Question

how to find the laplace of the H(1-t).

Answer 1

\[\int\limits_{0}^{ \infty } e ^ {-st} H (t-1) dt = \int\limits_{0}^{1} e ^ {-st} \times 0 \ dt + \int\limits_{1}^{\infty} e ^ {-st} \times 1 \ dt\]

Answer 2

can u explain how u take 0 and 1.

Answer 3

"can u explain how u take 0 and 1."

not totally sure i get you exact question but firstly this is the transform of the H function on its own.

secondly, H(t-1) is zero until t = 1 and then H = 1 thereafter.

so you can just split the interval of integration.

hope that helps

Answer 4

my question is H( 1-t) not H(t-1).

Answer 5

crumbs, i see :(

H (-x) = 1 - H(x)
H(1-t) = 1 - H(t-1)

http://math.stackexchange.com/questions/122604/negated-argument-of-the-heaviside-step-function

so switch the 0 and the 1 around in my incorrect solution.

i now get ( - e^(-s) + 1 ) / s

Answer 6

thank u so much.