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Help simplify the equation for ellipse

  1. Feb 11, 2013 #1
    This is really a simplification of an equation. It is just very long and complex and I need advice.

    For ellipse center at the origin with tilt angle [itex]\tau[/itex], the distance from origin to the ellipse is [itex]\rho(\varsigma)[/itex].

    It is given:
    [tex]\varsigma\;=\;\frac {y}{x}\;=\;\frac{2E_xE_y\cos\delta}{E_x^2-E_y^2}\;,\; a=\frac{1}{E_x^2\sin^2\delta}\;,\;b=\frac{2\cos \delta}{E_x E_y \sin^2\delta}\;,\;c=\frac {1}{E^2_y\sin^2\delta}[/tex]
    [tex]x^2=\frac{1}{c\varsigma^2-b\varsigma+a}[/tex]



    The equation of ellipse is:

    [tex] \vec E (0,t)\;=\;\hat x E_x\cos(\omega t)\;+\;\hat y E_y\cos(\omega t +\delta)[/tex]
    Where [itex] E_x,\;E_y, \; \delta[/itex] are all constant.
    [tex]x=E_x\cos(\omega t)\;\hbox{ and }\; y=E_y\cos(\omega t +\delta)[/tex]
    [tex]\rho^2(\varsigma)\;=\; x^2+y^2\;=\;x^2(1+\varsigma^2)\;=\;\frac {(1+\varsigma^2)}{c\varsigma^2-b\varsigma+a}[/tex]


    The major axis [itex]\rho_{max}[/itex]=OA is given by the book where:
    [tex]OA=\sqrt{\frac{1}{2}[E^2_x+E^2_y+\sqrt{E_x^4+E_y^4+2E_x^2E_y^2\cos(2 \delta)}}[/tex]

    I tried to substitute everything in, it get way complicated and no where close. It is just too long to type in my attempt. Can anyone suggest a way to simplify this?

    Thanks
     
    Last edited: Feb 11, 2013
  2. jcsd
  3. Feb 11, 2013 #2

    mfb

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    Neglecing redundant equations, I think you have more unknown parameters than equations.
    Which equations gives the ellipse? It should look like ##ax^2 + by^2 + cxy = d## with some parameters a,b,c,d. Based on them, you can calculate the major and minor axis.
    The distance of the ellipse from the origin is just the semi-minor axis.
     
  4. Feb 11, 2013 #3
    Thanks for the reply. The equation of the ellipse is:
    [tex]\rho^2(\varsigma)\;=\; x^2+y^2\;=\;x^2(1+\varsigma^2)\;=\;\frac {(1+\varsigma^2)}{c\varsigma^2-b\varsigma+a}[/tex]

    I should have said [itex] E_x,\;E_y, \; \delta[/itex] are all constant. So their is no problem of unknown. Also x and y are variable of t. It is the simplification that is my problem.

    As you can see, the equation is very long, I really worked on trying to simplify it, but I have no luck. I am just hoping there is a way to simplify it. It is really not a trick question, the book gave this equation as the major axis, I want to verify it as books make more mistake as people realize and I don't by default taking them on face value. In fact I have already found a mistake in the book already. Problem this is on polarization of electromagnetic wave and finding a book is not exactly easy, it is not common like the calculus books use in colleges where they are very well proof read, checked and corrected, that you don't find mistake or typos.
     
    Last edited: Feb 11, 2013
  5. Feb 11, 2013 #4

    mfb

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    If those are constant, y/x is constant. This is not an ellipse, it is a straight line.
     
  6. Feb 11, 2013 #5
    I modify the original post, I realize I left out a lot of stuff.

    [tex]x=E_x\cos(\omega t)\;\hbox{ and }\; y=E_y\cos(\omega t +\delta)[/tex]
     
  7. Feb 11, 2013 #6

    mfb

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    That looks more like an electromagnetic wave now.

    The time-dependent amplitude is ##x^2+y^2 = E_x^2 cos^2(\omega t) + E_y^2 cos^2(\omega t + \delta)##. This has a maximum for some t, which can be evaluated in the usual way (well, I would expect that the equations can be solved).
     
