Mathematical Methods

Solved Fourier transform exercises, with graphs

Twelve exercises with solutions: exponentials, delayed pulses, convolution, modulation, duality, Parseval, RC filter, in-band energy, aliasing, uncertainty principle.

ielettronica 25 October 2026 9 min read 0 comments

Exercises on the Fourier transform of the kind set in Mathematical Methods for Engineering and Signal Theory courses, in three levels: basic, with transforms from the definition; intermediate, with properties, convolution, duality and Parseval; exam, with filters, energy, sampling and the uncertainty principle. The result is given straight away; the full solution, with the graph, opens with a click. Convention: $F(\omega) = \int f(t)e^{-i\omega t}dt$ ; $\text{sinc}(x) = \frac{\sin x}{x}$ .

Basic level

Exercise 1. The two-sided exponential

Compute the transform of $f(t) = e^{-a|t|}$ , with $a \gt 0$ .

Result: $F(\omega) = \dfrac{2a}{a^2 + \omega^2}$ .

Solution

Split the integral over the two half-lines:

$F(\omega) = \int_{-\infty}^{0}e^{(a - i\omega)t}dt + \int_0^{\infty}e^{-(a + i\omega)t}dt = \frac{1}{a - i\omega} + \frac{1}{a + i\omega} = \frac{2a}{a^2 + \omega^2}$

The integrals converge because $a \gt 0$ . The transform is real and even, like the signal (a real even signal has a real even spectrum). The shape $\frac{1}{a^2 + \omega^2}$ is called a Lorentzian: it is the spectrum of the noise coming out of an RC filter.

Two-sided exponential and its transform — On the left the signal e to the minus a times the absolute value of t with a equal to 1, with a peak at the origin; on the right the transform, a Lorentzian of height 2.

Exercise 2. The one-sided exponential: magnitude and phase

Compute the transform of $f(t) = e^{-at}u(t)$ and write its magnitude and phase. What are they at $\omega = a$ ?

Result: $F = \dfrac{1}{a + i\omega}$ ; $|F| = \dfrac{1}{\sqrt{a^2 + \omega^2}}$ , $\arg F = -\arctan\dfrac{\omega}{a}$ ; at $\omega = a$ : $\dfrac{1}{a\sqrt 2}$ and $-45°$ .

Solution

$F(\omega) = \int_0^\infty e^{-(a + i\omega)t}dt = \dfrac{1}{a + i\omega}$ .

Multiplying by the conjugate: $F = \dfrac{a - i\omega}{a^2 + \omega^2}$ , so $|F| = \dfrac{1}{\sqrt{a^2 + \omega^2}}$ and $\arg F = -\arctan(\omega/a)$ .

At $\omega = a$ the magnitude is $\frac{1}{\sqrt 2}$ of its value at 0 (i.e. $-3$ dB) and the phase is $-45°$ . This is the cut-off frequency of the RC filter, with $a = 1/RC$ : the impulse response of the filter is exactly $\frac{1}{RC}e^{-t/RC}u(t)$ .

One-sided exponential and its spectrum — On the left the signal, which jumps to 1 at the origin and then decays; on the right the magnitude of the transform, in blue, and the phase, dashed red, which goes from plus 90 to minus 90 degrees through zero.

Exercise 3. A delayed pulse

Compute the transform of the pulse that is 1 for $2 \lt t \lt 4$ and 0 elsewhere. What changes compared with the pulse centred at the origin?

Result: $F(\omega) = 2\,\text{sinc}(\omega)\,e^{-3i\omega}$ ; the magnitude is the same, the phase becomes linear.

Solution

The pulse has duration $T = 2$ and is centred at $t_0 = 3$ : it is $\text{rect}_2(t - 3)$ . The centred pulse has transform $T\,\text{sinc}(\omega T/2) = 2\,\text{sinc}(\omega)$ (lesson), and by the delay property it is multiplied by $e^{-i\omega t_0}$ :

$F(\omega) = 2\,\text{sinc}(\omega)\,e^{-3i\omega}$

The magnitude $2|\text{sinc}(\omega)|$ does not change: the amplitude spectrum does not say when the signal occurs. The phase changes, adding $-3\omega$ : a straight line with slope equal to minus the delay. This is why a filter that must not distort a signal must have linear phase (all frequencies delayed by the same time).

