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)
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.
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.
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.
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.
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.