This document discusses using the Nyquist plot to determine the stability of a control system. It provides an example of a stable system where the Nyquist plot does not encircle the -1 point, indicating no zeros in the right half plane and stability. An unstable example is also given where the plot encircles the -1 point twice, indicating two zeros in the right half plane and instability. The behavior of the Nyquist plot at high frequencies is explained. MATLAB is referenced for generating Nyquist plots from transfer functions.

1. Academic Year : 2019 - 2020
II Year II- Semester
Control Systems
Faculty:
Mr.Siva Rama Rao
Associate Professor
SUBMITTED BY
Mr. PEDIREDLA SANJAY KUMAR

2. DETERMINATION OF STABILITY OF
A SYSTEM USING
NYQUIST PLOT
BY TAKING APPROPRIATE
EXAMPLE FROM MATLAB

3. CONTENTS :
■ Statement of the problem
■ The Nyquist path
■ Example
■ Nyquist path, no poles on jω axis, stable
Behavior of the Nyquist path when |s|→∞
 Nyquist path, no poles on jω axis, unstable
 MATLAB

4. Statement of the Problem
Given a single loop feedback system
To determine the stability of a system we:
■ Start with a system whose characteristic equation is given by "1+L(s)=0."
■ Make a mapping from the "s" domain to the "L(s)" domain where the path of "s" encloses the entire
right half plane.
■ From the mapping we find the number N, which is the number of encirclements of the -1+j0 point in
"L(s)."
Note: This is equivalent to the number of encirclements of the origin in "1+L(s)."
■ We can factor L(s) to determine the number of poles that are in the right half plane.
■ Since we know N and P, we can determine Z, the number of zeros of "1+L(s)" in the right half plane
(which is the same as the number of poles of T(s)).
■ If Z>0, the system is unstable.

5. The Nyquist Path (with no poles on the jω axis) :
The specification here is that the path of s should enclose the entire right half plane ,
To start, we assume that the function L(s) has no poles on the jω axis.
We define the path as starting at the origin, moving up the imaginary axis to j∞,
following a semicircle (in the clockwise direction), and then moving up the -jω
axis and ending at the origin.

6. Let us consider this example :
Consider a system with plant G(s), and unity gain
feedback (H(s)=1)
[ here L(s) is nothing but the transfer function]

7. If we map this function from "s" to "L(s)" with the variable s

8. If we zoom in on the graph in "L(s)"

9. The first thing we notice is the multiple arrowheads at the origin.
This is because as the path in "s" traverses a semicircle at ∞ the path in "L(s)" remains at the
origin, but the angle of L(s) changes.
More importantly, we can see that it does not encircle the -1+j0, so N=0.
We also know that P=0, and since N=Z-P, Z is also equal to zero. This tells us that the system is
stable.
And, if we close the loop, we find that the characteristic equation of the closed loop transfer
function is which has roots at -2±17.2j so the system is indeed stable.

10. Behavior of the Nyquist path when |s|→∞ :
A very large segment of the path in "s" occurs when |s|→∞, shown as the large semicircle in
the s-domain plot (i.e., the left plot). However, during this segment of the plot, the path in L(s) will
not move, assuming L(s) is a proper transfer function. The order of the numerator polynomial of
a proper transfer function is less than or equal to that of the denominator. Let's consider, first,
the case when the order of both polynomials is equal to n:

11. If the loop gain, L(s), is strictly proper (i.e., the order of the numerator
is less than that of the denominator)
then
and the path in L(s) is at the origin while |s|→∞.
If |s|→∞, then the highest order term of the polynomial
dominates and we get
So, when |s|→∞ the path in L(s) is at a single point.

12. Nyquist path, no poles on jω axis, unstable
Consider the previous system, with a sensor in the feedback loop

13. If we map this function from "s" to "L(s)" with the variable s following the
Nyquist path we get the following image

14. We can see that this graph encircles the -1+j0 twice in the clockwise direction so N=2.
We also know that P=0, and since N=Z-P, we can calculate that Z=2.
This tells us that the system is unstable, because the characteristic equation of the closed loop transfer
function has two zeros in the right half plane (or, equivalently, the transfer function has two poles there).
We can check this by closing the loop to get the characteristic equation of the closed loop transfer function:
which has roots at s=-7.5 and s=1.26±8.45j so the system is indeed unstable with two poles in the right half
plane.