Question 3 of 7: Fourier Transform of a Triangular Pulse

Question 3: Fourier Transform of a Triangular Pulse (20 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. A symmetric triangular pulse of peak height $1/a$ at $x=0$ and half-width $2a$ (zero outside $[-2a,2a]$), and the symmetric Fourier-transform convention above.

Find. The area under $f$; the closed-form $F(\omega)$; and the behaviour of both as $a\to0^+$.

Approach. The area is two equal right-triangles; the transform is a direct piecewise integral (the function is even, so only the cosine part survives); the $a\to0$ limit is read straight off the closed forms.

1. Area under the pulse. Each half is a triangle of base $2a$ and height $1/a$: $$\text{Area}=\int_{-2a}^{0}\frac1a\Bigl(1+\frac{x}{2a}\Bigr)dx+\int_{0}^{2a}\frac1a\Bigl(1-\frac{x}{2a}\Bigr)dx=\frac12(2a)\Bigl(\frac1a\Bigr)+\frac12(2a)\Bigl(\frac1a\Bigr)$$ $$\boxed{\text{Area}=2}\quad\text{for every }a\text{ (the pulse gets taller exactly as fast as it gets narrower)}$$
   Fig. 1 — the triangular pulse $f(x)$ for $a=1.0$ (peak $1$, half-width $2$) and $a=0.25$ (peak $4$, half-width $0.5$); both enclose the same area, $2$.
2. Fourier transform. $f$ is even, so $$F(\omega)=\frac{1}{\sqrt{2\pi}}\left[\int_{-2a}^{0}\frac1a\Bigl(1+\frac{x}{2a}\Bigr)e^{-i\omega x}dx+\int_{0}^{2a}\frac1a\Bigl(1-\frac{x}{2a}\Bigr)e^{-i\omega x}dx\right]$$ Carrying out the integration (the imaginary parts cancel by symmetry; the remaining real integral reduces to a standard $(1-\cos)/(\cdot)^2$ form) gives the closed form $$\boxed{F(\omega)=\sqrt{\frac{2}{\pi}}\left(\frac{\sin(a\omega)}{a\omega}\right)^{2}}$$ with the removable point at $\omega=0$ taken as its limit, $F(0)=\sqrt{2/\pi}\approx0.7979$ — consistent with $F(0)=\frac{1}{\sqrt{2\pi}}\int f\,dx=\frac{2}{\sqrt{2\pi}}=\sqrt{2/\pi}$ from Step 1. (Confirmed by symbolic integration of the two piecewise integrals and by direct numerical evaluation at several $(a,\omega)$ pairs.)
   Fig. 2 — $F(\omega)$ for $a=1.0$ (narrower main lobe) and $a=0.25$ (wider, lower-amplitude ripples spread further in $\omega$); both curves start at the same peak value $\sqrt{2/\pi}\approx0.798$.
3. Limit $a\to0^+$. The area stays fixed at $2$ while the base $4a\to0$ and the peak $1/a\to\infty$, so $f(x)\to2\,\delta(x)$, a unit-area-times-2 spike at the origin. Correspondingly, since $\sin(a\omega)/(a\omega)\to1$ for every fixed $\omega$ as $a\to0$, $$\boxed{F(\omega)\ \to\ \sqrt{2/\pi}\ \text{(a flat spectrum, independent of }\omega\text{)}}$$ This is the expected time–frequency trade-off: a pulse that becomes arbitrarily concentrated in $x$ has a transform that becomes arbitrarily spread out (flat) in $\omega$.