# find f(t) for L{f(t)}={((1-e^s)(1+e^2s))/s^2}?

• find f(t) for L{f(t)}={((1-e^s)(1+e^2s))/s^2}?

Answer #1 | 23/12 2013 19:59
I'll write it as L{f(t)} = 1/s^2 + e^(2s)/s^2 - e^s/s^2 - e^(3s)/s^2 In the Laplace Transform table in Pauls Online Math Notes, formula #27 says that e^(-cs) F(s) is the the Laplace transform of u(t-c) f(t-c), where u(t-c) is the unit step function at t=c. And formula #3 implies that 1/s^2 is the transform of f(t) = t. So in your problem, f(t) seems to be t + (t+2)u(t+2) - (t+1)u(t+1) - (t+3)u(t+3) It's not obvious what the meaning of the f values on the negative domain might be, since the Laplace transform operates on f(t) only from 0 to +infinity. The expression I wrote for f(t) is uniformly equal to -1 on the interval (0,infinity).
Positive: 46 %

Math 2280 - Assignment 10 ... L(f(t)) = 1 s2 − 2e−s s2 + e−2s s2 = 1− 2e−s +e−2s s2 = (1− e−s)2 s2. 11.
Positive: 46 %
F(s) = e 2s 1 s2 + s 2 ... Find the inverse Laplace transform of G(s) = 2e 2s(s 1) ... 1 e s Put this into the formula: L(f(t)) = 1 s (1 e s) 1 e 2s
Positive: 41 %
F(s) = 1 e 2s s2: We have L 1(F(s)) = L 1 ... e 2s s2) = t u 2(t)(t 2): Example 4. Find the inverse transform of G(s) = 1 s2 4s+ 5 ... November 1, 2013 15 ...