22-Elec-A3 Signals and Communications · May 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, May 2014 — 07-Elec-A3 Signals and Communications. Three hours, open book, any non-communicating calculator permitted. Ten questions of equal value (20 marks each); the rubric states that five questions constitute a complete paper and only the first five presented are marked. All ten are solved here, since the set is intended as a study resource. A table of Fourier-transform pairs and properties, a table of z-transform pairs, trigonometric identities and Bessel-function graphs/tables are supplied with the paper and are used freely below.
Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (sampling, PCM, AM/FM, matched filtering); B. P. Lathi, Linear Systems and Signals, 2nd ed. (Fourier analysis, LTI properties); A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (z-transform, regions of convergence); S. Haykin, Communication Systems, 5th ed. (digital transmission, eye diagrams).
Convention used throughout. This paper's own transform table defines the sinc function as $\operatorname{sinc}(x)=\sin x / x$ (Lathi's convention, not the normalised $\sin(\pi x)/(\pi x)$). Every bandwidth below follows from that table, in particular $2B\operatorname{sinc}(2\pi Bt)\leftrightarrow\operatorname{rect}(f/2B)$ and $B\operatorname{sinc}^2(\pi Bt)\leftrightarrow\Delta(f/2B)$.
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. $x(t) = 2\cos 2\pi f_0t$; memoryless non-linearity $y = x + x^2$; filter magnitude unity for $|f|\le 1.1f_0$, falling linearly to zero at $|f| = 1.5f_0$ and zero beyond; filter phase $+\pi/2$ for $f \lt -0.5f_0$, zero for $|f| \lt 0.5f_0$, and $-\pi/2$ for $f \gt 0.5f_0$.
[Figure not reproduced: Figure 6 (redrawn): a memoryless squaring non-linearity feeding the lowpass filter H(f). Only z(t) and x(t) are observable from outside the dashed box. See the official exam paper.]
Find. $X(f)$ and $Y(f)$; then $Z(f)$ and $z(t)$; then a justified verdict on linearity and distortionlessness; then a black-box test procedure.
Approach. Expand the squared cosine with the supplied identity to get the harmonic content, read the filter response at each harmonic, and reassemble the surviving components in the time domain.
The filter now decides which of these three components reaches the output, and what phase each acquires.
[Figure not reproduced: Figure 7 (redrawn): |H| is unity out to 1.1f0 and has fallen to zero by 1.5f0, so the 2f0 pair is annihilated. The phase is a constant ∓π/2 beyond ±0.5f0 and zero at DC. See the official exam paper.]
Part (c) — the verdict, argued from $x(t)$ and $z(t)$ alone.
(i) Is it linear? No. The input is a pure sinusoid at $f_0$ with no DC content, yet the output contains a constant term of 2 V. A linear time-invariant system can only scale and phase-shift the frequencies present at its input; it can never create energy at a frequency the input does not contain, and $f = 0$ is such a frequency here. That single observation settles it. The homogeneity test confirms it independently: the DC term arises from the squaring, so doubling the input amplitude to $4\cos 2\pi f_0t$ would quadruple the DC term to 8 while only doubling the fundamental — the output is not proportional to the input, so scaling fails.
(ii) Is it distortionless? No. A distortionless system satisfies $z(t) = k\,x(t-t_d)$ for constants $k$ and $t_d$: the output must be a faithful, merely scaled and delayed, copy of the input. Here the sinusoidal part alone would qualify, since $2\sin 2\pi f_0 t = 2\cos 2\pi f_0(t - t_d)$ with $t_d = 1/(4f_0)$ — unit gain and a constant delay of a quarter period. But the additive DC pedestal has no counterpart in $x(t)$ at any scale factor or delay, so no choice of $k$ and $t_d$ can reproduce $z(t)$ from $x(t)$. The system fails the distortionless test, and the failure is specifically non-linear distortion (a new spectral component) rather than amplitude or delay distortion.
Part (d) — black-box experiment design.
(i) Testing linearity. Linearity is the conjunction of homogeneity and additivity, so test each in turn. For homogeneity, apply a single tone $A\cos 2\pi f_1t$ and record the output spectrum; then repeat at $2A$, $4A$ and so on across the operating range. In a linear system every output component must scale by exactly the same factor as the input; any component that grows as $A^2$ (as the DC and second-harmonic terms do here) or that saturates immediately exposes non-linearity. For additivity, the sharper test is the two-tone intermodulation experiment: drive the box with $A\cos 2\pi f_1t + A\cos 2\pi f_2t$ where $f_1$ and $f_2$ are close together and not harmonically related, and inspect the output for energy at the intermodulation frequencies $f_2 \pm f_1$, $2f_1 - f_2$, $2f_2 - f_1$. A linear system returns energy at $f_1$ and $f_2$ and nowhere else; the appearance of sum, difference or third-order products is conclusive proof of non-linearity, and this test is preferred in practice because the third-order products fall close to the wanted tones and are therefore easy to observe even in a band-limited box. The rationale is that these frequencies cannot arise from any weighted sum of scaled inputs, so their presence cannot be reconciled with superposition.
(ii) Testing distortionlessness. A distortionless system requires $H(f) = ke^{-j2\pi ft_d}$ over the whole band the signal occupies: flat magnitude and linear phase. So perform a swept-sine measurement — apply a sinusoid of fixed amplitude, step its frequency across the band of interest, and at each step record the ratio of output to input amplitude and the phase difference. Distortionlessness then demands two things: the amplitude ratio must be constant with frequency (no amplitude distortion), and the phase must fall on a straight line through the origin, equivalently the group delay $\tau_g = -\frac{1}{2\pi}\,d\theta/df$ must be constant (no phase or delay distortion). A quicker qualitative version is to apply a narrow pulse or a square wave and compare the output with the input on a two-channel oscilloscope: a distortionless box returns the same shape, merely scaled and shifted in time, whereas ringing, rounding or asymmetry indicates amplitude or phase distortion. Note that the linearity test must come first, since the swept-sine transfer function is only meaningful once superposition is known to hold.
| Quantity | Result |
|---|---|
| $X(f)$ | $\delta(f-f_0)+\delta(f+f_0)$ |
| $y(t)$ | $2+2\cos 2\pi f_0t+2\cos 2\pi(2f_0)t$ |
| $Y(f)$ | Impulses: weight 2 at DC; weight 1 at $\pm f_0$ and $\pm 2f_0$ |
| Filter action | $|H|=1$ at DC and $f_0$; $|H| = 0$ at $2f_0$; phase $\mp\pi/2$ beyond $\pm 0.5f_0$ |
| $z(t)$ | $2+2\sin 2\pi f_0t$ |
| (c)(i) Linear? | No — DC created at a frequency absent from the input |
| (c)(ii) Distortionless? | No — the DC pedestal has no counterpart in $k\,x(t-t_d)$ |
| Delay of the surviving tone | $t_d = 1/(4f_0)$ |
