaying exponential and fails the absolute-integrability condition. The Laplace transform survives because the factor \(e^{-\sigma t}\) can be made to decay faster than \(h(t)\) grows, by choosing \(\sigma\) large enough. The Fourier transform, fixed at \(\sigma = 0\), has no such freedom — the \(j\omega\) axis lies outside the region of convergence.