Question 3 of 7: Fourier Transform of a Two-Sided Exponential Pulse

Question 3: Fourier Transform of a Two-Sided Exponential 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. $f(x)=\dfrac{K}{2a}e^{-|x|/a}$ (a symmetric two-sided exponential, i.e. a Laplace-shaped pulse), $K,a\gt 0$.

Find. (a) $\int_{-\infty}^{\infty}f\,dx$ and a plot of $f$ for $(K,a)=(20,2)$ and $(20,1)$; (b) closed-form $F(\omega)$; (c) a plot of $F(\omega)$ for the same $(K,a)$ pairs; (d) the limiting behaviour as $a\to 0$.

Approach. Split the area/transform integrals at $x=0$ using the even symmetry of $f$, integrate the exponential pieces directly, then read off the $a\to0$ limit from the closed forms.

1. (a) Area under $f(x)$. By symmetry, integrate each half and double is unnecessary since the two halves are already written separately: $$\int_{-\infty}^{0}\frac{K}{2a}e^{x/a}dx+\int_{0}^{\infty}\frac{K}{2a}e^{-x/a}dx = \frac{K}{2a}\cdot a + \frac{K}{2a}\cdot a = \frac{K}{2}+\frac{K}{2}.$$ $$\boxed{\text{Area} = K \quad(\text{independent of } a)}$$ For $K=20$: area $=20$. The peak height is $f(0)=K/2a$, so a smaller $a$ makes a taller, narrower pulse of the same enclosed area — visible in the graph below (a=1 peaks at $10$, a=2 peaks at $5$, both areas $20$).
2. (b) Fourier transform. Since $f$ is even, $F(\omega)=\dfrac{2}{\sqrt{2\pi}}\displaystyle\int_0^\infty f(x)\cos(\omega x)\,dx$ (the sine part cancels). With $f(x)=\frac{K}{2a}e^{-x/a}$ on $x\gt 0$: $$F(\omega) = \frac{K}{a\sqrt{2\pi}}\int_0^{\infty}e^{-x/a}\cos(\omega x)\,dx = \frac{K}{a\sqrt{2\pi}}\cdot\frac{1/a}{(1/a)^{2}+\omega^{2}} = \frac{K}{a\sqrt{2\pi}}\cdot\frac{a}{1+a^{2}\omega^{2}}.$$ $$\boxed{F(\omega) = \frac{K}{\sqrt{2\pi}\,\big(1+a^{2}\omega^{2}\big)}}$$ This is a Lorentzian in $\omega$ — the classical Fourier-transform pair of a two-sided exponential.
3. (c) Behaviour of $F(\omega)$. $F(0)=K/\sqrt{2\pi}$ is the same for every $a$ (both curves below start at the same height), but $F$ falls off as $1/\omega^2$ faster for larger $a$ (a wider time-domain pulse has a narrower frequency spectrum) — the usual time–bandwidth trade-off. For $K=20$: $F(0)\approx 7.979$ for both $a=2$ and $a=1$.
4. (d) Limit $a\to 0^{+}$. From (a) the area stays fixed at $K$ for every $a$, while the peak height $K/2a\to\infty$ and the pulse width $\sim a\to 0$: $f(x)$ collapses to an impulse of total weight $K$, i.e. $f(x)\to K\,\delta(x)$. Consistently, in (b) as $a\to0$, $a^2\omega^2\to0$ for every fixed $\omega$, so $$F(\omega) \to \frac{K}{\sqrt{2\pi}} \quad\text{(a flat, $\omega$-independent spectrum).}$$ A flat spectrum is exactly the Fourier transform of a scaled Dirac delta ($\mathcal F\{K\delta(x)\}=K/\sqrt{2\pi}$, constant in $\omega$) — the two limits are mutually consistent.