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How to get to this instantaneous power equation?

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madeza

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Hi everybody
As you can see, [1] are voltage and current with their respective phase angle.
[2] is the instantaneous power, which I could solve from [1]
But my question is: How can I get [3], if I read the next: instantaneous power is given by [3] where the voltage is resolved onto in phase and quadrature components to the current
Tips will be welcome...
Thanks for your support
 

Re: instantaneous power

p(t) = 1/2 VI cos(alfa-beta) + 1/2 VI cos(2wt+alfa+beta)

Let P = 1/2 VI cos(alfa-beta)
Then p(t) = P + 1/2 VI cos(2wt+alfa+beta)


But cos(2wt+alfa+beta) = cos(2wt+2beta + alfa-beta) = cos(alfa-beta)·cos(2wt+2beta) - sin(alfa-beta)·sin(2wt+2beta)


Then 1/2 VI cos(2wt+alfa+beta) = P·cos(2wt+2beta) - 1/2·VI·sin(alfa-beta)·sin(2wt+2beta)

Now let Q = 1/2·VI·sin(alfa-beta)


Finally p(t) = P + P·cos(2wt+2beta) - Q·sin(2wt+2beta)
 

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    madeza

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