结论「零输入响应Zero-input Response]线性时不变齐次系统X(t) = AX(t), X(to) = X(0)(2.2)的零输入响应，具有如下形式：(2.9)AtX(t) =X(0)e如果是更一般的初始条件X(t.)，系统(2.2)的零输入响应为eA(t-to)X(t) =X(to)(2.10)
结论 [零输入响应Zero-input Response] 线性时不变齐次系统 0 (2.2) X AX ( ) ( ), ( ) (0) t t X t X = = 的零输入响应，具有如下形式： X( ) X(0) At t = e (2.9) 如果是更一般的初始条件 , 系统(2.2) 的零输入响应为： ( ) ( ) 0 ( ) 0 t e t t t X X A − = (2.10) 0 X t( )
2.3Calculation of the Matrix Exponential Function矩阵指数函数的计算方法1（定义法）：1YAA't+=At+..=I+At+P2!3!优点：直观缺点：计算量大、级数有无穷多项
➢矩阵指数函数的计算 2.3 Calculation of the Matrix Exponential Function
方法2拉普拉斯变换法（Laplace transformmethod）Consider thehomogeneous state equationX(t) = AX(t)Taking the Laplace transformsX(s) - X(O) = AX(s)北(sI - A)X(s) = X(O)+X(s) = (sI - A)-1 X(O)Taking the inverse Laplace transformX(t) = L-"[(sI - A)-"JX(O)1eAt = L-'[(sI - A)-"]
方法2 拉普拉斯变换法（Laplace transform method） Consider the homogeneous state equation X(t) = AX(t) Taking the Laplace transform sX(s) − X(0) = AX(s) (sI − A)X(s) = X(0) ( ) ( ) (0) 1 X I A X − s = s − Taking the inverse Laplace transform ( ) [( ) ] (0) 1 1 X I A X − − t = L s − [( ) ] −1 −1 = I − A A e L s t
Example 2.1 Calculate e 4f by using the Laplace transform method重-32SolutionThecharacteristicmatrix and its inversematrix are calculated as(sI -A)S+31S+3and(sIAs(s + 3) +2-2S1$+3(s +1)(s +2)(s +1)(s + 2)-2S(s +1)(s + 2)(s +1)(s +2)
Taking the inverse Laplace transform, the matrix exponential functioncan be obtained aseAt = L-'[(sI -A)-I](s + 3)21+(s + 1)(s + 2)(s + 1)(s + 2)$+2s+2s+1S+1L-1= L-122-2-2-1sS+2S+2Ls+1s+1(s + 1)(s + 2)(s + 1)(s + 2) eae~2ts-a2e-2e- +2e-21-e-' +2e-21
1 1 [ ] at L e s a − = − Taking the inverse Laplace transform, the matrix exponential function can be obtained as [( ) ] −1 −1 = I − A A e L s t + + + + + + + − + + + − + + = + + + + − + + + + + = − − 2 2 s 1 -1 2 2 s 1 - 2 2 1 s 1 1 2 1 s 1 2 (s 1)(s 2) (s 1)(s 2) 2 (s 1)(s 2) 1 (s 1)(s 2) (s 3) 1 1 s s s s L s L − + − + − − = − − − − − − − − t t t t t t t t e e e e e e e e 2 2 2 2 2 2 2 2