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I would like to know what would be the relationship between $E_b/N_0$ and $\rm SNR$ when there is a coding rate of $R_c=K/N$. Normally when there is no channel coding according to this link.

$$\textrm{SNR(dB)} = \frac{E_b}{N_0}(\textrm{dB}) + 10\log10\left(\frac{R_b}{B}\right)$$

However what would happen if I have a linear block code or other coding schemes with code rate $R_c =1/2$ or other values, how to calculate the relationship between $\rm SNR$ and $E_b/N_0$ in this case ?

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2 Answers 2

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The relationship between SNR and $E_b/N_0$ is independent of the code rate. Note that $E_b$ is the energy per data bit (not the coded bits), and $R_b$ is the (uncoded) data bit rate. As long as you keep using these values, you can use the formula given in your question. Of course, when going from an uncoded system to a coded system, the values of $E_b$ and $R_b$ will usually change, so the resulting SNR will generally be different.

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  • $\begingroup$ Can you elaborate how Eb and Rb will be changing ? $\endgroup$
    – user59419
    Aug 19, 2016 at 15:27
  • $\begingroup$ @user59419: That depends on how you implement it. If you add a rate $1/2$ coder and if you keep the channel bit rate unchanged, then $R_b$ (the data bit rate) will be halved. If you spend the same energy per channel bit as without coding, then $E_b$ will be doubled. $\endgroup$
    – Matt L.
    Aug 19, 2016 at 17:25
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The answer to the question depends on the fact if the code rate implies a change in the bit rate f the system. The bit rate would be the rate of bits that come out of the system per second, and the code rate the rate between an uncoded length of data and the coded length of data.

If the coding of the data does not imply a change in the bit rate of the system, then the $SNR$ will remain the same, but the transmission in the sytem will be slower as you need more bits to transmit the same message.

However, if coding the data implied a change in the data rate so that the transmission velocity (in terms of data chunks) remains the same, then the bit rate of the system would change to $R_b'=\frac{R_b}{R_c}$, and the new $SNR$ would be calculated with the equation you posted but with $R_b'$.

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