控制理论基础-自控a_ch4

1,Chapter 4 The Stability of Linear Feedback Systems,4.1 The Concept of Stability4.2 Routh-Hurwitz Stability Criterion,2,4.1 The Concept of Stability,A stable system is dynamic system with a bounded response to a bounded input.Stability is the fundamental requirement of a control system.Absolute stability stable / not stableRelative stability the degree of stability,3,The response of a linear system to a stimulus has two component: (1). Steady state terms which are directly related to the input. (2). Transient terms which are either exponential or oscillatory with an envelope at exponential form. If the exponential terms decay as time increases then the system is said to be stable; otherwise the system is said to be unstable.,4,Stability means that with no input, the output of each integrator will decay to zero eventually.When the transfer function of system is T(s), the output Y(s) is:,5,According to inverse Laplace transform, there is:,the system is stable.,then total response ,so the system is unstable.,6,The conclusion will be gotten as following:,The system is stable only when all the closed-loop poles are located in the left-hand position (LHP) of the s-plane.The system becomes unstable as soon as one closed-loop pole is located in the right-hand position (RHP) of the s-plane.If a system has simple roots on the imaginary axis with all other roots in the LHP, it is called marginally stable.,7,4.2 Routh-Hurwitz Stability Criterion,Consider the following system:,8,A necessary but not sufficient criterion: For a stable system, all the coefficients of the polynomial q(s) must have the same sign and be nonzero.,The Routh-Hurwitz criterion is a necessary and sufficient criterion.,9,Routh array,10,b 1= ( an-1.an-2 - an. a n-3 )/ an-1 ,b 2= ( an-1 .an-4 - an. a n-5)/ a n-1, .etc.c 1= ( b1 .an-3- an-1. b 2 )/ b1 ,c 2= ( b1 .an-5- an-1. b 3 )/ b1 , .etc.,11,Routh Stability Criteria,For system stability the primary requirement is that all of the roots of the characteristic equation have negative real parts. All of the roots of the characteristic equation q(s) have negative real parts only if the elements in the first column of the Routh array are the same sign. The number of sign changes in the first column is equal to the number of roots of the characteristic equation q(s) with positive real parts.,12,Example 4.1 To test the stability of a system having a characteristic equation q(s) = s3 + 6 s2 + 12 s + 8 = 0 The Routh Array is constructed as follows：,The first column includes no changes of sign and therefore the roots of the characteristic equation have only negative real parts and the system is stable.,13,Example 4.2 The system characteristic equation is： q(s) = s 3 + 3 s 2 + 3s + K = 0.In order for the system to be stable there should be no sign change in column 1. To achieve this K must be greater than 0 and less than 9. Therefore for system stability, 0 K 余下的多项式重新做劳思表辅助多项式对s求导后系数代替全0行系数-继续劳思表,19,Example 4.4 Consider a closed loop control system with negative feedback which has an open loop transfer function.,The closed loop characteristic equation is,q(s) = s 4 + 2s 3 + 2s 2 + s + K = 0 .,The Routh Array is constructed as :,20,In this array row 3 becomes an all zero row if K = 3/4 and we obtain the auxiliary polynomial from row 2 :Which indicates that two roots are on the imaginary axis.By dividing q(s) by N(s) = (2 s2 + 1) the equation is ob