22-Elec-A3 Signals and Communications · May 2018

Question 3 of 5: Amplitude Modulation — Time Waveform, Spectrum, Envelope and Two Demodulators

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

Question 3: Amplitude Modulation — Time Waveform, Spectrum, Envelope and Two Demodulators (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. The time-domain AM expression and its plot; the exact line spectrum; the envelope with all its parameters; a component-level envelope detector; and a coherent-detector block diagram.

Approach. Normalise the message by its own peak so that $\mu$ has its textbook meaning, use the stated maximum peak-to-peak value to pin the carrier amplitude $A_c$, then expand the product $A_c[1 + \mu m_n(t)]\cos 2\pi f_c t$ into cosines to get the exact spectral lines. The envelope follows directly from the bracket, and the two detectors are the standard non-coherent and coherent receivers sized to this signal's numbers.

1. Normalise the message and write the AM expression. The message peaks at $|m|_{\max} = 1 + \tfrac12 = 1.5$ (both cosines peak together at $t = 0$; see the figure). Defining the normalised message $m_n(t) = m(t)/1.5$, which swings between $\pm 1$, the standard AM signal is $$s(t) = A_c\bigl[1 + \mu\,m_n(t)\bigr]\cos(2\pi f_c t) = A_c\left[1 + \frac{\mu}{1.5}\,m(t)\right]\cos(2\pi f_c t).$$ With $\mu = 1/2$ the bracket's scaling on $m(t)$ itself is $\mu/1.5 = 1/3$.
2. Pin the carrier amplitude from the peak-to-peak specification. The waveform reaches its greatest excursion when the envelope is greatest, so the maximum peak-to-peak value is twice the maximum envelope: $$V_{pp} = 2A_c(1+\mu) \quad\Rightarrow\quad A_c = \frac{V_{pp}}{2(1+\mu)} = \frac{4}{2(1.5)} = \frac{4}{3}\ \text{V}.$$ Substituting into step 1 gives the answer to part (a): $$\boxed{\,s(t) = \frac{4}{3}\left[1 + \frac{1}{3}m(t)\right]\cos\!\left(2\pi\,(10^{7})\,t\right)\ \text{V}\,}$$ with $A_c = 1.333$ V. The plot below is the message of the given figure ridden by the 10 MHz carrier: the envelope reproduces the message shape scaled and lifted, and the waveform's peaks trace it exactly.

3. Expand into spectral lines (part b). Write out the product using $\cos X\cos Y = \tfrac12[\cos(X-Y) + \cos(X+Y)]$. With $A_c\mu/1.5 = (4/3)(1/3) = 4/9$, $$s(t) = \underbrace{\tfrac{4}{3}\cos 2\pi f_c t}_{\text{carrier}} + \tfrac{4}{9}\cos(2\pi f_m t)\cos(2\pi f_c t) + \tfrac{2}{9}\cos(2\pi\,3f_m t)\cos(2\pi f_c t),$$ $$s(t) = \tfrac{4}{3}\cos 2\pi f_c t + \tfrac{2}{9}\bigl[\cos 2\pi (f_c\!-\!f_m)t + \cos 2\pi (f_c\!+\!f_m)t\bigr] + \tfrac{1}{9}\bigl[\cos 2\pi (f_c\!-\!3f_m)t + \cos 2\pi (f_c\!+\!3f_m)t\bigr].$$ So the spectrum is exactly five pairs of impulses. In terms of impulse weights $|S(f)|$ (each real cosine of amplitude $C$ contributes $C/2$ at $+f$ and $C/2$ at $-f$): $$\boxed{\,\text{lines at }9.988,\ 9.996,\ 10.000,\ 10.004,\ 10.012\ \text{MHz with weights } \tfrac{1}{18},\ \tfrac{1}{9},\ \tfrac{2}{3},\ \tfrac{1}{9},\ \tfrac{1}{18}\,}$$ The transmission bandwidth is $B_T = 2\times 3f_m = 24$ kHz, set by the third-harmonic component of the message and not by $f_m$.

