Stability This lecture we will concentrate on How to determine the stability of a system represented as a transfer function How to determine the stability of a system represented in state-space How to determine system parameters to yield stability
Definitions Remember that : The total response of a system is the sum of the forced and natural response, c(t) = c forced (t) + c natural (t) Using this concepts, we present the following definitions of stability, instability and marginal stability : Definition : A linear time invariant system is stable if the natural response approaches to zero as time approaches to infinity. A linear time invariant system is unstable if the natural response grows without bound as time approaches to infinity. A linear time invariant system is marginally stable if the natural response neither decays nor grows but remains constant or oscillates as time approaches to infinity.
BIBO Stability The definitions given on the previous slide leads to a more commonly used way of defining stability for LTI systems A system is stable if every bounded input yields a bounded output. We call this statement the bounded-input, bounded-output (BIBO) definition of stability.
Determining the Stability of Systems From our previous lectures, we learned that If the closed loop system poles are in the left-half of the s- plane and hence have a negative real part, the system is stable. Unstable systems have closed loop system transfer functions with at least one pole at the right half plane and/or poles of multiplicity greater than one on the imaginary axis. And marginally stable marginally stable systems have closed loop transfer function with only imaginary axis poles of multiplicity 1 and poles in the left half plane.
Routh-Hurwitz Criterion The Routh Hurwitz stability criterion is a mathematical test that is a necessary and sufficient condition for the stability of a linear time invariant (LTI) control system. This method enables us to investigate the stability information without the need to calculate for closed loop system poles Basicaly the test is composed of two steps : (1) Generate a data table called a Routh table and (2) 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 axis.
Routh Hurwitz Criterion for ch. polynomials of order 2, 3, and 4 In the following, we assume the coefficient of the highest order (e.g. in a second order polynomial) be positive. If necessary, this can always be achieved by multiplication of the polynomial with -1. For a second-order polynomial,, all the roots are in the left half plane (and the system with characteristic equation is stable) if all the coefficients satisfy. For a third-order polynomial, all the coefficients must satisfy, and For a fourth-order polynomial all the coefficients must satisfy, and and In general Routh stability criterion proclaims that all First column elements of Routh array is to be of same sign. System meeting the above criteria is said to closed loop stable otherwise unstable because there is sign changes in the first column elements.
Generation a Routh Table Consider the following closed-loop transfer function We are interested in system poles (the roots of the ch. eq.) therefore focus on the denominator Generate the Rought table Begin by labelling the rows with powers of s from the highest power of the denominator of the closed loop transfer function to s 0. Next start with the coefficient of the highest power of s in the denominator and list, horizontally in the first row, every other coefficient. In the second row list horizontally, starting with the highest power of s, every coefficient that was skipped in the first row. empty parts
Generation a Routh Table Each entry is a negative determinant of entries in the previous two rows divided by the entry in the first column directly above the calculated row. The left-hand column of the determinat is always the first column of the previous two rows, and the right-hand column is the elements of the column above and to the right. The table is complete when all of the rows are completed down to s 0
Example #1 Investigate the stabiliy of the system with the characteristic equation given as Solution : Form the Routh Table is positive and no sign changes The initial table Calculate the third row and simplify Complete the overall table System is STABLE
Example #2 Investigate the stabiliy of the system with the characteristic equation given as Solution : Form the Routh Table 4 sign changes, System is UNSTABLE
Example #3 (Zero coef. in 1st row) Consider the characteristic equation of the form Form the Rought table a 0 in the first column?? Case #1 insert a very small number instead Stability is dependent NOT STABLE!! (Why??)
Example #3 Case #2 Rewrite the polynomial in the form with Form the table now 2 sign changes NOT STABLE!!!
Example #4 (Zero Row in Table) Consider Routh Table zero row Define zero row (again) Define STABLE WHY?
Zero Row in the Table Looking further to the case of all zeros in a row : This can happen when a purely even or purely odd polynomial is a factor of the original polynomial Even polynomials only have roots that are symmetrical to the origin. This symmetry can happen only in 3 conditions The roots are symmetric and pure real The roots are symmetric and pure imaginary The roots are quadrantal.
Example #5 (variable gain) Consider determine the values of K in order for the system to be stable Form the Routh Table
Stability for Block Diagram Rep. Given comment on the stability of the plant Solution : Start with constructing the equivalent closed loop system representation. The closed loop block diagram representation and transfer function is in the form
Block Diagram Representation From the Routh Table Due to the sign change in the first column plant is UNSTABLE
Design with Block Diagram Rep. Find the range of K, for the system shown to be stable unstable and marginally stable Solution : Start with constructing the equivalent closed loop system representation. That is K T ( s) 3 2 s 18s 77s K Corresponding Rought Table
Since K assumed positive, we see that all elements in the first column are always positive except the s 1 row. This entry can be positive, zero, or negative, depending upon the value of K. If K<1386, all terms in the first column will be positive, and since there are no sign changes, the system will have three poles at the left half plane and be stable. If K>1386, the s 1 term in the first column is negative. There are two sign changes, indicating that the system has two right half plane poles and one left half plane pole, which makes the system unstable. If K=1386, we have an entire row of zeros, which colud signify jω poles. Returning the s 2 row and replacing K with 1386, we form the even polynomial P(s)=18s 2 +1386. Differetiating with respect to s, we obtain dp(s)/ds=36s.
Replacing the row of zeros with the coefficients of the last equation, we obtain the following Routh table. Since there are no sign changes from the even polynomial (s 2 row) down to the bottom of the table, the even polynomial has its two roots on the jω-axis of the unit multiplicity. Since there are no sign changes above the even polynomial, the remaining root is in the left half plane. Therefore, the system is marginally stable.
Stabiliy in State Space Given the system of the form The eigen values of the input matrix is also thepoles of the system therefore The chacteristic equation can be obtained as det( I A) 0 When the characteristic equaiton is obtained we can evaluate the stability of the system with Rought Hurwitz Criterion.
Example: For the system given below, find out how many poles are in the left half plane, in the right half plane, and on the jω-axis. Solution : Find
Form the Rought Table Since there is one sign change in the first column, the system has one right half plane pole and two left half plane poles. The system is unstable.