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I had a question about the properties of logarithms and exponents, that I need some assistance on:

If you are familiar with electrical engineering principles, you may be familiar with Shannon's Channel Capacity Theorem:

$$C = W\log_2(1 + \text{SNR})$$

Where $W$ represents the bandwidth and $C$ is the capacity of the channel. Well it can be found such that

$$\text{SNR} = \dfrac{E_b}{N_0}\cdot \dfrac{C}{W}$$

If we take the above, and replace it into the original capacity equation, we get:

$$C = W \log_2\biggl(1 + \dfrac{E_b}{N_0}\cdot \dfrac{C}{W}\biggr) = W \cdot 2^{\bigl(1 +\frac{E_b}{N_0} \cdot \frac{C}{W}\bigr)}$$

However, this is where I'm kinda stuck. I would like to solve for $E_b/N_0$, but I'm a bit rusty on my algebra. This might be a rather simple question, but are there some properties of exponents or logs that I can leverage to solve for $E_b/N_0$?

Thanks for your help in advanced! I really appreciate it!

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  • $\begingroup$ How did $\log_2 x$ become $2^x$? $\endgroup$
    – parsiad
    Feb 12, 2019 at 4:20
  • $\begingroup$ Is that not the proper way to remove a log? $\endgroup$ Feb 12, 2019 at 4:23
  • $\begingroup$ I assume the claim you are using is $\log_2 x = 2^x$ for any $x > 0$, which is not true. See my answer below. $\endgroup$
    – parsiad
    Feb 12, 2019 at 4:29

2 Answers 2

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Start with $$ C=W\log_{2}\left(1+\frac{E_{b}}{N_{0}}\frac{C}{W}\right). $$ Exponentiate both sides with base $2$ to get $$ 2^{C}=2^{W\log_{2}\left(1+\frac{E_{b}}{N_{0}}\frac{C}{W}\right)}. $$ Since $x\log_{y}z=\log_{y}(z^{x})$, it follows that $$ 2^{C}=2^{\log_{2}\left[\left(1+\frac{E_{b}}{N_{0}}\frac{C}{W}\right)^{W}\right]}. $$ Since $b^{\log_b x}=x$ for $x>0$, it follows that $$ 2^{C}=\left(1+\frac{E_{b}}{N_{0}}\frac{C}{W}\right)^{W}. $$ Now take roots of both sides to get $$ 2^{C/W}=1+\frac{E_{b}}{N_{0}}\frac{C}{W}. $$ You should be able to do the rest.

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You have $$C = W \log_2(1 + SNR)\tag 1$$ $$SNR = \frac{E_b}{N_0} \frac{C} W\implies1+SNR=1+\frac{E_b}{N_0} \frac{C} W\implies\log_2(1+SNR)=\log_2\left(1+\frac{E_b}{N_0} \frac{C} W \right)\tag 2$$ Combine both equations and write $$C=W\log_2\left(1+\frac{E_b}{N_0} \frac{C} W \right)\implies \frac C W=\log_2\left(1+\frac{E_b}{N_0} \frac{C} W \right)\implies 2^{\frac C W}=1+\frac{E_b}{N_0} \frac{C} W $$

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