Question 3 of 7: Fourier Transform of a Triangular Pulse

Question 3: Fourier Transform of a Triangular Pulse (a: 4 marks; b: 10 marks; c: 6 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 half-width $4\tau$ and peak height $1/\tau$: $f(x)=\frac1\tau\big(1+\frac{x}{4\tau}\big)$ on $[-4\tau,0)$, $f(x)=\frac1\tau\big(1-\frac{x}{4\tau}\big)$ on $[0,4\tau]$, zero elsewhere.

Find. (a) Area under $f$, with graphs for $\tau=0.5,0.25$. (b) Closed form of $F(\omega)$. (c) Graphs of $F(\omega)$ for the same $\tau$, and the limiting behaviour as $\tau\to0$.

Approach. The area is a triangle, computed directly. $f$ is even, so $F(\omega)$ is real: $F(\omega)=\frac{2}{\sqrt{2\pi}}\int_0^{4\tau}f(x)\cos(\omega x)\,dx$, evaluated by parts.

1. (a) Area under $f$. $f$ is a triangle of base $8\tau$ and height $1/\tau$: $$\text{Area}=\frac12\cdot(8\tau)\cdot\frac1\tau=\boxed{4}$$ independent of $\tau$ — the pulse gets narrower and taller as $\tau\downarrow0$, but it always encloses the same area. See Fig. 1 for the graphs at $\tau=0.5$ (base $\pm2$, peak $2$) and $\tau=0.25$ (base $\pm1$, peak $4$).
2. (b) Set up the transform integral. Since $f(-x)=f(x)$ (even), the transform reduces to a cosine integral: $$F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-4\tau}^{4\tau}f(x)e^{-i\omega x}dx=\frac{2}{\sqrt{2\pi}}\int_0^{4\tau}\frac1\tau\Big(1-\frac{x}{4\tau}\Big)\cos(\omega x)\,dx$$
3. Evaluate by parts. Using $\int_0^{a}(1-x/a)\cos(\omega x)dx=\dfrac{1-\cos(\omega a)}{a\omega^2}=\dfrac{2\sin^2(\omega a/2)}{a\omega^2}$ with $a=4\tau$: $$F(\omega)=\frac{2}{\sqrt{2\pi}}\cdot\frac1\tau\cdot\frac{2\sin^2(2\omega\tau)}{4\tau\,\omega^2}=\boxed{\dfrac{1}{\sqrt{2\pi}}\cdot\dfrac{\sin^2(2\omega\tau)}{\omega^2\tau^2}}$$ At $\omega=0$ (L'Hopital or the small-angle limit $\sin^2(2\omega\tau)\approx4\omega^2\tau^2$) this is continuous with value $F(0)=4/\sqrt{2\pi}=\text{Area}/\sqrt{2\pi}$, as it must be.
4. (c) Graph $F(\omega)$ and describe the $\tau\to0$ limit. See Fig. 2: both curves peak at $F(0)=4/\sqrt{2\pi}\approx1.596$ (independent of $\tau$) and oscillate in a $\sin^2/\omega^2$ envelope whose first zero is at $\omega=\pi/(2\tau)$ — smaller $\tau$ pushes that zero (and the whole envelope) further out, i.e. the transform spreads out and flattens as $\tau$ shrinks. As $\tau\to0$: the pulse $f(x)$ becomes an infinitely narrow, infinitely tall spike that still encloses area $4$ — i.e. $f(x)\to4\,\delta(x)$, a scaled Dirac delta. Correspondingly $\sin^2(2\omega\tau)/(\omega^2\tau^2)\to4$ for every fixed $\omega$ (Taylor-expand the sine), so $$F(\omega)\to\dfrac{4}{\sqrt{2\pi}}\quad\text{(a flat constant, for all }\omega\text{)}$$ which is exactly the Fourier transform of an impulse of strength $4$ — consistent with the uncertainty-principle trade-off: a signal concentrated in $x$ has a transform spread arbitrarily wide in $\omega$.