David said:

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

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

2) Change multiplication to addition.

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

"misleading"...

Robin