# Calculating $E_b/N_0$ from given SNR ratio

I have the following formula:

$$\frac{S}{N}=\frac{E_b \cdot R_b}{N_0 \cdot B }$$

Where $$S/N$$ is of course SNR in dB, $$E_b$$ is energy per information bit, $$R_b$$ is information bit rate in [bit/s], $$N_0$$ is noise spectral density and $$B$$ is bandwitch in [Hz].

So let's say the $$S/N$$ ratio = $$70\ \rm dB$$

• $$R_b = 250\ \rm kb/s$$
• $$B = 1\ \rm MHz$$

So:

$$70\ \mathrm{dB}=\frac{E_b \cdot 250 000\ \mathrm{b/s}}{N_0 \cdot 1000000\ \mathrm{MHz}}$$

$$70\ \mathrm{dB}=\frac{E_b \cdot 1\ \mathrm{b/s}}{N_0 \cdot 4\ \mathrm{MHz}}$$

Now, of course, I can't simply multiply because this 4 is not on a logarithmic scale. But can I just logarithmize 4 like that? Won't there be a problem with units? How to do it to get $$E_b/N_0$$ ratio in decibels, because I keep doing something wrong.

$$\frac{S}{N}=\frac{E_b \cdot R_b}{N_0 \cdot B }$$ With all units linear: $$\implies \frac{E_b}{N_0} = \frac SN\cdot \frac BR_b\tag{linear}$$ Now with $$S/N$$ in $$[\rm dB]$$ and $$R_b$$ and $$B$$ linear, you get $$E_b/N_0$$ in $$[\rm dB]$$ as follows: $$\implies \frac{E_b}{N_0} = \frac SN +10\log_{10}\left(\frac BR_b\right)\tag{in [dB]}$$