Know Your Eb/N0 - Part 3

This is the last part of our three-part series that explains the meaning of $E_b/N_0$ in digital communication context.

Calculating $E_b/N_0$

Having understood how Gaussian noise comes to affect transmitted symbols, it remains to define an appropriate average signal energy, and hence a signal-to-noise ratio (SNR). A commonly used definition of SNR is $E_b/N_0$, where $E_b$ is the average received energy per transmitted information bit, and $N_0$ is twice the two-sided power spectral density of the AWGN.

In digital transmission, the transmitter usually takes in information bits, encoding them (for the purposes of error detection and correction) with an encoder to produce a proportionate number of channel bits. The constant of proportionality, i.e., the ratio of the number of information bits to channel bits, is referred to as the encoder rate, and is denoted $R$. (For example, if every information bit on average produces two channel bits, the encoder rate is 1/2.)

The channel bits are then mapped (perhaps after interleaving or some other operation) to a signal constellation, producing a proportionate number of transmitted symbols. The constant of proportionality for a constellation of size $|\Omega|$ is $\log_2|\Omega|$ channel bits per symbol.

The overall transmission efficiency (or rate) is therefore given by

$R\log_2|\Omega|$ information bits per symbol.

(One can check that the units are correct: we have information bits per channel bit multiplied by channel bits per symbol, resulting in information bits per symbol.)

If the modem is capable of transmitting at a symbol rate of $B$ symbols per second$^1$, then the overall rate of information bit transmission is

$RB\log_2|\Omega|$ information bits per symbol.

Suppose that the transmitter selects from among the points of an $N$-dimensional constellation $\Omega$ with equal probability. The average symbol energy in $N$ dimensions is given by

$E_s(\Omega)=\frac{1}{\Omega}\sum_{x \in \Omega} ||x||^2$

Assuming energy is measured in Joules, $E_s(\Omega)$ can be thought of as having units of Joules per symbol.

It can be shown that the average symbol energy of a symmetric regular $M$-PAM signal constellation with minimum squared Euclidean distance $d^2_{min}$ min is given by

$E_s=\frac{d^2_{min}(M^2 - 1)}{12}$

Similarly, a square $M^2$-QAM constellation (which is really just two $M$-PAM systems in quadrature) has average symbol energy

$E_s=\frac{d^2_{min}(M^2 - 1)}{6}$

Now, if the transmission rate is $R\log_2|\Omega|$ bits per symbol, then the average information bit energy is given by

$E_b=\frac{E_s}{R\log_2|\Omega|}$

measured in Joules per information bit. (Again, one can verify that the units are correct: we have Joules per symbol divided by information bits per symbol, resulting in Joules per information bit.)
Now, knowing that the noise variance per signal space dimension is $\sigma^2 = N_0/2$, we may calculate $E_b/N_0$.

In summary,

$\frac{E_b}{N_0}=\frac{E_s}{2R\sigma^2\log_2|\Omega|}$

where
- $\Omega$ is an $N$-dimensional signal constellation,
- $E_s$ is the average energy (in $N$-dimensions) of the points of $\Omega$,
- $\sigma^2$ is the Gaussian noise variance per signal dimension,
- $R$ is the code rate.

Some Examples

1. A system uses a convolutional code of rate $R$, combined with 2-PAM constellation $\Omega$ = {−1, 1}. The noise variance per dimension is 2. What is $E_b/N_0$?
Solution: The noise variance per dimension is $\sigma^2 = N_0/2$; hence $N_0 = 2\sigma^2$. Each symbol has unit energy, i.e., $E_s$ = 1, and each symbol conveys $\log_2 2 = 1$ bit, so the energy per information bit is given by $E_b = 1/R$. It follows that

$\frac{E_b}{N_0}=\frac{1}{2R\sigma^2}$

2. The modulation of the previous example is modified to the 4-QAM constellation

$\Omega = {(1, 1), (1,-1), (-1, 1), (-1,-1)}.$

The noise variance per dimension is 2. What is $E_b/N_0$?
Solution: As above, $N_0 = 2\sigma^2$. Each symbol has energy $E_s = 2$. We have

$E_b=\frac{E_s}{R\log|\Omega|}=\frac{2}{2R}=\frac{1}{R}.$

It follows that

$\frac{E_b}{N_0}=\frac{1}{2R\sigma^2}.$

3. Suppose the constellation of the previous example is modified to $\Omega = {(1, 0), (-1, 0), (0, 1), (0,-1)}$. Now what is $E_b/N_0$?
Solution: Now $E_s = 1$, so $E_b = 1/(2R)$, and hence

$\frac{E_b}{N_0}=\frac{1}{4R\sigma^2}.$

4. The output of a rate 5/6 binary encoder is mapped to a 64-QAM constellation with $d^2_{min} = 4$. The noise variance per dimension is $\sigma^2$. What is $E_b/N_0$?
Solution: For $M^2$-QAM (here $M$ = 8), $E_s = d^2_{min}(M^2 - 1)/6 = 4(63)/6 = 42$. We have

$\frac{E_b}{N_0}=\frac{E_s}{2R\sigma^2\log_2|\Omega|}=\frac{42}{2(5/6)6\sigma^2}=\frac{4.2}{\sigma^2}$

$^1$ For example, an ideal QAM modem operating in a bandwidth $W$ can send up to $W$ symbols per second without intersymbol interference.

References:
[1] Bernard Sklar, Digital Communications: Fundamentals and Applications, Prentice Hall, 2 edition, January 2001.
[2] John Proakis, and Masoud Salehi, Digital Communications, McGraw-Hill, 5th edition, November 2007.
[3] John R. Barry, Edward A. Lee, and David G. Messerschmitt, Digital Communication, Springer, 3rd edition, September 2003.