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I am not able to find any source giving reason as to why the maximum bit rate in a noise less channel is twice bandwidth multiplied by log L base 2 where L is number of symbols used.

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What I came closest to is bit rate is twice bandwidth as a consequence of ISI and use of Nyquist pulse. But what about logL part.

Please help me out by giving a source to derivation proof or an intuitive reasoning too would do

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  • \$\begingroup\$ Instead of calculating bit rate, suppose you take bit rate and L as inputs and try to calculate the occupied bandwidth of your signal. What would your calculated bandwidth be? \$\endgroup\$ – scorpdaddy May 17 '16 at 13:42
  • \$\begingroup\$ I am not able to take L as input because to prevent ISI the output is sampled at instances of 1/Rb giving rise to rectangular pulse in frequency domain whose bandwidth is Rb/2 where Rb is bit rate of sampling \$\endgroup\$ – Yasith Mohim May 17 '16 at 13:51
  • \$\begingroup\$ \$\log_2(L)\$ is just the number of bits of information transmitted by a single symbol with \$L\$ possible (and equally likely) levels. \$\endgroup\$ – The Photon May 17 '16 at 16:05
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I'll add another answer into the mix.

So $$ \text{bitrate} = \text{number of bits transmitted per second} $$

$$ \implies \text{bitrate} = \text{(Symbols Per Second)} \times \text{(Bits Per Symbol)} $$

$$ \implies \text{bitrate} = f_s \times \log_2 L = 2B \log_2 L $$

The capacity tells us the maximum number of error free bits that can be transmitted over a channel. The capacity of a channel is given by

$$ C = \frac{f_s}{2} \log_2 \left ( 1 + S/N \right) $$

so for a noiseless channel (\$N = 0\$) we have \$C = \infty\$. So in a noiseless channel there is no upper limit on the number of error free bits we can transmit, any bitrate is achievable.

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The Shannon–Hartley theorem tells us this: -

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If your channel is noiseless then S/N (signal to noise ratio) is infinite hence channel capacity is potentially infinite OR, put another way, limited only by the bit depth.

Your formula appears to be like the "Hartley" part of the above i.e.: -

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This wiki page will probably help

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