> Only if you insist on carrying the signal on a single carrier.
> And, more correctly, it's the bandwidth, not the frequency that
> correlates with information content. The 10MHz of bandwidth
> between 0 and 10MHz has the same information carrying capacity as
> the 10MHz between 1GHz and 1.01GHz. The difference is in the
> practicalities of modulation/demodulation and transmission.
>> which law is this again? (remember reading it in Tannenbaum's network
> book). Also, this law only applies to "direct"/"binary" (what is the
> fancy computer term for this?) signalling methods, doesn't it? ie
> where there are 2 states for the signal such as rs232, old old
> 9600baud modems (i believe), 10baseT ethernet.. etc..
Shannon's law is:
C = Blog2(1 + S/N)
C is the information capacity (bps) of the channel
B is the bandwidth of the channel
S/N is the signal-to-noise power ratio of the channel
E.g. for 4kHz BW, 40dB SNR, you get
C ~= 53kbps (obviously not with a binary system)
As p4 of "Advanced Electronic Communications Systems" which I'm looking at
right now says, "Shannon's formula is often misunderstood"...
In fact I think it's being confused with the Nyquist criterion (or
Shannon-Whittaker sampling theorem) which states that to represent a signal,
one must sample at a rate at least twice that of the highest frequency
component in the signal. This is true but doesn't say anything about the
information contained transferred, for a binary transmission system, the
information transferred is the same as the symbol frequency but can be
higher for other schemes, up to the Shannon theorem limit. (e.g. imagine
that you have a 100dB S/N channel and put a 16-bit DAC at one end and an ADC
at the other - 16 bits at a time are sent across the channel at a time, so
your symbol rate is 1/16 of the bps rate)
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