Subtracting two electrical sinusoidal functions

[RonShap]

In an electrical transfer switch there are two sources. When measuring from phase A of one source to Phase A of the second source, there may be a phase angle difference and a resulting voltage difference.
I'm trying to workout the formula on excel to determine the value of one sin wave minus the other so I could determine the expected voltage difference based on varied phase angle differences.
In the US, the peak of the sinusoidal wave is approximately 391.737 (which is 277 * sqrt2). The frequency is 60 hertz.

The expected result for two sample points is a phase angle difference of 0 should result in a sinusoidal difference of 0 because they are on top of each other. A phase angle of 120 degrees should have a peak sinusoidal difference of 678.8225 (which is 480 * sqrt2).

Suggestions or corrections to this theory would be greatly appreciated.

 

[electricpete]

There are many ways to skin that cat. Here is one way using excel's complex functions (part of analysis toolpack).

Put into cells A1 and B1 the magnitude and angle (degrees) of your first number
Put into cells A2 and B2 the magnitude and angle (degrees) of your second number number
Put into C1 the formula =+COMPLEX(A1*COS(RADIANS(B1)),A1*SIN(RADIANS(B1)))
(that is the complex representation of the first vector)
Put into D1 the formula =+COMPLEX(A2*COS(RADIANS(B2)),A2*SIN(RADIANS(B2)))
(that is the complex representation of the second vector)
Put into A5 the formula =+IMSUB(A3,A4)
(that is the complex representation of the difference between first and second vectors)
Put into A6 the formula =+IMABS(A5)
(that is the magnitude of the difference)
Put into A7 the formula =+DEGREES(IMARGUMENT(A5))
(that is the phase angle of the difference)


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[RonShap]

Pete,
Thanks.
This is turning into quite the exercise for my brain. I hadn't had a reason to think about complex numbers in a very very long time.
I was trying to plot out the function in excel hoping it would help solidify what I can't seem to get my arms around. Of course the end result is to find the peak value of that resulting function (vector 2 minus vector 1) and then convert it to an RMS value.
Since you are so comfortable with this, other reference material would be helpful if possible.

 

[GregLocock]

You could do it as a timehistory which will at least allow you to plot the waveforms out directly.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[RonShap]

Pete,
Thanks, I got it!

 

[electricpete]

I guess you figured it out in spite of my major typo. Corrected in bold:

"Put into cells A1 and B1 the magnitude and angle (degrees) of your first number
Put into cells A2 and B2 the magnitude and angle (degrees) of your second number number
Put into A3 the formula =+COMPLEX(A1*COS(RADIANS(B1)),A1*SIN(RADIANS(B1)))
(that is the complex representation of the first vector)
Put into A4 the formula =+COMPLEX(A2*COS(RADIANS(B2)),A2*SIN(RADIANS(B2)))
(that is the complex representation of the second vector)
Put into A5 the formula =+IMSUB(A3,A4)
(that is the complex representation of the difference between first and second vectors)
Put into A6 the formula =+IMABS(A5)
(that is the magnitude of the difference)
Put into A7 the formula =+DEGREES(IMARGUMENT(A5))
(that is the phase angle of the difference)"

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[jghrist]

RonShap,

Do you have the magnitude and angle of each measurement or do you have instanteous sample points from a measurement?

If the latter, you could subtract the instantaneous points then use a cosine filter to determine the magnitude and angle of the difference.

 

[electricpete]

a simpler approach than my last would be law of cosines for adding two vectors:

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

Except you negate one of them to accomplish subtraction

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