Styx is correct, in the terms of units Power (Watts) are in joules/sec. Once you integrate over all frequency or over a period (see Parseval's Theorem) you get a number that represents the energy in the signal. Parseval's Theorem just states that there is a conservation of energy between time domain and frequency domain.Originally Posted by Styx
energy is just the sum of the power (or the integral over time of the power)
and what in a power of a signal :?:energy is just the sum of the power (or the integral over time of the power)
Thats what the power spectral density explains.
Just look at the integral. Its the integral of the absolute value squared of the transfer function multiplied by the fourier transform of the input function integrated over all frequency.
WHAT?
I thought the same :lol:WHAT?
It's probably easiest to visualise if you draw it out?.
Get a piece of graph paper, and draw a horizontal line across the middle (0V x-axis) - bear in mind, this now looks like a scope display!. Now draw a sine wave (or whatever wave you want) along the x-axis you just drew, going both above and below 0V. Next get a coloured pencil (or crayon) and fill it in, between the 0V line and the wave you drew, both above and below the 0V line - this will give you a series of coloured half-sine pulses, alternately positive and negative. This is the total amount of power contained in the signal.
Next, draw another sinewave, but this time DC shift it ABOVE the 0V line, so it doesn't cross 0V anywhere - then, as before, colour between the 0V line and the wave you drew (this time all the colour will be above the line). The coloured part again shows the amount of power in the signal.
In both cases (and all other ones) to measure the amount of power you need to calculate the surface area 'under the curve', which with graph paper can consist of counting the little squares :lol: Doing it electronically is FAR more complicated though, and requires 'true RMS' designs, which are essentially hardware multipliers.
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk
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.
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk
:? 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.
Len
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