Derivation of "arcsin" phase shift formula

[DeldotB]

Homework Statement

Good Day,

On an oscilloscope, when two incoming signals are out of phase, in an XY setting, an ellipse appears on the oscilloscope screen. The phase shift between the two incoming signals can be found by the formula:

[tex]sin^{-1}((Y_{max})/(Y_{int}))[/tex] where Y max is the ellipse's max value and Y int is the Y intercept.

I am asked to derive this equation, and I am not sure how to start.

Any help would be appreciated!

Homework Equations

I was not given any relevant equations - although the derivation shouldn't be too hard. My professor said this class is pretty UN math intensive.

The Attempt at a Solution

[/B]
I do not have an attempt- I don't know where to start!

 

[berkeman]

Homework Statement

Good Day,

On an oscilloscope, when two incoming signals are out of phase, in an XY setting, an ellipse appears on the oscilloscope screen. The phase shift between the two incoming signals can be found by the formula:

[tex]sin^{-1}((Y_{max})/(Y_{int}))[/tex] where Y max is the ellipse's max value and Y int is the Y intercept.

I am asked to derive this equation, and I am not sure how to start.

Any help would be appreciated!

Homework Equations

I was not given any relevant equations - although the derivation shouldn't be too hard. My professor said this class is pretty UN math intensive.

The Attempt at a Solution

[/B]
I do not have an attempt- I don't know where to start!

Start by sketching a few test cases. What does the display look like with zero degrees offset? With 90 degrees? With 45 degrees? With 30 degrees?

 

[DeldotB]

Hmm...well I know 90 degrees is a circle, zero degrees is the line y=x,180 degrees is the line y=-x and 45 degrees is an ellipse. Using this information how would I derive a formula?

 

[berkeman]

Hmm...well I know 90 degrees is a circle, zero degrees is the line y=x,180 degrees is the line y=-x and 45 degrees is an ellipse. Using this information how would I derive a formula?

And what are Ymax and Yint for each of those cases?

 

[DeldotB]

Well yint for 0 degrees and 180 degrees is zero (with no y max), I am not sure what y max and int are for 45 degrees or 90 degrees (or how to find them)
p.s sorry, I literally have no experience with an oscilloscope - this question was in a problem set my teacher gave me and I don't really have any "insight" about oscilloscopes.

 

[berkeman]

Well yint for 0 degrees and 180 degrees is zero (with no y max), I am not sure what y max and int are for 45 degrees or 90 degrees (or how to find them)
p.s sorry, I literally have no experience with an oscilloscope - this question was in a problem set my teacher gave me and I don't really have any "insight" about oscilloscopes.

Did the instructor say anything about how to find Yint and Ymax on the plots? That part is a little ambiguous to me...

http://77.162.55.232/usbscope/images/xyplot_animated.gif

BTW, in that animation above, the center is (2.5,2.5), not (0,0).

 

[DeldotB]

My professor did not say how to find ymax and ymin. Does it have to do with parameterizing an ellipse perchance?

[tex]x=Acos(t)[/tex] and [tex]y=Bsin(t)[/tex]?

 

[DeldotB]

After a little more research, this is called a "Lissajous Curve" https://en.wikipedia.org/wiki/Lissajous_curve
...still not sure how to derive it though...

edit: also question 13 on this page:http://www.allaboutcircuits.com/worksheets/ac-phase/
but it doesn't say how to do it...

 

[ehild]

Do you know how the oscilloscope works? You have two inputs, one for the x direction and one for the y direction. If there is zero signal at both inputs, you see a bright spot in the middle of the screen. Now add a signal X(t) to the X input, and Y(t) to the Y input. The bright spot is deflected both in the x and y direction, and it describes a curve, corresponding to x=X(t) and y=Y(t). You need to eliminate the parameter, t.
Take the case when X(t)=Asin(t), Y(t)=Asin(t+a) and eliminate t. You need the addition rule of the sine function.

 

[DeldotB]

Do you know how the oscilloscope works? You have two inputs, one for the x direction and one for the y direction. If there is zero signal at both inputs, you see a bright spot in the middle of the screen. Now add a signal X(t) to the X input, and Y(t) to the Y input. The bright spot is deflected both in the x and y direction, and it describes a curve, corresponding to x=X(t) and y=Y(t). You need to eliminate the parameter, t.
Take the case when X(t)=Asin(t), Y(t)=Asin(t+a) and eliminate t. You need the addition rule of the sine function.

When I eliminate t, I seem to not get the right answer...In your above example, is "a" the phase shift?

