Polar plot in control system
Table of Contents
Polar plot
A polar plot is a graph that connects magnitude and phase. Only normal values are used to represent the magnitudes in this case.
The polar form of G(jω)H(jω) is
G(jω)H(jω)=|G(jω)H(jω)|∠G(jω)H(jω)
The magnitude and phase angle of G(jω)H(jω) can be plotted using the Polar plot, which can be created by changing ω from zero to ∞. As seen in the following figure, the polar graph sheet.
Radial lines and concentric circles make up this graph sheet. So The magnitudes and phase angles are represented, respectively, by concentric circles and radial lines. These angles are represents in the clockwise direction by positive values. By rotating them clockwise, negative angles can be represented similarly. The angle -90 in clockwise direction, for instance, is equal to the angle 270 in anti-clockwise direction.
Rules for Drawing Polar Plot
Follow these rules for plotting the polar plots.
- Substitute, s=jω in the open loop transfer function.
- Write the expressions for magnitude and the phase of G(jω)H(jω).
- Find the starting magnitude and the phase of G(jω)H(jω) by substituting ω=0. So, the polar plot starts with this magnitude and the phase angle.
- Find the ending magnitude and the phase of G(jω)H(jω) by substituting ω=∞. So, the polar plot ends with this magnitude and the phase angle.
- Check whether the polar plots intersects the real axis, by making the imaginary term of G(jω)H(jω) equal to zero and find the value(s) of ω.
- Check whether the polar plot intersects the imaginary axis, by making real term of G(jω)H(jω) equal to zero and find the value(s) of ω.
- For drawing polar plots more clearly, find the magnitude and phase of G(jω)H(jω) by considering the other value(s) of ω.
Example
Consider the open loop transfer function of a closed loop control system.
G(s)H(s)=5/{s(s+1)(s+2)}
Let us draw the polar plot for this control system using the above rules.
Step 1 − Substitute, s=jω in the open loop transfer function.
G(jω)H(jω)=5/{jω(jω+1)(jω+2)}
The magnitude of the open loop transfer function is
M=5/{ω√(ω2+1)√(ω2+4)}
The phase angle of the open loop transfer function is
ϕ=−900−tan−1ω−tan−1ω/2
Step 2 − The following table shows the magnitude and the phase angle of the open loop transfer function at ω=0 rad/sec and ω=∞ rad/sec.
Frequency (rad/sec) | Magnitude | Phase angle(degrees) |
---|---|---|
0 | ∞ | -90 or 270 |
∞ | 0 | -270 or 90 |
The polar plots therefore begins at (∞, -90) and ends at (0, -270). Therefore, the first and second terms enclosed in brackets stand for the magnitude and phase angle, respectively.
Step 3 − So Based on the starting and the ending polar co-ordinates, this polar plot will intersect the negative real axis. The phase angle corresponding to the negative real axis is −1800 or 1800. So, by equating the phase angle of the open loop transfer function to either −1800 or 1800, we will get the ω value as √2.
By substituting ω=√2 in the magnitude of the open loop transfer function, we will get M=0.83. Therefore, the polar plot intersects the negative real axis when ω=√2 and the polar coordinate is (0.83,−1800).
Therefore, using the data above and the polar graph sheet, we can draw the polar plot.