Question 5 of 6: Envelope Detection of DSB-SC Signals

Question 5: Envelope Detection of DSB-SC Signals (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. Four baseband signals, each multiplied by a carrier $\cos\omega_c t$ with $\omega_c$ far above every baseband component, giving the transmitted signal $s(t)=x(t)\cos\omega_c t$.

Find. Which of the four can be demodulated by a simple envelope detector (diode, RC time constant), with justification.

Approach. An envelope detector recovers the magnitude of the slowly varying multiplier, i.e. $|x(t)|$, not $x(t)$ itself. So the single criterion is whether $x(t)$ ever changes sign.

1. State the criterion. Writing $s(t)=x(t)\cos\omega_c t=|x(t)|\cos\!\left(\omega_c t+\phi(t)\right)$ with $\phi=0$ where $x\gt 0$ and $\phi=\pi$ where $x\lt 0$, the envelope is $|x(t)|$. Hence $$\boxed{\ \text{envelope detection recovers }x(t)\iff x(t)\ge 0\ \text{for all }t\ }$$ Every sign reversal of the message appears at the detector as a rectified fold, which is irreversible distortion.
2. (a) $x=2\cos(2\pi f_1 t)$. The minimum is $-2$, so the message reverses sign twice per cycle. The envelope is $2|\cos(2\pi f_1 t)|$, a full-wave-rectified cosine at twice the message frequency. Not recoverable by envelope detection; coherent (synchronous) detection is required.
3. (b) $x=\cos(2\pi f_1 t)+\cos(2\pi f_2 t)$. Written as $2\cos\!\left(\pi(f_1-f_2)t\right)\cos\!\left(\pi(f_1+f_2)t\right)$, this is a beat waveform whose minimum is $-2$; it passes through zero at every beat null. Not recoverable.
4. (c) $x=2+\cos(2\pi f_1 t)$. The DC term dominates: $x_{\min}=2-1=1\gt 0$, so $|x(t)|=x(t)$ everywhere. Recoverable. The envelope reproduces the message with a DC pedestal, removed by the detector output capacitor. Strictly this is conventional AM with modulation index $\mu=\tfrac{1}{2}=50\%$ rather than true DSB-SC, since a carrier term $2\cos\omega_c t$ is transmitted.
5. (d) $x=2+2\cos(2\pi f_1 t)+\cos(2\pi f_2 t)$. The two tones can align in anti-phase with the pedestal, giving $x_{\min}=2-2-1=-1\lt 0$ whenever $f_1$ and $f_2$ are incommensurate (a numerical sweep over 600 001 samples confirms the minimum reaches $-1$). Because the pedestal is smaller than the summed tone amplitudes the carrier is over-modulated. Not recoverable — raising the DC term above 3 would fix it.