Consider the function ft 5 0 lessthanorequalto t lessthanor

Consider the function f(t) = {5, 0 lessthanorequalto t lessthanorequalto 2 -3, 2 lessthanorequalto t lessthanorequalto 5; -4, t greaterthanorequalto 5 Write the function in terms of unit step function f(t) =. (Use step(t-c) for u_c(t).) Find the Laplace transform of f(t) F(s) = .

Solution

1. in terms of unit step function :

f(t) = [ 5[u(t-0) - u(t-2)] -3[u(t-2) +u(t-5)] -4[ u(t-5)] ]

= 5u0t -5u2t -3u2t -3u5t -4u5t [ note : uot is u(t-0) , u5t is u(t-5) and so on .......]

=>5u0t - 8u2t -8u5t

2. now we know that

uot = 1/s

u2t = e^(-2s)/s

u5t = e^(-5s)/s

So the Laplace is

F(S) = 5u0t - 8u2t -8u5t

=> 5* 1/s -8e^(-2s)/s - 8e^(-5s)/s