- 2073 Bhadra (CS I) · 4+6 marks
Why AM wastes power and bandwidth; switching modulator
Short answer
In DSB‑FC AM the carrier carries no information yet uses most of the power, and both sidebands carry identical information so half the bandwidth is wasted. A switching modulator uses a diode switch driven by a large carrier; the switched output passes through a band‑pass filter to yield s(t) = (Ac/2)[1 + (4/(πAc))m(t)]cos ωct.
Why is conventional AM wasteful of power and bandwidth? Explain the method of conventional AM generation by using switching modulator.
Why conventional AM wastes power and bandwidth
Power: In DSB-FC AM, . The carrier carries no information, yet it takes most of the power. Even at , the carrier uses of the total and the sidebands only 33.3%. In practice on average, giving efficiency only.
Bandwidth: Both sidebands carry the same information (each is a mirror image of the other about ). Sending both needs bandwidth, while (one sideband) is enough. So half the bandwidth is wasted.
Switching modulator
A diode used as a switch, driven by a large carrier, generates DSB-FC AM.
m(t) --->(+)--- v1 ----|>|----+------> v2 --> BPF --> s(t)
^ D | (fc, 2W)
| R_L
Ac cos wc t |
(Ac >> |m(t)|) ---
Assumption: carrier amplitude , so the diode is ON during the positive half-cycles of the carrier and OFF during the negative half-cycles, regardless of .
Input: .
Output:
where is a unit-amplitude (0/1) pulse train at with 50% duty:
A BPF at with bandwidth keeps only the first two terms:
This is DSB-FC AM with .
v1: carrier + m(t) v2: positive half-cycles only
/\ /\ /\ /\ /\ /\
/ \/ \/ \/ / \_/ \_/ \_
(then BPF -> AM wave)
Advantage: works at high power levels with good efficiency, does not depend on an exact square-law characteristic.
✓ Answer checked by Claude Opus 5.5 (high effort)
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