 

[DeldotB]

How's this:

[tex]X(t)=V_{1}sin(\omega t)\: and\: Y(t)=V_{2}sin(\omega t+\delta )[/tex]

so Isolating t from X(t) equation:

[tex]t=\frac{1}{\omega }\left [ sin^{-1}\frac{X(t)}{V_{1}} \right ][/tex]

Putting that into the Y(t) equation I get:

[tex]Y(t)=V_{2}sin\left [ sin^{-1}(\frac{X(t)}{V_{1}} )+\delta \right ][/tex]

Now If I set Y to zero ( x intercepts) I get:[tex]0=V_{2}sin\left [ sin^{-1}(\frac{X(t)}{V_{1}} )+\delta \right ][/tex]

So,
[tex]\delta =-sin^{-1}\frac{X(t)}{V_{1}}[/tex] or just
[tex]\delta =sin^{-1}\frac{X(t)}{V_{1}}[/tex].

In the beginning, V1 was equal to the max amplitude Y max. But what about X(t)?

So I have:
[tex]\delta =sin^{-1}\frac{X(t)}{Y_{max}}[/tex]

Any help? Why would Y min = X(t)?

 

[SammyS]

How's this:

[tex]X(t)=V_{1}sin(\omega t)\: and\: Y(t)=V_{2}sin(\omega t+\delta )[/tex]
...

In the beginning, V1 was equal to the max amplitude Y max. But what about X(t)?
...

By the way: Isn't Y_max = V₂ ?

At what time(s) do you have x(t) = 0 ?

Can't you get the y-intercept directly from that ?

 

[DeldotB]

By the way: Isn't Y_max = V₂ ?

At what time(s) do you have x(t) = 0 ?

Can't you get the y-intercept directly from that ?

Yes...but how would I get a function for the phase in terms of Y max/min from this?

 

[DeldotB]

and Ok, say I do look at where X(t)= 0 (at t=0, pi/omega, 2pi/omega,...,(n)pi/omega) but how does this help?
I just don't understand how to get rid of the X(t) in my final answer for the phase angle. Thanks for any help

 

[SammyS]

and Ok, say I do look at where X(t)= 0 (at t=0, pi/omega, 2pi/omega,...,(n)pi/omega) but how does this help?
I just don't understand how to get rid of the X(t) in my final answer for the phase angle. Thanks for any help

Isn't the phase angle δ ?

Why bother with x(t) except to evaluate the y-intercept ?

 

[DeldotB]

Isn't the phase angle δ ?

Why bother with x(t) except to evaluate the y-intercept ?

Dont I have to "bother" with it? Its part of a pair of parametric equations. In order to eliminate the parameter "t" I need to solve one of the functions (I chose X(t)) for t so I could plug it into the Y(t) function...Do you see another way of obtaining the equation at the top of the page?

 

[SammyS]

...Do you see another way of obtaining the equation at the top of the page?

Yes.

Try what I have suggested above.

 

[ehild]

When I eliminate t, I seem to not get the right answer...In your above example, is "a" the phase shift?

Yes, "a" was the phase shift in my formula.
As the angular frequency is the same for both the X(t) and Y(t) signal, you can consider it unity. And you can assume that the amplitudes of the signals are equal. At least, you can set them equal on the screen of the oscilloscope.
You have the signals X(t)=Vsint and Y(t)=Vsin(t+δ). That is the parametric equation of a curve on the XY plane.
Applying the addition rule for the sine function, Y=Vsin(t+δ)= Vsin(t)cosδ+Vcos(t)sinδ, and you know that X=Vsin(t). Eliminating t, you get a relation between X and Y.
The maximum of that Y(X) curve is Ymax, and its y intercept is Yint. You should find the phase angle in terms of Ymax and Yint. You do not even need the y(x) curve. Follow Sammy's advice. You figured out that X=0 at t=0. What is Y (the y-intercept) then?

 

[DeldotB]

oh wow. That was too easy. Perfect example of over complicating a problem. Thanks for the help!

 

[Ondro]

Yes, "a" was the phase shift in my formula.
As the angular frequency is the same for both the X(t) and Y(t) signal, you can consider it unity. And you can assume that the amplitudes of the signals are equal. At least, you can set them equal on the screen of the oscilloscope.
You have the signals X(t)=Vsint and Y(t)=Vsin(t+δ). That is the parametric equation of a curve on the XY plane.
Applying the addition rule for the sine function, Y=Vsin(t+δ)= Vsin(t)cosδ+Vcos(t)sinδ, and you know that X=Vsin(t). Eliminating t, you get a relation between X and Y.
The maximum of that Y(X) curve is Ymax, and its y intercept is Yint. You should find the phase angle in terms of Ymax and Yint. You do not even need the y(x) curve. Follow Sammy's advice. You figured out that X=0 at t=0. What is Y (the y-intercept) then?

Eliminating t, you get a relation between X and Y. How did you do that? Thanks

 

[kuruman]

Eliminating t, you get a relation between X and Y. How did you do that? Thanks

You may have to wait a while for an answer. This thread is almost 6 years old.

 

[Ondro]

I've figured it out already, thanks