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EugP
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Magnitude and Phase Angle for Nyquist Plots
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]
[itex]
|G(j\omega)| \angle G(j\omega)
[/itex]
[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.
Homework Statement
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]
Homework Equations
[itex]
|G(j\omega)| \angle G(j\omega)
[/itex]
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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