1、Ch.3 Transient and Steady-State Response Analysis,Stability and Routh criterion,Basic concept of stability,stable,unstable,Definition of stability,All system will deviate from the original equilibrium under the disturbance. If the system can return to the equilibrium when all disturbance vanish, the

2、n it can be called stable system. Stability is intrinsic for a system. For a linear system, It relies not on the initial and external conditions, but on the structure, parameters.,Methods in analyze stability,Characteristic equation and characteristic roots(特征方程和特征根) Algebraic criterion(代數(shù)判據(jù)) Root l

3、ocus(根軌跡) Stability criterion in frequency domain(頻率穩(wěn)定判據(jù)),Since stability is the research of system after removing the disturbances, and has nothing to do with input, it can be represented by the pulse response function. If the pulse response convergences, the system is stable, and vice versa.,K rea

4、l characteristic roots,r pairs of conjugate roots,（1）if it will return to original equilibrium, theres oscillation because of complex roots; （2）if the response is exponential attenuation; （3）if or , when , its unstable; （4）if one of equals to 0, when , the system cannot return to the original equili

5、brium or shows even oscillation. Its also treated as unstable.,Sufficient and necessary conditions for stability,All roots of the characteristic equation have negative real parts. Namely all roots must lie on the left half of the s plane.,Marginally stable will be treated as unstable, since there al

6、ways exists approximation in system modeling. and the parameters are always changing.,If above method is adopted, all poles must be figured out for the closed-loop transfer function. It is nearly impossible for the system above 2-order. It is a natural demand for a substitute for this direct solutio

7、n. In 1877, Routh found one.,The Routh-Hurwitz Criterion The Routh criterion can ascertain the stability by determining the sign of the roots without solving the characteristic equation. The number of the roots with positive real parts is equal to the number of changes in sign of the first column of

8、 the Routh array. This is necessary and sufficient!,Subsidiary statements (1) For a stable system, the coefficients of the characteristic polynomial must be positive (have the same sign) and nonzero. This requirement is necessary but not sufficient.,(2) It is sufficient to assure the stability of th

9、e control system that all the entries in the first column of the Routh array are positive.,(3) It is sufficient to determine that the control system is unstable if there are positive and negative entries in the first column of the Routh array. (4) Theres no influence on Routh criterion if all entrie

10、s of one row in the Routh array multiply one positive number.,The Routh array is scheduled as:,How to schedule the Routh array consider the characteristic equation as:,Example 1,The system is unstable, and has 2 poles in the right half s-plane.,First order: If a1, a0 have the same sign, the system i

11、s stable.,Second order: If a1,a2,a0 have the same sign, the system is stable.,Third order: If a0,a1,a2,a3 are positive, and a1a2a3a0, the system is stable.,Several distinct cases: Case 1: No entry in the first column is zero. Case 2: There is a zero in the first column, and the row has at least one

12、nonzero entry. Case 3: There is a zero in the first column, and entries in the corresponding row are all zero.,Case 1,Example 2 ( with parameter ): the characteristic equation is,s3+s2+s+k=0; k is not 1or 0,Thus: The system is unstable when k is less than 0 and when k is greater than 1; Else, the sy

13、stem is stable.,Adjusting the parameter may change the stability!,Case 2,Example 3 The characteristic equation is:,s4+s3+s2+s+k =0,Let approaches to 0, there is the product of k and minus infinite in the first column. The system is always unstable.,Replace the zero element with a small positive ,Cas

14、e 3 Zero row,Example 4 The characteristic equation is:,s4+5s3+7s2+5s+6=0,2. Build the auxiliary polynomial using the above row. The order will be always even.,1. When there are roots symmetrically located about the origin, zero row will happen.,s2+1=0,Sufficient for unstable: If there is a zero row,

15、 the system is unstable!,s1,2=j,Example 5 welding control,The characteristic equation is:,Determine the range of k and a, to make the system stable.,Routh Array,Let k=40, then we need a0.639,Detection of the abundant degree of stability,The routh criterion can decide the abundant degree of stability

16、 as well. In this situation, the coordinate system can be moved and use the routh criterion again.,Example 6 The characteristic equation is: determine the range of k to make the system stable. If we require further the roots are all in the left side of s=-1, how to adjust the parameter k ?,The chara

17、cteristic equation can be rewritten as:,By Routh array, we find that the range of k to make the system stable is: 0k13,for the requirement that all the roots are in the left side of s=-1, substituting s=s1-1 into the characteristic equation, we have:,Routh array Letting all the elements in the first column to be positive yields more strict range:,Meathods to correct the structurally unstable system,No matter how to adjust the structural parameter, some systems are still unstable. They are called structurally unstable system.,Structurally unstable system,1. Modify the integral element,Employ