《现代控制理论》课程教学资源课程教学资源——教学课件_Chap2_2.4 Time Response of the LTI System

CHAPTER2 TIME RESPONSE OF THE LTISYSTEMCONTENT> 2.1 Time Response of the LTI HomogeneousSystem> 2.2 Calculation of the Matrix ExponentialFunction> 2.3 State Transition Matrix> 2.4 Time Response of the LTI System

CHAPTER2 TIME RESPONSE OF THE LTI SYSTEM • CONTENT  2.1 Time Response of the LTI Homogeneous System  2.2 Calculation of the Matrix Exponential Function  2.3 State Transition Matrix  2.4 Time Response of the LTI System

2.4 Time Response of the LTI SystemConsider the LTI system with the initial condition X(t.)= X(O) X(t) = AX(t) + Bu(t)Taking Laplace transform, we havesX(s) - X(O)= AX(s)+ BU(s)X(s) = (sI - A)-1X(O)+(sI - A)-BU(s)Note thatL[f, f(t)g(t-t)dt) =F(s)G(s)L[e4"]=(sI -A)-1So,the inverseLaplacetransformresults inthe solution['eA(t-t) .B .u(t) .dtX(t) = eAt X(0) + /

2.4 Time Response of the LTI System

2.4 Time Response of the LTI SystemConsider the LTI system with the initial condition X(t) = X(O)X(t) = AX(t) + Bu(t)A(t-t) . B . u(t) . dtX(t) = eAtX(0) +If the initial condition of the LTI system is the more general caseX(t) , the solution will be.eA(t-to)A(t-t)X(t)=x(t.) +B.u(t).dt2the solution can also be written as the more general caseX(t)= (t - t)X(t。)+ [ Φ(t - t). B . u(t) .dt

2.4 Time Response of the LTI System

2.4 Time Response of the LTI SystemX(t) =e4(-to) x(t)A(t-t) . B .u(t) .dtIt is clear that the solution of the LTI system is composed oftwoterms.The first term on the right-hand side is the solution of the LTIhomogenous system and it can be called the force-free responseor zero-input response of the LTI system

2.4 Time Response of the LTI System

2.4 Time Response of the LTI SystemeA(t-t) .B .u(t)·dtX(t) = e4(t-to) X(t) -一Let the initial state X(t,)= 0, in other words, the LTI system istaken onthe zero-statesituation.Based on the definition of the matrix exponential function, thederivation can be obtained asd-4(t-o) X(t) = -Ae-4(--o) X(t) +e-4(-t) X(t)dt= -e-A(t-to) AX(t) + e-A(t-to) X()= e-4(-to)[X(t) - AX(t)] = e-4(t-to) Bu(t)Furthermore,d[e-A(t-to) X()] = e-A(t-to) Bu(t)dt

2.4 Time Response of the LTI System

2.4 Time Response of the LTI System4(t-t) .B .u(t).dtX(t) = e4(t-to)X(t.) +Let the initial state X(t.) = O, in other words, the LTI system istaken on the zero-state situation. +d[e-A(t-to) X(t)] = e-4(t-to) Bu(t)dtTaking definite integral on both side, we have" d[e-4(-) X(t)] = e-A(-to) X(t) = [ e-A(r-o) Bu(t)d tToX(t)= [ eA(t-) Bu(t)dtTherefore,Obviously, it is the second term on the right-hand side and it canbe called the forced response or zero-state response of the LTIsystem

2.4 Time Response of the LTI System

2.4 Time Response of the LTI SystemConsider the LTI system with the initial condition X(t.)X(t) = AX(t) + Bu(t)the solution can also be written as the more general cased(t -t). B.u(t) .dtX(t) = Φ(t - t)X(t.)+zero-inputresponsezero-state response4(t-t) .B .u(t).dtX(t) = e4(t-to) X(t.) +

2.4 Time Response of the LTI System

Example 2.6 Determine the solution of the LTI systemdescribedby+[o]01X:X+t≥0X(0) =[x;(0)x2(0)]一u-2where u(t) =l(t) is the unit step functionSolutionFrom Example 2.1 we have obtained that2e'-e-2t2eyeΦ(t) = eAr -e'" + 2e-2tL-2e-+2e-2tTherefore, the solution of the LTI system can also be calculatedX(t) = @(t)X(O) + Φ(t -t)Bu(t)dt

Example 2.6 Determine the solution of the LTI systemdescribedby +[o0XX+t≥0X(0) =[x;(0)x2(0) ]u-2['Φ(t -t)Bu(t)dtSolutionX(t) = @() X(O) +2et - e-2te-t - e-2tx (0)-2e- +2e-2t-e-* + 2e-2 /Lx2(0)2e-(t-t) -e-2(t-t)0-2(t-z)0e-(t-t)dte-(t-t) +2e-2(t-t)-2(t-t)o-(t-t)+ 2e1201-1(2e-t - e-2t)x (0) + (e- - e-2)x2 (0)dt20-2(t-t-(-2e-t + 2e-2t)x (0) +(-e-t + 2e-2t)x, (0)

Example 2.6 Determine the solution of the LTI systemdescribedby[o0XX+t≥0X(0) =[x;(0)x2(0)]u-2'@(t -t)Bu(t)dtSolutionX(t) = @(t) X(O) +-2(t-t)(2e-t - e-2f)x (0) +(e-t -e-2f)x2 (0)(-2e- + 2e-2t)x (0)+(-e- +2e-2t)x (0)1(2e-t -e-2t)x (0)+(e-t -e-2t)x, (0)22(-2e-t + 2e-2t)x (0) +(-e-t + 2e-2t)x, (0)21e121+[2x (0) +x (0) -1]e-t -[x (0) +x (0) -2-[2x (0) +x2(0) -1]e-t +[2x (0) + 2x, (0) - 1]e-2t