Fourier transform of a triangular pulse: 04-BS-5, May 2015, Q3

Question 3: Fourier transform of a triangular pulse (a) 5 marks; (b) 10 marks; (c) 5 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $f(x)$ is an isosceles-triangle ("tent") pulse of half-width $a$ and peak height $1/a$ at $x=0$, zero outside $[-a,a]$.

Find. (a) area under $f$; (b) $F(\omega)$; (c) graphs at $a=1.0,0.5$ and the limiting behaviour as $a\to0^+$.

Approach. The area is elementary geometry (a triangle); $f$ is even, so its Fourier transform reduces to a real cosine transform, computed by integrating by parts twice over $[0,a]$.

Fig. Q3(a) — the triangular pulse $f(x)$ for $a=1.0$ (peak $1.0$, base $\pm1.0$) and $a=0.5$ (peak $2.0$, base $\pm0.5$); both triangles enclose unit area.

Area (part a). $f$ is a triangle of base $2a$ and height $1/a$: $$\text{Area}=\dfrac12\cdot(2a)\cdot\dfrac1a=1\qquad\text{for every }a\gt0$$ (confirmed by direct integration). The area is independent of $a$ — the pulse is a normalized (unit-area) window.
Set up the transform using evenness (part b). $f(-x)=f(x)$, so the $\sin(\omega x)$ part of $e^{-i\omega x}$ integrates to zero and $$F(\omega)=\dfrac{2}{\sqrt{2\pi}}\int_0^{a}\dfrac1a\left(1-\dfrac xa\right)\cos(\omega x)\,dx$$
Integrate by parts twice. $$\int_0^a\cos(\omega x)\,dx=\dfrac{\sin(a\omega)}{\omega},\qquad \int_0^ax\cos(\omega x)\,dx=\dfrac{a\sin(a\omega)}{\omega}+\dfrac{\cos(a\omega)-1}{\omega^2}$$ Combining $\tfrac1a\int_0^a\cos(\omega x)dx-\tfrac1{a^2}\int_0^ax\cos(\omega x)dx$, the $\sin(a\omega)/(a\omega)$ terms cancel, leaving $$\boxed{F(\omega)=\sqrt{\dfrac2\pi}\,\dfrac{1-\cos(a\omega)}{a^2\omega^2}\quad(\omega\ne0),\qquad F(0)=\dfrac{1}{\sqrt{2\pi}}\approx0.3989}$$ (the $\omega=0$ value is the removable-singularity limit, which equals $\text{Area}/\sqrt{2\pi}$).
Behaviour as $a\to0^+$ (part c). The base $2a\to0$ while the height $1/a\to\infty$, with area held at exactly $1$ — $f(x)$ converges (in the distributional sense) to the Dirac delta $\delta(x)$. Correspondingly, for any fixed $\omega$, $1-\cos(a\omega)\approx\tfrac12(a\omega)^2$ as $a\to0$, so $$F(\omega)\to\sqrt{\dfrac2\pi}\cdot\dfrac{\tfrac12(a\omega)^2}{a^2\omega^2}=\sqrt{\dfrac2\pi}\cdot\dfrac12=\dfrac1{\sqrt{2\pi}}\quad\text{for every }\omega$$ i.e. the spectrum flattens to the constant $1/\sqrt{2\pi}$ — the familiar fact that the Fourier transform of an impulse has infinite bandwidth (a flat spectrum).

Fig. Q3(c) — $F(\omega)$ for $a=1.0$ and $a=0.5$: both are sinc²-shaped lobes through $F(0)=1/\sqrt{2\pi}\approx0.3989$; the smaller-$a$ (narrower, taller) pulse produces the wider spectral lobe, illustrating the time–bandwidth trade-off.

Final results — Question 3
Quantity	Value
Area under $f(x)$	$1$ (all $a$)
$F(\omega),\ \omega\ne0$	$\sqrt{2/\pi}\,(1-\cos a\omega)/(a^2\omega^2)$
$F(0)$	$1/\sqrt{2\pi}\approx0.3989$ (all $a$)
$a\to0^+$ limit	$f\to\delta(x)$; $F(\omega)\to1/\sqrt{2\pi}$ (flat)

Topic: Fourier transform of a triangular (Bartlett) window; the uncertainty/time–bandwidth principle.

Key relations: $F(\omega)=\tfrac1{\sqrt{2\pi}}\int f(x)e^{-i\omega x}dx$; for an even real $f$, $F(\omega)=\tfrac{2}{\sqrt{2\pi}}\int_0^\infty f(x)\cos(\omega x)\,dx$ is real and even; a triangle in $x$ transforms to a $\text{sinc}^2$-type lobe in $\omega$ (equivalently, a triangle is the self-convolution of a rectangle, whose transform is $\text{sinc}$, and convolution in $x$ maps to multiplication in $\omega$, giving $\text{sinc}^2$).

Why this works: Because the pulse narrows and grows tall together (area pinned at 1), no energy is created or destroyed — only redistributed across frequency; as $a\to0$ the time-domain spread collapses to a point while the frequency-domain spread grows without bound, the standard time–bandwidth trade-off (an idealized limit of the uncertainty principle).

Common pitfall: forgetting that $f$ is even and attempting the full complex exponential integral (extra unnecessary algebra); mis-simplifying the $\omega\to0$ limit as "undefined" instead of recognizing the removable singularity and evaluating it via the Taylor expansion of $1-\cos$.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., §11.7–11.9 (Fourier Transform).

Question 3 of 7: Fourier transform of a triangular pulse

Question 3: Fourier transform of a triangular pulse (a) 5 marks; (b) 10 marks; (c) 5 marks