求函数的拉氏反变换:X(s)=(s+2)/[s·(s+1)^2·(s+3)]我的解法如下:X(s)=A/s+B/(s+1)^2+C/(s+3)A=(s+2)/[(s+1)^2·(s+3)] |(s=0) =2/3 ,B=(s+2)/[s·(s+3)] |(s=-1) =-1/2 ,C=(s+2)/[s·(s+1)^2] |(s=-3) =1/12 ,所以X(s)=(2/3) / s+(-
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![求函数的拉氏反变换:X(s)=(s+2)/[s·(s+1)^2·(s+3)]我的解法如下:X(s)=A/s+B/(s+1)^2+C/(s+3)A=(s+2)/[(s+1)^2·(s+3)] |(s=0) =2/3 ,B=(s+2)/[s·(s+3)] |(s=-1) =-1/2 ,C=(s+2)/[s·(s+1)^2] |(s=-3) =1/12 ,所以X(s)=(2/3) / s+(-](/uploads/image/z/11519818-34-8.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0%E7%9A%84%E6%8B%89%E6%B0%8F%E5%8F%8D%E5%8F%98%E6%8D%A2%EF%BC%9AX%EF%BC%88s%EF%BC%89%3D%28s%2B2%29%2F%5Bs%C2%B7%EF%BC%88s%2B1%EF%BC%89%5E2%C2%B7%EF%BC%88s%2B3%EF%BC%89%5D%E6%88%91%E7%9A%84%E8%A7%A3%E6%B3%95%E5%A6%82%E4%B8%8B%EF%BC%9AX%28s%29%3DA%2Fs%2BB%2F%28s%2B1%29%5E2%2BC%2F%28s%2B3%29A%3D%28s%2B2%29%2F%5B%28s%2B1%29%5E2%C2%B7%28s%2B3%29%5D+%7C%28s%3D0%29+%3D2%2F3+%2CB%3D%28s%2B2%29%2F%5Bs%C2%B7%28s%2B3%29%5D+%7C%28s%3D-1%29+%3D-1%2F2+%2CC%3D%28s%2B2%29%2F%5Bs%C2%B7%28s%2B1%29%5E2%5D+%7C%28s%3D-3%29+%3D1%2F12+%2C%E6%89%80%E4%BB%A5X%28s%29%3D%282%2F3%29+%2F+s%2B%28-)
求函数的拉氏反变换:X(s)=(s+2)/[s·(s+1)^2·(s+3)]我的解法如下:X(s)=A/s+B/(s+1)^2+C/(s+3)A=(s+2)/[(s+1)^2·(s+3)] |(s=0) =2/3 ,B=(s+2)/[s·(s+3)] |(s=-1) =-1/2 ,C=(s+2)/[s·(s+1)^2] |(s=-3) =1/12 ,所以X(s)=(2/3) / s+(-
求函数的拉氏反变换:X(s)=(s+2)/[s·(s+1)^2·(s+3)]
我的解法如下:
X(s)=A/s+B/(s+1)^2+C/(s+3)
A=(s+2)/[(s+1)^2·(s+3)] |(s=0) =2/3 ,
B=(s+2)/[s·(s+3)] |(s=-1) =-1/2 ,
C=(s+2)/[s·(s+1)^2] |(s=-3) =1/12 ,
所以
X(s)=(2/3) / s+(-1/2) / (s+1)^2+(1/12) / (s+3) .
故:
X(t)=2/3 - (1/2)t·e^(-t) + (1/12)·e^(-3t) .
目前为止我这个答案不错的,但是,跟参考答案就差了一个项:(-3/4)·e^(-t)
为什么呢?
求函数的拉氏反变换:X(s)=(s+2)/[s·(s+1)^2·(s+3)]我的解法如下:X(s)=A/s+B/(s+1)^2+C/(s+3)A=(s+2)/[(s+1)^2·(s+3)] |(s=0) =2/3 ,B=(s+2)/[s·(s+3)] |(s=-1) =-1/2 ,C=(s+2)/[s·(s+1)^2] |(s=-3) =1/12 ,所以X(s)=(2/3) / s+(-
首先,一开始就存在误区,s+1这个分母是二阶的,按照你的分解式,B应该是B=K(s+1)+H,其中K,H都是实数,(或者说你应该写成这样:X(s)=A/s+K/(s+1)+H/(s+1)^2+C/(s+3),)
其次,“B=(s+2)/[s·(s+3)] |(s=-1) =-1/2”这种算法,实际上计算出的是H=-1/2,而没有算出K,
而K=(X(s)-H/(s+1)^2)(s+1)|(s=-1)
然后你再做反变换就行了
关键要注意分母是二阶的
ps:实际上你在求A、B、C时是在做极限运算,例如求A
A=lim S->0(X(s)*s)),
而对于分母是二次方的式子,一定要考虑一阶的可能性,