# Laplace tranforms

EXPLAIN LAPLACE TRANSFORM OF UNIT STEP FUNTION???

i.e L{u(t)} = 1/s

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Cyosis
Homework Helper
The unit step function is defined as::

$$u(t)=\begin{cases} 0, & t < 0 \\ 1, & t \ge 0 \end{cases}$$

Now take the Laplace transform.

$$L[u(t)]=\int_0^\infty u(t) e^{-st} dt=\int_0^\infty 1*e^{-st} dt$$

Because on the interval $0 \leq x < \infty, u(t)=1$.

You should be able to work it out now.

The same holds for the two-sided Laplace transform, because on the interval $-\infty<x<0$ the unit step function is 0.

HallsofIvy
Homework Helper
Perhaps what you really want is L(u(t-a)).

$$u(t-a)=\begin{cases} 0, & t < a \\ 1, & t \ge a \end{cases}$$

Then
$$L(u(t-a))= \int_0^\infty u(t-a)e^{-st} dt= \int_a^\infty e^{-st}dt$$
Let v= t- a. Then t= v+ a, dv= dt, when t= a, v= 0, and when t= $\infty$, v= $\infty$. The integral becomes
$$L(u(t-a)= \int_0^\infty e^{s(v+a)}dv= \int_0^\infty e^{sv}e^{sa}dv$$
$$= e^{sa}\int_0^\infty e^{sv}dv= e^{sa}e^{-st}dt$$
and that last integral is the Laplace transform of 1.

That u(t-a) step function multiplying a function basically multiplies the Laplace transform of the function by $e^{as}$.