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Old 16th April 2003, 07:16 PM   #1
JDeV is offline JDeV  United Kingdom
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Question About dB's

I am bussy with a corespondence course in electronics and got the following question in an assignment. It sound a bit wrong to me (the uninformed one) and want to know if it make sense and what would the answer be? I don't grasp the dB thing yet, so please excuse silly questions (Maybe somebody can give me a "dB for Dummy's" explanation. )

Determine the dB level from a given sound power of 15dB, and a reference sound power of 20dB.

(Don't you measure sound power in Watts?)

Thanx for all help.
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Old 16th April 2003, 07:32 PM   #2
SY is offline SY  United States
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I think the text writer needs the "dB For Dummies" course more than you do. Or at least a writing course. Sound power is given in watts OR in dB referenced to some standard power level.

Here's the entire course:

For power ratios, dB = 10 log (P1/P2)

Since power = V^2/R, and log (x^y) = y log x, then for a voltage ratio, dB = 20 log (V1/V2)

That's all there is, there ain't no more. If the dB quoted in the question are both referenced to the same standard power level, the answer is -5 dB.
"You tell me whar a man gits his corn pone, en I'll tell you what his 'pinions is."
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Old 16th April 2003, 07:39 PM   #3
dhaen is offline dhaen  Europe
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dB's are a logarithmic scale. They are not absolute, and can be referenced to any point.
In this case they have told you the reference point (20dB).
The magic formulae I keep in my head are:
10*Log (P1/P2) (for power)
20*Log (V1/V2) for voltage (and other non squared values)
In other words, the dB figuer is describing the ratio between the 2 voltages, powers, or whatever...
You'll need to do some substitution...

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