Question 3 of 6: Exponential Fourier Series of a Triangular Wave

Question 3: Exponential Fourier Series of a Triangular Wave (20 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. Figure 2 shows a triangular wave of peak value $+1$ at $t=0$ falling linearly to $-1$ at $t=4$ and returning, repeating with zero crossings at $t=\pm2,\pm6,\dots$:

[Figure not reproduced: Figure 2 redrawn: even triangular wave, period T0 = 8 s, peak +1 at t = 0 and minimum -1 at t = 4. See the official exam paper.]

Find. (a) the exponential Fourier series $x(t)=\sum_{n}D_{n}e^{jn\omega_0 t}$; (b) the closed form of $D_n$ and stem plots of $|D_n|$ and $\arg D_n$ over $-7\le n\le 7$.

Approach. Exploit the symmetry first — even symmetry forces the coefficients to be real, and zero average forces $D_0=0$ — then evaluate the analysis integral over one period using integration by parts.

1. Fix the fundamental and the DC term. The waveform repeats every $T_0=8$ s, so $\omega_0=\dfrac{2\pi}{8}=\dfrac{\pi}{4}$ rad/s. The signal spends equal time above and below zero with equal triangular areas, hence $D_0=\dfrac{1}{T_0}\int_{-4}^{4}x\,dt=0$.
2. Write the analysis integral. $D_n=\dfrac{1}{T_0}\displaystyle\int_{-4}^{4}\left(1-\tfrac{|t|}{2}\right)e^{-jn\omega_0 t}\,dt$. Because $x(t)$ is even, the odd (sine) part of the exponential integrates to zero and the integral reduces to twice the half-range cosine integral: $$D_n=\frac{2}{8}\int_{0}^{4}\left(1-\tfrac{t}{2}\right)\cos\!\left(\tfrac{n\pi t}{4}\right)dt .$$
3. Do the integral. Write $a=\dfrac{n\pi}{4}$. Then $\displaystyle\int_{0}^{4}\cos(at)\,dt=\frac{\sin 4a}{a}=\frac{\sin n\pi}{a}=0$, and by parts $\displaystyle\int_{0}^{4}t\cos(at)\,dt=\left[\frac{t\sin at}{a}+\frac{\cos at}{a^{2}}\right]_{0}^{4}=\frac{\cos n\pi-1}{a^{2}}$. Combining, $D_n=\dfrac{1}{4}\left(0-\dfrac{1}{2}\cdot\dfrac{\cos n\pi-1}{a^{2}}\right)=\dfrac{1-\cos n\pi}{8a^{2}}$.
4. Simplify with $a^{2}=n^{2}\pi^{2}/16$. $D_n=\dfrac{16\,(1-\cos n\pi)}{8\,n^{2}\pi^{2}}=\dfrac{2\,(1-(-1)^{n})}{n^{2}\pi^{2}}$, so $$\boxed{\,D_n=\begin{cases}\dfrac{4}{\pi^{2}n^{2}},&n\ \text{odd}\\[6pt]0,&n\ \text{even (including }n=0).\end{cases}}$$ Every even harmonic vanishes — the signature of half-wave symmetry, $x\left(t+\tfrac{T_0}{2}\right)=-x(t)$.
5. Assemble the exponential series (part a). $$x(t)=\sum_{\substack{n=-\infty\\ n\ \text{odd}}}^{\infty}\frac{4}{\pi^{2}n^{2}}\,e^{jn\pi t/4}=\frac{8}{\pi^{2}}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{2}}\cos\!\left(\frac{(2k+1)\pi t}{4}\right),$$ where the second form pairs $n$ with $-n$ using $D_n=D_{-n}$. Evaluating at $t=0$ gives $\dfrac{8}{\pi^{2}}\cdot\dfrac{\pi^{2}}{8}=1$, the correct peak — a quick and complete sanity check.
6. Numerical coefficients and spectra (part b). $|D_{\pm1}|=\dfrac{4}{\pi^{2}}=0.4053$, $|D_{\pm3}|=0.04503$, $|D_{\pm5}|=0.01621$, $|D_{\pm7}|=0.008271$, and all even coefficients are zero. Since each $D_n$ is real and positive, $\arg D_n=0$ for every $n$ — the phase spectrum is identically zero, as it must be for a real and even signal.
7. Confirm by Parseval. $\sum_{n}|D_n|^{2}=2\sum_{n\ \text{odd}\gt 0}\left(\tfrac{4}{\pi^{2}n^{2}}\right)^{2}=0.3333$, while the mean square computed directly is $\dfrac{1}{8}\int_{-4}^{4}x^{2}dt=\dfrac{1}{3}$. The two agree, so no coefficient has been lost.