Question 3 of 7: Two-Sided Exponential and Its Fourier Transform

Question 3: Two-Sided Exponential and Its Fourier Transform (a) 5, (b) 9, (c) 3, (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. $f(x)=\dfrac{1}{4a}e^{-|x|/a}$ (the two branches combine into a single two-sided decaying exponential), $a\gt0$.

Find. (a) $\int_{-\infty}^{\infty}f(x)\,dx$ and its graph for $a=0.5,2.0$. (b) $F(\omega)$. (c) The graph of $F(\omega)$ for the same two $a$. (d) The limiting behaviour as $a\to\infty$.

Approach. Use the symmetry $f(-x)=f(x)$ to fold every integral onto $[0,\infty)$, then apply the standard Laplace-type integral $\int_0^\infty e^{-bx}\cos(\omega x)\,dx=\dfrac{b}{b^2+\omega^2}$ with $b=1/a$.

1. (a) Area under $f(x)$. Since $f(x)\ge0$ everywhere and is even, $$\int_{-\infty}^{\infty}f(x)\,dx=2\int_0^\infty\dfrac{1}{4a}e^{-x/a}\,dx=2\cdot\dfrac{1}{4a}\cdot a=\boxed{\dfrac12}$$ The area is exactly $1/2$ for every $a\gt0$ — changing $a$ trades peak height ($f(0)=1/4a$) for spread but leaves the total area fixed.
2. (a) Graph of $f(x)$. The figure below plots $f(x)$ for $a=0.5$ (narrow, tall peak, height $f(0)=0.5$) and $a=2.0$ (wide, low peak, height $f(0)=0.125$) — both curves enclose the same area $1/2$.
   $f(x)$ for $a=0.5$ and $a=2.0$; both enclose area $1/2$.
3. (b) Fourier transform. Because $f$ is even, the $\sin(\omega x)$ part of $e^{-i\omega x}=\cos(\omega x)-i\sin(\omega x)$ integrates to zero and only the cosine part survives: $$F(\omega)=\dfrac{2}{\sqrt{2\pi}}\int_0^\infty\dfrac{1}{4a}e^{-x/a}\cos(\omega x)\,dx=\dfrac{1}{2a\sqrt{2\pi}}\int_0^\infty e^{-x/a}\cos(\omega x)\,dx$$ With $b=1/a$, $\int_0^\infty e^{-bx}\cos(\omega x)\,dx=\dfrac{b}{b^2+\omega^2}=\dfrac{1/a}{1/a^2+\omega^2}=\dfrac{a}{1+a^2\omega^2}$, so $$F(\omega)=\dfrac{1}{2a\sqrt{2\pi}}\cdot\dfrac{a}{1+a^2\omega^2}=\boxed{\dfrac{1}{2\sqrt{2\pi}\,(1+a^2\omega^2)}}$$ $F(\omega)$ is real (as expected for an even, real $f$) and is a Lorentzian centred at $\omega=0$ with peak value $F(0)=1/(2\sqrt{2\pi})\approx0.1995$, independent of $a$.
4. (c) Graph of $F(\omega)$. For $a=0.5$ (narrow $f$) the transform is the wider Lorentzian; for $a=2.0$ (wide $f$) the transform is the narrower, more sharply peaked one — both share the same peak height $F(0)\approx0.1995$.
   $F(\omega)$ for $a=0.5$ and $a=2.0$; same peak height, reciprocal width.
   This is the time–frequency reciprocity (uncertainty) trade-off: the more spread out $f(x)$ is in $x$, the more concentrated $F(\omega)$ is in $\omega$, and vice versa.
5. (d) Limit $a\to\infty$. Pointwise, $f(x)=\dfrac{1}{4a}e^{-|x|/a}\to0$ for every fixed $x$ as $a\to\infty$ (the prefactor $1/(4a)\to0$), even though the area stays fixed at $1/2$ — the curve flattens out over an ever-widening range. Correspondingly $F(\omega)=\dfrac{1}{2\sqrt{2\pi}(1+a^2\omega^2)}\to0$ for every fixed $\omega\ne0$, but $F(0)=1/(2\sqrt{2\pi})$ stays exactly fixed for every $a$. So $F(\omega)$ collapses toward a narrow spike of fixed height $1/(2\sqrt{2\pi})$ concentrated at $\omega=0$ (width $\sim1/a\to0$) — the frequency-domain image of a time-domain function that spreads out and flattens without changing its total area.