Fourier Transform of a Two-Sided Exponential (Laplace) Pulse: 04-BS-5, December 2019, Q3

Question 3: Fourier Transform of a Two-Sided Exponential (Laplace) Pulse (a: 5 marks; b: 9 marks; c: 3 marks; d: 3 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. The two-sided exponential (Laplace-type) pulse $f(x)=\dfrac{K}{2a}e^{-|x|/a}$ (the two branches combine into this single even expression), $K,a>0$.

Find. (a) $\int_{-\infty}^{\infty}f(x)\,dx$ and a graph of $f$. (b) $F(\omega)$. (c) A graph of $F(\omega)$. (d) The limiting behaviour as $a\to0$.

Approach. Recognize $f$ is even, so split every integral at $x=0$ and double the $x>0$ half; evaluate the area and the Fourier transform as standard exponential integrals, then read off the $a\to0$ limit.

(a) Area under $f$. By symmetry, $$\int_{-\infty}^{\infty}f(x)\,dx=2\int_0^{\infty}\frac{K}{2a}e^{-x/a}\,dx=\frac{K}{a}\Big[-a\,e^{-x/a}\Big]_0^{\infty}=\frac{K}{a}\cdot a=\boxed{K}$$ Remarkably the area is independent of $a$ — for $K=20$ the area is $\mathbf{20}$ regardless of whether $a=2$ or $a=1$. The two curves (peak height $K/2a=5$ for $a=2$, $10$ for $a=1$) enclose the same total area but trade height for width.

Fig. 1 — $f(x)=\frac{K}{2a}e^{-|x|/a}$ for $K=20$: $a=2$ (blue, peak $5$) and $a=1$ (red, peak $10$). Both curves enclose the same area, $K=20$.

(b) Fourier transform. $f$ is even, so $\int_{-\infty}^{\infty}f(x)e^{-i\omega x}dx=2\int_0^{\infty}f(x)\cos(\omega x)\,dx$ (the odd $\sin$ part integrates to zero): $$F(\omega)=\frac{1}{\sqrt{2\pi}}\cdot\frac{K}{a}\int_0^{\infty}e^{-x/a}\cos(\omega x)\,dx$$ The standard Laplace-transform integral $\int_0^{\infty}e^{-x/a}\cos(\omega x)\,dx=\dfrac{1/a}{(1/a)^2+\omega^2}=\dfrac{a}{1+a^2\omega^2}$ gives $$\boxed{F(\omega)=\dfrac{K}{\sqrt{2\pi}\,\big(1+a^2\omega^2\big)}}$$ a Lorentzian (Cauchy-shaped) spectrum, peaking at $F(0)=K/\sqrt{2\pi}$ for every $a$.
(c) Graph of $F(\omega)$. For $K=20$: $F(0)=20/\sqrt{2\pi}\approx7.978846$ regardless of $a$; the $a=2$ curve (narrower time pulse… here the wider time pulse) falls off as $1/(1+4\omega^2)$ (narrower in $\omega$) while $a=1$ falls off as $1/(1+\omega^2)$ (wider in $\omega$) — the classic inverse relationship between time-domain and frequency-domain widths.

Fig. 2 — $F(\omega)=\dfrac{K}{\sqrt{2\pi}(1+a^2\omega^2)}$ for $K=20$: $a=2$ (blue, narrower) and $a=1$ (red, wider). Both peak at $F(0)=20/\sqrt{2\pi}\approx7.978846$.

(d) Limit $a\to0$. As $a\to0$, the peak height $K/2a\to\infty$ while the decay length $a\to0$, but the area stays fixed at $K$ for every $a$ (Step 1) — this is exactly the defining property of a Dirac delta impulse, so $$\boxed{f(x)\ \longrightarrow\ K\,\delta(x)\qquad\text{as } a\to0}$$ In the frequency domain, $$\lim_{a\to0}F(\omega)=\lim_{a\to0}\frac{K}{\sqrt{2\pi}(1+a^2\omega^2)}=\boxed{\frac{K}{\sqrt{2\pi}}}\ \text{(a constant, independent of }\omega\text{)}$$ A perfectly flat (all-frequencies-equal) spectrum — the well-known result that an impulse in time contains equal energy at every frequency.

Quantity	Result
Area under $f(x)$	$K=20$ (independent of $a$)
Peak $f(0)$	$K/2a$: $5$ ($a=2$), $10$ ($a=1$)
$F(\omega)$	$\dfrac{K}{\sqrt{2\pi}(1+a^2\omega^2)}$
$F(0)$	$20/\sqrt{2\pi}\approx7.978846$ (both $a$)
$a\to0$ limit	$f(x)\to K\delta(x)$; $F(\omega)\to K/\sqrt{2\pi}$ (flat)

Topic: Fourier transform of the two-sided (Laplace/double) exponential pulse; the delta-function limit.

Key relations: $\int_0^\infty e^{-x/a}\cos(\omega x)dx=\frac{a}{1+a^2\omega^2}$; even function $\Rightarrow$ real, even transform; $\int f\,dx$ (area) $=\sqrt{2\pi}\,F(0)$.

Why this works: splitting the even integrand at $x=0$ reduces both the area and the transform to the same elementary exponential integral; the Lorentzian shape of $F(\omega)$ is the hallmark transform pair of an exponential decay, and its width $1/a$ is inversely proportional to the time-domain decay length $a$ — a concrete instance of the time–bandwidth (uncertainty) trade-off.

Common pitfall: forgetting the $x\lt0$ branch and only integrating $x>0$ (halving the area); or assuming $F(\omega)$'s peak height changes with $a$ when in fact $F(0)=K/\sqrt{2\pi}$ is independent of $a$ — only the width changes.

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

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

Question 3: Fourier Transform of a Two-Sided Exponential (Laplace) Pulse (a: 5 marks; b: 9 marks; c: 3 marks; d: 3 marks)