# Derivation of the Shannon Limit

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:

### Background:

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 $$ln(\frac{c}{w})$$.

### Question:

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?

• 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. Feb 12, 2019 at 5:24
• I'll definitely have to check it out! However, it doesn't quite answer the derivation that I have... Feb 12, 2019 at 23:18

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