# More on Shannon’s limit

Shannon’s limit does not depend on BER. Shannon’s limit tell us the minimum possible Eb/No required to achieve an arbitrarily small probability of error as M->∞. (M is the number of signaling levels for the modulation technique, for BPSK M=2, QPSK M=4 , and so on..).

It gives the minimum possible Eb/No that satisfies the Shannon’s-Hartley law. In otherwords, it gives the minimum possible Eb/No to achieve maximum transmission capacity (R=C, where, R is the rate of transmission and C is the channel capacity). It will not specify what BER you will get at that limit. It also will not specify which coding technique to use to achieve the limit. As the capacity is approached, the system complexity will increase drastically. So the aim of any system design is to achieve that limit. LDPC coding “nears” the Shannon’s limit but it cannot achieve it.

Shannon’s limit depends on the type of channel that is being investigated. Here the channel is assumed to be AWGN. The method to calculate the shannon’s limit for an AWGN channel is as follows..

Given B and S/N, assume we are transmitting R bits/second and we wish to ensure that R is less than the Shannon-Hartley limit C (if R>C BER will increase drastically and the entire system will fail). Then:

$latex R \leq B \times log_{2} \left( 1+ \frac{S}{N}\right)&s=2&fg=0000A0$

Now assume that we wish to transmit average energy/bit of Eb (Joules per bit) and the AWGN noise has 2-sided power spectral density N_{0} /2 Watts (normalised to 1 Ohm) per Hz. It follows that the signal power S = RE_{b} and the noise power N = BN_{0} Watts. Therefore:

$latex \frac{R}{B} \leq log_{2} \left( 1+ \frac{RE_{b}}{BN_{0}}\right)&s=2&fg=0000A0$

$latex \frac{R}{B}$ is called the bandwidth efficiency in units of bit/second/Hz.

rearranging ….

$latex 2^{\frac{R}{B}}\leq 1+ \frac{RE_{b}}{BN_{0}}&s=2&fg=0000A0$

substituting $latex \frac{Eb}{N_{0}}$ with $latex \left( \frac{Eb}{N_{0}} \right)_{min}$ and rearranging,

$latex \left( \frac{E_{b}}{N_{0}} \right)_{min} \geq \frac{ \left( 2^{\frac{R}{B}} – 1 \right)}{\left( \frac{R}{B} \right)}&s=3&fg=0000A0$

This gives the minimum possible normalised energy satisfying the Shannon-Hartley law. If we draw a graph of $latex \left( \frac{Eb}{N_{0}} \right)_{min}$ against bandwidth efficiency (R/B) we observe that $latex \left( \frac{Eb}{N_{0}} \right)_{min}$ never goes less than about 0.69 which is about -1.6 dB in logarthmic scale.Therefore if our normalised energy per bit is less than -1.6dB, we can never satisfy the Shannon-Hartley law and the designed communication system will fail completely.

### See also:

[1] Shannon’s Limit and Channel Capacity

[2] Performance of Channel Codes and coding metrics

[3] Introduction to Channel Coding