A simple question about dB calculation.

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theasus

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Hi,

I have read definiton of desibel at Wikipedia. But I couldn't understand one point.

When a signal attenuate 3db , how to power of signal change?
 

theasus said:
I have read definiton of desibel at Wikipedia.
Click to expand...
Really?
Paragraph 3 of that Wikipedia article answers your question.
So does the table at the top of the page.
 

"A change in power ratio by a factor of two is approximately a 3 dB change."

I put this in formula but I didn't understand.
 

3dB increase gives double the power.
3dB decrease gives half the power.
 
The decibel is a dimensionless number that can express the ratio between two power levels. If we have, for instance P1 and P2

Power ratio(dB)=10*Log(P2/P1)

if the power P2 is the double of the power P1, that is P2=2*P1, then:

Power ratio(dB)=10*Log(2*P1/P1)=10*Log(2)≈3 dB

It works also using absolute dB unit (that means number that have dimension of power), such dBW, dBm and so on.
For instance dBm is the power referred to 1mW:

P(dBm)=10*Log[P(W)/1mW)

if we have, like before P2=2*P1, then the difference between the two levels, in dB will be:

Power ratio(dB)=10*Log[P2/1mW)-10*Log[P2/1mW)=10*Log[2*P1/1mW)-10*Log[P2/1mW)

using the Log properties:

Power ratio(dB)=10*[Log(2)+Log(P1)-Log(1mW)-Log(P1)+Log(1mW)]=10*Log(2)≈3 dB
 
3 dB is 3 dB. No matter for what - powers or magnitudes relationship. For power this just means 2 times, for magnitude this means sqrt(2) times.
But the thread starter just did not apply his mind to resolve it by himself, I think.
 

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