Frequency and Power

1. Mar 6, 2015

We all know that the electrical power = V * I * cos Q. and I know how this formula is derived. but yet intuitively I was expecting to see the frequency in the formula. I just need intuitive explanation how the frequency is not appearing in the power formula while we know in physics that higher frequency equal to higher power !!

2. Mar 6, 2015

milesyoung

This is the average, or real, power over a period, where V and I are the RMS values of the voltage and current, respectively.

If you convert the energy E over a period T, the average power is E/T.

If you now, for instance, double the frequency, you convert half the energy in half the time, so the average power is E/2/(T/2) = E/T, and, in general, you'll find that the average power is independent of frequency.

Hmm, only thing that comes to mind is how the energy of a photon is directly proportional to its frequency, but if that's what you mean, then you definitely shouldn't confuse that with what you're asking here.

3. Mar 6, 2015

Thank you for the clarification... after all I am after intuitive clarification.

when double frequency I assume double energy is converted or same energy in half the time, I think this is the source of my confusion.

4. Mar 6, 2015

milesyoung

Here's a power curve for a resistive load (I just grabbed the first image on Google Images):

The area under the power curve (integral in general) is equal to the amount of energy you have converted in whatever period of time you're considering. If you focus on a single period of the voltage/current waveform, can you see how the area under the power curve must decrease as you "squeeze" the period, i.e. as you increase the frequency?

Last edited by a moderator: May 7, 2017
5. Mar 6, 2015

jim hardy

Maybe this will help...

If volts = sin(X)
and we'll make R = 1 ohm for simplicity
So current also = sin(X)

power = volts X amps
power = sin^2(X)

it's a trig identity that sin^2(x) = 1/2 - 1/2 cos(2x) (see http://www.math.com/tables/trig/identities.htm )
which = 0.5 X ( 1-cos(2x))
1. It's always positive, because cos(anything) only can be between -1 and +1 so it cycles between 0 and 1 (edit - fixed arithmetic mistake jh)

2. Its cyclic nature has doubled in frequency
as shown in MilesYoung's most excellent picture.

Instantaneous power contains a frequency term ( 1-cos(2x))
but remember what's the whole purpose of RMS ?
It's to calculate equivalent heating value of any arbitrary waveform, equivalent to DC(no frequency)

RMS gives the average of instantaneous power over a cycle....
We use sinewaves and RMS so exclusively that we tend to forget from whence we came...

old jim

Last edited: Mar 6, 2015
6. Mar 8, 2015

Thank You for trying to help confused man

Can I summarize as the following, and please correct me if I am wrong:
Because the power is sine wave in shape then the RMS value only depend on the peak and not the frequency, but if the shape is something like square wave or sawtooth then definitely it will depend on frequency

Thanks.

7. Mar 8, 2015

jim hardy

No, that's about as wrong as it can be.
Frequency has nothing to do with it.
Sine wave is so prevalent that we forget it is a mathematical oddity in that its shape does not change when you differentiate it.

Force a sine wave current through an inductor and you get voltage that's a cosine wave. It looks identical.
Force a triangle wave current through same inductor and you'll get voltage that's a square wave , looking very different .

Look up your definition of RMS: http://en.wikipedia.org/wiki/Root_mean_square

The key to understanding RMS is this:
It's just a sequence of three arithmetic steps. We do them backward, over one complete cycle , whatever be its shape or frequency:
first, Square every instantaneous value
second, take Mean(the average) of those squared instantaneous values over one cycle whatever be its shape or frequency
third, take square Root of that mean
Pictorially here's what it does for a sine wave:

http://www.physics.udel.edu/~watson/phys345/fall1998/class/6-rms.html

See the three steps? The kindly Professor Watson numbered them for us .....
and you can apply them to waves of any shape or frequency.

"Mean" or "Average" is just a constant with same area beneath the curve, the black horizontal line in that image
You should go through the three steps for square and rectangular waves, because those shapes are easily solved by geometry.
That'll replace the faulty word-based image in your brain with a math based one. Then see if you think there's better words to describe it...

Study that link i think it's well written
again it's http://www.physics.udel.edu/~watson/phys345/fall1998/class/6-rms.html

8. Mar 8, 2015

milesyoung

The shape of the waveform doesn't matter. Following Jim's lead, if we take the basics:

The RMS value of some $T$ periodic function $f$ is given by:
$$\sqrt{\frac{1}{T}\int_0^T[f(t)]^2dt}$$
The $\frac{1}{T}\int_0^T[f(t)]^2dt$ part is just the average of $[f(t)]^2$, so ask yourself if the average of a periodic function depends on its frequency?

If it doesn't, the RMS value of any periodic function must be independent of its frequency, and the expression you posted for average power must then also be independent of frequency.

What is it, intuitively, that tells you it must depend on frequency? If you tried to explain it, maybe it will help us help you.

Edit: Jim beat me to it.

9. Mar 8, 2015

jim hardy

no, miles wrote an elegant paragraph while i fumbled for words .

Nice job !

10. Mar 8, 2015

What made me post the previous is this equation:
Positive square wave with duration T and frequency f :
Vrms= Vpk*sqrt(f*T)
Vavg= f*T*Vpk
from http://www.daycounter.com/Calculators/RMS-Calculator.phtml

As you can see frequency is mentioned in both although this is for voltage I assumed the same for Power ( I may be wrong as usual)

I will spend sometime to study your post and I am sure that your help will adjust my thoughts. Your time is very much appreciated. Thank

11. Mar 8, 2015

milesyoung

It's a bit confusing the way they present it, but T is the "on-time" of the pulse train, so if Tprd is the period of the waveform, then f*T = T/Tprd. This is the fraction of the period where it's "on", also called its duty cycle, which is independent of frequency.

12. Mar 8, 2015

jim hardy

I'd say that webpage made a mistake putting frequency in the equation.
He has not defined "duration"
and he's only inferred a square wave moving between zero and some positive value.
if by "duration T" he means "T is the fraction of time that the square wave is positive not zero"; well that is duty cycle.

and frequency does NOT enter the equation. Putting it in there undefined was IMHO poor technical writing at best...........

It should be obvious that for a wave that's positive half the time and zero half the time, ie T=0.5, its average is half its positive value.

If instead he means that T is the number of (micro? pico? femto?)seconds per cycle that the wave is positive, and that f is the number of cycles per second(observe not necessarily an integer), then F X T is the fraction of any second during which the wave is positive not zero. And that's duty cycle too.

So even if he's not technically incorrect, his ambiguity certainly misled you to a wrong conclusion .

Beware of references from the internet . Always corroborate between at least two of them.
Make sure one of them is an institute of higher learning -
it's not that academics know more than practicing science folks , just they are trained to explain things unambiguously and define their terms up front.
Hence the old saying "If you want to really learn a subject, Teach It ! "

We working engineers are by and large notoriously bad at exposition. That's why engineering curricula include a few credit hours of Literature - in hope some of the writing skills will rub off.

old jim

edit - as usual miles said it in 1/10 as many words...

Last edited: Mar 8, 2015