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Newbie struggling to understand why a larger chunk of signal bandwidth is needed to carry more information. For example, why do you need more bandwidth to carry video as opposed to voice or CW? Apparently missing some fundamental concept here. Have a sneaking suspicion this will lead to information theory. Thanks in advance for any help!

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Have a sneaking suspicion this will lead to information theory.

Well, it won't lead to information theory so much as it is information theory.

In non-technical terms, if you have a chunk of signal, in order to pack more information into it you either need to be able to distinguish between fine differences in the signal (low noise, or high signal energy) or you need to pack more separate values (symbols) into that chunk of signal.

As data rates get higher (CW to voice to video), you need to pack more data into the signal per second. There's only so far you can go to reduce the noise (or boost the power); once you've done your best there then you can only increase the symbol rate. More symbols per second mean a more rapidly-changing signal, and that means more bandwidth.

What I've just given you is about as much of the Shannon channel capacity theorem as I can give you without math or pictures. Beyond that, yes, it's all information theory (of which the Shannon channel capacity is one of, or depending on who you ask, the, foundational theorem(s)).

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  • $\begingroup$ Thanks! I'll have to look into Shannon's Channel Capacity theorem. I agree that the answer might lie there. However, with regard to "More symbols per second mean a more rapidly-changing signal, and that means more bandwidth." I think that is essentially the question I was asking, that is, why do more signals per second require more bandwidth? I'm failing to grasp some fundamental relationship that I think might be intuitive for other folks. $\endgroup$
    – georgers
    Aug 20 at 14:46
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One perspective: greater bandwidth $\Leftrightarrow$ lesser subsampleability. That is, lesser lossless subsampleability; (opposite case) if fewer data points can represent the same signal, then it's more compressible/reducible/redundant, or, the original is "more of the same thing".

If we have 128 samples and 16 DFT bins, -8 to 7, are nonzero, then we can subsample without aliasing by 8 (128/2/max(8,7)). This follows from subsampling in time $\Leftrightarrow$ folding in frequency - example unaliased subsampling by 2, in frequency:

$$ [a, b, 0, 0, 0, 0, 0, 0, c, d] \rightarrow [a, b, 0, c, d] $$

it's right half + left half. Aliased:

$$ [a, b, A, B, 0, 0, C, D, c, d] \rightarrow [a, b + C, A + D, B + c, d] $$

This is "bad" since, if $b + C = 6$, we don't know if that's $2 + 4$ or $7 - 1$.

Note, aliasing is sensitive to where the bandwidth is, but information isn't: we can always shift the bandwidth to center around DC ("baseband") in subsampling, then undo the shift in upsampling. Indeed, aliasing != lossy.

Intuitively, greater bandwidth $\Leftrightarrow$ greater range of variation. One might think the location in frequency matters, i.e. higher freq = more info since it's more wiggles per unit time, but if they're the only wiggles, as in a high freq pure sine, then these wiggles can't tell us much, nor do we need much to describe them. What matters is how many kinds of other wiggles there are - like putting legos together.

This is also evident in STFT, which, despite extreme aliasing with large hop_size, can remain perfectly invertible. Wider windows have greater maximum hop_size, and of course wider in time $\Leftrightarrow$ narrower in frequency, meaning greater subsampleability, and hop_size is subsampling of STFT along time.

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    $\begingroup$ Thanks for the detailed answer. A bit over my head just now but I'm working on it. $\endgroup$
    – georgers
    Aug 20 at 14:55
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The concept of channel capacity is easiest to understand in a wired serial link like RS232, Ethernet, or USB. If you want to send more data per second, you have two options:

  • Increase the symbol rate, also called baud rate or bandwidth. This means changing the voltage on the wire more frequently. At some point, you run into problems with parasitic capacitance and inductance of the wire, which prevent you from changing the voltage all too quickly.

  • Go from binary signaling to 4, 8, or 256 distinct voltage levels. The receiver needs a good analog-to-digital converter and you need to keep noise to a minimum (shielded cables, etc.). At some point, you run into the noise limit and the receiver will decode the wrong symbol if the voltage difference is too small.

If you use 16 distinct voltage levels (4 bits/symbol) and switch to a different voltage 1'000'000 times per second (1 MHz bandwidth), you get a data rate of 4 Mbit/s.

If you encode the data directly in voltage levels, that's called baseband communication and it occupies the entire spectrum up to your symbol rate. The alternative is to pick a carrier frequency and encode data in amplitude or phase of that signal. This is called carrier frequency communication and occupies the part of the spectrum that falls within carrier ± symbol rate. If you change the amplitude of a 2.4 GHz carrier up to 1'000'000 times per second, you get a voltage that fluctuates with 2.4 GHz ± 1 MHz depending on the exact amplitude pattern, but you can still only encode 4 Mbit/s with 16 distinct amplitude levels, despite the high carrier frequency. Most radio links and some wired links use carrier frequency communication. Most wired links use baseband communication.

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  • $\begingroup$ "Increase the symbol rate... ...means changing the voltage on the wire more frequently" But wouldn't that imply that, say, 1 MHz of bandwidth on a higher frequency portion of the spectrum would carry more data that a 1 MHz portion at a lower frequency, which I understand isn't correct? $\endgroup$
    – georgers
    Aug 20 at 14:54
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    $\begingroup$ Bandwidth is the number of symbols per second. Each symbol carries a fixed amount of data depending on the number of distinct voltage/amplitude/phase/whatever levels. The same bandwidth carries the amount of data regardless of carrier frequency. You don't even need a carrier frequency. $\endgroup$
    – Rainer P.
    Aug 20 at 16:31
  • $\begingroup$ I think I understand this a little better. For FM the larger the chunk of spectrum gives you a wider range of frequencies to use in composing your symbols. However, it seems like with AM, since only the amplitude changes, it would seem that in that case a higher frequencies would allow for more information throughput. So does the rule about portions of similar bandwidth from different parts of the spectrum being equal in terms of bandwidth not really apply to AM? $\endgroup$
    – georgers
    Aug 20 at 17:20
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    $\begingroup$ The thing is that amplitude modulation distorts the carrier enough to change its frequency. That sounds weird, but it's just math: sin(x) sin(y) = cos(x-y)/2 - cos(x+y)/2. You end up with two cosines of carrier+modulation and carrier-modulation. (@georgers) $\endgroup$
    – Rainer P.
    Aug 20 at 17:37

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