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Question on inverse tangent.

  1. Feb 17, 2013 #1
    [tex]\Psi=\tan^{-1}\left(\frac{\cos\omega t}{\cos(\omega t+\delta)}\right)[/tex]

    I want to find out whether ##\Psi## increase or decrease with time t, if ##\delta## is positive and if ##\delta## is negative.

    [tex]\Psi=\tan^{-1}\left(\frac{\cos\omega t}{\cos(\cos\omega t \cos \delta+\sin\omega t \sin\delta)}\right)\;=\;\tan^{-1}\left(\frac {1}{(\cos\delta + \tan(\omega t)\sin \delta)}\right)[/tex]

    [tex]\Rightarrow\;\Psi=\tan^{-1}\left(\frac{1}{K_1+K_2\tan \omega t}\right)[/tex]
    where ##K_1## is a constant and ##K_2## is a constant following the sign of ##\delta##.

    This is as far as I know how to go, what can I do to simplify it. I want to find out whether ##\Psi## increase or decrease with time if ##\delta## is positive and if ##\delta## is negative.
     
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  3. Feb 17, 2013 #2

    Simon Bridge

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    You'll have to use the known limits to the problem and the properties of the tangent function.

    I would be thinking: if ##\Psi## is increasing, what happens to ##\tan\Psi## ?
    What are the allowed values for ##\omega t## ?
     
  4. Feb 17, 2013 #3
    Thanks for the reply. I think I come up with a way and see whether you agree or not:

    Let ##\theta=\frac {\pi}{2}-\Psi##. So if ##\tan \Psi=\frac y x\;\Rightarrow \;\tan \theta = \frac x y##

    So ##\theta \;=\;\tan ^{-1} \left(\frac{\cos(\omega t +\delta}{\cos \omega t}\right)##

    This will be easy to solve and find out the direction. Then we know if θ increase with t, then ##\Psi## decrease with t as ##\theta=\frac {\pi}{2}-\Psi##.

    You think this make sense?
     
  5. Feb 17, 2013 #4

    epenguin

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    Sounds too complicated. This is your first step.


    Or maybe easier for you if x is increasing what is tan-1x doing?
     
  6. Feb 17, 2013 #5
    Tangent function of ##\Psi## is discontinue ( to +ve ∞) at 90 degree, then it start at -∞ from 90+##\delta## to 0 at 180. And so on. That's the reason I am trying to get into

    ## B=\tan^{-1} ( \tan A)## type of format so B=A and B increase and decrease with A. This is a continuous function.

    Particular in my very problem I am facing is to find the direction of rotation of the vector where ## \vec E (0,t)=\hat x \cos \omega t +\hat y \sin \omega t## where ##\Psi## is the angle made by A respect to x-axis. Thereby

    ##\Psi=\tan^{-1}(\tan(\frac{\cos \omega t}{\sin \omega t})=\omega t##

    In this case ##\Psi## increase with time t.
     
    Last edited: Feb 17, 2013
  7. Feb 17, 2013 #6
    Thanks for the reply, I don't see your first step.

    But my concern is described in post #5.
     
  8. Feb 17, 2013 #7

    vela

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    I'd start by plotting the argument of arctan for various values of ##\delta## to get a feel for what the answer should be.
     
  9. Feb 17, 2013 #8

    Simon Bridge

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    I can see why you want to try the approach you want - however, I am suggesting you look instead at:$$B=\tan \Psi=\frac{1}{K_1+K_2 \tan \omega t }$$From the behavior of ##B## you can perhaps deduce the behavior of ##\Psi##

    As vela says, though, it will help you to plot the functions for some sample values to see how they work and use that to guide you in your approach.

    However - since this was a phasor, then what was the original phasor representation?

    It looks like you have a phasor of form ##z=\cos(\omega t + \delta) + i\cos\omega t##
    You are trying to investigate the phase angle dependence on time to see which way it rotates ... this would be given by the $$\theta(t)=\tan^{-1}\left [ \frac{\cos\omega t}{\cos(\omega t + \delta)} \right ]$$If I have this right, then wouldn't you want it in form ##z(t)=f(t)\big [\cos(At)+i\sin(At)\big ]=f(t)e^{iAt}## so that the phase angle is given by At and you will be able to tell, by inspection, which way the phasor is rotating.
     
