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Argand Diagram help

  1. May 16, 2009 #1
    9abk87.jpg

    For this I started with the 1st interval [0,pi/2] and said

    arctan [4/ ... ] = [0 , pi/2]

    but studying the tan graph this is all positive values of the y-axis

    so surely the range of values for this is (0 , inf)?

    If I've done that right, how do I do the other interval [-pi/2 , pi]?

    Because using that method gives the result (-inf , 0), but then that would also be true for the quadrant [0 , -pi/2] surely, which brings me back to my first answer also being true for [pi/2 , pi]..

    How do I attempt this because this is obviously wrong..

    This was under the chapter about Laplace / Fourier Transforms

    And if you could shed any light on how to maximise too I'd be greatful, because my notes havn't touched on this..

    The y(t) = ... was formed from solving a differential equation using laplace transforms, I didn't include that because I didn't think it was relevant. =)
     
  2. jcsd
  3. May 17, 2009 #2

    tiny-tim

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    Hi Firepanda! :smile:

    (have an omega: ω and a phi: φ and a pi: π :wink:)
    I don't really follow what you're doing :confused:

    to get arg, you separate 4/… into the form x + iy (not 1/(x + iy) !), and then use tan-1 :wink:
     
  4. May 17, 2009 #3
    Oh shoot I read the question wrong =)

    I mean for [-π/2 , 0]

    So I guess that's the values for (-inf , 0)..

    AH yeah I see now I was doing something wrong in my method too

    So how I'm interpreting the question is this:

    Where does the angle given by arg[4/-ω2+5ωi+6] lie in the bottom right quadrant [-π/2 , 0] of an argand diagram?

    So do I need to change [4/-ω2+5ωi+6] into the form x + iy?
     
  5. May 17, 2009 #4

    tiny-tim

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    No, it's asking what are the values of ω for which 4/3 lies in the bottom right (4th) quadrant? (and then same question for the 3rd quadrant).
    Yes!! :smile: (or you'll end up in the wrong quadrant :rolleyes:)
     
  6. May 17, 2009 #5
    Great ok,

    I canceled it all down and got my

    arctan 5ω/(ω2-6) (lets assume this is right, forget my working too much to type out :P)

    then 5ω/(ω2-6) has to lie in (-inf,0) for my arctan 5ω/(ω2-6) to equal the interval (-pi/2 , 0)

    so how do I find ω from this?
     
  7. May 17, 2009 #6

    tiny-tim

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    This is getting unnecessarily complicated …

    you don't actually need to consider tan-1 since all you are interested in is the quadrant, which means all you need to consider is the signs of x and y (in x + iy) …

    just draw the four quadrants, and mark them ++ +- -+ and -- :wink:
     
  8. May 17, 2009 #7
    now I got it! ty. Much simpler, and I have myself an answer

    Any idea of how to maximise? because that part has me stumped
     
  9. May 18, 2009 #8

    tiny-tim

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    just maximise x2 + y2 :smile:
     
  10. May 18, 2009 #9
    So I use my x and y, and add up their squares..

    Could you give me the basic method on how to do this, just so it clicks with me and I'm able to =)
     
  11. May 18, 2009 #10

    tiny-tim

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    erm … I don't see what you think the difficulty is :confused:

    add up the squares … what do you get?
     
  12. May 18, 2009 #11

    tiny-tim

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    yes :smile:

    (except I think the question is asking for any ω)
     
  13. May 18, 2009 #12
    wouldnt that just be infinity then?

    or do you mean I have to find another range of values?
     
  14. May 18, 2009 #13

    tiny-tim

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    Nooo … what formula are you using for Aω? :confused:
     
  15. May 18, 2009 #14
    Well I was using my x+iy values

    I then found x^2 + y^2

    and I got ω^4 -12ω^2 +61

    So I'm trying to find the value of ω where this is largest, and by the graph it zooms off the screen so I was assuming inf and -inf for my ω
     
  16. May 18, 2009 #15

    tiny-tim

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    No, it's 1/(that) …

    (except you've used 25 instead of 25ω2)
     
  17. May 19, 2009 #16

    tiny-tim

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    Hi Firepanda! :smile:

    I'm getting confused :confused:

    |1/A| = 1/|A|, so isn't it just 1/(ω4+13ω2+36)?

    (and you can factor that :wink:)
     
  18. May 19, 2009 #17
    Well you said this earlier

    So that's why I was doing that

    So I thought I was to compute x2 + y2, then find the maximum on the graph
     
  19. May 19, 2009 #18

    tiny-tim

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    I meant I don't see where your fraction (ω4-12ω2+61)/(ω4+13ω2+36) comes from :confused:
     
  20. May 19, 2009 #19
    EDIT: my working out was riddled with mistakes so ill jsut put the answer!

    I seeeeee!!!!!

    Woohoo

    16/(ω4+36+13ω2)

    I have the roots of the denominator, what do i do with those?
     
    Last edited: May 19, 2009
  21. May 19, 2009 #20

    tiny-tim

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    he he :biggrin:

    now you see you were going round in circles … |1/A| = 1/|A| ? :smile:
    hmm :rolleyes: … on second thoughts, factoring doesn't help … the roots are all imaginary, which is a teeny bit irrelevant since ω is real :redface:

     
    Last edited by a moderator: Apr 24, 2017
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