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I understand that similar questions like this one have been asked before on this site, listed below. However, I am confused about the answers. If I explain what I think I understand, can somebody please point out where i'm wrong?

I'll start with what I do know:

Shannon Law gives the theoretical upper limit

$$C_{noisy}=B*log_{2}(1+\frac{S}{N})$$

if S = N, then C = B

As N→∞, C→0

As N→0, C→∞

Nyquist Formula says approximately how many levels are needed to achieve this limit

$$C_{noiseless}=2*B*log_{2}M$$

(If you do not use enough logic levels you can not approach the shannon limit, but by using more and more levels you will not exceed the shannon limit)


My problem is that I'm having a hard time understanding why bandwidth relates to bit rate at all. To me it seems like the upper limit of the frequency that can be sent down the channel is the important factor.

Here's a very simplified example: No noise at all, 2 logic levels (0V and 5V), no modulation, and a bandwidth of 300 Hz (30 Hz - 330 Hz). It will have a Shannon Limit of ∞, and a Nyquist Limit of 600bps. Also assume that the channel is a perfect filter so anything outside of the bandwidth is completely dissipated. As I double the bandwidth, I double the bit rate etc.

But why is this? For two level digital transmission With a bandwidth of 300 Hz (30 Hz - 330 Hz), the digital signal of "0V's" and "5V's" will be a (roughly) square wave. This square wave will have the harmonics below 30 Hz and above 330 Hz dissipated, so it will not be perfectly square. If it has a fundamental frequency at the minimum 30 Hz, (so the "0V's" and "5V's" are switching 30 times a second), then there will be a good amount of harmonics and a nice square wave. If it has a fundamental frequency at the max 330 Hz, the signal will be a pure sine wave as there are no higher order harmonics to make it square. However, as there is no noise the receiver will still be able to discriminate the zeros from the ones. In the first case the bit rate will be 60 bps, as the "0V's" and "5V's" are switching 30 times a second. In the second case the bit rate will be a maximum of 660bps, (if the threshold switching voltage of the receiver is exactly 2.5V), and slightly less if the threshold voltage is different.

However this differs from the expected answer of 600 bps for the upper limit. In my explanation it is the upper limit of the channel frequency that matters, not the difference between the upper and lower limit (bandwidth). Can somebody please explain what have I misunderstood?

Also when my logic is applied to the same example but using FSK modulation (frequency shift keying), I get the same problem.

If a zero is expressed as a 30 Hz carrier frequency, a one is expressed as a 330 Hz carrier frequency, and the modulation signal is 330 Hz, then the max bit rate is 660 bps.

Again, can somebody please clear up my misunderstanding?

Also why use a square wave in the first place? Why cant we just send sine waves and design the receivers to have a switching threshold voltage exactly in the middle between the max and min value of the sin wave? This way the signal would take up much less bandwidth.

Thanks for reading!

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  • \$\begingroup\$ Apologies for the very poor formatting, I didn't preview before I posted. I have fixed this now. \$\endgroup\$ – Blue7 Mar 30 '14 at 22:08
  • \$\begingroup\$ @Ignacio Vazquez-Abrams, oh okay, that surprises me; I assumed that it would simplify my example. 5 or so Harmonics usually gives a pretty decent square wave though so why would you need frequencies outside the bandwidth to avoid distortion? \$\endgroup\$ – Blue7 Mar 30 '14 at 22:13
  • \$\begingroup\$ Instead of thinking about what happens with a passband of 30-300 Hz, imagine what would happen if your passband was 1.0 to 1.3 kHz, for example. \$\endgroup\$ – The Photon Mar 30 '14 at 22:30
  • \$\begingroup\$ @ThePhoton: I suppose in this case you will not be able to have any higher order harmonics, because when the fundamental frequency is 1kHz, the 1st harmonic is 3KHz, which is way outside the passband. But this still leave me confused. What would be the harm in just transmitting the fundamental frequency? \$\endgroup\$ – Blue7 Mar 30 '14 at 22:47
  • \$\begingroup\$ first some terminology. The fundamental is the same thing as the first harmonic. If the fundamental is 1 kHz,then 3 kHz is the third harmonic. \$\endgroup\$ – The Photon Mar 31 '14 at 0:10
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It's a subtle point, but your thinking is going astray when you think of a 330-Hz tone as somehow conveying 660 bits/second of information. It doesn't — and in fact, a pure tone conveys no information at all other than its presence or absence.

In order transmit information through a channel, you need to be able to specify an arbitrary sequence of signaling states that are to be transmitted, and — this is the key point — be able to distinguish those states at the other end.