  10. Feb 18, 2013 #9
    Thanks for your time Simon.

    It is an electromagnetic plane wave propagating in either +ve or -ve z direction. The equation at z=0 is

    [tex]\vec E(0,t)\;=\; Re[\hat x E_{x0}e^{j\omega t-\delta}+\hat y E_{y0}e^{j\omega t}]\;=\;\hat x E_{x0}\cos( \omega t-\delta)+\hat y E_{y0}\cos \omega t[/tex]
    Whereby:
    [tex]\Psi\;=\;\tan^{-1}\left(\frac{y(t)}{x(t)}\right)\;=\;\tan^{-1}\left(\frac{\cos \omega t}{\cos( \omega t-\delta)}\right)[/tex]
    Assuming ##E_{x0}=E_{y0}## to simplify the equation.

    I know drawing it out or put in number will work, but as I said, this is part of the polarization of EM wave and I want to get a general equation of determining the direction of rotation. This is only one part as I specifically left of the +ve or -ve z direction of propagation. All the books only give in specific direction. That's the reason I want to derive an equation of the polarity of ##\frac {\partial \Psi}{\partial t}## and then determine the direction of rotation when the direction of propagation is known AND the polarity of ##\delta##. The material about this topic is not cover very well ( to put it politely) in All the books that I check ( over 8 or 9 already). They are full of contradictions, that's the reason I want to derive my own equation.
     
    Last edited: Feb 18, 2013
  11. Feb 18, 2013 #10

    DrClaude

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    [tex]
    \frac{d \Psi}{dt} = \frac{2 \omega \sin \delta}{2 + \cos (2 \omega t) +
    \cos(2 (\omega t + \delta))}
    [/tex]
    With the denominator clearly always positive, I think you have your answer.
     
  12. Feb 19, 2013 #11
    I don't know how you arrive with the equation. Are you using:
    [tex]y=tan^{-1}x\;\Rightarrow \tan y\;=\;x\;\Rightarrow \;d\tan y=\sec^2y\ dy=dx\;\Rightarrow\;\frac {dy}{dx}= \frac {1}{1+\tan^2y}[/tex]

    In my situation, x is a long equation, so ##x'=dx(t) dt##.

    Let me follow this way:

    [tex]\Psi\;=\; \tan^{-1}\left( \frac{\\cos \omega t}{\cos(\omega t+\delta)}\right)\;\Rightarrow\; \tan \Psi\;=\; \frac{\\cos \omega t}{\cos(\omega t+\delta)}[/tex]
    [tex]\Rightarrow \; d(\tan \Psi)\;=\left[\frac{-\omega \sin \omega t \cos(\omega t+\delta)\;+\;\omega \sin\omega t \cos(\omega t +\delta)}{\cos^2(\omega t +\delta)}\right]dt\;\Rightarrow\; \frac {d\Psi}{dt}\;=\; \left[\frac{-\omega \sin \omega t \cos(\omega t+\delta)\;+\;\omega \sin\omega t \cos(\omega t +\delta)}{\cos^2(\omega t +\delta)}\right]\frac {1}{1+\tan^2\Psi}[/tex]

    I don't see this is an easier way compare to what I used in post #5. What I did is just to use either ##\Psi## or ##\frac {\pi}{2}-\Psi## in order to keep the denominator as ##\cos \omega t## and put the complicate stuff in the numerator. I think for my purpose of only need the increase or decrease of ##\Psi## with t, the way I did it in post 5 is as simple as it gets..... by making it arctangent of a tangent.
     
    Last edited: Feb 19, 2013
  13. Feb 19, 2013 #12

    DrClaude

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    Mathematica
     
  14. Feb 19, 2013 #13

    epenguin

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    All I am saying is, as you did note somewhere, tan-1(x) simply increases with x everywhere from - ∞ to + ∞. So if x is the function inside the brackets in the first equation of your first post, the overall function will be increasing as long as the expression in brackets is, and decreasing wherever that does, you just don't have to worry about the tan-1.

    At least that was how I understood your original problem as you stated it.

    Also I saw tan-1(tan(something)) in one of your posts (brackets incompleted) .
    But tan-1(tan(something)) is just that original something.
    f-1(f(something)) is that original for any function f (at least if it is single-valued both ways).
     
  15. Feb 19, 2013 #14

    Simon Bridge

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    I think I got sidetracked away from this post for some reason:
    I can see why that's a problem for what you stated - but your function is not as simple as the one below:
    ... it looks to me that your vector is ##\vec{E}=\hat{x}\cos(\omega t + \delta) +\hat{y}\cos\omega t## ... is that correct?

    Why not express this as a complex phasor?

    Anyway - the way I'm reading the problem, what you care about is which way the vector is rotating. Is this correct?

    In which case, you don't need to know the details of the relationship - only the signs.
    This is where you use the properties of the functions to help you.

    i.e. for ##u>0##, then ##\tan^{-1}u > 0## and the problem becomes "what is the sign of ##u##? ... and #u## is everything inside the inverse tan.
     
