Something to think about:

[rattus]

A square wave with amplitude Vp has an RMS value of Vp, right?

Now let this square wave drive a sinusoidal current with amplitude,Ip, through a filter. The RMS value is 0.707xIp, right?

How does one compute the power?

 

[chris kennedy]

How does one compute the power?

I would try using a computer.

 

[nakulak]

A square wave with amplitude Vp has an RMS value of Vp, right?

I would call this an "rms equivalent" value, since the area under the square wave is easily calculated.

Now let this square wave drive a sinusoidal current with amplitude,Ip, through a filter. The RMS value is 0.707xIp, right?

How does one compute the power?

a

wouldn't the area under the I curve simply be a scalar multiple of V ?

 

[quogueelectric]

A square wave

A square wave

Will not create a sinusoidal current unless you are driving a nonlinear load. It will create a square current wave minus the clipped edges. Be prepared for harmonic havoc.

 

[crossman]

You are talking about the power at the load, right? I mean, you are not counting any power generated by heat or other losses in the filter?

Here is my guess, and I am assuming we are measuring voltage, current, and power at the load. To create the sine wave current, the voltage at the load will be a sine wave voltage, therefore proceed accordingly:

Irms of sine wave = Ipeak x .707

Erms of sine wave = Epeak x .707

Multiply these two together, Power at load = 1/2(Ipeak x Epeak)

Now on the other hand, if you are looking at the entrie amount of power put out by the source, we would need more information on exactly how the filter is accomplishing its task.

And there you have my educated guess.

 

[rattus]

Nuh-uh:

Nuh-uh:

Will not create a sinusoidal current unless you are driving a nonlinear load. It will create a square current wave minus the clipped edges. Be prepared for harmonic havoc.

We are not speaking of rectifiers here, just a linear filter between the source and the load. This filter blocks all frequencies other than the fundamental.

 

[nakulak]

We are not speaking of rectifiers here, just a linear filter between the source and the load. This filter blocks all frequencies other than the fundamental.

assuming that he could build a sine generating current source, I still maintain that the output would be a scalar multiple of the easily calculated original function (the square wave voltage), and the power is easily calculated by the product without any rms calculations.

jmo

 

[rattus]

Crossman, assume that the filter is ideal--perfect filtering and no losses. The load is not specified, therefore we cannot compute the power there. All we have is the square wave voltage and sinusoidal current.

 

[rattus]

assuming that he could build a sine generating current source, I still maintain that the output would be a scalar multiple of the easily calculated original function (the square wave voltage), and the power is easily calculated by the product without any rms calculations.

jmo

We don't need to assume anything other than the conditions stated in the problem. Now give us a formula for the power.

 

[crossman]

Oh, now I got ya! Hang on asecond! Don't run off!

 

[nakulak]

if the area under the current wave form was indeed a scalar of the voltage, then power = k * voltage squared. the scalar k would presumably be the efficiency of the sine wave generator/transformer(?)

 

[crossman]

(1/2)(Ipeak x Epeak)cos theta

where cos theta = phase angle between voltage and current

 

[quogueelectric]

Eliminating all of the hypothetical

Eliminating all of the hypothetical

You have a square wave driving exactly what to create a sinusoidal current???

 

[dezwitinc]

If sinusoidal depleneration is the reason for this question, I believe that the following link may provide the means required to eliminate this problem.

http://www.youtube.com/watch?v=TD1LHejil6M

 

[rattus]

You have a square wave driving exactly what to create a sinusoidal current???

It is a filter which passes only the fundamental of the square wave. Doesn't matter how it is made or even if it can be made. It is hypothetical--something to exercise your mind.

Let's say also that the current is in phase with the fundamental of the square wave.

 

[nakulak]

if they are in phase then my formula above is correct. if not, it would require correction, but the math is simple (subtracting rectangles )

 

[crossman]

Let's say also that the current is in phase with the fundamental of the square wave.

Then we can omit the cos theta so:

1/2(Ipeak x Epeak)

which is equivalent to:

Irms x .707Epeak

 

[winnie]

http://en.wikipedia.org/wiki/Square_wave

This is a job for Fourier analysis. Simply take the square wave, analyze it as a sum of sine functions, and then do a separate analysis of the circuit for each sine function.

For example, you have a square wave voltage applied to a load which consists of an ideal filter and then a resistor. This imaginary load essentially has infinite impedance for all voltage components other than fundamental, and is a resistor for the fundamental voltage component.

You use Fourier analysis to figure out the voltage of the fundamental component, and then just use Ohm's law to figure out the power delivered through the circuit by the fundamental component.

If you take a +-1V 50% square wave and feed it through an _ideal_ filter that will only pass fundamental, then you will end up with a +-4/pi sine function. Read that again The peak voltage of the fundamental component of the a square wave is greater than the peak voltage of the source square wave. Its right there in the Fourier series, and has real application in VFD design.

VFD aside: the way that the VFD works, the output is synthesized by using IGBT devices to switch an internal DC supply to the output. The peak voltage of the output waveform is set by the internal DC voltage (usually, when a half bridge topology is used, output peak to peak voltage is equal to the DC voltage). A very common trick is to _intentionally_ add a bit of third harmonic to the output. This lets the fundamental peak to peak voltage exceed the DC rail voltage. The composite waveform output is limited to the DC rail voltage, but the fundamental component can be slightly greater. A three phase motor will not pass third harmonic, and thus acts like a near perfect blocking filter for this component. The net result is that the motor windings 'see' a voltage that is greater than the inverter can supply.

-Jon

 

[rattus]

How about this?

How about this?

First assume the current is in phase with the fundamental of the square wave, and let the instantaneous current be,

i(t) = Ip x cos(wt)

Then the instantaneous power is,

p(t) = Ep x Ip x |cos(wt)| [added absolute value symbols|

Then, the average power is,

P = Ep x Ip x 0.636

Where 0.636 is the average value of the cosine wave.

The point is that the power is not equal to the product of the RMS values in this hypothetical case.

This approach differs from Winnie's approach using the fundamental of the square wave. Both yield the same results however! [Corrected my earlier misstatement!]

 

[crossman]

First assume the current is in phase with the fundamental of the square wave, and let the instantaneous current be,

i(t) = Ip x cos(wt)

Then the instantaneous power is,

p(t) = Ep x Ip x cos(wt)

I'm not catching this, rattus. If the filter is blocking out all the voltage except the fundamental, and then we want to know the voltage at any given instant, don't we need to adjust Ep by multiplying it by cos(wt)?

What I am seeing, if the filter is perfect, then it is creating a sine wave voltage from the square wave. This new sine wave voltage is what causes the current to flow.

To me, the square wave/filter combo isn't any different than a regular generator producing a sine wave voltage.

What am I missing?