4. Envelope and its parameters (part c). Because $1 + \mu m_n(t) > 0$ everywhere (the bracket ranges over $1 \pm \mu$), the envelope is simply that bracket scaled by $A_c$: $$a(t) = A_c\left[1 + \tfrac13 m(t)\right],\qquad a_{\max} = A_c(1+\mu) = 2\ \text{V},\qquad a_{\min} = A_c(1-\mu) = \tfrac{2}{3} = 0.667\ \text{V}.$$ $$\boxed{\,a_{\max}=2\ \text{V},\quad a_{\min}=0.667\ \text{V},\quad \text{mean }A_c=1.333\ \text{V}, \quad \text{period }1/f_m = 250\ \mu\text{s}\,}$$ The envelope's shape is a scaled, DC-shifted copy of $m(t)$; its swing is $a_{\max} - a_{\min} = 2\mu A_c = 1.333$ V, and the modulation index can be read back off the plot as $\mu = (a_{\max}-a_{\min})/(a_{\max}+a_{\min}) = 1.333/2.667 = 0.5$, confirming the given value. Because $a_{\min} > 0$ there is no over-modulation and envelope detection is legitimate. The power efficiency — sideband power divided by total power — is $\eta = 0.0617/(0.889+0.0617) = 6.49\%$, typical of the wasteful but cheap full-carrier AM format.

5. Envelope-detector circuit (part d). The classic non-coherent demodulator is a series diode feeding a parallel $RC$ load. The diode conducts on positive carrier peaks and charges $C$; on the falling carrier the diode blocks and $C$ discharges through $R$. For the capacitor to hold between carrier cycles but still track the envelope down its steepest slope, the time constant must satisfy $$\frac{1}{f_c} \ll RC \le \frac{\sqrt{1-\mu^{2}}}{2\pi\mu W},$$ where $W$ is the highest message frequency present, here $3f_m = 12$ kHz (not $f_m$). Numerically $1/f_c = 0.1\ \mu$s and $$RC_{\max} = \frac{\sqrt{1-0.25}}{2\pi(0.5)(12\times10^{3})} = \frac{0.8660}{3.770\times10^{4}} = 22.97\ \mu\text{s}.$$ Choosing a value comfortably inside the window, $$\boxed{\,R = 10\ \text{k}\Omega,\quad C = 1\ \text{nF}\ \Rightarrow\ RC = 10\ \mu\text{s} \quad (0.1\ \mu\text{s} \ll 10\ \mu\text{s} \le 23\ \mu\text{s})\,}$$ A Schottky diode is specified because its low forward drop (≈ 0.3 V) matters against an envelope whose minimum is only 0.667 V. A series DC-blocking capacitor at the output removes the $A_c$ pedestal and recovers $m(t)$ alone.

6. Coherent detector (part e). The synchronous receiver multiplies the received signal by a locally generated carrier of the same frequency and phase, then low-pass filters: $$s(t)\cdot 2\cos(2\pi f_c t) = A_c\left[1+\tfrac13 m(t)\right]\bigl[1 + \cos(4\pi f_c t)\bigr],$$ and the low-pass filter (cut-off just above 12 kHz, far below $2f_c = 20$ MHz) discards the double-frequency term, leaving $A_c[1 + \tfrac13 m(t)]$. A DC block then removes the constant $A_c$ and delivers $\tfrac{A_c}{3}m(t) = 0.444\,m(t)$. The local carrier must be phase-locked — a phase error $\phi$ scales the output by $\cos\phi$, and a frequency error produces a slow beat — so a phase-locked loop or Costas loop recovers the carrier from the strong transmitted carrier line, which for full-carrier AM is easy because that line carries 93.5% of the total power. $$\boxed{\,\text{multiplier} \to \text{LPF (12 kHz)} \to \text{DC block},\ \text{LO phase-locked to } f_c\,}$$