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A channel has bandwidth 3 KHz, signal to noise ratio is 40 dB find maximum data rate for 16 level encoding schemes.

There are two ways to find channel capacity. First one is with Nyquist theorem and second one is with shanon capacity. For Nyquist theorem I found 24 kbps and for Shanon capacity I found 39.84 kbps. So which one would be preferable? Is there any other method to do?

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  • \$\begingroup\$ What was your first approach? \$\endgroup\$ – Matt Young Mar 23 '13 at 4:33
  • \$\begingroup\$ what do you mean by first approach ? \$\endgroup\$ – Åbħísħêk Đêsħkăr Mar 23 '13 at 4:33
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    \$\begingroup\$ What did you try? We're not here to do your homework for you. You need to show us some effort. \$\endgroup\$ – Matt Young Mar 23 '13 at 4:36
  • \$\begingroup\$ Lol ! OK. Actually there are two ways to find channel capacity ! First one is with Nyquist theorem and second one is with shanon capacity ! For Nyquist theorem I found 24 kbps and for Shanon capacity I found 39.84 kbps. So which one would be preferable ? Is there any other method to do ? \$\endgroup\$ – Åbħísħêk Đêsħkăr Mar 23 '13 at 4:40
  • \$\begingroup\$ I edited your question so it will be more likely to get an answer, and less likely to get down voted. \$\endgroup\$ – Matt Young Mar 23 '13 at 4:50
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The difference between the two formulas arises from the fact that the Nyquist formula uses the number of encoding levels that was explicitly given (16 levels implies 4 bits/baud), while the Shannon formula is the theoretical maximum based on the SNR of the channel (40 dB implies about 6.64 bits/baud).

3000 Hz × 2 baud/cycle × 4 bits/baud = 24000 bits/sec

3000 Hz × 2 baud/cycle × 6.64 bits/baud = 39840 bits/sec

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I'm not going to do your homework for you. Instead of trying to plug numbers into some formula, go back to what Nyquist and Shannon were really saying.

You want to send data encoded with 16 different analog levels. Divide the voltage range of your signal up so that 16 levels are maximally spaced apart. That means each level will be 1/15 of the range from the next. This means each level has 1/32 noise margin. In other words, you can add up to 1/32 noise to each signal and still be able to distinguish which level it is at the other end.

The question now becomes, what is the time for the worst case step to settle to within 1/32 of its final value. That is the absolute minimum time the transmitter has to dwell on each level for it to be distinguishable at the other end. I'll let you work out the 1/32 settling time of a step that is limited to 3 kHz. In the end, you have to remember you are sending Log2(16) = 4 bits at a time.

That was the case without noise. What noise does is add a certain amount of error that never settles away. Convert the noise level to the fraction of full scale. The signal needs to settle to within the 1/32 level minus the noise. Without noise, your minimum settling level is 1/32 = .03125. If you think the maximum noise is .01, for example, then you have to wait for settling to within .03125 - .01 = .02125. Conversely, you can do this calculation in logarithmic scale, typically in units of dB.

There is more you can do if you are allowed to make some assumptions about the noise. Actually, we already made one assumption above, which is that the noise has a maximum voltage excursion. That may not be true from just a dB figure. If, for example, you know the noise will average out over time (doesn't cover the full 3 kHz bandwidth, in this case not including low frequencies), then you can eventually recover a level even if the noise amplitude is larger than the error band around each level. Effectively though, you are reducing the bandwidth of your channel because you are adding low pass filtering at the receiver. By the way, this is actually how GPS receivers decode some of the signals. The satellite signals are so weak that they are something like 20 dB below the noise floor. They are recovered by some fancy math, with one way to look at it being that a lot of filtering is applied at the receiver, effectively reducing the frequency range of the channel to where the signal does exceed the noise.

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  • \$\begingroup\$ What is important to realize is that the \$ 1/15 \$ is a "zone" within which you signal must travel and not have excursions into the adjacent "zone". The signals are not just simply subtracted. And acceptable level to ensure that one signal level does not interfere is to use a 6 sigma level so that level confusion is vanishing small. So that \$ 1/16 \$ level maps to a actual acceptable noise level of \$ 1/(16*6) = 1/96 \$ which in log terms is very close to 40 dB. Additionally noise sources add as RSS (Root sum of squares). \$\endgroup\$ – placeholder Mar 23 '13 at 16:13
  • \$\begingroup\$ GPS operates using SSDS (Spread spectrum direct sequence) and the noise operation is due to the encoding side and the matched filter on the receiving side. The reduction in transmitted BW is regained in something called process gain which can be applied against, in this case, noise penetration. But equally can be used in CDMA systems for increased BW with multiple codes. \$\endgroup\$ – placeholder Mar 23 '13 at 16:15
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Nyquist theorem gives your the bit rate achieved with a certain modulation such as 16-QAM (4bits/symbol). You can go higher and higher by selecting even more complex modulation schemes e.g. with 64-QAM you can achieve 36kbps and with 256-QAM 48kbps, even higher than Shannon Capacity. Is this possible? Short answer NO.

Shannon capacity which is calculated as 39.86kbps in the above case gives the maximum error free bit rate that is only achievable with advanced coding schemes such as Turbo Codes. So Shannon basically says you cannot achieve 48kbps using 256-QAM because there would be just too many errors in transmission (due to noise) that you will not be able to correct at the receiver.

A recent development (actually not that recent) is MIMO which allows us to go beyond channel capacity of SISO channel by transmitting multiple parallel streams between the transmit and receive antennas.

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