Fourier Transform of a Raised-Cosine Pulse: 04-BS-5, May 2017, Q3

Question 3: Fourier Transform 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 lobe $f(x)=a\cos^2(ax)$ supported on $|x|\le\pi/(2a)$ and zero elsewhere, with $a\gt0$. The pulse height is $a$ (at $x=0$) and its total width is $\pi/a$, so height and width trade off inversely as $a$ varies.

Find. (a) the enclosed area and the graphs for $a=1,2$; (b) the Fourier transform $F(\omega)$ in the stated normalisation; (c) the limiting behaviour of both $f$ and $F$ as $a\to\infty$.

Part (a): the pulse for $a=1$ (solid) and $a=2$ (dashed). Doubling $a$ doubles the height and halves the width, leaving the shaded area unchanged at $\pi/2$.

Approach. Use the power-reduction identity $\cos^2\theta=\frac12(1+\cos2\theta)$ to turn both the area integral and the transform integral into elementary cosine integrals, exploiting the evenness of $f$ to discard the imaginary part of $e^{-i\omega x}$.

Part (a) — area under the pulse. Substituting $u=ax$ (so $du=a\,dx$) maps the support onto $|u|\le\pi/2$ and cancels the height factor $a$ entirely: $$\text{Area}=\int_{-\pi/2a}^{\pi/2a}a\cos^2(ax)\,dx=\int_{-\pi/2}^{\pi/2}\cos^2u\,du=\Big[\dfrac{u}{2}+\dfrac{\sin 2u}{4}\Big]_{-\pi/2}^{\pi/2}=\boxed{\dfrac{\pi}{2}}$$ The area is independent of $a$ — the parameter merely reshapes the pulse, it does not change how much it encloses. This invariance is what makes part (c) interesting, and it is drawn to scale in the figure above.
Part (b) — reduce the transform integral. Because $f$ is even, $\int f(x)\sin\omega x\,dx=0$ and only the cosine part of $e^{-i\omega x}$ survives, so $F(\omega)$ is real. Applying $\cos^2(ax)=\frac12\big[1+\cos(2ax)\big]$ and writing $L=\pi/(2a)$, $$F(\omega)=\dfrac{1}{\sqrt{2\pi}}\int_{-L}^{L}\dfrac{a}{2}\big[1+\cos(2ax)\big]\cos(\omega x)\,dx$$
Evaluate the two elementary integrals. The first is immediate, and the second is split with the product-to-sum identity $\cos\alpha\cos\beta=\frac12[\cos(\alpha+\beta)+\cos(\alpha-\beta)]$: $$\int_{-L}^{L}\cos\omega x\,dx=\dfrac{2\sin\omega L}{\omega},\qquad \int_{-L}^{L}\cos(2ax)\cos(\omega x)\,dx=\dfrac{\sin\big[(2a+\omega)L\big]}{2a+\omega}+\dfrac{\sin\big[(2a-\omega)L\big]}{2a-\omega}$$ With $L=\pi/(2a)$ put $\theta=\omega L=\pi\omega/(2a)$; then $(2a\pm\omega)L=\pi\pm\theta$, and since $\sin(\pi+\theta)=-\sin\theta$ while $\sin(\pi-\theta)=+\sin\theta$, the second integral collapses to $$-\dfrac{\sin\theta}{2a+\omega}+\dfrac{\sin\theta}{2a-\omega}=\dfrac{2\omega\sin\theta}{4a^2-\omega^2}$$
Combine into the transform. Substituting both results, $$F(\omega)=\dfrac{a}{2\sqrt{2\pi}}\left[\dfrac{2\sin\theta}{\omega}+\dfrac{2\omega\sin\theta}{4a^2-\omega^2}\right]=\dfrac{a\sin\theta}{\sqrt{2\pi}}\cdot\dfrac{(4a^2-\omega^2)+\omega^2}{\omega(4a^2-\omega^2)}$$ The $\omega^2$ terms cancel in the numerator, leaving $$\boxed{F(\omega)=\dfrac{4a^3\,\sin\!\big(\dfrac{\pi\omega}{2a}\big)}{\sqrt{2\pi}\;\omega\,(4a^2-\omega^2)}}$$ Numerical spot-checks against direct quadrature agree to ten decimal places: for $a=1,\omega=0.7$ both give $0.5786897430$; for $a=2,\omega=1.3$ both give $0.5851171730$; for $a=1.5,\omega=5.0$ both give $0.0583021377$.
Interpret the two apparent singularities. Both are removable. As $\omega\to0$, $\sin\theta\to\pi\omega/(2a)$ and $$F(0)=\dfrac{1}{\sqrt{2\pi}}\cdot\dfrac{\pi}{2}=\dfrac{\pi}{2\sqrt{2\pi}}\approx0.6267$$ which is just the area of Step 1 divided by $\sqrt{2\pi}$, as the definition demands. At $\omega=\pm2a$ the denominator vanishes but so does $\sin(\pm\pi)$, and the limit is finite, $F(\pm2a)=\pi/(4\sqrt{2\pi})\approx0.3133$. The spectrum is therefore continuous everywhere.

