Question 3 of 7: Fourier Transform of a Trapezoidal Pulse
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 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, with internal splits as marked); 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) — Sturm–Liouville eigenproblems, Fourier series and the Fourier transform (Ch. 11), least-squares curve fitting, Lagrange/Newton interpolation, Romberg integration, and root-finding by bisection/Newton/fixed-point iteration (Ch. 19), Cholesky factorization (Ch. 20); Strang, Introduction to Linear Algebra (6th ed., Wellesley-Cambridge) — symmetric positive-definite systems and Cholesky factorization.
Question 3: Fourier Transform of a Trapezoidal Pulse (20 marks)
Given. The even, continuous trapezoidal pulse $f(x)$: unit plateau on $[-1,1]$, linear ramps down to zero on $[-2,-1]$ and $[1,2]$, zero outside $[-2,2]$; the transform convention $F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-i\omega x}\,dx$.
Find. $F(\omega)$ for all $\omega$.
The trapezoidal pulse f(x): a unit-height plateau on |x|≤1 ramping linearly to zero at |x|=2.
Approach. $f$ is even, so the transform reduces to a real cosine integral, $F(\omega)=\dfrac{2}{\sqrt{2\pi}}\int_0^2f(x)\cos(\omega x)\,dx$; split the integral at the corner $x=1$ (plateau vs. ramp) and integrate the ramp piece by parts.
Use evenness to drop the imaginary (sine) part. $\sin(\omega x)$ is odd and $f$ is even, so their product integrates to zero over $[-2,2]$:
$$F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-2}^{2}f(x)e^{-i\omega x}\,dx=\frac{2}{\sqrt{2\pi}}\int_0^{2}f(x)\cos(\omega x)\,dx$$
Integrate the plateau, $x\in[0,1]$.
$$\int_0^1 1\cdot\cos(\omega x)\,dx=\frac{\sin\omega}{\omega}$$
Integrate the ramp, $x\in[1,2]$, by parts. With $u=2-x,\,dv=\cos(\omega x)dx$,
$$\int_1^2(2-x)\cos(\omega x)\,dx=\frac{\cos\omega-\cos2\omega}{\omega^2}-\frac{\sin\omega}{\omega}$$
Add the two pieces — the $\sin\omega/\omega$ terms cancel exactly.
$$\int_0^2f(x)\cos(\omega x)\,dx=\frac{\sin\omega}{\omega}+\left[\frac{\cos\omega-\cos2\omega}{\omega^2}-\frac{\sin\omega}{\omega}\right]=\frac{\cos\omega-\cos2\omega}{\omega^2}$$
Assemble $F(\omega)$ and check the removable point $\omega=0$.
$$\boxed{F(\omega)=\sqrt{\dfrac{2}{\pi}}\;\dfrac{\cos\omega-\cos2\omega}{\omega^2}\quad(\omega\ne0)}$$
As $\omega\to0$, l'Hopital (or the Taylor expansion of the two cosines) gives the removable value $F(0)=\tfrac32\sqrt{2/\pi}$, which equals $\sqrt{2/\pi}$ times the area under $f$ (the plateau area 1 plus the two half-triangle ramps, total $3/2$) — a useful check.
