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