1. Chapter 6. Stability
Youngjune, Han

2. Introduction C(t)=Cforced(t)+Cnatural(t)
For a bounded input, depend upon natural response, the system can be stable, unstable or marginally stable
A system is stable if every bounded input yields a bounded output (BIBO)
If input is bounded but the total response is not bounded, the system is unstable

3. Stability
A system is stable if the natural response approaches zero as time approaches infinity.
A system is marginally stable if the natural response neither decays nor grows, but remains constant or oscillates.

4. Stability Marginally stable
Note that the roots are on the imaginary axis

5. Closed-loop poles and response

6. Routh-Hurwitz Criterion
Routh Table Criterion
The number of roots of the polynomial that are in the right half-plane is equal to the number of sign changes in the first column
The method requires two step
Generate a data table(Routh table)
Interpret the Routh table to tell how many closed-loop system poles are in the left half-plane, the right half-plane, and on the jw-axis

7. Routh-Hurwitz Criterion
Generating a Basic Routh Table
Equivalent Closed-loop
Transfer function
If no sign changes, stable

8. Routh-Hurwitz Criterion
Example 6.1

9. Routh-Hurwitz Criterion
Example 6.1
Two sign changes, two nstable poles

10. Routh-Hurwith Criterion: Special Case
Zero Only in the first Column
Two sign changes; two unstable poles

11. Routh-Hurwith Criterion: Special Case
Zero Only in the first Column  A polynomial that has the reciprocal root of the original polynomial has its roots distributed the same

12. Routh-Hurwith Criterion: Special Case
Entire Row is zero
P(s)
All zero
P’(s)
P(s)=s4+6s2+8, P’(s)=4s3+12s

13. Pole distribution via Routh table with row of zeros
An entire row of zeros will appear in the Routh table when a purely even or purely odd polynomial is a factor of the original polynomial.
Even polynomials only have roots that are symmetrical about the origin.

14. Pole distribution via Routh table with row of zeros
Since jω roots are symmetric about the origin, if we do not have a row of zeros, we cannot possibly have jω roots.
Everything from the row containing the even polynomial down to the end of the Routh table is a test of only the even polynomial.

15. Pole distribution via Routh table with row of zeros
All zero

16. Pole distribution via Routh table with row of zeros
Total 8 roots exist because it is 8th order
P(s)=s4+3s2+3, no sign change after this means no real pair  4 roots on j-axis

17. Pole distribution via Routh table with row of zeros
Example 6.8

18. Pole distribution via Routh table with row of zeros
Example 6.8
P(s)
P(s)=s6+8s4+32s2+64; two rhs (two sign changes),
so two lhs, the remaining two on j-axis

19. Stability design via Routh-Hurwitz
Example 6.9

20. Stability design via Routh-Hurwitz
If no sign change at the first column; unstable
If there is a zero row, jw is possible
Stable 0<K<1386

21. Stability design via Routh-Hurwitz
Routh table for Example 6.9 with K = 1386
P(s)=18s2+1386
P’(s)=36s
As there is no sign change below, there are two jw poles
Marginally stable

22. Stability in State Space
System poles are equal to the eigen values of the system matrix A
Ax=lx
(lI-A)x=0
x= (lI-A)-10
x= [adj(lI-A) /det (lI-A)]0
det (lI-A)=0
Use det (sI-A)=0

23. Stability in State Space
Example 6.11
s(s-8)(s+2)+30+10
+10(s-8)+5s-6(s+2)
=s3-6s2-7s-52

24. Stability in State Space
Example 6.11
One rhs, two lhs; unstable