Nyquist Bandwidth

Nyquist Bandwidth

To begin, let us consider the case of a channel that is noise free. In this environment,
the limitation on data rate is simply the bandwidth of the signal. A formulation of

this limitation, due to Nyquist, states that if the rate of signal transmission is 2B, then
a signal with frequencies no greater than B is sufficient to carry the signal rate. The
converse is also true: Given a bandwidth of B, the highest signal rate that can be carried
is 2B. This limitation is due to the effect of intersymbol interference, such as is
produced by delay distortion.3 The result is useful in the development of digital-toanalog
encoding schemes.
Note that in the preceding paragraph, we referred to signal rate. If the signals
to be transmitted are binary (take on only two values), then the data rate that can be
supported by B Hz is 2B bps. As an example, consider a voice channel being used,
via modem, to transmit digital data. Assume a bandwidth of 3100 Hz. Then the
capacity, C, of the channel is 2B = 6200 bps. However, as we shall see in Chapter 6,
signals with more than two levels can be used; that is, each signal element can represent
more than one bit. For example, if four possible voltage levels are used as
signals, then each signal element can represent two bits. With multilevel signaling,
the Nyquist formulation becomes
C = 2B logz M
where M is the number of discrete signal elements or voltage levels. Thus, for
M = 8, a value used with some modems, a bandwidth of B = 3100 Hz yields a
capacity C = 18,600 bps.
So, for a given bandwidth, the data rate can be increased by increasing the
number of different signal elements. However, this places an increased burden on
the receiver: Instead of distinguishing one of two possible signal elements during
each signal time, it must distinguish one of M possible signals. Noise and other
impairments on the transmission line will limit the practical value of M.