Delayed pulse and its spectrum — On the left the pulse between 2 and 4; on the right the magnitude of the transform, in blue, equal to that of the centred pulse, and the phase, dashed red, which falls linearly and is wrapped between minus pi and pi.

Exercise 4. The triangle as a convolution

Show that the convolution of $\text{rect}_1$ with itself is the triangle $\text{tri}_1$ (base from $-1$ to 1, height 1) and derive the transform of the triangle.

Result: $\mathcal F\{\text{tri}_1\} = \text{sinc}^2(\omega/2)$ .

Solution

$(\text{rect}_1 * \text{rect}_1)(t) = \int\text{rect}_1(\tau)\,\text{rect}_1(t - \tau)\,d\tau$ is the length of the overlap between the interval $[-\frac{1}{2}, \frac{1}{2}]$ and the same interval shifted by $t$ . It is 1 for $t = 0$ and decreases linearly to 0 for $|t| = 1$ : $1 - |t|$ , the triangle.

By the convolution theorem the transform is the product: $\left(\text{sinc}(\omega/2)\right)^2$ .

Computing it directly from the definition would take two integrations by parts: convolution is a powerful shortcut.

Triangle and its spectrum — On the left the triangle with base from minus 1 to 1; on the right its transform, sinc squared, always positive, compared with the magnitude of the plain sinc, dashed.

Intermediate level

Exercise 5. Amplitude modulation

A signal $x(t)$ has a triangular spectrum, zero for $|\omega| \gt 2$ . It is transmitted as $y(t) = x(t)\cos(8t)$ . Sketch the spectrum of $y$ and explain how $x$ is recovered.

Result: $Y(\omega) = \frac{1}{2}X(\omega - 8) + \frac{1}{2}X(\omega + 8)$ : two halved copies centred at $\pm 8$ .

Solution

$\cos 8t = \frac{1}{2}\left(e^{i8t} + e^{-i8t}\right)$ , and by the modulation property each exponential shifts the spectrum: $Y(\omega) = \frac{1}{2}X(\omega - 8) + \frac{1}{2}X(\omega + 8)$ .

The copies occupy the bands $6 \lt |\omega| \lt 10$ and do not overlap because $8 \gt 2$ . To recover $x$ (demodulation) multiply again by $\cos 8t$ : this gives $\frac{1}{2}x(t) + \frac{1}{2}x(t)\cos 16t$ , and a low-pass filter removes the term at angular frequency 16. This is the scheme of every AM radio and, with small variations, of digital transmission.

Spectrum before and after modulation — On the left the spectrum of the baseband signal; on the right the modulated spectrum, with two copies of half height centred at plus and minus omega zero.

Exercise 6. Duality

Knowing that $\mathcal F\{e^{-|t|}\} = \frac{2}{1 + \omega^2}$ , compute the transform of $f(t) = \frac{1}{1 + t^2}$ without integrating.

Result: $\pi\,e^{-|\omega|}$ .

Solution

Duality: if $g(t) \leftrightarrow G(\omega)$ , then $G(t) \leftrightarrow 2\pi g(-\omega)$ . With $g = e^{-|t|}$ , $G = \frac{2}{1 + \omega^2}$ :

$\mathcal F\left\{\frac{2}{1 + t^2}\right\} = 2\pi e^{-|\omega|} \quad\Rightarrow\quad \mathcal F\left\{\frac{1}{1 + t^2}\right\} = \pi e^{-|\omega|}$

Computing it directly would require the residue theorem (lesson). Check: $F(0) = \pi = \int\frac{dt}{1 + t^2}$ , the area of the signal.

Note that the signal is smooth (infinitely differentiable) and the spectrum decays exponentially; the spectrum has a peak at 0 and the signal decays only as $1/t^2$ . Smoothness on one side, decay on the other.

The function 1 over 1 plus t squared and its transform — On the left the signal, smooth and with slow tails; on the right the transform divided by pi, the exponential e to the minus the absolute value of omega, with a peak at the origin.

