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

Question 3: Fourier Transform of a Two-Sided Exponential (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. The odd, two-sided decaying exponential $f(x)$ with rate parameter $a\gt0$: amplitude $2/a$ at $x=0^{\pm}$, decaying/growing with time constant $a$.

Find. (a) Total area between $f$ and the $x$-axis, with sketches for $a=1,0.5$. (b) $F(\omega)$ in closed form. (c) Sketches of $F(\omega)$ for $a=1,0.5$. (d) The limiting behaviour as $a\to0$.

Approach. Confirm $f$ is odd; integrate $|f|$ over each half-line for the area; split the Fourier integral at $x=0$ into two convergent one-sided exponential integrals for $F(\omega)$; then use the closed form to describe the $a\to0$ limit.

1. (a) Total (unsigned) area. By symmetry $f(-x)=-f(x)$ (odd), so the signed integral is zero, but the area bounded by the curve and the axis is the unsigned area: $$A=\int_0^{\infty}\dfrac{2}{a}e^{-x/a}\,dx+\int_{-\infty}^{0}\dfrac{2}{a}e^{x/a}\,dx=2+2=\boxed{4}$$ independent of $a$ — a smaller $a$ makes the spike taller and narrower but the enclosed area is conserved.
2. (b) Split the transform at $x=0$. $$F(\omega)=\dfrac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^{0}\!\!\left(-\dfrac2a e^{x/a}\right)e^{-i\omega x}dx+\int_{0}^{\infty}\!\dfrac2a e^{-x/a}e^{-i\omega x}dx\right]$$ Each piece is a convergent one-sided exponential integral: $\displaystyle\int_0^\infty\dfrac2a e^{-x(1/a+i\omega)}dx=\dfrac{2}{1+i\omega a}$ and $\displaystyle-\int_{-\infty}^0\dfrac2a e^{x(1/a-i\omega)}dx=-\dfrac{2}{1-i\omega a}$.
3. Combine over a common denominator. $$\dfrac{2}{1+i\omega a}-\dfrac{2}{1-i\omega a}=\dfrac{2(1-i\omega a)-2(1+i\omega a)}{1+\omega^2a^2}=\dfrac{-4i\omega a}{1+\omega^2a^2}$$
4. Box the transform. $$\boxed{F(\omega)=\dfrac{-4ia\omega}{\sqrt{2\pi}\,\big(1+a^2\omega^2\big)}}$$ $F(\omega)$ is purely imaginary and odd in $\omega$ (as expected: the Fourier transform of a real odd function is purely imaginary and odd).
5. (c) Magnitude and sketch. $|F(\omega)|=\dfrac{4a|\omega|}{\sqrt{2\pi}(1+a^2\omega^2)}$; the sketches below plot $\mathrm{Im}\{F(\omega)\}$ (the non-zero part) for $a=1$ and $a=0.5$ — both are odd, single-humped curves through the origin.
6. (d) Locate the peak of $|F(\omega)|$. Differentiating $|F|$ with respect to $\omega$ and setting the result to zero gives $\omega_{\text{peak}}=1/a$, at which $$|F(1/a)|=\dfrac{4a(1/a)}{\sqrt{2\pi}\,(1+1)}=\dfrac{2}{\sqrt{2\pi}}\approx0.797885$$ — a height that is the same for every $a$. As $a\to0$: for any fixed $\omega$, $F(\omega)=-4ia\omega/[\sqrt{2\pi}(1+a^2\omega^2)]\to0$, but the peak location $\omega_{\text{peak}}=1/a\to\infty$ while the peak height stays fixed at $2/\sqrt{2\pi}$. So in the time domain $f(x)$ sharpens into an increasingly tall, narrow odd pair of spikes at the origin (a "doublet") while conserving the total area found in (a); correspondingly its transform does not vanish but spreads its energy over an ever wider band of frequencies, exactly the time–frequency trade-off (uncertainty principle) expected when a signal concentrates in $x$.