Part (b): the spectrum $F(\omega)$ for $a=1$ and $a=2$. Every curve passes through the same DC value $F(0)=\pi/(2\sqrt{2\pi})$; increasing $a$ stretches the spectrum outwards, which is the reciprocal-spreading behaviour discussed in part (c).

Part (c) — the limit $a\to\infty$. In the $x$-domain the support $|x|\le\pi/(2a)$ shrinks to the single point $x=0$ while the peak height $a$ grows without bound, yet Step 1 showed the enclosed area stays pinned at $\pi/2$. A family of non-negative pulses that narrows to a point at constant area is by definition a delta sequence, so $$f(x)\;\longrightarrow\;\dfrac{\pi}{2}\,\delta(x)\qquad(a\to\infty)$$ In the $\omega$-domain, for any fixed $\omega$ the argument $\pi\omega/(2a)\to0$, so $\sin(\pi\omega/2a)\to\pi\omega/(2a)$ and $4a^2-\omega^2\to4a^2$, giving $$F(\omega)\;\longrightarrow\;\dfrac{4a^3}{\sqrt{2\pi}}\cdot\dfrac{\pi\omega/(2a)}{\omega\cdot4a^2}=\boxed{\dfrac{\pi}{2\sqrt{2\pi}}\approx0.6267\quad\text{for every }\omega}$$ The spectrum flattens to a constant: the first zero crossing, at $\omega=2a$, marches off to infinity, so the pulse’s bandwidth grows in exact proportion as its duration shrinks. This is the reciprocal-spreading (uncertainty) property of the Fourier transform, and it is consistent with the standard result that the transform of $\delta(x)$ is the constant $1/\sqrt{2\pi}$ — here scaled by the fixed area $\pi/2$.

Quantity	Result
Area under $f$ (part a)	$\pi/2$, independent of $a$
Fourier transform (part b)	$F(\omega)=\dfrac{4a^3\sin(\pi\omega/2a)}{\sqrt{2\pi}\,\omega(4a^2-\omega^2)}$
$F(0)$	$\pi/(2\sqrt{2\pi})\approx0.6267$ for every $a$
$F(\pm2a)$	$\pi/(4\sqrt{2\pi})\approx0.3133$ (removable singularity)
Limit as $a\to\infty$ (part c)	$f\to\tfrac{\pi}{2}\delta(x)$; $F(\omega)\to\pi/(2\sqrt{2\pi})$, a flat spectrum

Topic: Fourier transform of a compactly supported pulse; delta sequences and reciprocal spreading.

Key relations: $F(\omega)=\frac{1}{\sqrt{2\pi}}\int f(x)e^{-i\omega x}dx$; power reduction $\cos^2\theta=\frac12(1+\cos2\theta)$; product-to-sum $\cos\alpha\cos\beta=\frac12[\cos(\alpha+\beta)+\cos(\alpha-\beta)]$; $F(0)=\frac{1}{\sqrt{2\pi}}\int f$.

Why this works: squaring a cosine doubles its frequency and adds a constant, so power reduction converts an awkward $\cos^2$ integrand into two integrals that any table handles. Evenness is the second lever: it kills the sine half of $e^{-i\omega x}$ outright, which is why the answer is purely real and why no complex arithmetic is needed. The apparent poles at $\omega=0$ and $\omega=\pm2a$ are artefacts of writing the answer as a single quotient — a compactly supported, absolutely integrable function always has a continuous (indeed entire) transform, so any pole in the formula must be cancelled by a zero of the numerator, and checking those limits is a free correctness test on the algebra.

Common pitfall: forgetting the normalising $1/\sqrt{2\pi}$ that this paper’s stated definition carries (other texts put $2\pi$ in the inverse transform instead, which changes $F(0)$ by a factor of $\sqrt{2\pi}$), and reporting the pulse as vanishing in the limit $a\to\infty$ because its width goes to zero — the area is what is conserved, so the correct limit is a scaled delta, not the zero function.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch. 11 §11.9 (Fourier Transform) and §11.7 (Fourier Integral).

Question 3 of 7: Fourier Transform of a Raised-Cosine Pulse

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