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259 points zdw | 7 comments | | HN request time: 0.001s | source | bottom
1. davekeck ◴[] No.41832998[source]
I always assumed it was because FM station bandwidths (200kHz) are much wider than AM (10kHz). AM's 10 kHz chops off a lot of human-hearable frequencies.
replies(1): >>41833153 #
2. ndndjdjdn ◴[] No.41833153[source]
AM doesn't use the frequency for modulation though so it shouldn't matter.
replies(3): >>41833331 #>>41833354 #>>41835233 #
3. t-3 ◴[] No.41833331[source]
AM does use the frequency, it just doesn't need as much and uses it differently than FM. If it was all at a single frequency, there just be a single tone getting louder and softer.
replies(1): >>41833375 #
4. kragen ◴[] No.41833354[source]
When you amplitude-modulate a carrier wave with an audio signal, you spread it out into a bunch of sum and difference frequencies, as you can see if you use the trigonometric angle-sum formula to factor cos(85000·2πt) · (2 + cos(440·2πt)), a 440-hertz flute being transmitted on 85-kilohertz AM. These so-called "sidebands" mean that the bandwidth of AM does matter, and consequently, using a too-narrow bandpass filter on your AM radio station will result in low-pass filtering your demodulated audio signal.
5. ndndjdjdn ◴[] No.41833375{3}[source]
Thanks. I just learned that doing a rabbit hole about sidebands! Still getting my head around it.
replies(1): >>41834791 #
6. YZF ◴[] No.41834791{4}[source]
Once you change the amplitude of a sine wave (modulate it) it's no longer a side wave. It spreads in the frequency domain. Take the fourier transform of that and you can see the frequency components.
7. KK7NIL ◴[] No.41835233[source]
Other comments gave a nice explanation of why AM does need a bandwidth, but here's the information theory explanation: https://en.m.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theo...

TL;DR: the information one can reliably send through a noisy channel (C) is proportional to the bandwidth of that channel.