Question 3 of 7: Area, Fourier Transform and Limiting Behaviour of a Raised-Cosine Pulse

Question 3: Area, Fourier Transform and Limiting Behaviour of a Raised-Cosine Pulse (20 marks: (a) 5, (b) 9, (c) 6)

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 single raised-cosine pulse of height $a$ supported on the symmetric interval $|x| \le \pi/(2a)$ and identically zero outside it, with $a \gt 0$ and the transform convention supplied by the question.

Find. The area under the pulse, its graph for the two requested values of $a$, a closed form for $F(\omega)$, and the limiting behaviour of both $f$ and $F$ as $a\to\infty$.

Approach. Substitute $u=ax$ for the area, which removes $a$ entirely; then use the power-reduction identity $\cos^{2}\theta = \tfrac12(1+\cos 2\theta)$ and a product-to-sum step to reduce the transform integral to elementary cosine integrals over the finite support; finally take $a\to\infty$ in the closed forms.

1. Area under the pulse (part a). Only the support contributes. Putting $u=ax$, $du = a\,dx$ maps the limits to $\mp\pi/2$ and cancels the factor $a$ altogether: $$\text{Area} = \int_{-\pi/(2a)}^{\pi/(2a)} a\cos^{2}(ax)\,dx = \int_{-\pi/2}^{\pi/2}\cos^{2}u\,du = \Big[\frac{u}{2}+\frac{\sin 2u}{4}\Big]_{-\pi/2}^{\pi/2} = \boxed{\;\frac{\pi}{2} \approx 1.5708\;}$$ The area is independent of $a$: the family is a one-parameter set of pulses of fixed strength. For the graphs, $f$ peaks at $f(0)=a$ and falls to zero exactly at the ends of the support, so $a=2$ gives a pulse of height $2$ spanning $|x| \le \pi/4$ and $a=4$ a pulse of height $4$ spanning $|x| \le \pi/8$, as plotted above.
2. Reduce the transform integrand (part b). Because $f$ is even, the imaginary part of the transform integral vanishes and $$F(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-L}^{L} a\cos^{2}(ax)\cos(\omega x)\,dx ,\qquad L = \frac{\pi}{2a} .$$ Applying $\cos^{2}(ax) = \tfrac12\big(1+\cos 2ax\big)$ splits this into a plain cosine integral and a product of two cosines.
3. First piece. Writing $s \equiv \sin\!\big(\pi\omega/(2a)\big)$ and using $\sin(\omega L) = s$, $$\frac{a}{2}\int_{-L}^{L}\cos(\omega x)\,dx = \frac{a}{2}\cdot\frac{2\sin(\omega L)}{\omega} = \frac{a\,s}{\omega} .$$
4. Second piece. The product-to-sum identity $\cos(2ax)\cos(\omega x) = \tfrac12\big[\cos((2a+\omega)x) + \cos((2a-\omega)x)\big]$ gives two more elementary integrals. Since $(2a\pm\omega)L = \pi \pm \pi\omega/(2a)$ and $\sin(\pi\pm\theta) = \mp\sin\theta$, they evaluate to $-s$ and $+s$ respectively, so $$\frac{a}{2}\int_{-L}^{L}\cos(2ax)\cos(\omega x)\,dx = \frac{a\,s}{2}\left(\frac{1}{2a-\omega}-\frac{1}{2a+\omega}\right) = \frac{a\,s\,\omega}{4a^{2}-\omega^{2}} .$$
5. Combine over a common denominator. Adding the two pieces, the $\omega^{2}$ terms cancel in the numerator: $$\frac{a s}{\omega} + \frac{a s \omega}{4a^{2}-\omega^{2}} = a s \cdot \frac{4a^{2}-\omega^{2}+\omega^{2}}{\omega\,(4a^{2}-\omega^{2})} = \frac{4a^{3}s}{\omega\,(4a^{2}-\omega^{2})} ,$$ and therefore $$\boxed{\;F(\omega) = \frac{1}{\sqrt{2\pi}}\cdot\frac{4a^{3}\,\sin\!\big(\pi\omega/(2a)\big)}{\omega\,\big(4a^{2}-\omega^{2}\big)}\;}$$ The transform is real and even, as it must be for a real even pulse.
6. Check the two apparent singularities. At $\omega=0$ the expansion $\sin\theta \approx \theta$ gives $F(0)=\dfrac{1}{\sqrt{2\pi}}\cdot\dfrac{\pi}{2} = 0.6267$, which agrees with $F(0)=\dfrac{1}{\sqrt{2\pi}}\times(\text{area}) = \dfrac{\pi/2}{\sqrt{2\pi}}$ from part (a). At $\omega=\pm 2a$ the numerator vanishes with the denominator and the limit is $\dfrac{1}{\sqrt{2\pi}}\cdot\dfrac{\pi}{4} = 0.3133$. Both singularities are removable, so $F$ is smooth everywhere.
7. Limit as $a\to\infty$ (part c). The pulse keeps its area $\pi/2$ while its height $a$ grows without bound and its width $\pi/a$ shrinks to zero, so $f$ concentrates at the origin and tends, in the distributional sense, to the scaled impulse $\tfrac{\pi}{2}\delta(x)$. On the transform side, for any fixed $\omega$ we have $\pi\omega/(2a) \to 0$, hence $\sin\!\big(\pi\omega/(2a)\big) \approx \pi\omega/(2a)$ and $$F(\omega) \;\longrightarrow\; \frac{1}{\sqrt{2\pi}}\cdot\frac{4a^{3}\cdot \pi\omega/(2a)}{\omega\cdot 4a^{2}} = \frac{1}{\sqrt{2\pi}}\cdot\frac{\pi}{2} = 0.6267 ,$$ a constant. The spectrum flattens out: the first zero of $F$, at $\omega = \pm 4a$, runs off to infinity, so the pulse acquires unlimited bandwidth. This is the reciprocal spreading (time–bandwidth) property of the Fourier transform — narrowing a signal in $x$ necessarily broadens it in $\omega$ — and it is consistent with the impulse limit, whose transform is the constant $\tfrac{\pi}{2}/\sqrt{2\pi}$.