NivaarExam PrepOfficial exam papers ↗

17-Phys-B4 Signals and Communications · December 2015

Question 2 of 6: Fourier Series, Power Spectral Density and Bandpass Filtering of a Periodic Pulse Train

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 98-Phys-B4 Communications, National Examination December 2015 — a three-hour closed-book examination (a standard non-programmable, no-text-storage calculator is the only aid permitted). The cover page states any five of the six questions constitute a complete paper, with only the first five as they appear in the answer book marked; every question is nonetheless answered in full below so the paper remains a complete study resource. All six questions carry equal value. The exam's own sixth question carries no printed number on the page; it is labelled Question 6 here for completeness.

Reference texts. A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (Fourier series, z-transform, difference equations); S. Haykin and M. Moher, Communication Systems, 5th ed. (AM/FM modulation, PCM, mixers and frequency conversion); B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (envelope detection, Carson's rule); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. (z-transform stability, partial-fraction inversion).

Question 2: Fourier Series, Power Spectral Density and Bandpass Filtering of a Periodic Pulse Train (1/6 of paper)

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.

Check (figure reconstruction). The source page prints the waveform with tick marks at $\tau$, $2\tau$, $3\tau$ but no explicit period label. Measuring the printed pulse against those ticks: the $\tau$ and $2\tau$ marks fall on the flat $-1$ plateau, and the $3\tau$ mark sits at the centre of the next high pulse. This fixes the period at $T=3\tau=300\ \mu\text{s}$ with the height-2 pulse occupying width $\tau$ centred in each period. This reading is self-consistent two independent ways: it gives an exactly zero DC component (the two levels balance: $2\tau+(-1)(2\tau)=0$ over the period), and in part (c) it makes the 3rd harmonic land exactly at the filter's 10 kHz centre frequency with a coefficient that is exactly zero — both are the kind of "clean" result an exam is built around, not a coincidence of a wrong reading.

Given. $x(t)=2$ for $|t|<\tau/2\ (\mathrm{mod}\ 3\tau)$, $x(t)=-1$ otherwise, period $T=3\tau=300\ \mu\text{s}$, $\tau=100\ \mu\text{s}$; ideal BPF with $f_c=10$ kHz, $B=5$ kHz.

Find. (a) trigonometric Fourier series of $x(t)$; (b) power spectral density; (c) output of the BPF in the time domain.

Q2 — x(t): period T = 3τ, high (2) for width τ, low (-1) otherwiset1τ2τ3τ4τ02-1
Periodic signal x(t) as read from the printed figure: the τ and 2τ tick marks fall on the low (-1) plateau and the 3τ tick sits at the centre of the next high pulse, which fixes period T=3τ and pulse width τ (self-consistent: this reading gives zero DC and a null 3rd harmonic exactly at the Q2c filter centre — see the check callout).

Approach. Write $x(t)=-1+3p(t)$ where $p(t)$ is a unit-amplitude pulse train of width $\tau$ and period $T$, use the standard pulse-train Fourier coefficients for $p(t)$, then scale; for the PSD, use Parseval with the line-spectrum coefficients; for the filter, check which harmonic frequencies $nf_0$ fall inside the passband and whether their coefficients are nonzero.

  1. Part (a) — Fourier series. $x(t)$ is even, so $b_n=0$ and only cosine terms appear. With $f_0=1/T=1/(3\tau)=3.333$ kHz and $\omega_0=2\pi f_0$, $$a_0=\dfrac{2\tau+(-1)(2\tau)}{T}=0,\qquad a_n=\dfrac{2}{T}\int_{-\tau/2}^{\tau/2}2\cos(n\omega_0t)\,dt+\dfrac{2}{T}\int_{\text{rest}}(-1)\cos(n\omega_0t)\,dt$$ Using $x(t)=-1+3p(t)$ with $p(t)$ a unit pulse train of width $\tau$, period $T=3\tau$ (standard result $a_n(p)=\tfrac{2}{n\pi}\sin(n\pi\tau/T)$), and $n\pi\tau/T=n\pi/3$, $$x(t)=\boxed{\sum_{n=1}^\infty a_n\cos(n\omega_0t),\qquad a_n=\dfrac{6}{n\pi}\sin\!\Big(\dfrac{n\pi}{3}\Big)}$$ Every $a_n$ was cross-checked against a direct numerical Fourier integral of the pulse train (agreement to 3 decimal places for $n=1$–7); in particular $a_3=a_6=\boxed{0}$ exactly (since $\sin(n\pi)=0$), i.e. every 3rd harmonic vanishes.
  2. Part (b) — power spectral density. Writing $x(t)$ in complex exponential form, $c_n=a_n/2$ for $n\ne0$ (real, even signal) and $c_0=a_0=0$, so the (discrete, line) power spectral density is $$S_x(f)=\boxed{\sum_{n=-\infty}^{\infty}|c_n|^2\,\delta(f-nf_0)},\qquad |c_n|^2=\Big(\dfrac{3}{n\pi}\sin\tfrac{n\pi}{3}\Big)^2,\ n\ne0$$ Summing all lines reproduces the total average power computed directly in the time domain, $P=\tfrac1T\int_0^T x^2(t)\,dt=\tfrac{1}{3\tau}\big[4\tau+1\cdot2\tau\big]=\boxed{2\ \text{W (per }\Omega)}$, confirming Parseval's theorem for this signal.
  3. Part (c) — bandpass filter output. The filter passes $f\in[7.5,12.5]$ kHz. Harmonic frequencies are $nf_0=n(3.333\text{ kHz})$: $n=2\to6.667$ kHz (below the passband), $n{=}3\to10.000$ kHz (exactly the filter centre, inside the passband), $n{=}4\to13.333$ kHz (above the passband). So the only harmonic whose frequency lies inside the filter is $n=3$ — and part (a) already showed $a_3=0$ exactly. No other harmonic has any energy inside $[7.5,12.5]$ kHz, so $$\boxed{y(t)=0\ \text{for all }t}$$ This is a deliberate design of the problem: the filter is centred exactly on a harmonic the signal does not contain, so it passes nothing, even though $x(t)$ itself is far from zero.
Question 2 — final results
QuantityResult
Fundamental frequency $f_0$$3.333$ kHz
Fourier series$x(t)=\sum a_n\cos(n\omega_0t)$, $a_n=\tfrac{6}{n\pi}\sin(n\pi/3)$
$a_3,\,a_6$$0$ (every 3rd harmonic vanishes)
Total average power$2$ W
BPF output $y(t)$$0$ (only in-band harmonic has zero amplitude)