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Derivation of "arcsin" phase shift formula

  1. Aug 27, 2015 #1
    1. The problem statement, all variables and given/known data
    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 Im not sure how to start.

    Any help would be appreciated!

    2. Relevant equations

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

    3. The attempt at a solution

    I do not have an attempt- I dont know where to start!
     
  2. jcsd
  3. Aug 27, 2015 #2

    berkeman

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    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? :smile:
     
  4. Aug 27, 2015 #3
    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?
     
  5. Aug 27, 2015 #4

    berkeman

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    And what are Ymax and Yint for each of those cases?
     
  6. Aug 27, 2015 #5
    Well yint for 0 degrees and 180 degrees is zero (with no y max), Im 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 dont really have any "insight" about oscilloscopes.
     
  7. Aug 27, 2015 #6

    berkeman

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    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
    xyplot_animated.gif

    BTW, in that animation above, the center is (2.5,2.5), not (0,0).
     
  8. Aug 27, 2015 #7
    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]?
     
  9. Aug 27, 2015 #8
    Last edited: Aug 27, 2015
  10. Aug 27, 2015 #9

    ehild

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    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.
     
  11. Aug 28, 2015 #10
    When I eliminate t, I seem to not get the right answer....In your above example, is "a" the phase shift?
     
  12. Aug 28, 2015 #11
    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)?
     
  13. Aug 28, 2015 #12

    SammyS

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    By the way: Isn't Ymax = V2 ?

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

    Can't you get the y-intercept directly from that ?
     
  14. Aug 28, 2015 #13
    Yes....but how would I get a function for the phase in terms of Y max/min from this?
     
  15. Aug 28, 2015 #14
    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 dont understand how to get rid of the X(t) in my final answer for the phase angle. Thanks for any help
     
  16. Aug 28, 2015 #15

    SammyS

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    Isn't the phase angle δ ?

    Why bother with x(t) except to evaluate the y-intercept ?
     
  17. Aug 28, 2015 #16
    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?
     
  18. Aug 28, 2015 #17

    SammyS

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    Yes.

    Try what I have suggested above.
     
  19. Aug 28, 2015 #18

    ehild

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    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?
     
    Last edited: Aug 28, 2015
  20. Aug 28, 2015 #19
    oh wow. That was too easy. Perfect example of over complicating a problem. Thanks for the help!
     
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