Exercise 7. An integral with Parseval

Use Parseval’s theorem to compute $\displaystyle\int_{-\infty}^{\infty}\text{sinc}^2(t)\,dt$ .

Result: $\pi$ .

Solution

The pulse $\text{rect}_2$ (1 for $|t| \lt 1$ ) has transform $2\,\text{sinc}(\omega)$ . Parseval:

$\int|\text{rect}_2|^2dt = \frac{1}{2\pi}\int 4\,\text{sinc}^2(\omega)\,d\omega$

The left-hand side is 2 (the duration). So $\int\text{sinc}^2(\omega)\,d\omega = 2\cdot\frac{2\pi}{4} = \pi$ .

An integral that is hard in time becomes trivial in frequency. With the same idea, using the transform itself instead of Parseval (area = value of the transform at 0), one also finds $\int\text{sinc}(t)\,dt = \pi$ .

The square of sinc — The shaded area under the curve equals pi; almost all of it is in the central lobe between minus pi and pi.

Exercise 8. Differentiation in frequency

Compute the transform of $f(t) = t\,e^{-at}u(t)$ using the property $\mathcal F\{t\,g(t)\} = iG'(\omega)$ .

Result: $\dfrac{1}{(a + i\omega)^2}$ .

Solution

With $g = e^{-at}u(t)$ , $G = \frac{1}{a + i\omega}$ and $G'(\omega) = -\frac{i}{(a + i\omega)^2}$ . So $F = i\cdot\left(-\frac{i}{(a + i\omega)^2}\right) = \frac{1}{(a + i\omega)^2}$ .

Magnitude: $\frac{1}{a^2 + \omega^2}$ , which decays as $1/\omega^2$ : the signal is continuous at 0 (it is 0 on both sides), but its derivative jumps from 0 to 1. This is exactly the decay rule of the in-depth study.

The signal t times e to the minus t and the magnitude of its transform — On the left the signal, which starts from zero without jumps, rises to its maximum at t equal to 1 and decays; on the right the magnitude of the transform, which decays as 1 over omega squared.

Exam level

Exercise 9. A pulse through an RC filter

A rectangular pulse of amplitude 1 and duration $T = 2$ enters an RC filter with $\tau = RC = 0.5$ . Compute the output in time with convolution and explain the result in frequency.

Result: $y(t) = 1 - e^{-t/\tau}$ for $0 \lt t \lt T$ , $y(t) = (1 - e^{-T/\tau})e^{-(t - T)/\tau}$ for $t \gt T$ ; $y(T) = 0.982$ .

Solution

Impulse response $h(t) = \frac{1}{\tau}e^{-t/\tau}u(t)$ . The output is $y(t) = \int_0^{\min(t, T)}\frac{1}{\tau}e^{-(t - s)/\tau}ds$ .

For $0 \lt t \lt T$ : $y = e^{-t/\tau}\left[e^{s/\tau}\right]_0^t = 1 - e^{-t/\tau}$ . For $t \gt T$ : $y = e^{-t/\tau}(e^{T/\tau} - 1) = (1 - e^{-T/\tau})e^{-(t - T)/\tau}$ . At the end of the pulse $y(T) = 1 - e^{-4} = 0.982$ .

In frequency. $Y(\omega) = \frac{2\,\text{sinc}(\omega)e^{-i\omega}}{1 + i\omega\tau}$ . The filter attenuates the components beyond $1/\tau = 2$ rad/s: they are the ones that make up the steep edges, which indeed become exponential. The main lobe of the pulse spectrum (up to $\pi$ rad/s) passes almost entirely, and the output gets close to level 1. With $\tau$ much larger than $T$ the output would stay small: the filter would “not have time” to follow the pulse.

Response of an RC filter to a rectangular pulse — The input pulse, dashed, and the output, in blue, which rises exponentially during the pulse and then discharges.

Exercise 10. How much energy is in the band

For the signal $f(t) = e^{-at}u(t)$ , what fraction of the energy lies in the band $|\omega| \lt a$ ?

Result: 50 %.

Solution

Total energy in time: $\int_0^\infty e^{-2at}dt = \frac{1}{2a}$ .

