I keep seeing $$\frac{E_b}{N_0} = \frac{E_\text{s}}{\rho N_0}; ~~ \rho=\log_2{M}$$ But my own calculation is:

$$\frac{E_b}{N_0} = \frac{E_s}{N_0}\frac{1}{k}$$

Where $k = \frac{\rho n}{ms}$, where $\rho = msr$ is the spectral efficiency, $m$ is the modulation efficiency, $s$ is the percentage of carriers that carry data from the actual coded bitstream, $r$ is the coding rate $\frac{k}{n}$, $n = sNm$ is the total number of coded data bits in the symbol and $k$ is the total number of information bits in the symbol and $N$ is the number of subcarriers.

Wikipedia shows the following:

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There are multiple issues with this

  1. It uses $\rho$, which is spectral efficiency, and actually refers to it as spectral efficiency but then substitutes it with $\log_{2}M$ which actually is the modulation efficiency
  2. It says that 'this is the energy per bit, not the energy per information bit', to follow up its use of modulation efficiency now instead, but there's a contradictory equation that is using the net bitrate i.e. the information rate $f_b$ in the section above
  3. The contradictory equation, which is correct, produces my formulation, and the formulation that I've seen on a few sources i.e. this one: https://uk.mathworks.com/help/comm/ug/awgn-channel.html

$$\frac{E_b}{N_0} = \frac{\frac{P_C}{f_b}}{\frac{P_N}{B}} = \frac{P_C}{P_N}\frac{B}{f_b}$$ and $$\frac{E_s}{N_0} = \frac{\frac{P_C}{F}}{\frac{P_N}{B}} = \frac{P_C}{P_N}\frac{B}{F}$$

Which shows how $E_b/N_0$ differs in that it has a $f_b$ instead of an $F$ term. Therefore $F$ needs to be timesed by something that produces $f_b$, and that thing is $k$. $Fk$, the baud rate times the number of information bits in the symbol, is the information rate.

  1. The equation I keep seeing across multiple sources as well as wikipedia is dividing the energy of a symbol by modulation efficiency rather than the number of bits in the symbol, which does not make semantic sense if the symbol has multiple subcarriers, and if it doesn't, this would only give the energy per gross bitrate transmission bit.

My guess as to what they have done wrong is calling $\rho$ $\log_{2}M$ or using $\log_{2}M$ in these equations at all, because AFAIK $E_b/N_0$ is with respect to the information rate. If they use actual spectral efficiency, then it is correct IF the symbol has one carrier. If it has multiple then they'd have to times the single carrier spectral efficiency $\rho = mr$ by $sN$ to get $\rho sN$ i.e. $k$

  • $\begingroup$ I think the mismatch comes from the understanding of "symbol". By $E_s$ we mean the energy of a QAM symbol which is modulated in one subcarrier (you are referring to multicarrier modulation e.g. OFDM, right?). Therefore, $E_s = E_b \times log_2 M$. Indeed it makes sense because we want to compare this energy to noise density $N_0$ which affects each subcarrier. $\endgroup$
    – AlexTP
    Oct 15, 2020 at 13:02
  • $\begingroup$ @AlexTP I added a last paragraph and I'm pretty sure now that that's whats going on and it's calling spectral efficiency modulation efficiency because there is no coding (assumption 1), and there is only one subcarrier being modulated (assumption 2) $\endgroup$ Oct 15, 2020 at 13:04
  • $\begingroup$ "$E_b/N_0$ is with respect to the information rate" not necessary. It depends on what you mean by "bit", which could be either before or after channel coding. For the former, yes, $E_s = E_b \times \log_2 M \times k$ where $k$ is code rate. I have never seen ratio $E_s/N_0$ with $E_s$ being the energy of multiple subcarriers. $\endgroup$
    – AlexTP
    Oct 15, 2020 at 13:10
  • $\begingroup$ @AlexTP You're right that it makes sense to talk about spectral efficiency in terms of a data subcarrier itself, and possibly makes more sense than referring to the spectral efficiency of the whole OFDM symbol (where you multiply by the the $sN$ factor of 48/80 * 80, because there are 80 subcarriers in OFDM 802.11a theoretically as there are 80 samples with the CP). I came up with the convention of using $E_c/N_0$ for the coded data bit (which means per chip in CDMA) and $E_b/N_0$ for the information bit $\endgroup$ Oct 15, 2020 at 13:49

1 Answer 1


It's just a matter of finding out the assumptions behind the equation. Sometimes, unfortunately, those assumptions are not made explicit.

For the first equation you present, the one you have issues with, the assumptions are (off the top of my head):

  • Orthogonal, ideal sinc pulses are used as pulse shape.
  • The signals involved are strictly baseband, or, if passband, then the upconversion is suppressed-carrier, and the entire process is ideal.
  • The transmitted signal is narrowband; there are no multicarriers.
  • The channel is AWGN; there is no signal distortion, not even attenuation.
  • There is no coding and the transmitted bits are uncorrelated.
  • The received signal is matched-filtered. The matched filter is ideal (i.e. an infinite sinc pulse).

Under these assumptions, that equation is correct.

If you change those assumptions, then the relationship changes, as you found. I didn't check every one of you equations carefully but in general it seems like you are on the right track.


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