(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Maginutude and Phase Angle for Nyquist Plots

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