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I recently used an oscilloscope in X-Y mode to draw the phase ellipse of two voltages. I then used the formula $\phi = \arcsin(2y/B)$ where $y$ is the value of the ellipse at $x = 0$ and $B$ is the total distance from the highest point of the ellipse to the lowest.

I really want to know why this works.

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I guess You speak of an oscilloscope, don't You? –  Georg Oct 2 '11 at 10:54
Yes, sorry. I will update the question. –  Althalos Oct 2 '11 at 12:00

1 Answer 1

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The first thing to notice is where $2y$ comes from. You have two values at $x=0$. But for symmetry reasons, they differ only in sign. Hence, they're $+y$ and $-y$, and their difference is $2y$. Similarly, your Y signal has an amplitude A, and thus can have range from $-A$ to $+A$. The difference is $2A$, which you call $B$. Therefore, $2y/B$ is the ratio of the Y signal at $x=0$ and the max Y signal.

But why is this important? Let's look at some special cases. If the phase shift is zero, your ellipsis becomes degenerate. It's a line, and $y=0$. If the phase shift is $\pi/2$, it's a circle and $y=B/2$. Clearly the shape matters, and $y/B$ is a measure of that shape.

Not let's put in numbers:

$$x = A \sin(ft)$$ $$y = A \sin(ft+\phi)$$

$x=0$ means $\sin(ft) = 0$ and therefore $y = A \sin(\phi)$. Substituting $B$, you get $$x=0 \implies y = B/2 \sin(\phi) \implies \sin(\phi) = B/2y\qquad \qquad\Box$$

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I think the XY-mode of an oscilloscpe works by some switches, which connect the second channel amplifier to the horizontal plates in the Braun tube, instead of the time sweep circuit which is normally connected to those plates. –  Georg Oct 8 '11 at 17:07
@Georg: Modern scopes are a bit more complex, but that's indeed how it used to work. –  MSalters Oct 10 '11 at 7:47

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