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:

\(R \leq B \times log_{2} \left( 1+ \frac{S}{N}\right)\)

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 N0 /2 Watts (normalised to 1 Ohm) per Hz. It follows that the signal power S = REb and the noise power N = BN0 Watts. Therefore:

\(\frac{R}{B} \leq log_{2} \left( 1+ \frac{RE_{b}}{BN_{0}}\right)\)

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

rearranging ….

\(2^{\frac{R}{B}}\leq 1+ \frac{RE_{b}}{BN_{0}}\)

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

\(\left( \frac{E_{b}}{N_{0}} \right)_{min} \geq \frac{ \left( 2^{\frac{R}{B}} – 1 \right)}{\left( \frac{R}{B} \right)}\)

This gives the minimum possible normalised energy satisfying the Shannon-Hartley law. If we draw a graph of \(\left( \frac{Eb}{N_{0}} \right)_{min}\) against bandwidth efficiency (R/B) we observe that \(\left( \frac{Eb}{N_{0}} \right)_{min}\) 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

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  • Wirless Communication Student

    yeah that is the point, when I have a vectore of SNR values e.g. EbN0= [1,2,3…,10], Which SNR value of this vector should I use to calculate the bandwidth of the channel?

    • Mathuranathan

      Striking a balance between signal Power ( translates to operating Eb/N0) and Bandwidth is like walking on a razor.
      In this articles , the shannon’s limit is derived for Gaussian Channel. The shannon’s-Hartley equation is specific for specific channel. If your channel assumption is different from Gaussian Channel, then the Shannon’s Hartley equation will change.

      Assuming your channel is Gaussian, you can rearrange the given Shannon Hardley equation as,

      $latex \frac{E_b}{\eta} = \frac{B}{C} \left( 2^{\frac{C}{B} – 1} \right) &s=2 $

      Here $latex \eta $ is the noise spectral density (see link: )
      Then plot $latex \frac{E_b}{\eta}$ on the x axis and B/C on Y axis. Check for asymptote. The asymptote gives your the minimum $latex \frac{E_b}{\eta}$ achievable.
      From this you can calculte the minimum Eb/N0 giving the Shannon’s limit.

  • Wirless Communication Student

    I am implementing a system, where I use a code rate of 0.75, I just want to know how do we decide on the Bandwidth when we find the Eb/N0 for the shananon limit. Thanks

    • Mathuranathan

      Given a channel with certain bandwidth and noise characteristics, Shannon’s limit specifies the maximum rate at which data can be sent over the channel without any error. This maximum rate is called channel capacity or Shannon’s limit.

      Shannon’s limit depend on two parameters only : 1) Bandwidth of the channel and 2) Noise characteristics – Translated to Signal to Noise Ratio (SNR) or Eb/N0

      Shannon’s limit does not depend on code rate. It is the inherent capacity of a given channel.

      So your problem can be restated as : Given a channel, you want to transmit bits (without errors) at given maximum rate R at some Eb/N0, what will be bandwidth of the channel to achieve this?

      1) Since you know the operating Eb/N0, you have to convert it to SNR (details given here : )

      2) Then rearranging the equation

      $latex R = B log_2 \left( 1+SNR\right) &s=2$

      $latex B = \frac{R}{log_2 \left( 1+SNR \right)}&s=2$

      3) This gives the bandwidth of the channel required to achieve the given Shannon’s limit. Sometimes , the channel may not be able to provide this bandwidth, in such case you have to increase your operating Eb/N0 to satisfy the relationship