1、Lecture 910Analysis in the time domain (III)North China Electric Power UniversitySun Hairong9/1/20221Topics of this class Second-order systems: (Reading Module 6) （Disturbance rejection and Rate feedback）Higher-order system: (Reading Module 7)Rouths Method; (Reading Module 9)System type and steady-s

2、tate errors (Reading Module 8)21. Second-order systems: Disturbance rejection and Rate feedback The benefits of utilizing feedback control systemOne, to improve the dynamic response beyond that available with an open loop system;Two, to ignore unwanted input, often called its disturbance rejection c

3、apability;Three, to improve the dynamic performance with adjusting some parameter. 32. Higher-order systems a. Evans, or root locus, form . Example: b. Bode form. Example: 2.1 Reduction to lower-order systemsA higher-order system may be reduced to lower-order system through the identification of dom

4、inant poles. However, some qualifications need to be specified.Two forms of transfer function,4If the input isThen the steady-state output is If the reduced system is If the reduced system is The Bode form must be used when reducing higher-order system in order to ensure the steady-state output the

5、same. cf.5Given nth-order linear time-constant systemThe characteristic equation yields,The closed-loop system may have both real and complex poles.Assuming the system has q real poles, and r pairs of complex poles ( ), the output is given by2.2 Effect of a closed-loop pole62.4 Effect of a closed-lo

6、op zero(See page 114117)Fig. 7.6 Effect of zero on overshoot2.3 Third order systems(See page 112114)Fig. 7.3 PO for general third-order system7Then the closed-loop transfer function is given by Clearly, the poles of the closed-loop are decided by the characteristic equation.And the zeros are the zer

7、os of the forward path transfer function and the poles of the feedback transfer function.2.5 Occurrence of closed-loop zeros and polesTaking a feedback system8Assuming the system has q real poles, and r pairs of complex poles( ), the output is givenGiven nth-order linear time-constant systemThe char

8、acteristic equation yields,The closed-loop system may have both real and complex poles.3. Rouths Method 3.1 Review: 9Rouths criterion states that, The number of closed-loop poles in the right-hand half complex plane is equal to the number of sign changes of the elements of the first column of Rouths

9、 array.The Rouths criterion is both necessary and sufficient condition of a stable system.3.2 Rouths method10Example 1: The characteristic equation is given by D(s)=3s4+10s3+6s2+40s+9=0Determine if the system is stable.Write Rouths array,11Example 2:Consider the feedback system shown in following fi

10、gure. Determine the range of the gain K on the condition that the system is stable. 12DefinitionThe feedback system is known asError is defined as two forms(a). E(s) = R(s)-C(s)H(s)(b.) E(s) = R(s)-C(s)Calculate the steady-state error subjected to definition (a)The steady-state error is The steady-s

11、tate error can also calculate by using the final-value theorem as (assuming the poles of sE(s) are all located on the left-hand complex plane)4. System type and steady-state errors13Where Kk is called the open-loop gain. (Bode form). The system type is defined by the value of n. 4.1 Step inputFor th

12、e type zero systemFor the type one systemFor the type two systemFirstly, A generalized open-loop transfer function will be defined as ,the constant Kp is called the position error constant.144.2 Ramp input, the constant Kv is called the velocity error constant.For the type one systemFor the type two systemFor the type zero system154.3 Acceleration input, the constant Ka is call