Chapter 6. Stability Youngjune, Han young@ssu.ac.kr

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

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.

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

Closed-loop poles and response

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

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

Routh-Hurwitz Criterion Example 6.1

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

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

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

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

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.

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.

Pole distribution via Routh table with row of zeros All zero

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

Pole distribution via Routh table with row of zeros Example 6.8

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

Stability design via Routh-Hurwitz Example 6.9

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

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

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

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

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