Question 7 of 7: AM with a triangular message: time expression, spectrum, envelope and detectors (20 marks)

Question 7 — AM with a triangular message: time expression, spectrum, envelope and detectors (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 plot; (b) the line spectrum out to the fourth message harmonic, with the computation procedure stated; (c) the envelope with all parameters; (d) an envelope-detector circuit with component values; (e) a coherent-detector block diagram.

Approach. Fix the carrier amplitude from the stated peak-to-peak value, write the standard AM expression, expand the triangular message in a Fourier cosine series to generate the sideband lines, and then size the detector time constant against both the carrier period and the envelope’s steepest slope.

1. Fix the carrier amplitude. For standard AM, $s(t) = A_c\bigl[1 + a\,m(t)\bigr]\cos\omega_c t$ with $|m|_{\max} = 1$. The signal reaches $+A_c(1+a)$ and $-A_c(1+a)$, so $$s_{pp} = 2A_c(1 + a) = 4\ \text{V} \qquad\Longrightarrow\qquad A_c = \frac{4}{2(1.8)} = 1.111\ \text{V}.$$
2. Write the time-domain expression. With $f_c = 10\ \text{MHz}$, $$\boxed{\;s(t) = 1.111\bigl[1 + 0.8\,m(t)\bigr]\cos\!\left(2\pi \times 10^{7} t\right)\ \text{V}\;}$$ where $m(t)$ is the unit-amplitude triangular wave of period $200\ \mu\text{s}$ shown above. Since $a = 0.8 \lt 1$ the bracket never becomes negative, so the carrier never reverses phase and the signal is envelope-detectable — the fact parts (c) and (d) depend on.

(The carrier is drawn at forty cycles per message period so that both the carrier and the envelope are visible; the true ratio is $f_c/f_m = 2000$.)

3. Expand the message in a Fourier series — the computation procedure. The message is real and even about $t = 0$, so its series contains cosines only and every coefficient is real (all sideband lines are in phase, and no phase plot is needed). For a unit-amplitude triangular wave, $$m(t) = \sum_{n\ \text{odd}} c_n\cos(2\pi n f_m t), \qquad c_n = \frac{8}{\pi^{2}n^{2}} .$$ Even harmonics vanish identically because of the wave’s half-wave symmetry, so $c_1 = 8/\pi^{2} = 0.8106$, $c_2 = 0$, $c_3 = 8/9\pi^{2} = 0.0901$, $c_4 = 0$. (Check: $\sum_{n\,\text{odd}} c_n = 1 = m(0)$, and the mean-square value is $1/3$.)
4. Generate the line spectrum. Substituting the series into $s(t)$ and applying the product-to-sum identity, each message line $c_n\cos 2\pi n f_m t$ produces a pair of sidebands at $f_c \pm nf_m$ of one-sided amplitude $A_c a c_n/2$, alongside the carrier line at $A_c$: $$s(t) = A_c\cos\omega_c t + \frac{A_c a}{2}\sum_{n\ \text{odd}} c_n\left[\cos 2\pi(f_c + nf_m)t + \cos 2\pi(f_c - nf_m)t\right].$$ With $A_c a/2 = 0.4444$: $$\boxed{\;A_{\text{carrier}} = 1.111\ \text{V};\quad A_{\pm1} = 0.360\ \text{V at } 10\ \text{MHz} \pm 5\ \text{kHz};\quad A_{\pm3} = 0.0400\ \text{V at } 10\ \text{MHz} \pm 15\ \text{kHz}\;}$$ Neglecting harmonics above the fourth, as instructed, leaves exactly these five lines — the second- and fourth-harmonic sidebands are zero — so the outermost non-zero lines sit at $10\ \text{MHz} \pm 15\ \text{kHz}$ and the occupied bandwidth is $6f_m = 30\ \text{kHz}$, within the $8f_m = 40\ \text{kHz}$ that the instruction permits.

Because only 17.6 % of the total power lies in the sidebands (the carrier carries the rest, as the $a^{2}\overline{m^{2}}/(2 + a^{2}\overline{m^{2}})$ ratio with $\overline{m^{2}} = 1/3$ confirms), standard AM is spectrally inefficient here — the price paid for a two-component detector.

5. Describe the envelope. Since $1 + a\,m(t) \gt 0$ throughout, the envelope is simply $$E(t) = A_c\bigl[1 + 0.8\,m(t)\bigr],$$ a triangular wave of the same $200\ \mu\text{s}$ period as the message, riding on a d.c. pedestal: $$\boxed{\;E_{\max} = A_c(1+a) = 2.000\ \text{V},\quad E_{\min} = A_c(1-a) = 0.222\ \text{V},\quad E_{\text{avg}} = A_c = 1.111\ \text{V}\;}$$ Its peak-to-peak swing is $1.778\ \text{V}$ and its slope on each ramp is $\pm A_c a\,(4f_m) = \pm 17.8\ \text{kV/s}$. As a cross-check, the index recovered from the envelope, $(E_{\max}-E_{\min})/(E_{\max}+E_{\min}) = 1.778/2.222 = 0.8$, returns the given value of $a$.

6. Design the envelope-detector time constant. A series diode followed by a parallel $RC$ load is the classical circuit. The capacitor must hold its charge between carrier peaks but discharge fast enough to follow the falling envelope, which for the triangular message falls at the constant rate $A_c a\,(4f_m)$. The two-sided condition is $$\frac{1}{f_c} \ll RC \le \frac{E_{\min}}{A_c a\,4f_m} = \frac{1-a}{4af_m} = \frac{0.2}{4(0.8)(5000)} = 12.5\ \mu\text{s}.$$ Choosing $R = 10\ \text{k}\Omega$ and $C = 1.0\ \text{nF}$ gives $$\boxed{\;RC = 10\ \mu\text{s},\qquad 0.1\ \mu\text{s} \ll 10\ \mu\text{s} \lt 12.5\ \mu\text{s}\;}$$ so the detector smooths the 10 MHz carrier by a factor of 100 while still tracking the envelope without diagonal clipping. A germanium or Schottky diode is preferred for its low forward drop, since $E_{\min} = 0.222\ \text{V}$ is comparable with a silicon diode’s 0.7 V threshold; a series d.c.-blocking capacitor at the output removes the $A_c$ pedestal.

7. Give the coherent detector. Multiplying the received signal by a phase-locked replica $2\cos\omega_c t$ gives $$s(t)\cdot 2\cos\omega_c t = A_c\bigl[1 + a m(t)\bigr]\bigl(1 + \cos 2\omega_c t\bigr),$$ and a low-pass filter of bandwidth 20 kHz (enough for the retained harmonics, far below $2f_c = 20\ \text{MHz}$) removes the double-frequency term, leaving $A_c[1 + a\,m(t)]$. A d.c. block then strips the $A_c$ pedestal, giving $A_c a\,m(t) = 0.889\,m(t)$. The local carrier is derived from the received carrier line by a phase-locked loop; a static phase error $\phi$ multiplies the output by $\cos\phi$, so the loop must hold lock to within a few degrees.

For this signal the envelope detector of part (d) is the practical choice — it is cheaper, needs no carrier recovery, and works precisely because $a \lt 1$. The coherent detector of part (e) is nonetheless superior in noise, being unaffected by the threshold effect that degrades envelope detection at low carrier-to-noise ratios.