The relationship between channel capacity C (in bits per second, bps)
and bandwidth B (in Hertz, Hz):
The Nyquist formula applies if there is no noise. This formula says:
=
where M is the number of signal (often voltage) levels
Example: What is the channel capacity if the bandwidth is 1000 Hz and there are 8 signal
levels?
=
Note that
=
3
=
=
since
=
Example: Clearly, doubling the bandwidth B in this formula results in the doubling of the
channel capacity. If we have a bandwidth of 3000 Hz and double the number of signal levels
from 8 to 16, what effect does this have on the channel capacity?
=
=
=
=
=
=
Thus the channel capacity increased by one-third; it did not double. The percentage
increase depends on the initial value of M.
The Shannon-Hartley law applies in the presense of thermal noise. It says:
=
where SNR is the signal-to-noise ratio, the ratio between the signal power and
the noise power level.
Note that
=
is often used in these problems.
This is the decibel version of the signal-to-noise ratio.
Example: Suppose that the spectrum of a communications channel is 3 MHz to 4 MHz.
Suppose also that the decibel version of the signal-to-noise ratio is:
=
24 dB
a) Find the channel capacity (in the presence of noise, of course):
=
=
1 MHz
By definition,
=
Thus
=
=
Apply the inverse function for the log function, the exponential function base 10,
to both sides:
=
=
By calculator:
=
Now we use the Shannon-Hartley law:
=
We use the change of base formula to find log
base 2:
=
=
=
Then log base 10 is done on a calculator.
=
=
(M stands for mega, million, which is 10 to the sixth power)
b) If there were no noise, how many signal levels would be needed to achieve the above
channel capacity of 8 Mbps?
Use the Nyquist formula:
=
=
=
=
M
=
=
16 signal levels are needed
