# Maginutude and Phase Angle for Nyquist Plots

1. Dec 12, 2009

### EugP

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.

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

2. Relevant equations

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

3. The attempt at a solution

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

From that I know that the magnitude is found like this:
$|G(j\omega)| = \frac{k\sqrt{(\omega^2+4)}}{\sqrt{(\omega^2+1)}\sqrt{(\omega^2+9)}}$

Now, the solution says that the phase angle is:

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

What I don't understand is why there is a $180^o$ shift in the last term.
Any help would be greatly appreciated.

Last edited: Dec 12, 2009
2. Dec 16, 2009

### AqA

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