logs in the shannon-hartley law?

Question

i've been doing the shannon hartley law for my networking classes and i've come across this example in a book, but it doesn't explain how they got the answer.

the example is:
assuming a circuit has a bandwidth of 3000Hz and a signal to noise ratio of 20db, determine the maximum theoretical info (data) rate that can be achieved

SNR = 10 log10 (S/N)

therefore 20 = 10 log 10(S/N)
hence: S/N=100

so far i have gotten how they got this.

then it goes onto this:

now: C = B log2(1+S/N)

therefore: C = 3000 * log2 (1+ 100)
= 19,963 bps

its the log 2 bit that i am stuck on (the 2 is supposed to be a little 2 btw). when i type this in my calculator i get this answer: 91212.08869.

i don't have a log2 button on my calculator. so is there another way of working it out, (my uni is strict with what calc model we can use so i cant change it) or am i just missing something.

thanks

Answer 1

You convert between logarithms in any two bases a and b using the following rule:

Log_a(x) = Log_b(x) / Log_b(a)

where I have used the notation "Log_b(x)" to indicate "Logarithm to the base b of the value x,"

So if you use the Log_10 button on your calculator, then Log_2(x) = Log_10(x) / Log_10(2). If you use the Ln button on your calculator (which means Log_e in the notation I have been using) then Log_2(x) = Ln(x) / Ln(2).

How did you ever get out of grammar school algebra without learning this?

Answer 2

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Answer 3

First of all, Shannon Hartley is working with the power S/N ratio and not the log of it. So your S/N ratio is 100 as far as power goes. The log 2 is a binary equation. It means what power of two gives me the value around 101 in your case. Your answer looks good.

Answer 4

Log_2(x) = Log_10(x)/Log_10(2) = log_10(x)/.3

Example: log_2(8) = .9/.3 = 3