Chapter 14

14.3 Solved Examples for Polar Plots

14.3.1 Example

Consider the following transfer function:

G(s)=200s3+11s2+38s+4G(s)=200s3+11s2+38s+4

Consider its frequency response, G(jω)G(jω), at a specific frequency of ω=1ω=1 rad/sec. Show its rectangular and polar forms.

G(jω)=200(jω)3+11(jω)2+38(jω)+4G(jω)=200(jω)3+11(jω)2+38(jω)+4

=200j(ω)311(ω)2+38(jω)+4==200j(ω)311(ω)2+38(jω)+4= 200(411ω2)+jω(38ω2)200(411ω2)+jω(38ω2)

G(j1)=2007+j37=G(j1)=2007+j37= 200(7j37)49+1369=0.9873j5.2186200(7j37)49+1369=0.9873j5.2186

The polar form of this function is:

G(j1)∣=(0.9873)2+(5.2186)2=5.3112G(j1)=(0.9873)2+(5.2186)2=5.3112

G(j1)=1.7578rad=100.71G(j1)=1.7578rad=100.71

G(jω)=∣G(jω)e(jG(ω)G(jω)=G(jω)e(jG(ω)

G(j1)=5.3112ej1.7578G(j1)=5.3112ej1.7578

G(j1)=5.3112ej1.7578=0.9873j5.2186G(j1)=5.3112ej1.7578=0.9873j5.2186

14.3.2 Example

Consider a simple first order system, with one real pole:

G(s)=110s+1G(s)=110s+1

G(jω)=G(jω)= 110jω+1=110jω+1= 1(10ω)2+1tan1(10ω)1(10ω)2+1tan1(10ω)

M(jω)=1(10ω)2+1M(jω)=1(10ω)2+1

Φ(jω)=tan1(10ω)Φ(jω)=tan1(10ω)

Now consider the rectangular representation of the same frequency response function G(jω)G(jω):

G(jω)=11+10jω)=1(110jω)(1+10jω)(110jω)=11+100ω2j10ω1+100ω2G(jω)=11+10jω)=1(110jω)(1+10jω)(110jω)=11+100ω2j10ω1+100ω2

Re(ω)=11+100ω2Re(ω)=11+100ω2

Im(ω)=10ω1+100ω2Im(ω)=10ω1+100ω2

The standard frequency response plot (Bode Plot) with magnitude in decibels and phase in degrees is shown below. For the Polar Plot, crossovers with the Imaginary and Real axes can be calculated analytically by setting first the Real, then the Imaginary part to zero, and solving for frequency. In this example:

Re(ω)=0Re(ω)=0 ω=ω=

Im()=0Im()=0

Im()=0Im()=0  ω=0,ω=0, ω=ω=

Re(0)=1Re(0)=1

Re()=0Re()=0

 

 

This indicates that the polar plot starts at (1, j0) location for ω=0ω=0 (DC condition), and ends at (0, j0) for ω=ω= . The sense of increasing frequency ωω should always be shown on the polar plot. The polar plot of the system G(s)G(s) is shown.

To do plot polar plots in MATLAB, use subroutine Nyquist – see below. The second plot (on the following page) shows a so-called Nyquist contour, which will be discussed in detail later. The Nyquist contour consists of the polar plot for positive frequencies,0<ω<+0<ω<+, and its mirror image for negative frequencies, <ω<0<ω<0.  

 

14.3.3 Example

A process transfer function is described as follows: G(s)=10s3+4s2+6s+8.G(s)=10s3+4s2+6s+8. Frequency plots of G(s)G(s) are shown. Sketch a polar plot for G(s)G(s).

Solution:   It is helpful to construct a table with the important coordinates:

Frequency Phase Magnitude
ω=0ω=0 rad/s ϕ=0ϕ=0 1.251.25
ω=1.4ω=1.4 rad/s ϕ=90ϕ=90 1.771.77
ω=2.4ω=2.4 rad/s ϕ=180ϕ=180 0.6250.625
ω=+ω=+ rad/s ϕ=270ϕ=270 00

The resulting polar plot can be also plotted using MATLAB subroutine Nyquist.

 

 

14.3.4 Example

Consider a unity feedback control system under Proportional Control. The process transfer function is described as follows:

G(s)=5s(s+1)(s+5)G(s)=5s(s+1)(s+5)

Frequency plots of G(s)G(s) are shown. It is helpful to construct a table with the important coordinates read off the plot. Note that this is a Type I system, with an integrator, and therefore its polar plot will begin with an infinite gain at the DC level.

Frequency Phase Magnitude in dB Magnitude in Volt/Volt
ω=0ω=0rad/s ϕ=90ϕ=90 ++ ++
ω=224ω=224rad/s ϕ=180ϕ=180 15.5615.56 dB 0.16670.1667 Volt/Volt
ω=+ω=+rad/s ϕ=270ϕ=270 00

 

 

The resulting polar plot is shown.

 

License

Icon for the Creative Commons Attribution-NonCommercial 4.0 International License

Introduction to Control Systems Copyright © by Malgorzata Zywno is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.