A question arising from some misunderstanding involving the Laplace Transform of the Heaviside function

Question

I am trying to compute the Laplace Transform of the Heaviside Step-function. I define the Heaviside step-function with the half-maximum convention:

$H(t-t_0) = 0$ for $t < t_0$ ; $H(t-t_0) = 1/2$ for $t=t_0$ ; $H(t-t_0) = 1$ for $t > t_0$.

Performing the integrations necessary to find the Laplace transform of the Heaviside step-function, I found that: (I use the symbol $L$ to denote the laplace transform.)

$L[ 0 ] = 0 $ ; $L[1/2] = \frac{1}{2s} $ ; $L[1] = \frac{1}{s} $ .

I thought that this is the correct Laplace Transform for the different intervals on which the Heaviside Step-function is defined. However, my book mentions that: $$L [ H(t - t_0) f(t - t_0) ] = e^{-t_0 s } F(s) , $$ where $F(s) = L [f(t) ] $. So if we look at the interval in which $t > t_0$, we find that $$L [ H(t - t_0) f(t - t_0) ] = \frac{ e^{-t_0 s } }{s} .$$

This doesn't agree with my calculations. Can you please point who goes wrong, and where and how?

Answer 1

The unilateral Laplace transform $F(s)$ of a function $f(t)$ is defined by

$$F(s)=\int_{0}^{\infty}f(t)e^{-st} \,dt$$

For the shifted step function $H(t-t_0)$ this gives

$$F(s)=\int_{0}^{\infty}H(t-t_0)e^{-st} \,dt=\int_{t_0}^{\infty}e^{-st} \,dt=\frac{e^{-st_0}}{s}$$

So I guess your book is correct.

Answer 2

Your assumption about the value o the Laplace transform is wrong. Because the step function is not nonZero until $t=t_0$, the integral defining the LT does not have its lower limit at $0$, but rather $t_0$.