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I try to understand this question since a while and I think I came to an understanding. But I need someone to confirm it. At first, I thought the bitrate was dependent on the frequency (more sines per sec = more data per second). Now I understand that that was completely wrong.

Here is my understanding now:

Let's say I have a carrier frequency of 1000 HZ

  • So my 1000 HZ frequency represents one state A.
  • Now, if I have a change in that frequency (modulation) to 1001, that would be state B.
  • Another change to 1002, that would be state C.
  • And one more change to 1003, that would be state D.

Now let's look at the bandwidth:

  • If I look at frequencies 1000 & 1001 only, I would have a bandwidth of 2, and I could represent 2 states (A and B, or 0 and 1)
  • If I look at all 4 frequencies (1000-1003) I would have a bandwidth of 4, and I could represent 4 states (A, B, C, D or 00, 01, 10, 11)

So in case 1, with bandwidth of 2 I have 1 bit of information that I can carry and in case 2 I can carry 2 bit per time unit.

I know this is way to simple, but is it in the right direction?

Update: ok, I think I got something wrong. This helped me:

I just found those slides about FSK https://www.slideshare.net/jessierama/frequencyshift-keying

There is this image: enter image description here

So in this case, with FSK there is a bitrate of 5 bps. Because the bandwidth allows to switch the states 5 times per second. When I have a higher bandwidth, I can switch more often (faster) the states, so I have more bits per second. Is that assumption right?

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    \$\begingroup\$ There's nothing saying you can't have a state E with frequency 1000.5 Hz. Frequency is a continuous variable, not discrete. The important thing to realize is that it takes around (not exactly) 1 second for a receiver circuit to tell the difference between 1000 and 1001 Hz. But it takes around 2 s to tell the difference between 1000 and 1000.5 Hz. \$\endgroup\$
    – The Photon
    Dec 15 '20 at 21:12
  • \$\begingroup\$ Also you could send two voltages like 1V for 1 bit and 0V for 0 bit and tranmit 1000 bits per second. But you could also add voltages and use 16 voltages to send 4 bits at a time to have 4000 bits per second, by still using baud rate of 1000. \$\endgroup\$
    – Justme
    Dec 15 '20 at 21:24
  • \$\begingroup\$ I don't think the AB thing holds up. If you switch AB at rate X you need some amount of bandwidth. Now if you switch at a rate 2x as fast then you need more bandwidth. \$\endgroup\$
    – Aaron
    Dec 15 '20 at 21:25
  • \$\begingroup\$ ok, maybe the problem is, that I don't have the right background knowledge about this stuff....but your answers give me a headache ;). So basically you say my understanding is still totally wrong? \$\endgroup\$
    – DanielG
    Dec 15 '20 at 22:20
  • \$\begingroup\$ @ThePhoton : hmm, I think that's the main part I'm missing. Why does it take longer to tell the difference between 1000 and 1000.5 Hz? \$\endgroup\$
    – DanielG
    Dec 15 '20 at 22:22
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What you describe is called "FSK", or https://en.wikipedia.org/wiki/Frequency-shift_keying

Also important to note that

  1. you need more bandwidth than just the spread of the states, in order to be able to transition between them. How much depends on how fast you want to switch.
  2. how long you have to hold a frequency before you switch to another depends on how far the frequencies are apart.

All this matters if you are transmitting into a band limited channel, meaning that only a certain amount of bandwidth is available, either by regulatory limits or by physical limits.

Consider the "Talking Tube" or "Speaking Tube" used long ago as intercoms on ships, in houses and now in playgrounds. The longer the tube the less bandwidth is available to you and the slower you have to talk. This is separate from attenuation, which requires you to speak louder over longer tubes.

These tubes are a great example of a physically bandlimited communications channel, with lots of reflections and dispersion!

https://en.wikipedia.org/wiki/Speaking_tube

The bandwidth of the tube determines what individual frequencies get through, but also -and this should not be overseen- how long it takes for a switched frequency to be discernible from others. That limits how fast you can switch from one to another.

If you transmit tones (of around 1kHz), the longer the tube the longer you have to hold the tone before the receiver can determine what frequency was sent. Through longer tubes it takes time for a switched sinusoid at the input to start looking like a sinusoid at the output. This is not just a matter of tube delay (pure delay wouldn't limit your transmission rate) but rather due to the reflections inside the tube.

For more details on the bandwidth for FSK see here:

enter image description here

Notice that the bandwidth is "blobbier" than just two spikes at two frequencies, and the total width is more than the just the span between the two frequencies.

https://www.testandmeasurementtips.com/analog-and-digital-modulation-and-modulation-measurements/

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  • \$\begingroup\$ Ok, but for FSK my simple approach is true, isn't it? Or am I missing something? For FSK (or M-FSK), the more frequencies I have, the more data can be transported. I mean, it is clearly more complicated, but the basic idea is correct? Basically, I want to understand, why bandwidth is important for bitrate and not frequency. So, let's say I have 2 frequencies: 1MHz and 2MHz and for both I use the same bandwidth, then the bitrate should be similar. That's what I want to understand. \$\endgroup\$
    – DanielG
    Dec 16 '20 at 8:58
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    \$\begingroup\$ @DanielG, and to your other point "I can switch more often (faster) the states, so I have more bits per second. Is that assumption right?" -- Yes. \$\endgroup\$
    – P2000
    Dec 16 '20 at 15:39
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    \$\begingroup\$ @DanielG "let's say I have 2 frequencies: 1MHz and 2MHz and for both I use the same bandwidth, then the bitrate should be similar." The bandwidth needs to span 1MHz to 2MHz plus, on either end, side-lobes that have a width depending on the switching rate. If you are independently switching the carriers on/off, then it's not FSK but OOK (on-off-keying) and the bandwidth around both depends on the rate, and you can say "for both I use the same bandwidth, then the bitrate should be similar". \$\endgroup\$
    – P2000
    Dec 16 '20 at 17:47
  • \$\begingroup\$ @DanielG You can of course transmit slower. If you transmit faster, there is a maximum speed and if you go even faster you won't be able to properly receive what you transmitted any more. If you want to transmit even faster and actually receive it, you'll need to increase the bandwidth. \$\endgroup\$
    – user253751
    Mar 22 at 17:39
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I know this is way to simple, but is it in the right direction?

Sorry, that's not really how it works. This is all well captured in the Shannon-Hartley Theorem, which is refreshingly simple: Take a look at https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem

There are basically two things that determine the amount of information that you can send: bandwidth and signal-to-noise ratio. One way to think about this is "symbols". A communication stream sends a bunch of "symbols" from the transmitter to the receiver. The higher the bandwidth, the faster the signal can change and the more symbols per second you can send. The higher the signal-to-noise, the more bits you can stuff into one symbol.

Shannon worked out the math behind it: turns out you can send no more than two symbols per Hz of bandwidth. I.e. if your channel has 1000 Hz of bandwidth you can send up to 2000 symbols per second.

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