Q & A on Nyquist's Theorem Q: > >> 2.(p44, end of 5.2) The formula 2BT for the data rate of phase > >> modulation, should, I believe, not have the constant 2. It should > >> just be BT. On p36, where you present Nyquist's Theorem, B is the bandwidth. The baud rate, in the context of p36, is 2B. But on p44 B is the baud rate, and thus no coefficient is needed to adjust from bandwidth to baud rate. At a common sense level, if you can change your signal B times per second and you can encode T bits per change, you get BT bits/sec. Part of the problem is that Nyquist's Theorem, in the simplest form, applies most directly to digital transmission of bits, and that is usually how it is presented in network texts. When the jump is made to analog encodings of data, the application of the theorem is less obvious. Phase modulation will only provide a baud rate equal to the bandwidth, and amplitude modulation will also provide the same. The intuitive understanding that I glean from various texts (and this may be taken with a grain of salt) is that since amplitude modulation and phase modulation are orthogonal, they are each providing 1/2 of the theoretical limit given by Nyquist's Theorem. When combined, which they are in all modems, they do indeed provide a baud rate equal to twice the bandwidth of the signal (although most texts don't call it a doubling of the baud rate, but rather look at it as each signal change is a 2-dimensional change). A: I'm not a hardware expert, so you could be right. The part of you argument that bothers me is the statement: ``Phase modulation will only provide a baud rate equal to the bandwidth.'' I don't understand why you would say that. What am I missing?