I have read in Data Communications and Networking, 5th Edition, by Forouzan that the formula for calculating the bandwidth for QAM modulation is \$B=(1+d)S\$ where \$d\$ is factor that depends on filtering process and is between 0 and 1, and \$S\$ is symbol rate.

My question is why is that formula used instead of e.g. Nyquist or Shannon formula for capacity to calculate bandwidth?


Because they are computing different things, used for different purposes.

In the formula you quote, bandwidth means amount of 'occupied bandwidth' in the RF spectrum. This is the amount of channel width that must be dedicated to transmitting the signal. Theoretically, given enough transmission time and computing power, d could approach zero. In practice, these are traded off with d varying between 0.2 and 0.5 for different systems. It's independent of the type of modulation used, whether 4QAM or 256QAM.

Nyquist uses bandwidth with respect to the sampling theorem. In a sense, the factor d is a measure of how far 'above Nyquist' the bandwidth is, so the relationship is quite close.

Shannon has to do with the capacity of a channel when noise is taken into account, which is a different purpose to your formula. Here the different between 4QAM and 256QAM is profound.

  • \$\begingroup\$ Could you please explain the differences in bandwidths in a little more detail with respect to Nyquist. \$\endgroup\$ – Quirik Jul 31 '16 at 10:49
  • \$\begingroup\$ We are only using Nyquist capacity formula when we want do convert analog to digital signal. For example, if we sample voice with bandwidth 0-4kHz with 8 bits, the capacity would be 64kbps. Then, we can use QAM 16 to transmit this digital signal at occupied bandwidth of 64.000 bps/4 bits = 16kHz. Is this correct? \$\endgroup\$ – Quirik Jul 31 '16 at 11:27
  • \$\begingroup\$ No. We are only using Nyquist when we want to convert continuous signal (eg from an analogue microphone) to a sampled signal, usually so it can be digitised. For example, if we sample voice with bandwidth 0-4kHz, then we must sample above a minimum rate of 8kHz, but preferrably a bit more, or use a smaller bandwidth. Note telephony uses 3400Hz as the upper voice frequency, and 8kHz sample rate, so is sampling at 17% above the minimum Nyquist rate. It has nothing to say about the bit width of the samples, or the gross data rate. \$\endgroup\$ – Neil_UK Jul 31 '16 at 11:47

The Nyquist rate tells you how frequently you have to sample an analog signal to retain all the information it contains. It doesn't particularly apply to a communication system where the inputs and outputs are both digital.

The Shannon formula gives the minimum bandwidth needed to achieve a certain error-free data transmission rate. It doesn't tell you the bandwidth of any actual implementation of a communication system.

You can compare the bandwidth used by your actual system to the Shannon bandwidth to see how "efficiently" it operates. That is, how close it gets to the Shannon limit. But you cannot use the Shannon formula to predict the performance of a system.

  • \$\begingroup\$ E.g. for a 1 Mbps and SNR = 1000, Shannon would give the minimum required bandwidth of aprox. 100 kHz. The formula for occupied bandwidth for QAM 4 would yield 240 kHz. How to interprete this? In addition, a "more efficient" system would be the one closer to the Shannon limit? \$\endgroup\$ – Quirik Jul 31 '16 at 17:42
  • \$\begingroup\$ @Navi, correct. \$\endgroup\$ – The Photon Aug 1 '16 at 16:25

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