22-Elec-A3 Signals and Communications · May 2015

Question 3 of 7: Conventional AM — Power Budget, Spectrum, Envelope and Detection

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

Question 3: Conventional AM — Power Budget, Spectrum, Envelope and Detection (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 expression with numerical carrier amplitude, the line spectrum, the envelope and its parameters, block diagrams for both detectors, and a reasoned comparison.

Approach. Write the standard tone-modulated AM expression, use the power split between carrier and sidebands to pin the carrier amplitude \(A_c\), then read the spectrum and envelope extremes directly from that expression.

1. Time-domain expression and the power budget. For conventional AM with a cosine message, $$s(t) = A_c\left[1 + \mu\cos(2\pi f_m t)\right]\cos(2\pi f_c t)$$ Its average power splits into carrier and sideband parts, $$P_T = \underbrace{\frac{A_c^2}{2}}_{\text{carrier}} + \underbrace{\frac{A_c^2\mu^2}{4}}_{\text{sidebands}} = \frac{A_c^2}{2}\left(1 + \frac{\mu^2}{2}\right)$$ Solving with \(P_T = 3\) W and \(\mu = 0.9\): $$A_c = \sqrt{\frac{2P_T}{1+\mu^2/2}} = \sqrt{\frac{6}{1.405}} = \boxed{\,2.067\ \text{V}\,}$$ so \(s(t) = 2.067\left[1 + 0.9\cos(2\pi\cdot 20{,}000\,t)\right]\cos(2\pi\cdot 10^{7}t)\ \text{V}\).
2. Plot of the modulated waveform. The carrier fills in between an upper and lower envelope; with \(f_c/f_m = 500\) the drawing below compresses the carrier for legibility.
   Figure Q3.1 — AM waveform (blue) with its envelope (dashed): Ac = 2.067 V, μ = 0.9, envelope 0.21 to 3.93 V. Carrier compressed for legibility.
3. Spectrum. Expanding the product gives three lines on each side of the origin. On the positive-frequency axis the carrier line has amplitude \(A_c/2 = 1.033\) V at 10 MHz and each sideband has amplitude \(A_c\mu/4 = 0.465\) V at \(10\ \text{MHz} \pm 20\ \text{kHz}\), so the transmission bandwidth is $$B_T = 2f_m = \boxed{\,40\ \text{kHz}\,}$$
   Figure Q3.2 — positive-frequency line spectrum: carrier 1.033 V at 10 MHz, sidebands 0.465 V at 10 MHz ± 20 kHz; BT = 40 kHz.
4. Envelope and its parameters. Since \(\mu \lt 1\) the bracket never changes sign, so the envelope is simply $$E(t) = A_c\left[1 + \mu\cos(2\pi f_m t)\right]$$ a 20 kHz cosine riding on a DC pedestal. Its parameters are \(E_{\max} = A_c(1+\mu) = 3.926\) V, \(E_{\min} = A_c(1-\mu) = 0.207\) V, mean level \(A_c = 2.067\) V, and period \(1/f_m = 50\ \mu\text{s}\). As a check, \(\mu = (E_{\max}-E_{\min})/(E_{\max}+E_{\min}) = 3.719/4.133 = 0.9\). The dashed curve in the step-2 figure is exactly this envelope.
5. Transmission efficiency. Only the sidebands carry information: $$\eta = \frac{A_c^2\mu^2/4}{P_T} = \frac{0.865}{3} = \boxed{\,28.8\%\,}$$ The carrier consumes the remaining 2.135 W — the well-known price of conventional AM, and the reason DSB-SC or SSB is used where power matters.
6. Envelope detector (part d). A diode rectifies the received signal and an RC network follows the peaks; the time constant must be slow compared with the carrier but fast compared with the message, \(1/f_c \ll RC \ll 1/f_m\), i.e. \(0.1\ \mu\text{s} \ll RC \ll 50\ \mu\text{s}\) — a value near \(5\ \mu\text{s}\) is a sound choice here. A series capacitor then removes the DC pedestal.
   Figure Q3.3 — envelope detector: diode rectifier, RC follower, DC block.
7. Coherent detector (part e). Multiplying by a locally generated carrier of the same frequency and phase and low-pass filtering recovers the message: $$s(t)\cdot 2\cos(2\pi f_c t) = A_c\left[1+\mu\cos(2\pi f_m t)\right]\left[1 + \cos(4\pi f_c t)\right]$$ The low-pass filter removes the \(2f_c\) term, leaving \(A_c[1+\mu\cos(2\pi f_m t)]\); a DC block strips the constant. The local carrier must be phase-locked, typically by a phase-locked loop or a Costas loop, because a phase error \(\phi\) scales the output by \(\cos\phi\) and nulls it completely at \(\phi = 90^\circ\).
   Figure Q3.4 — coherent (synchronous) detector: multiplier driven by a phase-locked local carrier, low-pass filter, DC block.

Part (f) — which detector is preferred. For this signal the envelope detector is preferred. It is essentially free — one diode, one resistor, one capacitor, no oscillator and no synchronisation — and it works correctly here precisely because the transmission satisfies the conditions for envelope detection: a large unmodulated carrier is present and \(\mu = 0.9 \lt 1\), so the envelope is a faithful copy of \(1 + \mu m(t)\) and never folds over. That cheapness is the entire commercial argument for conventional AM broadcasting: complexity and power are pushed into the single transmitter so that millions of receivers can be trivial.

The coherent detector is nevertheless used when the envelope carries no usable information or when performance is critical. It is mandatory for DSB-SC and SSB, where there is no carrier to define an envelope, and for over-modulated signals (\(\mu \gt 1\)) where the envelope is a distorted, rectified version of the message. It also performs better at low signal-to-noise ratio: the envelope detector exhibits a threshold effect below roughly 10 dB carrier-to-noise ratio, where the output degrades far faster than the input, whereas the coherent detector degrades gracefully. Its cost is the need to regenerate a carrier of exactly the right frequency and phase.