rms

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To make things a bit clearer, I've drawn a couple of pictures, in both case the horizontal black line is zero volts, and the red area is the amount of power supplied. Example1 is an AC coupled sinewave (centred on zero volts), example2 is the same sinewave, but with a positive DC offset.
 

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:? it implies that power of a signal is area under the curve

and

energy of a signal is ........... :!: [power * time]?
 
Well no,

What Nigel has plotted is a voltage waveform.
The area under that graph is just Volt-seconds (which has its uses elsewhere)


IF that sinewave was a power sinewave (ie sinusoidal amps * sinusoidal voltage => resistive load)

Then those graphs (black line) represent instantanious power for a given time, the RMS gives you the "average" (note average in quotes) power

The read area under the "power" curve gives you the energy for that time
 
I did the calculus for the general case, ie. the RMS value of a sinewave signal superimposed on a DC level.

If V = a + b sin(wt) then Vrms = sqrt(a^2 + b^2/2)

Check

For a DC level, Vrms = sqrt(a^2 + 0) = a as expected.

For a sinewave without a DC offset, Vrms = sqrt(0 + b^2/2)= b/ sqrt(2) as expected.

So if a = 3 and b = 1, then Vrms = 3.0822 approx. As _3iMaJ showed in his post.
 
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