# Lesson 30 Flashcards — Demodulation Click a question to reveal the answer.
1. Which part of an AM signal actually carries the message?

Only the sidebands. The carrier term carries none of it, which is why modulation efficiency matters.

2. Give the modulation efficiency formula.

\(\eta = \alpha^2/(\alpha^2+2)\) — exact for a single-tone message, and the course's estimate for any other. At \(\alpha=1\) this is \(1/3 = 33.3\%\), the best an envelope-detectable signal can do.

3. As α decreases, what happens to efficiency?

It decreases — more power is wasted in the carrier.

4. If under-modulation wastes power, why does AM broadcast use it?

Because it makes the receiver a diode and a few resistors and capacitors. One transmitter serves millions of receivers, so spending power at the expensive end to make the cheap end cheaper is the right trade.

5. What are the two demodulation methods and when does each work?

Envelope detection works only for \(\alpha \le 1\) and is cheap. Synchronous detection works for all AM signals but is expensive and complex.

6. Describe the envelope detector block chain.

Rectifier → low-pass filter → high-pass filter. The rectifier keeps the positive half, the LPF removes everything from \(f_c - f_{m,\max}\) up (the carrier-frequency content), and the HPF strips the DC level, leaving the message's shape. When other channels share the band, a band-pass filter tuned to this channel goes first.

7. Describe the synchronous detector block chain.

Multiply by a local oscillator at exactly \(f_c\), then LPF then HPF. Multiplying shifts the sidebands back to baseband and creates an image at \(2f_c\) which the LPF discards; the HPF removes the DC term the carrier leaves behind.

8. What does the mixer in a synchronous detector do to each component of the received spectrum?

It scales the component by \(A_\text{LO}/2\) and moves it to \(f - f_c\) and \(f + f_c\). The two sidebands land on the same baseband frequency and add, so a message tone returns at \(A_cA_\text{LO}A_m/2\). The carrier lands on 0 Hz once: a DC term of \(A_cA_\text{LO}B/2\), which does not double.

9. What is \(A_\text{LO}\), and what else might a problem call it?

The amplitude of the detector's local oscillator. A quiz may call it the carrier signal amplitude, \(A_c\) — meaning the demodulating carrier inside the detector, not the transmitter's carrier.

10. Why does over-modulation not defeat a synchronous detector?

Because recovering the message never depended on the envelope shape — only on the frequency shift. The cost is needing the local oscillator to match \(f_c\) in frequency and phase.

11. How do you specify the two filters in a demodulator design?

LPF cutoff just above \(f_{m,\max}\); HPF cutoff at about 10 Hz — enough to strip DC without touching the lowest message tone. Any LPF cutoff from \(f_{m,\max}\) up to below the lowest unwanted component is a correct design: that component starts at \(f_c - f_{m,\max}\) for an envelope detector and at \(2f_c - f_{m,\max}\) for a synchronous one.

12. A signal has max(message) = 4 and bias 3. Which detector must you use?

\(\alpha = 4/3 = 1.333 > 1\), so it is over-modulated and only a synchronous detector will work. Its efficiency is \((4/3)^2/((4/3)^2+2) = 8/17 = 47.1\%\).