In frequency, $|F|^2 = \frac{1}{a^2 + \omega^2}$ , and the energy in the band is

$\frac{1}{2\pi}\int_{-a}^{a}\frac{d\omega}{a^2 + \omega^2} = \frac{1}{2\pi}\cdot\frac{2}{a}\arctan 1 = \frac{1}{2\pi}\cdot\frac{2}{a}\cdot\frac{\pi}{4} = \frac{1}{4a}$

Fraction: $\frac{1/(4a)}{1/(2a)} = 50$ %. To contain 90 % one would need $|\omega| \lt a\tan(0.45\pi) = 6.3\,a$ : the slow tails of the Lorentzian make the bandwidth of a signal with a jump wide.

Energy spectral density of the one-sided exponential — The curve is the squared magnitude of the transform; the shaded area, between minus a and a, is half of the total area.

Exercise 11. Aliasing

A signal $\cos(2\pi\cdot 3000\,t)$ is sampled at 5 kHz. What frequency does the processor “see”? And if it were sampled at 8 kHz?

Result: 2 kHz with 5 kHz sampling; 3 kHz (correct) with 8 kHz.

Solution

With $f_c = 5$ kHz the Nyquist limit is 2.5 kHz, and 3 kHz exceeds it. The samples are $\cos(2\pi\cdot 3000\cdot n/5000) = \cos(1.2\pi n)$ . Since $\cos(1.2\pi n) = \cos(2\pi n - 0.8\pi n) = \cos(0.8\pi n)$ , they are identical to those of a cosine at $0.8\pi\cdot 5000/(2\pi) = 2000$ Hz.

In general the apparent frequency is the distance from the nearest multiple of $f_c$ : $|3 - 5| = 2$ kHz. The processor has no way of telling them apart: the information is lost at the moment of sampling.

At 8 kHz the limit is 4 kHz and the 3 kHz signal is represented correctly.

Aliasing of a 3 kHz cosine sampled at 5 kHz — The blue curve is the 3 kHz cosine, the dashed red one a 2 kHz cosine: the black dots, the samples taken every 0.2 ms, lie on both.

Exercise 12. The uncertainty principle for the Gaussian

For $f(t) = e^{-t^2/(2\sigma^2)}$ , compute the duration $\Delta t$ and the bandwidth $\Delta\omega$ defined as in the in-depth study, and check that the product is $\frac{1}{2}$ .

Result: $\Delta t = \frac{\sigma}{\sqrt 2}$ , $\Delta\omega = \frac{1}{\sigma\sqrt 2}$ , product $\frac{1}{2}$ .

Solution

$|f|^2 = e^{-t^2/\sigma^2}$ is a Gaussian with variance $\frac{\sigma^2}{2}$ (compare with $e^{-t^2/(2s^2)}$ : $s^2 = \sigma^2/2$ ). The weighted mean of $t^2$ is the variance: $\Delta t^2 = \frac{\sigma^2}{2}$ .

In frequency $|F|^2 \propto e^{-\sigma^2\omega^2}$ , a Gaussian with variance $\frac{1}{2\sigma^2}$ : $\Delta\omega^2 = \frac{1}{2\sigma^2}$ .

Product: $\Delta t\,\Delta\omega = \frac{\sigma}{\sqrt 2}\cdot\frac{1}{\sigma\sqrt 2} = \frac{1}{2}$ , the minimum possible. No other signal is as concentrated in time and frequency: this is why Gaussian pulses are used in radar and communications (GMSK modulation in GSM, Bluetooth).

Gaussians and their transforms — Narrowing the Gaussian in time widens it in frequency, and the product of the two widths stays constant.

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Basic level

Exercise 1. The two-sided exponential

Exercise 2. The one-sided exponential: magnitude and phase

Exercise 3. A delayed pulse

Exercise 4. The triangle as a convolution

Intermediate level

Exercise 5. Amplitude modulation

Exercise 6. Duality

Exercise 7. An integral with Parseval

Exercise 8. Differentiation in frequency

Exam level

Exercise 9. A pulse through an RC filter

Exercise 10. How much energy is in the band

Exercise 11. Aliasing

Exercise 12. The uncertainty principle for the Gaussian

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