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2.3 Stability of Feedback Control Systems

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Lecture 2.3
Stability of
Feedback Control Systems
CH158P Process Dynamics and Control
Contents
• Stability
• Concept
• Definition
• Criterion
• Routh Test for Stability
2
Stability
3
Concept of Stability
The first- and second-order systems discussed so far were inherently stable—the responses were
bound.
We now consider the problem of stability in a control system that is only slightly complicated than
those studied before.
Consider the proportional control of two stirred-tank heaters (in series) with a measurement lag:
We are going to consider only the servo
problem (U = 0, set-point tracking).
The overall transfer function for this is:
4
Concept of Stability
In terms of the individual transfer functions,
It can be seen that the denominator is a
third-order polynomial.
5
Concept of Stability
For a unit-step change in R, the response in the Laplace domain is:
Inversion of the previous equation provides the equation for the response in the time domain.
This requires obtaining the roots of the third-order polynomial in the original transfer function,
which entails lengthy algebraic methods or the use of math software, such as MATLAB.
It is, however, apparent that the roots will depend on the values of 𝐾𝐢 , 𝜏1 , 𝜏2 , and 𝜏3 . These roots
determine the nature of the transient response (constant, oscillatory, exponentially increasing,
decaying, etc.).
6
Concept of Stability
Assuming the time constants 𝜏1 , 𝜏2 , and 𝜏3 are constant, and only 𝐾𝐢 is varied, the step responses of
the closed-loop system will look like these:
As 𝐾𝐢 is increased, the response becomes
more oscillatory. It can be seen, however, that
at high enough values of 𝐾𝐢 , the oscillations
become unbounded (e.g. for 𝐾𝐢 = 12).
When the response grows rather than decays,
it is deemed unstable.
In the design of control systems, we are
interested in being able to determine quickly
the values of 𝐾𝐢 that give unstable responses.
7
Definition of Stability
For our purposes, we define stable control systems as follows:
A bounded input function is a function of time that always falls within certain bounds during the
course of time (e.g., step, sinusoidal). The ramp function 𝑓 𝑑 = 𝐴𝑑 is an example of an unbounded
function.
This definition is only true in the mathematical sense. In reality, physical bounds or constraints are
always exhibited, such as the fully closed and fully open positions of a valve. We describe such
limitations using the term saturation.
A physical system, when unstable, may not follow the response of its linear mathematical model
beyond certain physical bounds, but rather may saturate.
We still need to predict system stability since operation is always within the saturating boundaries;
operation at saturation implied unsatisfactory control.
8
Stability Criterion
We now translate the definition of stability into a simpler criterion, one that can be used to
ascertain the stability of control systems of the form shown:
The response of this closed-loop system to changes in both U and R is:
9
Stability Criterion
To determine under what conditions this system is stable, it is necessary to test the response to a
bounded input.
Suppose a unit-step change in set point is applied. The response is:
where π‘Ÿ1 , π‘Ÿ2 , … , π‘Ÿπ‘› are the n roots of the characteristic equation:
and 𝐹 𝑠 is a function that arises in the rearrangement to the single-term form of the response.
10
Stability Criterion
For example, for the control system being considered earlier, the step response is:
which may be rearranged to:
11
Stability Criterion
Comparing this final result to the original form of the response,
For this case, the function 𝐹 𝑠 is:
and the characteristic equation, whose roots are π‘Ÿ1 , π‘Ÿ2 , and π‘Ÿ3 , is:
12
If there are any roots in the
right half of the complex plane,
the response will contain a term
that grows exponentially in
time—the system is unstable.
If there is an 𝑠 π‘š in the
characteristic equation, where
π‘š ≥ 2, there is more than one
root at the origin and the
response is again unbounded.
If there is a repeated pair of
conjugate roots on the
imaginary axis, the contribution
is a sinusoid with an amplitude
that increases with time.
The right-half plane,
including the imaginary axis,
is the unstable region of
roots.
13
Stability Criterion
The mathematical definition of stability is therefore:
Note that the characteristic equation of a control system is same for set point or load changes.
Furthermore, this definition is applicable for any input—not only for the step input.
