# Thread: Max & Min for 2 Sine Waves out of Phase

1. ## Max & Min for 2 Sine Waves out of Phase

Hey. I'm taking some power classes for my electrical engineer degree, and I need to know the maximum and minimum for instantaneous power for an alternating voltage source. Is there any easy way to determine it without going through the derivative, finding what value makes it zero(with all those damn trig functions), and then plugging it back in to see which one is a min and a max?

example: $P(t) = VI\sin (wt + a) \sin (wt + b)$

Or if that is the only show-your-work kind of way to do it, could someone run through it with me? I find solving for zero with that many trig functions impossible (and i find solving for the periodic nature of the answer impossible) I resorted to graphing & using the max/min function to find the answer. I do not know if that will fly on a power test, though.

I have, through some manipulation, made a function that could be easier to solve, but still, it seems hopeless. If we are interested in the max & min without any specification for when they occur, we can say that the second sine wave is at an angle of $wt$ and the first is at an angle of $wt + a - b$
then
$P(t) = VI\sin (wt + a - b) \sin (wt)$
and the voltage's phase angle is $a$ and the current's is $b$
so $a - b = c$, the power factor angle.
$P(t) = VI\sin (wt + c) \sin (wt)$
then
$\sin (U + V) = \sin (U) \cos(V) + \cos(V) \sin(U)$
so
$P(t) = VI[\sin (wt) \cos(c) + \cos(wt) \sin(c)] \sin (wt)$
$P(t) = VI[\sin^2 (wt) \cos(c) + \cos(wt) \sin(c) \sin(wt)]$
and
$sin^2(U) = \frac{1 - cos(2U)}{2}$
so
$P(t) = VI[\frac{1 - cos(2wt)}{2} \cos(c) + \cos(wt) \sin(c) \sin(wt)]$
and
$\sin (U) cos(V) = \frac{\sin(U + V) \sin(U-V)}{2}$
so
$P(t) = VI[\frac{1 - cos(2wt)}{2} \cos(c) + \sin(c) [\frac{\sin (2wt) + \sin (0)}{2}]]$
simplifying:
$P(t) = \frac{VI}{2}[\cos(c) - \cos(c) cos(2wt) + \sin(c) \sin (2wt)]$

I think I made an error or something. I'm so tired of thinking about this problem. Can someone chime in? I believe there is a way to transform the sin(wt - a) sin(wt) into a constant plus(or minus) 1 sin(or cos) wave. That would be much easier to handle in this task.

2. Use the formula $\sin A\sin B = \tfrac12\bigl(\cos(A-B) - \cos(A+B)\bigr)$. Then $P(t) = \tfrac12VI\bigl(\cos(a-b) - \cos(2\omega t+a+b)\bigr)$. So $|P(t)|$ has a maximum value $\tfrac12VI\bigl(1 + |\cos(a-b)|\bigr)$, attained when $\cos(2\omega t+a+b) = \pm1$ (depending on whether $\cos(a-b)$ is positive or negative).

3. Originally Posted by Opalg
Use the formula $\sin A\sin B = \tfrac12\bigl(\cos(A-B) - \cos(A+B)\bigr)$. Then $P(t) = \tfrac12VI\bigl(\cos(a-b) - \cos(2\omega t+a+b)\bigr)$. So $|P(t)|$ has a maximum value $\tfrac12VI\bigl(1 + |\cos(a-b)|\bigr)$, attained when $\cos(2\omega t+a+b) = \pm1$ (depending on whether $\cos(a-b)$ is positive or negative).
Thank you!
$v(t) = 113.1371 sin(377t + 13^{\circ})$
$i(t) = 56.5685 sin(337t - 7^{\circ})$
we will say then this:
$v(t) = 113.1371 sin(377t)$
$i(t) = 56.5685 sin(337t - 20^{\circ})$
$P(t) = 113.1371 * 56.5685sin(377t)sin(337t - 20^{\circ})$
$P_{max}(t) = \frac{113.1 * 56.6}{2} ( \cos (20^{\circ}) + 1) = 6207 W$

Yes!
and the stuff above with - 1 instead of + 1 yields the right min! yes!