Laplace Transforms: Explaining Unit Step Function

[aspiring_gal]

EXPLAIN LAPLACE TRANSFORM OF UNIT STEP FUNTION?

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

 

[Cyosis]

The unit step function is defined as::

<br /> u(t)=\begin{cases} 0, &amp; t &lt; 0 \\ 1, &amp; t \ge 0 \end{cases}<br />

Now take the Laplace transform.

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

Because on the interval 0 \leq x &lt; \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&lt;x&lt;0 the unit step function is 0.

 

[HallsofIvy]

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

u(t-a)=\begin{cases} 0, &amp; t &lt; a \\ 1, &amp; 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}.