Question

F(s)=s/(s+1)(s²+4s+13)求f(t)

Answer 1

问：F(s)=s/(s+1)(s²+4s+13)求f(t)

答：你好，答案是：f(s)=2s+1/2*(1/(s+4)-1/(s+6)

问：过程有吗

答：设s/(s+1)²(s+2)=a/(s+1)+b/(s+1)²+c/(s+2)将右边通分，分子为(a+c)s²+(3a+b+2c)s+(3a+2b+c)比较两端系数，有a+c=03a+b+2c=12a+2b+c=0把c=-a代入得a+b=1a+2b=0解得a=2，b=-1，c=-2所以F(s)=2/(s+1)-1/(s+1)²-2/(s+2)自己去套拉氏反变换。