Lesson 29 Flashcards — AM Bandwidth and Multiplexing#
Click a question to reveal the answer.
1. What single question decides everything about an AM signal?
Does the envelope ever cross zero? As long as it stays above zero its shape is the message. That threshold is exactly \(\alpha = 1\).
2. What is the AM envelope, and how do you read \(\alpha\) from it?
The envelope is \(A_c[B + v_m(t)]\). For a single tone it runs from \(A_c(B - A_m)\) up to \(A_c(B + A_m)\), and \(\alpha = (\max - \min)/(\max + \min)\).
3. Describe under-modulation.
\(\alpha < 1\). The envelope never reaches zero, so a cheap envelope detector recovers the message — but a lot of power sits in the carrier doing nothing (Lesson 30 puts a number on it).
4. Describe full modulation.
\(\alpha = 1\). The envelope just touches zero at the message minimum. This is the most power-efficient case an envelope detector can still handle (Lesson 30 puts a number on it).
5. Describe over-modulation and why it is not merely 'worse'.
\(\alpha > 1\). The envelope crosses zero and folds, so its outline no longer traces the message. An envelope detector recovers garbage — it breaks the cheap receiver entirely.
6. Give the bandwidth of an AM signal.
\(BW = 2f_m\) for a single tone, or \(BW = 2f_{m,\max}\) for a multi-tone message.
7. Does the carrier frequency affect AM bandwidth?
No. Bandwidth is set by the message. The carrier only decides where that block of spectrum sits.
8. In an AM output spectrum, where does the bias appear, and what is \(f_{low}\)?
The bias rides at \(f_c\) with height \(A_cB\), not at \(f = 0\). So \(f_{low} = f_c - f_{m,\max}\): Lesson 15's rule that a DC term makes \(f_{low} = 0\) is about the message, not the AM output.
9. What does a multi-tone message produce in the spectrum?
Each tone produces its own pair of sidebands, \(A_c/2\) times that tone's amplitude, so the spectrum is symmetric about the carrier. Total bandwidth is set by the highest tone.
10. How do you find max(message) for a multi-tone message?
For cosines that all peak together, add the tone amplitudes: tones of 3 V, 2 V, and 1 V give \(\max(\text{message}) = 3 + 2 + 1 = 6\) V. With a bias of 8 V, \(\alpha = 6/8 = 0.75\).
11. How is a non-sinusoidal message handled?
Any signal decomposes into a sum of sinusoids (Objective 2.1), so its spectrum is a continuous curve and every point on it follows the same rule: the upper sideband is the message's shape shifted up by \(f_c\) and scaled by \(A_c/2\), the lower sideband is its mirror image, and \(BW = 2f_{m,\max}\). With no bias there is no carrier line.
12. What is frequency division multiplexing?
Giving each channel its own carrier and spacing the carriers at least \(BW\) apart so the blocks do not overlap. Real systems add a guard band because filters are not ideal.
13. Three 4 kHz voice channels are multiplexed from 100 kHz. Where do the carriers go?
\(BW = 8\) kHz each, so carriers at 100, 108, and 116 kHz, occupying 96–120 kHz in total.