I am doing the derivation of the Shannon Limit, and I think I ran into a snag, and am in need of a bit of assistance:


From information theory, we know that the Shannon Limit is: $$C = W *log_2(1 + SNR)$$ Where W represents the bandwidth and C is the capacity of the channel. It can be found that (assuming that data rate is equal to the theoretical capacity):

$$SNR = E_b/N_0 * C/W$$

Plugging the SNR in, and doing a bit of algebra we get the equation: $$E_b/N_0 = \frac{2^\frac{c}{W} - 1}{\frac{c}{w}}$$

Computing the value of $$\frac{Eb}{N_0}$$ in the limit as C/W -> 0, we get the Shannon Limit. However, when I take the limit of the above equation, I get that the limit is $$ln(\frac{c}{w})$$.


From Wikipedia (https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem) and other internet sources, the "Ultimate Shannon Limit" is ln(2). My question is, am I correct with my mathematics and the work that I have done? How can I derive an expression for the required Eb/N0 as a function of capacity and bandwidth (assuming that data rate is equal to the theoretical capacity), to achieve the Shannon limit?

Thanks in advanced for your help!

  • \$\begingroup\$ It's pretty easy to just find Shannon's original paper online (I think it's on the AT&T Labs website) if you google for it. It's not too hard to read. \$\endgroup\$
    – The Photon
    Feb 12, 2019 at 5:24
  • \$\begingroup\$ I'll definitely have to check it out! However, it doesn't quite answer the derivation that I have... \$\endgroup\$ Feb 12, 2019 at 23:18

1 Answer 1


Your reasoning is almost correct.

In the step where you take the limit of the \${E_b}/{N_0}\$ equation, you're making a mistake when applying L'Hôpital's rule. It should go as follows:

$$ \lim_{C/W \to 0^+} \frac{E_b}{N_0} = \lim_{C/W \to 0^+} \frac{2^{C/W} - 1}{C/W} = \lim_{C/W \to 0^+} \ln {(2)} 2^{C/W} = \ln{(2)} $$

Your problem arises when you take the derivative of \$2^{C/W} - 1\$, just remember that

$$ \frac{d(a^{x})}{dx} = \ln{(a)} a^{x} $$

Hope it helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.