利用拉氏变换的性质求下列函数的拉氏变换
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/13 00:44:25
利用拉氏变换的性质求下列函数的拉氏变换
(1)=L[t^2]+3L[t]+2L[1]
=2/s^3+3/s^2+2/s
(2)=L[1]-L[e^(-t)*t]
=1/s-F[s+1]
F(s)=L[t]=1/s^2
=1/s-1/(s+1)^2
(3)=5L[sin2t]-3L[cos2t]
=5/2*F1(s/2)-3/2*F2(s/2)
F1=L[sint]=1/(s^2+1)
F2=L[cost]=s/(s^2+1)
=(5/2)[1/((s/2)^2+1)]-(3/2)[(s/2)/((s/2)^2+1)]
(4)=L[e^(at)*(1-cos2t)/2]
=(1/2){L[e^(at)]-L[e^(at)cos2t]}
=(1/2){1/(s-a)-F(s-a)}
F(s)=L[cos2t]=s/(s^2+4)
=(1/2)[1/(s-a)-(s-a)/[(s-a)^2+4]]
(5)我只看出来用级数法
e^(-at)=sum (-at)^n/n!
(1-e^(-at))/t=-sum (-a)^n t^(n-1)/n!
L[(1-e^(-at))/t]=sum [(-1)^(n-1)a^n/n!]L[t^(n-1)]
=sum [(-1)^(n-1)a^n/n!]*(n-1)!/s^n
=sum (-1)^(n-1)[(a/s)^n/n]
=ln(a/s)
=2/s^3+3/s^2+2/s
(2)=L[1]-L[e^(-t)*t]
=1/s-F[s+1]
F(s)=L[t]=1/s^2
=1/s-1/(s+1)^2
(3)=5L[sin2t]-3L[cos2t]
=5/2*F1(s/2)-3/2*F2(s/2)
F1=L[sint]=1/(s^2+1)
F2=L[cost]=s/(s^2+1)
=(5/2)[1/((s/2)^2+1)]-(3/2)[(s/2)/((s/2)^2+1)]
(4)=L[e^(at)*(1-cos2t)/2]
=(1/2){L[e^(at)]-L[e^(at)cos2t]}
=(1/2){1/(s-a)-F(s-a)}
F(s)=L[cos2t]=s/(s^2+4)
=(1/2)[1/(s-a)-(s-a)/[(s-a)^2+4]]
(5)我只看出来用级数法
e^(-at)=sum (-at)^n/n!
(1-e^(-at))/t=-sum (-a)^n t^(n-1)/n!
L[(1-e^(-at))/t]=sum [(-1)^(n-1)a^n/n!]L[t^(n-1)]
=sum [(-1)^(n-1)a^n/n!]*(n-1)!/s^n
=sum (-1)^(n-1)[(a/s)^n/n]
=ln(a/s)