Question 3 of 7: A Cosine Pulse and Its Fourier Transform
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2013 — 04-BS-5 Advanced Mathematics. Three-hour, closed-book exam (one double-sided 8.5"×11" aid sheet permitted; approved Casio/Sharp calculator allowed). Format: seven questions of equal value (20 marks each); any five constitute a complete paper and only the first five appearing in the answer book are marked. All seven are solved below for completeness.
Reference texts: Kreyszig, Advanced Engineering Mathematics (10th ed., Wiley) — power-series solutions of ODEs (Ch. 5), Fourier series and the Fourier transform (Ch. 11), least-squares curve fitting, Lagrange interpolation, Romberg integration and root-finding (Ch. 19); Strang, Introduction to Linear Algebra (6th ed., Wellesley-Cambridge) — the Cayley–Hamilton theorem and matrix inversion.
Question 3: A Cosine Pulse and Its Fourier Transform (a) 5, (b) 9, (c) 6 marks
Given. The finite-support cosine pulse $f(x)=2a\cos(ax)$ on $\left(-\tfrac{\pi}{2a},\tfrac{\pi}{2a}\right)$, zero elsewhere, with $a\gt 0$ a parameter.
Find. (a) The area under $f$; graphs for $a=1.0,2.0$. (b) $F(\omega)$. (c) Graphs of $F(\omega)$; the limiting behaviour of $f$ and $F$ as $a\to\infty$.
Approach. (a) Direct definite integral. (b) $f$ is even, so $F(\omega)$ reduces to a real cosine integral; use the product-to-sum identity to evaluate it in closed form. (c) Examine the closed-form $F(\omega)$ as $a\to\infty$.
(a) Compute the area.
$$\text{Area}=\int_{-\pi/2a}^{\pi/2a}2a\cos(ax)\,dx=\Big[2\sin(ax)\Big]_{-\pi/2a}^{\pi/2a}=2\Big(\sin\tfrac\pi2-\sin(-\tfrac\pi2)\Big)=2(1-(-1))$$
$$\boxed{\text{Area}=4}\quad\text{for every }a\gt 0\text{ (independent of }a\text{).}$$
$f(x)=2a\cos(ax)$ for $a=1.0$ and $a=2.0$: same area (4), taller and narrower as $a$ grows.
(b) Use evenness to drop the sine part. Since $f$ is even, $\int f(x)\sin(\omega x)\,dx=0$, so
$$F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\pi/2a}^{\pi/2a}2a\cos(ax)\cos(\omega x)\,dx.$$
(b) Product-to-sum, then integrate. With $\cos(ax)\cos(\omega x)=\tfrac12[\cos((a-\omega)x)+\cos((a+\omega)x)]$ and $L=\pi/2a$: both $(a\mp\omega)L=\tfrac\pi2\mp\tfrac{\omega\pi}{2a}$ have the SAME sine, $\cos\!\big(\tfrac{\omega\pi}{2a}\big)$ (since $\sin(\tfrac\pi2\pm\theta)=\cos\theta$), so
$$\int_{-L}^{L}\cos(ax)\cos(\omega x)\,dx=\cos\!\Big(\dfrac{\omega\pi}{2a}\Big)\left[\dfrac1{a-\omega}+\dfrac1{a+\omega}\right]=\cos\!\Big(\dfrac{\omega\pi}{2a}\Big)\dfrac{2a}{a^2-\omega^2}$$
(b) Assemble $F(\omega)$.
$$F(\omega)=\dfrac{2a}{\sqrt{2\pi}}\cdot\cos\!\Big(\dfrac{\omega\pi}{2a}\Big)\dfrac{2a}{a^2-\omega^2}=\boxed{\dfrac{4a^2}{\sqrt{2\pi}}\cdot\dfrac{\cos\!\big(\omega\pi/2a\big)}{a^2-\omega^2}}$$
(the apparent singularity at $\omega=\pm a$ is removable — numerator and denominator vanish together, and l'Hopital's rule gives a finite value there.)
$F(\omega)$ for $a=1.0$ and $a=2.0$: wider, lower-amplitude lobes for the narrower/taller time-domain pulse.
(c) Take the limit $a\to\infty$ at fixed $\omega$. $\cos(\omega\pi/2a)\to\cos0=1$ and $4a^2/(a^2-\omega^2)\to4$, so
$$\boxed{F(\omega)\ \longrightarrow\ \dfrac{4}{\sqrt{2\pi}}\approx1.5958}\quad\text{(a constant, independent of }\omega\text{).}$$
Meanwhile the pulse grows without bound in height ($2a\to\infty$) and shrinks in width ($\pi/a\to0$) while its area stays fixed at 4 — i.e. $f(x)\to4\delta(x)$ — and correspondingly $F(\omega)$ flattens into the constant $4/\sqrt{2\pi}$, the transform of an impulse of strength 4. This is a concrete instance of the time–bandwidth trade-off: a signal concentrated in time has a spectrum spread out in frequency, and vice versa.