  8. Feb 11, 2013 #7
    Actually this is where I got this:

    http://www.ece.mcmaster.ca/faculty/nikolova/antenna_dload/current_lectures/L05_Polar.pdf

    The formula I want to verify is (5.8) in page 8. I worked through and verified the correctness in Appendix 1 on page 15 to page 17. You can see the last line in the last page that you can derive (5.8) by using A-4. That's where I got lost!!! even though this is physics, but my problem is a simple math problem I cannot solve!!!

    (5.8) is used in EM book by Balanis. I have a few issue with this book, but sadly Balanis and Kraus are the two books that even get into this polarization in detail. Kraus uses Poincare sphere representation which is different. It uses spherical trig. which is another can of worm!!! Balanis is even wrong on the equation of the tilt angle [itex]\tau[/itex], that's how bad this kind of books are. I just cannot trust what is given in the book unless I can verify them. Painful is an under statement. Please help.

    I looked into partial fraction and completing the square. It just didn't look like they're going to get me there. Any suggestion.

    Thanks

    Alan
     
    Last edited: Feb 11, 2013
  9. Feb 11, 2013 #8

    mfb

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    A-5 to A-8 are needed, too. This is just the method I described in post 6.
     
  10. Feb 11, 2013 #9
    I tried evaluate A-5, I am going nowhere either. I don't see how A-5 help as it gives the [itex]\varsigma[/itex] and it's not simpler than from A-8. Then still have to substitute back into A-4.
     
  11. Feb 11, 2013 #10
    I am going to try from this angle as I am not going anywhere with the original way:

    ##\rho^2= x^2+y^2 = E_x^2 cos^2(\omega t) + E_y^2 cos^2(\omega t + \delta)##

    Let ##\omega t=A\;,\;\delta=B##

    To get max and min, ##\frac {d\rho^2(A)}{dA}= -2E_x^2 \cos A \sin A\;-\;2E^2_y \cos(A+B)\sin(A+B)\;=\;0##

    ##E_x^2 \cos A \sin A =-E_y^2 \cos (A+B)\sin (A+B)\;=\; -E_y^2(\cos A\cos B-\sin A\sin B)(\sin A \cos B+\cos A \sin B)\;=\;-E_y^2[\cos 2B(\sin A \cos A)+\cos 2A( \sin B \cos B)]##

    ##\Rightarrow\;\frac {\cos 2A}{\sin A \cos A}\;=\;2\cot 2A\;=\;-\frac {E_x^2+E_y^2 \cos 2B}{E_y^2 \sin B \cos B}##

    ##A\;=\;\omega t\;=\; \frac {1}{2} \cot^{-1}\left[-\frac {E_x^2+E_y^2 \cos 2B}{2E_y^2 \sin B \cos B}\right]##

    Is this the right way? But still don't seem like I am going to get (5.8).
     
    Last edited: Feb 12, 2013
  12. Feb 12, 2013 #11

    mfb

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    The idea looks good. If you plug this into the original equation for x^2+y^2, you might get the correct result.
     
  13. Feb 12, 2013 #12
    How do you solve ##\cos(\cot^{-1}A) \;,\; \cos ( \cot{-1}A + \delta)##

    Sorry to keep asking, it's been a while I study calculus, math. I am very rusty in these.

    Thanks
     
  14. Feb 12, 2013 #13

    Dick

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    Draw a right triangle with an angle cot^(-1)(A). The cot of that angle would be A. So you put the opposite side to be 1 and the adjacent side to be A. The hypotenuse is sqrt(1+A^2). What's the cos of the angle?
     
  15. Feb 12, 2013 #14
    Thanks for the reply. I know it is easy to draw and easy if there are concrete numbers to plug into the equation. But I am trying to derive and verify the equation for OA, so I am stuck with constants like ##E_x\;,\;E_y\;,\;\delta## and only A to show:

    ##A\;=\;\omega t\;=\; \frac {1}{2} \cot^{-1}\left[-\frac {E_x^2+E_y^2 \cos 2B}{2E_y^2 \sin B \cos B}\right]##

    To plug into the this equation:

    ##x^2+y^2 = E_x^2 cos^2A + E_y^2 cos^2(A+ \delta)##

    To get

    [tex]OA=\sqrt{\frac{1}{2}[E^2_x+E^2_y+\sqrt{E_x^4+E_y^4+2E_x^2E_y^2\cos(2 \delta)}}[/tex]
     
    Last edited: Feb 12, 2013
  16. Feb 12, 2013 #15
    Anyone?
     
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