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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.

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HINT:

$$ \frac{S}{N}=\frac{E_b \cdot R_b}{N_0 \cdot B } $$ 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: $$\implies \frac{E_b}{N_0} = \frac SN +10\log_{10}\left(\frac BR_b\right)\tag{in [dB]}$$

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  • $\begingroup$ Thank you so much! So Eb/N0 = 70dB + 6.021dB; Eb/N0 = 76.021dB $\endgroup$ – MagicMan Jan 23 at 18:51
  • $\begingroup$ You're welcome. :) $\endgroup$ – Gilles Jan 23 at 18:54

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