So why square the function (wave) if all that's needed is taking its absolute value? Is squaring it a necessary part of calculating its integral? (My guess is "no", but my calculus is a bit ... rusty.)
Hi again carbon,
What i meant was that the 'graphic' was not correct. It showed a graphic that was really the abs() when the true calculation requires a squaring operation, not abs value. The required squaring operation makes a functions hills look more narrow than a simple absolute value operation. If compared by overlapping the graphics of the abs() and the f^2 functions even if f^2 is scaled so that the peaks are the same in both graphics, the f^2 function hills will be more narrow than the abs() function hills. That's one way to tell the difference between abs() and the true squaring functions when examining the graphics. In fact, you can use Paint or some other program to copy and paste the original function right on top of the second graphic in that diagram, and it overlaps perfectly on the positive cycles, and if you flip the function graph vertically it fits perfectly on the negative half cycles too, so it's just the abs() not the squaring function. But that's just the graphic. The theory requires squaring.
A totally different issue was the stated function itself:
|f(t)|^2
which is not really incorrect, but the abs value signs are not necessary when the function is squared because everything squared is always positive anyway. That's why i think they mixed up F(jw) with f(t) and used the abs value signs for f(t) when they are only required for F(jw). This is a different issue though. Both forms however do require squaring...that's a must.
The reason for the squaring is quite simple really. When we calculate power we can use either:
V^2/R
or:
I^2*R
where it's easy to recognize that either V or I is being squared, but which one do we use? Using V^2/R leads to a different result than usign I^2*R (obviously one is divided by R and one is multiplied by R so they cant be the same). But there's one little interesting exception to that problem, and that is when we make R exactly equal to 1 ohm. When we do this, we get:
V^2/R=V^2/1=V^2
or we get:
I^2*R=I^2*1=I^2
and now either one leads to a squaring of the variable.
So anyway, we square it in order to get the power that would develop in a 1 ohm resistor, that's why the total result is called the "1 ohm energy", but often just called the 'energy' with the 1 ohm being implied by the context.
On a related topic, we have slightly bigger motor A and different slightly smaller motor B. We try to use both motors to lift a block make up of 10 million cubic meters of pure lead. Neither motor can lift the block even a small fraction of a millimeter. Which motor is more powerful (ie has more power)?