Question 3 of 7: Fourier Transform of a Finite Cosine Window

Question 3: Fourier Transform of a Finite Cosine Window (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 raised-cosine window $f(x)=\tfrac{a}{4}\cos(ax)$ on $|x|\lt\pi/(2a)$ and zero outside — nonnegative throughout its support since $ax\in(-\pi/2,\pi/2)$ there.

Find. (a) The area under $f$, with sketches for $a=4,8$. (b) $F(\omega)$ in closed form. (c) Sketches of $F(\omega)$ for $a=4,8$, and the limiting behaviour as $a\to\infty$.

Approach. Integrate $f$ directly over its finite support for the area; use evenness of $f$ to reduce the transform to a real cosine integral, apply a product-to-sum identity, integrate term by term, then examine the $a\to\infty$ limit of the closed form.

1. (a) Area under the curve. Since $f\ge0$ on its support, the area bounded by $f$ and the $x$-axis is simply the signed integral: $$A=\int_{-\pi/2a}^{\pi/2a}\dfrac{a}{4}\cos(ax)\,dx=\dfrac14\Big[\sin(ax)\Big]_{-\pi/2a}^{\pi/2a}=\dfrac14\big[\sin(\tfrac\pi2)-\sin(-\tfrac\pi2)\big]=\dfrac14(1-(-1))$$ $$\boxed{A=\dfrac12}\quad\text{independent of }a$$ A larger $a$ narrows the window (width $\pi/a$) but raises the peak (height $a/4$) so as to conserve the enclosed area — visible in the sketch below (Figure 3a): the $a=8$ curve is half as wide and twice as tall as the $a=4$ curve.
2. (b) Reduce the transform to a real cosine integral. Because $f$ is even and real, and the support is symmetric, the imaginary (sine) part of $e^{-i\omega x}=\cos(\omega x)-i\sin(\omega x)$ integrates to zero: $$F(\omega)=\dfrac{1}{\sqrt{2\pi}}\int_{-\pi/2a}^{\pi/2a}\dfrac{a}{4}\cos(ax)\cos(\omega x)\,dx$$ Apply the product-to-sum identity $\cos(ax)\cos(\omega x)=\tfrac12\big[\cos((a-\omega)x)+\cos((a+\omega)x)\big]$ and integrate each cosine term over the symmetric interval (valid for $\omega\ne\pm a$): $$\int_{-\pi/2a}^{\pi/2a}\cos(kx)\,dx=\dfrac{2}{k}\sin\!\Big(\dfrac{k\pi}{2a}\Big)$$
3. Combine the two pieces. With $k=a-\omega$ and $k=a+\omega$, and using $\sin\!\big(\tfrac{(a-\omega)\pi}{2a}\big)=\cos\!\big(\tfrac{\omega\pi}{2a}\big)=\sin\!\big(\tfrac{(a+\omega)\pi}{2a}\big)$ (a half-angle complementary-angle identity, since both differ from $\pi/2$ by $\pm\tfrac{\omega\pi}{2a}$), both pieces reduce to the same $\cos(\pi\omega/2a)$ factor. Assembling and simplifying gives the closed form $$\boxed{F(\omega)=\dfrac{a^2\cos\!\big(\tfrac{\pi\omega}{2a}\big)}{2\sqrt{2\pi}\,(a^2-\omega^2)}},\qquad \omega\ne\pm a$$ (at $\omega=\pm a$ the $0/0$ form resolves to a finite value by l'Hopital's rule, so $F$ is actually smooth there — there is no real singularity, only a removable one in this formula.) $F(\omega)$ is real, as expected for the transform of a real even function.
4. (c) Peak height and graph. At $\omega=0$: $F(0)=\dfrac{a^2\cos(0)}{2\sqrt{2\pi}\,a^2}=\dfrac{1}{2\sqrt{2\pi}}\approx0.19947$, independent of $a$ — both the $a=4$ and $a=8$ curves peak at the same height (Figure 3b), but the $a=8$ curve is visibly wider in $\omega$ (first zero near $\omega=8$) than the $a=4$ curve (first zero near $\omega=4$): narrowing the window in $x$ widens its transform in $\omega$, the Fourier uncertainty principle.
5. Limit as $a\to\infty$. The window width $\pi/a\to0$ while the peak height $a/4\to\infty$, but part (a) showed the enclosed area stays fixed at $\tfrac12$ for every $a$ — so $f(x)$ approaches an impulse of strength $\tfrac12$ at the origin, $f(x)\to\tfrac12\delta(x)$. Correspondingly, for any fixed $\omega$, $$\lim_{a\to\infty}F(\omega)=\lim_{a\to\infty}\dfrac{a^2\cos(\pi\omega/2a)}{2\sqrt{2\pi}(a^2-\omega^2)}=\dfrac{1}{2\sqrt{2\pi}}$$ $$\boxed{F(\omega)\ \to\ \dfrac{1}{2\sqrt{2\pi}}\approx0.19947\ \text{for every }\omega}$$ i.e. the spectrum flattens to a constant over all frequencies — exactly the transform of $\tfrac12\delta(x)$, and the expected limiting case of the uncertainty principle: a perfectly time-localized pulse has a perfectly flat (frequency-unlimited) spectrum.