# Convert Decibels (Db) to Signal to Noise (S/N)?

Discussion in 'Electronic Basics' started by KP, Nov 17, 2003.

1. ### KPGuest

Hi All

I am abit stuck on converting a number in decibels (Db) to signal/noise
ratio.

I know the equation for converting into DB: -

SN (Db) = 10 Log [Signal Power / Noise Power]

The way which I have managed to conver the Decibel to S/N is using the
following formula: -

Inverse Log [XXXX db / 10]

So using the above formula, 30 Db will work out to be 1000 s/n which is
correct.

The problem is, explaining how I got this formula. Any ideas or better ways
of working this out would be appreciated.

Regards

Karl

2. ### SteveGuest

um, rearrange the formula? It's easier than you think.

S/N[dB] = 10 Log (S/N ratio) - divide both sides by ten
S/N[dB] / 10 = log (S/N ratio) - take antilog of both sides
alog (S/N[dB] / 10) = S/N ratio - et voila!

3. ### KPGuest

Thanks Steve.... I ended up with the same formula as you got, just I couldnt
explain how I got there.

Thanks again for the help... Its very much appreciated.

Karl

4. ### DavidGuest

Hi Steve

Thanks for helping my son, now I am stuck on another equation (like fathe,
like son).

If I have been asked to calculate the signal power at a point in the system
where the s/n ratio is 23Db and the Noise Power is 3mW, what would be first
step be?

I have looked through loads of books, and my son has only just got into this
kind of stuff, so not much help!

Any ideas?

David

5. ### Ben WeaverGuest

Signal is going to be 23dB bigger than the noise. You just need to find
the ratio form of 23dB and multiply it by the noise power.

Whereas 30dB was a factor of 1000, 23dB is a factor 200. I know that
because 20dB is 100 (dB's drop 10 for every power of ten change) and 3dB
is a factor of 2. Now for an addition in dB you just multiply the
factors. 100 x 2 = 200.

Now just multiply the noise power by that factor. 3mW x 200 = 600mW.

Or... You could covert 3mW into dBm by doing 10xlog(3)=4.77dBm. Then you
can just add the 23dB to it. The answer is 27.77dBm. And if you really
want, you can convert that back into a "proper" number by dividing by
ten and doing 10^(answer). You get the same result.

Ben
~~~

6. ### Guest

10 dB in terms of power is ten times greater.
3 dB in terms of power is well nigh double.

so

10 + 10 + 3 = 23dB is the same as:-
10 * 10 * 2 = 200 times

This is the *real* reason for using dB i.e.:-
1) Reduce the range to "human" proportions e.g. not many things are >
120 dB

If you are playing with some power levels that change from say 1 to
1000000 you could re-label them 0dB to 60dB. This then allows you to
stop at some level say 45.7 dB and, from that, double it (45.7 + 3 =
48.7 dB) or halve it (42.7dB) (by doing the sums in your head).

If you make a piece of equipment and label everything in dB then it is
easier for the operator to do make such adjustments. E.g.

"set output to the onset of clipping then back off 3dB". The
alternative would have been :-

"set output to the onset of clipping then back off until it becomes
half of what it was."

This might sound trivial (it is!) but it makes the *ratios* of things
truly appear the same (when they *are*) e.g. +3dB for a power
amplifier might mean 100 watts whereas +3dB for a front end might be a
pico watt and a pico watt is certainly different to watts which is

Robin

7. ### DavidGuest

Thank you very much for both of your help, its very much appreciated.

While im here...... No, im only joking, you have all helped me more than
enough!

Thanks again

David and Karl  