# Link Capacity and the Shannon-Hartley Theorem

I'm reading Computer Networks: A Systems Approach by Peterson and Davies. One of the examples demonstrates the relationship between link capacity and the Shannon-Hartley Theorem.

We can find the channel capacity by the formula:

$$C = B \log_2 \left( 1+\frac{S}{N} \right)$$

In the example of the book, they define bandwidth of the channel to be 3000Hz and the signal to noise ratio to be 30 dB, which they say would imply that S/N = 1000.

$$C = 3000 \times \log_2 (1001)$$

However, I don't understand how a signal to noise ratio of 30 dB is equivalent to 1000? How is this worked out? It's not explained in the example.

• See Decibel 30db is 1000, and it's explained there. – gbulmer Oct 3 '14 at 20:45
• Oh I see, so its the power ratio - since decibels is logarithmic? – user54505 Oct 3 '14 at 20:55
• @gbulmer, with a couple more sentences of explanation, your comment could be an answer. – The Photon Oct 3 '14 at 20:57
• @ThePhoton - I am still new here. However, I thought ee.se was not about collecting answers that are already satisfied by trivial web searches, i.e. type one word into wikipedia and get a definitive answer. I was only being helpful in my comment. I had expected this question to be closed because it fell into that category of "very well-answered on the web with minimal effort already" question. Am I wrong? – gbulmer Oct 4 '14 at 11:32
• @gbulmer, you're right it's not a great question. But it's borderline and hasn't gotten any close votes yet. So it's better we get an answer posted and accepted to keep the question from re-appearing on the front page. – The Photon Oct 4 '14 at 15:46

    Example: 3000Hz voice bandwidth, s/n = 30 dB. or a ratio of 1000.