14
Stability Criterion
Exercise 2.3.1
In terms of the figure shown, a control system has the
transfer functions
Find the characteristic equation and its roots, and
determine whether the system is stable.
15
Stability Criterion
Exercise 2.3.1
16
Stability Criterion
Exercise 2.3.1
17
Routh Test for Stability
18
Routh Test
The Routh test is a purely algebraic method for determining how many roots of the characteristic
equation have positive real parts: from this it can also be determined whether the system is stable.
If there are no roots with positive real parts, the system is stable.
The test is limited to systems with polynomial characteristic equations (without an exponential
terms, such as those for systems with transportation lag).
The procedure for examining the roots is to write the characteristic equation in the form:
where π‘Ž0 is positive.
In this form, it is necessary that all the coefficients π‘Ž0 , π‘Ž1 , π‘Ž2 , … , π‘Žπ‘›−1 , π‘Žπ‘› be positive if all the roots
are to lie in the left half-plane. If any coefficient is negative, the system is definitely unstable, and
the Routh test is not needed to determine stability.
If all the coefficients are positive, the system may stable or unstable.
19
Routh Test
If the coefficients are all positive, the procedure for the Routh test is as follows:
1.
Arrange the coefficients of the characteristic equation into the first two rows of the Routh array.
The array shown has been filled in for n = 7. In general, there are 𝑛 + 1 rows. For n even, the
first row has one more element than the second row.
2.
The elements in the remaining rows are found from
the formulas
During the computation of the Routh array, any row can
be divided by a positive constant without changing the
results of the test.
20
Routh Test
The following theorems are then applied to determine stability.
THEOREM 1
The necessary and sufficient condition for all the roots of the characteristic equation to have
negative real parts is that all elements of the first column of the Routh array (π‘Ž0 , π‘Ž1 , 𝑏1 , 𝑐1 , etc.) be
positive and nonzero.
THEOREM 2
If some of the elements in the first column are negative, the number of roots with a positive real
part (in the right half-plane) is equal to the number of sign changes in the first column.
THEOREM 3
If one pair of roots is on the imaginary axis, equidistant from the origin, and all other roots are in
the left half-plane, then all the elements of the nth row will vanish and none of the elements of the
preceding row will vanish. The location of the pair of imaginary roots can be found by solving the
equation 𝐢𝑠 2 + 𝐷 = 0, where the coefficients C and D are the elements of the array in the (n-1)st
row as read from left to right, respectively.
21
Routh Test
Exercise 2.3.2
Given the characteristic equation 𝑠 4 + 3𝑠 3 + 5𝑠 2 + 4𝑠 + 2 = 0, determine the stability by the Routh
criterion.
22
Routh Test
Exercise 2.3.3
1
1
Using 𝜏1 = 1, 𝜏2 = , and 𝜏3 = , determine the values of 𝐾𝐢 for which the control system is stable.
2
3
For the value of 𝐾𝐢 for which the system is on the threshold of stability, determine the roots of the
characteristic equation with the help of Theorem 3.
23
Routh Test
Exercise 2.3.3
1
2
𝜏1 = 1, 𝜏2 = , 𝜏3 =
1
3
24
Routh Test
Exercise 2.3.3
1
2
𝜏1 = 1, 𝜏2 = , 𝜏3 =
1
3
25
Routh Test
Exercise 2.3.3
1
2
𝜏1 = 1, 𝜏2 = , 𝜏3 =
1
3
26
Routh Test
Exercise 2.3.4
1
2
Determine the stability of the system for which a PI controller is used. Use 𝜏1 = 1, 𝜏2 = , 𝜏3 =
1
, 𝐾𝐢
3
= 5, and 𝜏𝐼 = 0.25.
27
Routh Test
Exercise 2.3.4
1
2
Determine the stability of the system for which a PI controller is used. Use 𝜏1 = 1, 𝜏2 = , 𝜏3 =
1
, 𝐾𝐢
3
= 5, and 𝜏1 = 0.25.
28
Lecture 2.3 Problems
Solve the following problems from the textbook (LeBlanc & Coughanowr, 3E):
1.
Problem 13.1
2.
Problem 13.9
3.
Problem 13.10
29
Lecture 2.3
Stability of
Feedback Control Systems
CH158P Process Dynamics and Control
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