Question 7 of 7: AM Signal — Time Domain, Spectrum, Envelope and Demodulation

Question 7: AM Signal — Time Domain, Spectrum, Envelope and Demodulation (1/5 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.

Given. $a=0.8$; AM signal peak-to-peak $=4$ V; $m(t)$ = normalized ($\pm1$ peak) triangular wave, $f_m=5$ kHz; $f_c=10$ MHz.

Find. (a) $s(t)$ in terms of $m(t)$, with $A_c$ evaluated numerically, and its time-domain plot; (b) the line spectrum (harmonics of $m(t)$ up to the 4th only); (c) the envelope, with all parameters; (d) an envelope-detector circuit with component values; (e) a coherent-detector block diagram.

Approach. Solve for $A_c$ from the stated peak-to-peak value and $a$; write $s(t)$ in the standard AM form; get the message's Fourier series (a triangular wave has only ODD harmonics, decaying as $1/n^2$) to build the line spectrum; the envelope follows directly from $|1+a\,m(t)|$; size the envelope-detector RC to avoid both carrier ripple and diagonal-clipping distortion.

1. Part (a) — carrier amplitude and s(t). The AM envelope peaks at $A_c(1+a)$ and troughs at $-A_c(1+a)$ (a full-signal peak-to-peak swing of $2A_c(1+a)$), so $$2A_c(1+a)=4\ \text{V}\quad\Rightarrow\quad A_c=\dfrac{4}{2(1+0.8)}=\boxed{\dfrac{10}{9}\approx1.111\ \text{V}}$$ $$s(t)=\boxed{A_c\big[1+0.8\,m(t)\big]\cos(2\pi\times10^7t),\quad A_c=10/9\ \text{V}}$$

2. Part (b) — line spectrum. $m(t)$ is an odd, zero-mean triangular wave rising through the origin, whose Fourier series contains only ODD harmonics: $$m(t)=\dfrac{8}{\pi^2}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^2}\sin\!\big(2\pi(2k+1)f_mt\big)=c_1\sin(2\pi f_mt)+c_3\sin(2\pi\cdot3f_mt)+\dots$$ with $c_1=8/\pi^2=0.8106$ and $c_3=-8/(9\pi^2)=-0.0901$. Since the question says to neglect message harmonics above the 4th, only $n=1$ and $n=3$ survive (the triangular wave has no 2nd or 4th harmonic at all). Each surviving $m(t)$ harmonic $c_n\sin(2\pi nf_mt)$ produces a symmetric sideband PAIR at $f_c\pm nf_m$ of height $(a\,A_c/2)|c_n|$, alongside an unmodulated carrier line of height $A_c$ at $f_c$: $$\boxed{\text{lines at }f_c\ (\text{height }A_c),\quad f_c\pm f_m\ \big(\text{height }\tfrac{aA_c}{2}c_1\big),\quad f_c\pm3f_m\ \big(\text{height }\tfrac{aA_c}{2}|c_3|\big)}$$ — computation procedure: take the message's Fourier series, scale each surviving sideband pair by $aA_c/2$, and place the carrier line at $f_c$ with height $A_c$.

3. Part (c) — envelope. Since $a=0.8<1$ (no overmodulation), the envelope tracks $m(t)$ directly and never crosses zero: $$e(t)=\boxed{A_c\big[1+0.8\,m(t)\big],\quad e_{\max}=A_c(1+a)=2.00\ \text{V},\quad e_{\min}=A_c(1-a)=0.222\ \text{V}}$$ with the same period as the message, $T_m=1/f_m=200\ \mu$s, and the same triangular shape (the envelope is a scaled-and-shifted copy of $m(t)$ itself).

4. Part (d) — envelope-detector circuit. A series diode into a parallel RC recovers the envelope directly, provided the RC time constant sits between two bounds: fast enough to follow the carrier's half-cycles ($RC\gg1/f_c$, so ripple stays small) yet slow enough not to induce diagonal-clipping distortion at the message's steepest slope ($RC\ll\sqrt{1-a^2}/(a\,\omega_{3f_m})$, using the HIGHEST kept message harmonic, $3f_m=15$ kHz, as the worst case): $$\dfrac{1}{f_c}=100\ \text{ns}\ \ll\ RC\ \ll\ \dfrac{\sqrt{1-0.8^2}}{0.8\times2\pi(15\,\text{kHz})}=7.96\ \mu\text{s}$$ Choosing $\boxed{R=2.2\ \text{k}\Omega,\ C=470\ \text{pF}\ \Rightarrow\ RC=1.03\ \mu\text{s}}$ sits comfortably inside both bounds (10× the carrier period, 7.7× below the clipping limit).

5. Part (e) — coherent detector. Multiplying $s(t)$ by a phase-synchronous local carrier $\cos(2\pi f_ct)$ and low-pass filtering recovers $m(t)$ directly (this method works regardless of $a$, including $a\ge1$, unlike envelope detection): $$s(t)\cos(2\pi f_ct)=\dfrac{A_c}{2}\big[1+0.8\,m(t)\big]+\dfrac{A_c}{2}\big[1+0.8\,m(t)\big]\cos(4\pi f_ct)$$ the LPF (cutoff between $3f_m=15$ kHz and $2f_c$) removes the $2f_c$ term, and a final DC-blocking stage removes the residual $A_c/2$ carrier term, leaving a signal $\boxed{\propto m(t)}$.