Why when doing bode plots 20* logs is used?

[mince]

Question about dB

When you want to convert a gain to dB, you say 10*log(A), but when doing bode plots, you use 20*log instead. Why is it 20*log when doing bode plots but 10*log everywhere else?

 

[flatulent]

Re: Question about dB

Power gain is 10, voltage gain and levels are 20.

 

[pardeep]

Re: Question about dB

In 10×log(A), the A is the power gain. Since power is proportional to square of the voltage so it becomes 20×log(V),where V is voltage gain.
Hope you got the point.
pardeep

 

[netsearcher]

Re: Question about dB

in bode plots we plot magntiude of the varibale so for finding magnitude we calculate second power of the varibale and ...

10 * Log ( |G(jw)| ^2 ) = 20 * Log ( |G(jw)| )

it's the answer .

 

[sxg]

Re: Question about dB

because P=U^2/R
so 10*log(P)=10*log(U^2/R)=20*log(U)-10*log(R),where R is impedance always is
a constant resistance.

 

[Jack// ani]

Question about dB

Power is proportional square of current/voltage. Now try to recall the rule of log, power terms, gets multiplied with 10 to give you 20.

 

[Borber]

Re: Question about dB

But when drawing Baude plot you can use power relation or voltage relation according to both formulaes and you will get equal curves, plots. Because dB is dB and it does not matter how it is calculated.