Question 3 of 7: Windowed-Cosine Pulse — Area and Fourier Transform

Question 3: Windowed-Cosine Pulse — Area and Fourier Transform (a: 5, b: 9, 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 finite cosine pulse $f(x)=\tfrac{Ma}{2}\cos(ax)$ windowed to $|x|\le\pi/2a$ (one half-cycle wide), zero elsewhere; $M,a\gt0$.

Find. (a) the area under $f$ for $(M,a)=(10,0.5)$ and $(10,1)$; (b) $F(\omega)$ in closed form; (c) the behaviour of $f$ and $F$ as $a\to\infty$.

Approach. Integrate the cosine window directly for the area; for $F(\omega)$ use evenness to reduce the transform to a real cosine integral, apply a product-to-sum identity, and evaluate at the window edges $x=\pm\pi/2a$.

1. Area under $f(x)$. $$\text{Area}=\int_{-\pi/2a}^{\pi/2a}\frac{Ma}{2}\cos(ax)\,dx=\frac{Ma}{2}\cdot\frac{2}{a}\sin\!\left(\frac{\pi}{2} \right)=\boxed{M}$$ — independent of $a$. For $M=10$, both $a=0.5$ and $a=1$ give Area $=10$ (the pulse gets taller and narrower as $a$ grows, but always encloses the same area).
2. Set up the transform. $f$ is even, so $F(\omega)=\frac{1}{\sqrt{2\pi}}\cdot\frac{Ma}{2} \int_{-L}^{L}\cos(ax)\cos(\omega x)\,dx$ with $L=\pi/2a$ (the $\sin$ part of $e^{-i\omega x}$ integrates to zero by odd symmetry). Using $\cos A\cos B=\tfrac12[\cos(A-B)+\cos(A+B)]$, $$\int_{-L}^{L}\cos(ax)\cos(\omega x)\,dx=\frac{\sin[(a-\omega)L]}{a-\omega}+\frac{\sin[(a+\omega)L]}{a+\omega}.$$
3. Evaluate at $L=\pi/2a$. $(a\mp\omega)L=\tfrac{\pi}{2}\mp\tfrac{\pi\omega}{2a}$, and $\sin(\tfrac\pi2\mp\theta)=\cos\theta$ for both signs, so both sine terms collapse to the SAME factor $\cos(\pi\omega/2a)$: $$\int_{-L}^{L}\cos(ax)\cos(\omega x)\,dx=\cos\!\left(\frac{\pi\omega}{2a}\right)\left[\frac{1}{a-\omega}+ \frac{1}{a+\omega}\right]=\frac{2a\cos(\pi\omega/2a)}{a^{2}-\omega^{2}}.$$ Multiplying by $\frac{1}{\sqrt{2\pi}}\cdot\frac{Ma}{2}$ gives the closed form $$\boxed{F(\omega)=\frac{Ma^{2}\cos(\pi\omega/2a)}{\sqrt{2\pi}\,(a^{2}-\omega^{2})}}$$ (verified against direct numerical integration of the defining transform at several $(M,a,\omega)$ triples).
4. Resolve the removable singularity at $\omega=\pm a$. The closed form is a $0/0$ form there; applying L’Hopital’s rule to numerator and denominator in $\omega$ gives the finite limit $$F(\pm a)=\boxed{\frac{M\sqrt{2\pi}}{8}}$$ — also independent of $a$ (e.g. $M=10\Rightarrow F(\pm a)\approx3.1333$ for both $a=0.5$ and $a=1$), and $F(\omega)$ is continuous through $\omega=\pm a$ once this limiting value is used.
5. Behaviour as $a\to\infty$. The window width $\pi/a\to0$ while the peak height $Ma/2\to \infty$, but Step 1 showed the area stays fixed at $M$ — so $f(x)\to M\,\delta(x)$, a Dirac impulse of weight $M$. Correspondingly, in the closed form, $\cos(\pi\omega/2a)\to1$ and $a^{2}-\omega^{2}\to a^{2}$ for every fixed $\omega$, so $$F(\omega)\ \longrightarrow\ \boxed{\frac{M}{\sqrt{2\pi}}}\quad\text{(a flat, $\omega$-independent spectrum).}$$ This matches $\mathcal F\{M\delta(x)\}=M/\sqrt{2\pi}$ exactly — the time-domain pulse collapsing to an impulse and the frequency-domain spectrum flattening out are the same limit, viewed two ways (time–frequency duality / uncertainty principle: an infinitely narrow pulse has infinitely broad, flat frequency content).