22-Elec-A3 Signals and Communications · December 2017

Question 3 of 5: Amplitude Modulation with a Triangular Message

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

Question 3: Amplitude Modulation with a Triangular Message (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.

Find. (a) $s(t)$ and its waveform; (b) the line spectrum out to the fifth message harmonic and the recipe for computing it; (c) the envelope with every parameter; (d) a practical envelope detector with component values; (e) a coherent-detector block diagram.

Approach. Convert the peak-to-peak specification into the carrier amplitude $A_c$ using the definition of the modulation index, write the standard AM expression, expand the triangular message in a cosine Fourier series to place the sidebands, and then design the two demodulators against the resulting bandwidth.

1. (a) Extract the carrier amplitude. Conventional AM with a normalised message ($|m(t)|_{\max} = 1$) has the form $s(t) = A_c[1 + a\,m(t)]\cos 2\pi f_c t$, so its envelope swings between $A_c(1 - a)$ and $A_c(1 + a)$. The maximum peak-to-peak excursion of the modulated waveform is twice the maximum envelope value: $$V_{pp,\max} = 2A_c(1 + a) \;\Longrightarrow\; A_c = \frac{V_{pp,\max}}{2(1+a)} = \frac{4}{2(1.8)} = \boxed{1.1111\ \text{V}}$$
2. Write the AM signal. Substituting the numbers, $$\boxed{\,s(t) = 1.1111\,\bigl[\,1 + 0.8\,m(t)\,\bigr]\cos\!\bigl(2\pi \times 10^{7}\,t\bigr)\ \text{V}\,}$$ The waveform is plotted below. Because the carrier at 10 MHz is 2000 times the 5 kHz message frequency, the figure uses a reduced carrier-to-message ratio so that the modulation remains visible; the essential feature — a triangular envelope that never collapses to zero — is preserved.
Figure 3.2 — The AM waveform (carrier drawn at a reduced frequency ratio so the modulation is visible). Dashed red: the envelope, peaking at 2 V and dipping to 0.222 V.
3. (b) Expand the triangular message. The message is real and even about $t = 0$, so its Fourier series contains cosine terms only and no DC: $$m(t) = \sum_{n\ \text{odd}} a_n \cos(2\pi n f_m t), \qquad a_n = \frac{8}{\pi^{2}n^{2}} \ (n\ \text{odd}), \qquad a_n = 0 \ (n\ \text{even})$$ giving $a_1 = 0.8106$, $a_3 = 0.09006$ and $a_5 = 0.03242$. A useful check is that $\sum_{n\ \text{odd}} a_n = (8/\pi^{2})(\pi^{2}/8) = 1$, which is the peak value at $t = 0$, and the even harmonics vanish because a triangular wave is odd-harmonic only (it possesses half-wave symmetry).
4. Place the AM lines. Substituting the series into $s(t)$ and applying the product rule $\cos X \cos Y = \tfrac12[\cos(X-Y) + \cos(X+Y)]$ converts every message harmonic into a pair of sidebands straddling the carrier. This is the computation procedure the question asks for: $$s(t) = A_c\cos 2\pi f_c t + \frac{A_c a}{2}\sum_{n\ \text{odd}} a_n\Bigl[\cos 2\pi (f_c + nf_m)t + \cos 2\pi (f_c - nf_m)t\Bigr]$$ so the carrier line has amplitude $A_c = 1.1111$ V at 10 MHz and each sideband at $f_c \pm nf_m$ has amplitude $$\frac{A_c a\,a_n}{2}: \quad 0.36025\ \text{V at } \pm 5\ \text{kHz}, \quad 0.04003\ \text{V at } \pm 15\ \text{kHz}, \quad 0.01441\ \text{V at } \pm 25\ \text{kHz}$$ Truncating at the fifth harmonic as instructed, the transmission bandwidth is $B_T = 2 \times 5f_m = \boxed{50\ \text{kHz}}$, i.e. 9.975 MHz to 10.025 MHz.
Figure 3.4 — Line spectrum about the carrier (positive-frequency half). Only odd harmonics of the triangular message appear; components beyond the 5th are neglected.
5. (c) The envelope and its parameters. Since $a = 0.8 \lt 1$ the bracket $1 + a\,m(t)$ never changes sign, so the envelope is simply $$e(t) = A_c\bigl[1 + a\,m(t)\bigr]$$ a scaled and DC-shifted copy of the triangular message. Its parameters are: maximum $e_{\max} = A_c(1+a) = 2.000$ V; minimum $e_{\min} = A_c(1-a) = 0.2222$ V; mean (DC pedestal) $A_c = 1.1111$ V; peak-to-peak swing $1.7778$ V; waveform triangular; period $200\ \mu$s (5 kHz). As a consistency check, recovering the index from the envelope extremes, $$a = \frac{e_{\max} - e_{\min}}{e_{\max} + e_{\min}} = \frac{2.0000 - 0.2222}{2.0000 + 0.2222} = 0.800 \ \checkmark$$
Figure 3.3 — The envelope: a triangular wave riding on a DC pedestal Ac, identical in shape to m(t) because a < 1.
6. Power budget (a useful by-product). Using normalised powers, the carrier carries $A_c^{2}/2 = 0.6173$ W while all six sidebands together carry $0.1310$ W, so the modulation efficiency is $0.1310/(0.6173 + 0.1310) = 17.5\ \%$. The rest is spent on the carrier — the price paid for allowing the simple envelope detector of part (d) instead of a coherent receiver.
7. (d) Envelope detector. Because the envelope never dips to zero, a diode followed by an RC low-pass will track it directly. The time constant must be long enough to hold charge between carrier cycles yet short enough to follow the fastest message component: $$\frac{1}{f_c} \ll RC \le \frac{\sqrt{1-a^{2}}}{a\,2\pi W}$$ With $W = 5f_m = 25$ kHz (the highest retained message harmonic) the upper bound is $$RC \le \frac{\sqrt{1-0.64}}{0.8 \times 2\pi \times 25\,000} = \frac{0.6}{125\,664} = 4.775\ \mu\text{s}$$ Choosing $\boxed{R = 4\ \text{k}\Omega,\ C = 1\ \text{nF}}$ gives $RC = 4\ \mu$s, comfortably inside the bound and forty times the carrier period $1/f_c = 0.1\ \mu$s. A series capacitor after the detector removes the 1.1111 V DC pedestal, leaving a scaled replica of $m(t)$.
Figure 3.5 — Diode envelope detector. A series DC-blocking capacitor after the output terminals removes the Ac pedestal.
8. (e) Coherent detector. Multiplying the received signal by a locally generated carrier of exactly the same frequency and phase gives $$s(t)\cdot 2\cos 2\pi f_c t = A_c[1 + a\,m(t)]\bigl[1 + \cos 4\pi f_c t\bigr]$$ The term at $2f_c = 20$ MHz is removed by a low-pass filter with cut-off just above 25 kHz, leaving $A_c[1 + a\,m(t)]$; a DC block then strips the pedestal and recovers $m(t)$. The local oscillator must be phase-locked to the carrier — a phase error $\phi$ scales the output by $\cos\phi$ and kills it completely at $\phi = 90^{\circ}$ — which is why a Costas loop or a pilot-derived reference is used in practice.
Figure 3.6 — Coherent (synchronous) detector. The local carrier must match the received carrier in both frequency and phase.