Something to think about:

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Well Holy Cr@p!

Well Holy Cr@p!

Rattus! :grin:

This is what has been bashing me in the head for the past four hours!!! I knew there was a connection somewhere, I finally figured it out. And you made reference to it up there when you said something to the order of "my way and winnie's way are the same thing. So you were just hinting around, wondering if I would ever figure it out for myself?:cool: I would have done this earlier, but I was trying to pin you down about whether the input and output powers were the same.

Check it out:

You say that the average power of the input to the filter is:

Pt = .636 x Ep x Ip

We're good on that.

I said that the average power at the load would be:

.5 x Ep x Ip ; this comes from .707 x Ep x .707 x Ip

I didn't recognize or have a clue (and still have my doubts:roll: ) that the peak voltage of the fundamental which would pass through the filter would be 4/pi Ep.

So, if we include the increased output voltage of the fundamental in my formula for average power at the load, we get:

4/pi x .5 x Ep x Ip

and since the filter is perfect with no losses, Input power = output power:

Your equation for input power = my equation for output power:

.636 x Ep x Ip = 4/pi x .5 x Ep x Ip

do the algebra, Ep and Ip cancel,

.636 = 4/pi x .5

and it certainly does. That is dadgum beautiful and unexpected. :smile:

So, I will conclude that the "perfect" filter needed for the job above must as a necessity produce a sine wave voltage which has a peak which is a tad bit larger than the peak of the square wave input.

Edit: I just did this with something else you posted:

Vrmsf = Vp x 4/(sqrt(2) x pi) = 0.9Vp

If I substitute this as the voltage in my equation for average power at the output of the filter, everything works out also. Makes sense since it just algebra with the numbers.

Edit again: Winnine, thank you also for your informative posts. I hope you will continue monitoring this thread because I have at least 5 questions I need cleared up.
 
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Back to the Fourier analysis:

1. So does the fundamental sine wave actually exist in the square wave? Or did the "filter" create the fundamental sine wave?

2. The fourier series for the square wave consists of only odd harmonics. Does that mean that there are no even harmonics contained in the square wave?
 
Fourier Analysis is a mathematical tool. It may be used to describe waveforms as possibly infinite sums of sinusoids. Clearly, what you really physically have is a single square wave, not a pile of sine waves. You could take that square wave and describe it as a different sum, for example as a sum of complex exponentials or as a sum of impulse functions or any number of other mathematical descriptions.

So in this sense, the harmonics are 'not really there'.

However, in a very real and important sense, the harmonics _are_ really there.

There are zillions of different mathematical tools. Their usefulness depends upon the particular mathematical tool correctly describing the physical situation to which they are applied.

If you want to understand how a particular complex system will respond to a complex waveform, it turns out that you can use Fourier Analysis to 'decompose' the complex waveform into a sum of sinusoids. Then you answer the question: how does the system respond to each of those sinusoids. Then you add up the result for each harmonic. For a large number of electrical systems, this sum will be a very very good approximation to the true response of the system.

Consider a sensor designed to pick up particular frequencies. This is a physical sensor that turns on a light if the set frequency is present. Take a complex waveform, do the harmonic analysis, and if one of the harmonics that is present is the frequency that you've tuned to, then the sensor will respond. If you tune it to 180 Hz and 'play' a 60Hz square wave, then the device will respond...but if you tune it to 120Hz and play a 60Hz square wave, then the device will not respond. The fact that a sensor tuned to the harmonic frequencies will detect something is a clear indication that in very real physical sense, those harmonics are _really_ there.

In the electric motor example that I gave, a complex waveform is produced by the inverter. This complex waveform can be analyzed as consisting of fundamental and third harmonic, along with a whole bunch of harmonics associated with the switching frequency. The physical response of the motor to the different harmonics is different, and the net result is that the motor 'sees' a higher fundamental voltage than the inverter output, and almost doesn't see the switching frequency at all. The current flow through the motor is nearly pure fundamental.

-Jon
 
Also:

Also:

The harmonic content of rectangular waves is evidenced by the radio noise they generate. Think fluorescent lamps and dimmers.
 
My point:

My point:

The point I was trying to make is that the product of an RMS voltage and an RMS current does not always provide average power as one might expect.
 
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