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Tan and ArcTan calculation

  1. Jun 24, 2012 #1
    Hello
    Could you help me to solve this problem? i want to find w
    71_1340528592.gif
    asnwer is 151660
    but i want to know the approach to solve it.
     
  2. jcsd
  3. Jun 24, 2012 #2

    HallsofIvy

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    Well, what have you done on this? I can see a few obvious things to do to start- like multiplying through by -1! Have you done that? And taking the tangent of both sides seems obvious. What is tan(180)? Do you know the formula for tan(a+b)? Can you extend that to tan(a+ b+ c)?
     
  4. Jun 24, 2012 #3
    Yes, i do 2 ways but none of them work fine...could you write the equation because its a start point of G.M(Gain margin) and Phase margin ampilfier.
     
  5. Jun 24, 2012 #4

    Curious3141

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    Multiply throughout by -1.

    Bring one of the arctan terms over to the RHS (to the same side as the 180 degree term). Suggest the one with 104 in the denominator.

    Take tangent of both sides. LHS can be resolved with angle sum formula. RHS is easy.

    That should be enough to start you out. Show your work (in detail), then you'll get more help.
     
  6. Jun 24, 2012 #5
    Is 180 in degrees?
    It cannot be in degrees because the range of arctan lies between -90 and +90 (degrees) and even if you simplify the LHS, you get tan-1x=180, (x is anything which remains after applying identities) and 180 degrees is out of range. Is my reasoning correct?
     
    Last edited: Jun 24, 2012
  7. Jun 24, 2012 #6

    pcm

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    range of phi is -1.5pi to 1.5 pi.so obviously 180 is degree
     
  8. Jun 24, 2012 #7
  9. Jun 24, 2012 #8

    pcm

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    what i meant was phi cant be greater than 1.5pi and cant be less than -1.5pi.see there are 3 arctan terms.
     
    Last edited: Jun 24, 2012
  10. Jun 24, 2012 #9
    Ah, thanks for rectifying. I was just confused by the solution i was doing for this problem.
     
  11. Jun 24, 2012 #10

    SammyS

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    Suppose you have three arctangents, all with equal argument, A. Then you have:

    [itex]-tan^{-1}(A)-tan^{-1}(A)-tan^{-1}(A)=-180^\circ\ .[/itex]

    Then you have tan-1(A) = 60° .

    No problem with that!
     
  12. Jun 24, 2012 #11

    SammyS

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    If you let u = ω/105, then ω/104 = 10u and ω/(2∙105) = u/2 .

    That leaves you with

    [itex]\displaystyle -\tan^{-1}(10u)-\tan^{-1}(u)-\tan^{-1}(u/2)=-180^\circ\ .[/itex]

    That looks a bit less intimidating.

    WolframAlpha gives an exact solution for u, but doesn't show a method.
     
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