With your 30-330 Hz channel, you can specify 660 states per second, but it will turn out that 9% of those state sequences will violate the bandwidth limitations of the channel and will be indistinguishable from other state sequences at the far end, so you can't use them. This is why the information bandwidth turns out to be 600 b/s.

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  • \$\begingroup\$ Actually, by sending only 30 symbols of 2 states per second, the data rate is 30bps. Nyquist tells us about the upper data rate limit given a bandwidth and number of states per symbol. The chosen FSK encoding does not come close to this limit as the chosen frequencies are not optimum. Nyquist says that we can choose better frequencies. \$\endgroup\$ – le_top Jun 11 '16 at 12:17
  • \$\begingroup\$ @le_top: Just to be clear, I was not talking about FSK modulation, even though the OP mentioned it in his question. I'm talking about straight baseband signaling (e.g., two voltage levels). I don't think anything I wrote was worth a downvote. Can you explain what you think is wrong with what I wrote? \$\endgroup\$ – Dave Tweed Jun 11 '16 at 13:50
  • \$\begingroup\$ * The presence or absence of the 330Hz tone does convey information as its presence can be interpreted as a 1 and its absence as a 0. The modulation is on/off. * So the 330Hz could convey 660bps information in the absence of the 30Hz tone. Which would be noise in Shannon's formula. * The confusion still existed after reading this. * It is not explained that the 9% loss is explained by the Nyquist sampling theorem indicating that the signal is perfectly reconstructed from precisely 2B samples per second. \$\endgroup\$ – le_top Jun 11 '16 at 15:12
  • \$\begingroup\$ * If you try to do more, you have aliasing effects, hence the limitation 2B symbols equal to the number of samples. * 2B symbols of each 1 bit (2 states) is 600bps with B=300. * 660 states are possible if symbols represent at least 2.2 states. \$\endgroup\$ – le_top Jun 11 '16 at 15:12
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    \$\begingroup\$ @le_top: I really don't understand where you're going with this. We have not been talking about sampled (discrete-time) systems at all, so the question of aliasing never arises. What is your point relative to the question at hand? \$\endgroup\$ – Dave Tweed Jun 11 '16 at 23:09
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This is only a partial answer, but hopefully it gets at the main points you're misunderstanding.

My problem is that I'm having a hard time understanding why bandwidth relates to bit rate at all. ...

If a zero is expressed as a 30 Hz carrier frequency, a one is expressed as a 330 Hz carrier frequency, and the modulation signal is 330 Hz, then the max bit rate is 660 bps.

If you switch down to 30 Hz for a zero, you need to have about 1/60 s or so to really know you got 30 Hz and not 20 Hz or 50 Hz or something. Really in this case you are just on-off keying your 300 Hz carrier, and the 30 Hz signal that's sent for 1/660 s during the zeros is just confusing things.

To talk about FSK, let's take a more realistic example. Say you use 1 MHz for the zero and 1.01 MHz for the one. It turns out you need to measure the signal for about \$1/2\Delta{}f\$, in this case 1/20,000 s, to be able to reliably distinguish those two frequencies. If you just measured the signal for 1 us, you wouldn't really be able to tell the difference between a 1 MHz signal and a 1.01 MHz signal (although in an ideal, noise-free scenario you could do it, just as Shannon's formula says you can transmit infinite data with zero bandwidth when SNR goes to infinity)

So in this example the bit rate you can send is about 20 kHz, corresponding to 2x the difference between your 1 and 0 frequencies, just as the Nyquist formula leads you to expect for a 2-level code.

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Your questions are valid and the path to a proper understanding of what the theory means ;-).

To the question how more bandwidth means a higher bit rate, the explication may look simple but be bad at the same time.

Here is a "bad" explication which looks ok. It is a start to understanding why bigger bandwith is more data though. Suppose that I have a first WiFi channel number 1 running at 1Mb/s given the power and encoding conditions. Then I take another WiFi channel number 2 which has the same bandwidth, power and encoding conditions. It is also running at 1Mb/s. When I sum the two together, I have doubled the bandwidth (two different channels) and double the data throughput (2x1Mb/s).

If you think that this looks like a perfect explication, you forget that we also doubled the power. So is the double data throughput due to the doubled power or due to the doubled bandwidth. It is a bit of both actually.

If I maintain the total power the same while doubling the bandwidth, I need to compare a first WiFi channel running at 1Mb/s with the sum of two other WiFi channels running each at half the received power. I am not going to check the datasheets of WiFi modems, but this would be an interesting excercise to compare with the following theoretical approach. Shannon helps us in prédicting what will happen more or less if the encoding adapts itself to the power levels (which is the case for WiFi). If the encoding does not adapt, the data rate remains constant until the reception level is too low at which time it drops to 0.

