2.求下列函数的Laplace逆变换(象原函数),并用另一种方法加以验证.-|||-(1) (s)=dfrac (1)({s)^2+(a)^2} ;-|||-(2) (s)=dfrac (s)((s-a)(s-b)) ;-|||-(3) (s)=dfrac (s+c)((s+a){(s+b))^2} ;-|||-(4) (s)=dfrac ({s)^2+2(a)^2}({({s)^2+(a)^2)}^2} :-|||-(5) (s)=dfrac (1)(({s)^2+(a)^2)(s)^3} ;-|||-(6) (s)=dfrac (1)(s(s+a)(s+b)) ;-|||-(7) (s)=dfrac (1)({s)^4-(a)^4} ;-|||-(8) (s)=dfrac ({s)^2+2s-1}(s{(s-1))^2} =-|||-(9) (s)=dfrac (1)({s)^2((s)^2-1)} ;-|||-(10) (s)=dfrac (s)(({s)^2+1)((s)^2+4)} -

求函数 F(s)= (s)/((s+1)(s+2)) 的拉普拉斯逆变换A. 以上都不对B. $2e^{-2t} - e^{-t}$C. $e^{-2t} - e

已知象函数 F(s)= (s+1)/(s(s+2))，则原函数 f(t)= L^-1(F(s)) 为 ()。A. $\frac{1}{2} + \frac{1}

单边拉氏变换F(s)=(e^-s)/(s^2)+1的原函数为（）A. sin(t-1)u(t-1)B. sin(t-1)u(t)C. cos(t-1)u(t-1

[题目]求 (s)=(s+4)/(2(s)^2+3s+1) 的拉普拉斯-|||-反变换-|||-(s)=dfrac (s+4)(2{s)^2+3s+1}

(B) (n-1)(S)^2+(overline {X)}^2 (C) (S)^2+(overline {X)}^2. (D) dfrac (n-1)(n)(S

已知 (s)=dfrac (1)((s-5)(s+4))则它的拉普拉斯逆变换为( )已知 (s)=dfrac (1)((s-5)(s+4))已知 (

f(t)的波形如图所示, 则F(s)=_____。1 ---|||-t-|||-(}^2[ dfrac {1)(2)-dfrac (1)(2)(e)^-2s-s

脱吸因数法（平衡线为直线）：_(OG)=dfrac (1)(1-S)ln [ (1-S)dfrac ({Y)_(1)-({X)_(2)}^3}({Y)_(2)-

dfrac (1)(2)(S)_(8)^2-+(e)^-arrow (S)_(4)^2- (Na)^++dfrac (x)(4)(S)_(4)^2-+2(1-d

统的输出;m,和m 2为质量,y1(t )为m1位移,k1、k2为弹簧刚度系数,-|||-2-3 求下列函数的拉氏反变换。-|||-(1) (s)=dfrac