The Stability of linear feedback systems

1
The Stability of linear feedback systems

• Dr. Fouad AL-Sunni
• SE302

2
Outline

• The concept of stability
• The Routh-Hurwitz Stability criterion
• Examples

3
Stability

• Absolute stability (yes or no)
• Relative stability (stable to which degree)
• A stable system is a dynamic system with a
  bounded response to a bounded input (BIBO)

4
Concept of stability
5
Stability time characterization
6
Necessary and sufficient condition for stability

• A necessary and sufficient condition for a
  feedback system to be stable is that all the
  poles of the system transfer function have
  negative real parts.
• One way to characterize stability is by computing
  the poles.

7
Pole of the closed transfer function
Characteristic equation
8
Steps to analyze stability

• Find all the poles
• Ex1 (s1)(s2)0
• Ex2(s3)(S26)0
• Ex3s33s25s40
• If at least one of the poles has positive real
  part the system is unstable.

9
Routh-Hurwitz Criterion

• Need to characterize stability without having to
  find the poles.
• Tool
• Routh-Hurwitz stability criterion

10
Idea

• Stability means all the poles are in the LHS of
  the s-plane

11
Necessary conditions

• All the coefficients of the characteristic
  equation have the same sign.
• None of the coefficient is zero.
• Conditions are necessary but not sufficient.
• Ex s3s22s8

12
R-H criterion is a necessary and sufficient
condition

• Steps
• Put the coefficients in a special table form
• Compute cofactors
• Observe the change of sign of the entries in the
  first column
• Suppose

13
Step 1 Table form
14
Example
15
Step 2 compute cofactors
16
Example
17
Theorem

• The roots of the equation are all in the left
  half of the s-plane if all the elements of the
  first column of the Routh-Hurwitz table are of
  the same sign.
• The number of changes of sign in the elements of
  the first column equals the number of roots with
  positive real parts.

18
Difficulties

• The following difficulties may occur when
  formulating the Routh-Hurwitz table
• The first element in any one row of the
  Routh-Hurwitz table becomes zero, but the others
  are not.
• The elements in one row become all zero.

19
First element of a row is zero

• In the first case, if a zero appears in the first
  element of a row, the elements in the next row
  will all become infinite.
• To remedy this situation, we replace the zero
  element in the first column by an arbitrary small
  positive number e, and then proceed with the
  Routh-Hurwitz table.

20
Example
e
Take limit e-gt0
21
Example I

• Two changes in sign
• System is unstable with two roots having positive
  real parts

22
Example II

• Consider the CE
• What values of K will make the CE stable?
• The RH array is

23
Example II

• For any K greater than zero, the system is
  unstable
• The last entry in the first column is K. This
  means a positive value of K will result in an
  unstable system.
• We can not find any K to make the CE stable.

24
Case two All the row is zero

• Indicates at least one of the cases
• The equation has at least one pair of real roots
  with equal magnitude but opposite sign.
• The equation has one or more pairs of imaginary
  roots.
• The equation has pairs of complex-conjugate roots
  forming symmetry about the origin.

25
Example
26
Steps

• 1- Auxiliary equation (above row of zeros)
• Long division

27
Steps (continue)
28
Conclusion

• System marginally stable and has 2 pure imaginary
  conjugate poles.

29
Example I

• Consider the CE
• The RH array gives
• System is stable for 0ltKlt8
• What happens when K 0?
• What happens when K 8?

30
K8..a zero row!!

• How to deal with this?
• We form an equation called the auxiliary
  Polynomial (AE) from the row preceding the zero
  row.
• The AE is a cofactor of the original polynomial
  we are testing

31
K8..a zero row!!

• The first two row..
• The AE is

32
Dividing q(s) /AE
33
Repeated roots on the jw-axis

• If the polynomial has repeated roots on the
  jw-axis, the RH will not be able to detect this.
  It will fail!!
• Another example Ex

34
Example II

• Consider
• The AE is
• Two roots on the jw-axis.
• The remaining part of q(s)

35
Examples. Continuations

• Apply the RH on the remaining part
• The system has two unstable poles, two poles on
  the jw-axis, and one stable pole.

36
Repeated roots on the jw-axis

• Another example Ex

37
Another example

• Consider the welding machine dynamics
• The Closed loop CE
• Or
• Use the RH test to study stability

38
Using the RH

• The RH
• For stability
• a K positive
• b3 positive implies Klt60
• c3 positive implies
• An exampleK 40, a0.639

39
Tabulated results

• For the general polynomial
• The RH test gives

40
Another example

• The system
• The CE

41
The RH table

• Table..
• Conditions on the parameters

42
Region of stability
43
Design

• Select K and a so that the system has steady
  state error for a ramp input less than or equal
  to 24 of the magnitude of the ramp
• For a ram r(t) At
• So,

44
Design

• If aK 42, see if ss error is 23.8 of A
• We can choose K 70, a0.6 as shown in the figure.

45
Class assignment and HW

• CA Page 339- The example in section 6.7
• HW
• E6.1-6.9, 6.16
• P6.1 (b,d,e), 6.4, 6.11, 6.12,
• AP 6.1

46
Relative stability

• We want to examine if a system have its poles to
  the left of a given value.

47
Relative stability

• Consider
• Do we have roots to the left of S-1?
• Make the substitution sns1 or ssn-1 to get
• Apply the RH test
• We have roots on the new jw-axis.