So shannon says: C=B∗log2(1+S/N) . When keeping total power, but doubling bandwidth, C2=2*B*log2(1+(S/2)/N) where C2 is the potential data rate. Filling in actual numbers we could suppose that S=2xN so that log2(1+2)=1.58 and log2(1+1)=1. So C=B*1.58 and C2=B*2 . In other words, when my signal level at the biggest bandwidth equals the noise level, the potential data rate is bout 26% higher than the same total power emitted in half the bandwidth. So theoretically, ultra narrow band can not be more efficient than ultrawide band based on Shannon's theorem. And doubling the bandwidth with the same total power level does not double the bandwidth as our WiFi example suggested. But the bandwidth is higher. If we can neglect the "1" term in the log2 of the Shannon expression, then you can easily see that it is more interesting to increase bandwidth than to increase the power (which is subject to a log2, lowering its impact on the data rate).

However, as I mentionned, the encoding must adapt, it must be optimised to the actual power and bandwidth available. If the encoding stays the same, I simply go from operational to dysfunctionnal.

Switching into your second question, if I have an FSK signal changing at 30Hz with two frequencies, then I can only emit at 30bps because I am emitting 30 symbols per second each corresponding to a bit of 1 or 0. If I introduce 4 states (=4 frequencies) by introducing two frequencies in between the previous ones because my noise level allows it, then I emit at 4x30bps=120bps. With FSK, I do not think that bandwidth remains constant when increasing the number of states this way, but one can surely find a way to keep it more or less constant (considering the 3dB limits because the theoretical frequency spectrum is unlimited).

Why use a square wave for the "modulating" signal? This is a choice in this encoding which makes it "easier" to decode as on the receiver side you simply have to have a bandpass filter for each frequency. You are still emitting "sine waves" - if you are emitting only "1" values, you have just one frequency. However the frequency shifts imply the presence of "harmonics" that allow/accompany these frequency shifts. Other encodings have other advantages and disadvantages. For instance, Direct Sequence Spread Spectrum allows having a signal below the noise level (and therefore have lower antenna power requirements for a similar bitrate in many other encodings), but it is more difficult to decode (and hence require more (compute) power and complexity in the decoding circuit).

Whatever the chosen encoding is, it has to respect the Shannon theorem which fixes the upper limit. You can not just apply Shannon to an encoding like FSK if you do not adjust the power level, number of states, and other parameters of the FSK signal as the noise level or signal level (distance) changes. Shannon allows you to check the absolute minimum power for a given bandwidth and data rate. The encoding method will increase the minimum power limit. And when the power levels exceed this limit, the bit rate will simply remain constant. Applying Shannon there is simply incorrect if you want to explain that more bandwidth means a higher bitrate. The WiFi example might very well apply in practice for an explication there, but it is not the general anwser based on Shannon's theorem.

Edit: rereading your question, "In the second case the bit rate will be a maximum of 660bps". Actually I do not fully understand how you get to 660bps as your frequency changes only 30 times each second and you encode on two frequencies which is 1 bit. Hence my 30bps above. This encoding allows for one full period at 30Hz and 22 full periods at 660Hz for each symbol. But 22 periods does not change the fact that there is only one symbol. It looks like something is missing or that the reasoning is wrong.

Edit2: I got it - you are comparing with the nyquist limit. This nyquist limit tells you the upper limit of the data rate given a bandwidth and the number of states per symbol. Here, the selected FSK encoding is not optimimum. You are using 30Hz and 660Hz. The Nyquist limit says that 30bps=2*B*log2(2), therefore, the bandwith must be at least B=15Hz. Without checking into detail, it says more or less that setting the FSK frequencies to 645Hz and 660Hz would be a good optimisation of the bandwidth (if FSK is otherwise an optimum encoding and without checking the precise bandwidth due to harmonics - the 15Hz may be too low for FSK).

Edit 3 - Explication following after further analysis to further explain source of confusion with other anwser and original question.

  • The Nyquist formula is based on the sampling theorem indicating that a signal with a bandwidth B is perfectly reconstructed from precisely 2B samples per second.
  • Hence the 2B samples can each represent a symbol (the intensity can determine which symbol).
  • A signal with bandwidth of 300Hz can be reconstructed with 600 symbols - no more no less.
  • This is why "aliasing" exists - bandwidth limitation can make two different signals look the same after sampling.
  • If each symbol only represents 2 states, then only 600 bps is possible.
  • The FSK from 30Hz to 330Hz can represent more than 600 bps, but then you need to consider more than 2 states per symbol. But is no longer an FSK demodulation because one can not only consider the frequency.
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