What is the energy and power os signal mean in real...??

[MrAl]

Hi,

Yes there is definitely a little 'squared' sign as post #39 clearly shows using Latex.

 

[Jugurtha]

Energy is the ability to do work.
Power is like energy that hasnt done anything yet.

I've always liked to see it as "Power is the INTENSITY with which ENERGY is EXERTED" .. The word "intensity" suggest the notion of time in that concept.

Say it takes X amount of ENERGY to lift an elevator from Basement to the Fifth floor.

Motor A is able to lift the elevator to the fifth floor in 5 seconds.

Motor B is able to lift the elevator to the fifth floor in 10 seconds..

We don't say Motor A has more ENERGY .. Because it's the SAME amount of ENERGY it takes to BOTH to take the elevator from Basement to the Fifth Floor..

We say Motor A is MORE POWERFUL than B. Because it does the job in half the time.

Energy: Joule. Power: Watt or Horsepower. A 75 HP and a 150 HP .. They CAN exert the same energy, it's just the TIME at which it's done that DIFFERS.

 

[ericgibbs]

The definition of Power is explained here.

**broken link removed**

 

[koolguy]

OK, i got it.....
Thanks

 

[carbonzit]

Hmm, I looked at your linked page ("Signal Energy vs. Signal Power"). Interesting.

1. In the first illustration (finding the energy of a signal), it looks as if all they did was take the absolute value of the signal and find its integral, not the square of the signal.

2. What's with all those "[Math Processing Error]"s all over the page? Javascript error messages?

I don't know, the page is displayed without error here. And it's actually a squared signal, maybe there's a display error that kept the tiny ² from showing up Sneaky little ².

No, I know there's a little squared "2" in the label, but as I said, it looks as if they simply took the absolute value (I'm talking about the waveform illustrations). It looks like they just flipped over the negative-going parts of the wave, rather than squaring the wave.

Of course, it's just a stupid little drawing, so it's hard to tell, but still ...

 

[MrAl]

Hi carbon,

I checked the graphic and yes it is not graphically correct. It is the abs value, not squared, but if you do square it and rescale you get something that looks like the second graphic except the mountains would look more narrow. So it's an approximation of a squared function.

You'll also notice that they state
|f(t)|^2

which is a mix of ideals. There's no need for the abs value signs if the function is squared. In other words
|f(t)|^2=f(t)^2

in all cases. In the frequency domain it's another story because the abs value signs really do signify a special operation, not truly abs() although sometimes called that.

 

[carbonzit]

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.)

 

[Jugurtha]

Hello again.. (This answer is for memo as this thread is really not about stability or stuff like this)

I was just reading about this last night in my engineering textbook. (I should point out that most of this stuff is waaaaay over my head!)

You might look into exploring root-locus analysis (developed by W.R. Evans in the 1940s), like this page and this page. It has everything to do with determining the stability of a system, using the locus (location) of the system's poles and zeroes in the s-plane (which, as I said, I don't yet fully understand; I'm guessing that the O.P. doesn't either).

In some other methods (Nyquist, Black & Nichols for example) the focus is on "pulsation" . In Root Locus, the focus is on "gain".


Grosso modo, Root Locus is about gain (Evans Gain 'Kev') .. You "travel" your gain and you "observe" your system for different values of this gain (stability, limit of stability, intersection with imaginary axis and unit circle, etc)...



If you've got some questions about it, if you want me to develop it further with examples of the 12 steps, feel free to open a thread and hit me up.

(This or State Space Representation (digital or else), z domain, digital correctors (R.S.T Polynoms in z for example) feel free to ask, I might be able to help on some questions..

 

[MrAl]

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)?

 

[carbonzit]

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)?

Hmm, do I detect a trick question here, perhaps?

I'll bite: my guess is that the power actually expended by each motor is zero, since no actual work was done, there's no difference between them.

(Does heat generated by a stalled motor count as power?)

 

[MrAl]

Hi carbon,

Well maybe sort of It's just that we have two motors each presumably of a different power TRYING to lift a weight, yet we can't calculate the power of either motor because the weight didnt move. It's almost like the two motors have the potential power to move something, but may not be able to do so. So power cant always be calculated based on something that shows an obvious energy expenditure.
Power is work done per unit of time, but what if we use a 1000 watt motor to try to lift something that doesnt move? Bringing the electrical resistance into the picture isnt going to help this situation, because we can say we have a theoretically perfect motor made of super conductors. The motor is normally rated at 1000 watts, but it actually doesnt do anything when applied to lift a 10 million cubic meter lead block (with a normal gearing ratio). The motor applies a force, but that doesnt mean it moves anything.
So it's interesting that we can have things rated in watts when power is defined as the work done per unit of time, when there might not actually be any work done.
Using the 'solid' analogy that i was suggesting, we can see power as a cross section of a solid, which simply means that we can view power as something of a lower dimension than energy, which might help.