# A question on Shannons Formula Simplification

1. Aug 12, 2011

### dragonflare

Hello all, i encountered a question which i coudnt really solve, was hoping someone could help me. Its regarding Shannons Formula for finding capacity of a channel.

So it goes like this:

C = B log (base2 ) [1 + SNR]

KEY:
C = capacity
B = bandwidth
SNR = SIgnal Noise Ratio

when the SNR (Signal Noise Ratio) is high, the above formula is not very reliable and we can use this formula instead:

*Also it states that if the SNR is high we can ignore the 1.

C = B * [SNR(subscript db)/3]

Question: How do we get the second formula after simplifying the first?

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This is what i have tried till now.

SNR(subscript db) = 10 log (base 10) SNR

We make SNR subject of formula :

10^(SNRdb/10) = SNR

Therefore substituting SNR in the text with the SNR (subscript dB)gives you

*We ignore the 1

C = B * log (base 2) [ 10 ^ (SNRdb / 10)]

Now im given to understand that to find log base 2 of a number we can do this:

C = [B * log (base 10) [ 10 ^ (SNRdb / 10)]] / [log (base 10) 2]

....

I am really confuse how to simplify after that.. i dont even know if im right in the first place -.-

Any help would be appreciated.

Thanks

How do we go to this simplified formula from the one given above??

2. Aug 13, 2011

### dynamicsolo

It looks like you basically have this, if not in a form you are recognizing: you use the "change-of-base" formula for logarithms to write

log(2) SNR = [ log(10) SNR ] / [ log(10) 2 ] .

Now from your formula 10^(SNRdb/10) = SNR , we have

SNRdb/10 = log(10) SNR .

Upon substituting this into the earlier equation, we get

log(2) SNR = [ SNRdb/10 ] / [ log(10) 2 ] = [ SNRdb ] / [ 10 log(10) 2 ]

The common log of 2 is 0.3010... , so 10 log(10) 2 is real close to 3 .

We can now put these pieces together to get, for SNR >> 1 ,

C = B log(2) [1 + SNR] ≈ B log(2) SNR ≈ B * [ SNRdb ] / 3 .

3. Aug 14, 2011

### dragonflare

thanks for the reply :) i better understand my mistake.. i forgot to make it into the value 0.3 ..... silly me :|

thanks anyway :D