  16. Feb 20, 2013 #15
    Thanks for both of your detail reply, let me digest these a little bit and comeback to this hopefully by tonight.

    Many many thanks.

    Alan
     
    Last edited: Feb 20, 2013
  17. Feb 20, 2013 #16
    I notice I make a big typo in post #5, it should be

    ##\Psi=\tan^{-1}(\frac{\sin \omega t}{\cos \omega t})=\omega t##

    I am looking for the direction of the change of ##u## with time t, not just the sign of ##u##. That is even if ##u## is negative, as long as it goes less negative with increase of time t, then the change is in positive direction. more like looking at the sign of ##\frac{du}{dt}##.

    My whole approach is to make it as ##\Psi=\tan^{-1}\left(K_2\tan\omega t\right)## or ##\Psi=\tan^{-1}\left(K_1+K_2\tan \omega t\right)##, which means
    [tex] \tan\Psi=\left(K_1+K_2\tan \omega t\right)\;\Rightarrow\; \sec^2\Psi d\Psi=\omega K_2 \sec^2\omega t dt\;\Rightarrow\; \frac{d\Psi}{dt}=\omega K_2\frac{\sec^2\omega t}{\sec^2\Psi}[/tex]
    Since both ## sec^2\Psi## and ##\sec^2\omega t## are positive, whereby, ##\frac{d\Psi}{dt} = +ve (##\Psi## increase with t if ##K_2## is positive), and ##\frac{d\Psi}{dt}## = -ve (##\Psi## decrease with t if ##K_2## is negative).

    So if ##\Psi## increase with t, the rotation is CCW on the xy plane. If ##\Psi## decrease with t, the rotation is CW.

    Does this make sense?

    Thanks
     
    Last edited: Feb 20, 2013
  18. Feb 20, 2013 #17
    That's a good question, let's look at this:
    [tex]\vec E(0,t)\;=\; Re[\hat x E_{x0}e^{j\omega t}+\hat y E_{y0}e^{j\omega t+\delta}][/tex]
    The first thing I see is that E(0,t) is real, the complex vector inside the Re[] is complex. Can you even do this?

    Second, let's suppose you can, so

    [tex]\Psi=Re[\tan^{-1}\left(\frac{E_{y0}e^{j\omega t +\delta}}{E_{x0} e^{j\omega t}}\right)]=Re[\tan^{-1}\left(\frac{E_{y0}e^{j\delta}}{E_{x0}}\right)][/tex]

    The t is gone!!!
     
  19. Feb 21, 2013 #18

    Simon Bridge

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    Careful how you apply Euler's formula...
    ##\Re[e^{j\theta}] = \cos\theta##, and ##\Im [e^{j\theta}]=\sin\theta##.
    ##2\cos(\omega t+\delta) = e^{j(\omega t + \delta)} + e^{-j(\omega t +\delta)}##

    But - I meant,
    you have a phasor which is basically an arrow rotating.
    You chose to represent it in R^2 ...
    why not represent it on the complex plane instead?

    Thus: ##E = \cos\omega t + j\cos(\omega t + \delta)## and try to get it in form ##E = f(t)e^{j\Omega t}## ... then you don't need to find the argument, just read-off the angular frequency directly.
     
  20. Feb 21, 2013 #19
    I don't get how to use complex plane for a vector in this situation. I don't get what you are trying to do with ##E = \cos\omega t + j\cos(\omega t + \delta)##. How does that relate to ##\vec E(0,t)=Re[\hat x E_{x0}e^{j\omega t}+\hat y E_{y0}e^{j(\omega t +\delta)}]##?

    Can you take a look at post #16, I edited it and it is very simple to me already, I just need to confirm that it is correct and I think I am done.

    Thanks

    Alan
     
    Last edited: Feb 21, 2013
  21. Feb 21, 2013 #20

    Simon Bridge

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    Well that would be a good reason not to use it then :)
    http://en.wikipedia.org/wiki/Phasor
    ... in effect it replaces your x-y axis with the complex plane and exploits the properties of complex numbers to represent the rotation.

    But back to post #16 ...
    I though you wanted to find the way the phase angle changes with time. The angle will go clockwise or anticlockwise right? That would be a positive or negative angular velocity right?

    What I was hoping to do was to get you to focus on the properties of the functions, rather than on finding the exact relationship, as the way forward - which is what you seem to be doing. Well done.

    I cannot judge if you get the correct answer or not because that would involve actually solving the problem myself - and nobody is paying me ;) Anyway - it is good practice for you to get used to judging your own work.

    Have fun.
     
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