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Maginutude and Phase Angle for Nyquist Plots

  1. Dec 12, 2009 #1
    Magnitude and Phase Angle for Nyquist Plots

    1. The problem statement, all variables and given/known data
    The assignment is to draw a Nyquist Plot of a certain transfer function. The problem is that I can't figure out how they got the angle for the plot.

    [itex]

    G(s) = \frac{k(s+2)}{(s+1)(s-3)}

    [/itex]


    2. Relevant equations

    [itex]

    |G(j\omega)| \angle G(j\omega)

    [/itex]


    3. The attempt at a solution

    [itex]

    G(j\omega) = \frac{k(j\omega+2)}{(j\omega+1)(j\omega-3)}


    [/itex]

    From that I know that the magnitude is found like this:
    [itex]

    |G(j\omega)| = \frac{k\sqrt{(\omega^2+4)}}{\sqrt{(\omega^2+1)}\sqrt{(\omega^2+9)}}

    [/itex]

    Now, the solution says that the phase angle is:

    [itex]

    \angle G(j\omega) = \angle \tan ^{-1} (\frac{\omega}{2})-\tan ^{-1} (\frac{\omega}{1})-(180 - \tan ^{-1} (\frac{\omega}{3}))

    [/itex]

    What I don't understand is why there is a [itex]180^o[/itex] shift in the last term.
    Any help would be greatly appreciated.
     
    Last edited: Dec 12, 2009
  2. jcsd
  3. Dec 16, 2009 #2

    AqA

    User Avatar

    If you consider (jw-3) alone, its y coordinate is 'w' and its x coordinate is '-3'.
    Thus (jw-3) as a directed vector from the origin makes an angle of 180-tan-1(w/3) with the positive x axis.
    The angle with the positive x axis is to be considered, not just tan-1(y/x).
     
    Last edited: Dec 16, 2009
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