Determine the Laplace transform for the following:
$$ f(t) = u(t) $$ If you are rusty (unfamiliar) with the unit step function (Heaviside function), it is defined as follows:
$$ u_c(t) = \left\{ \begin{array}{ll} 0 & t \lt c \\ 1 & t \ge c \end{array} \right. \qquad (Definition \; 1) $$
and is somtimes defined via the following form (which is equivalent to definition #1):
$$ u(t-t_0) = \left\{ \begin{array}{ll} 0 & t \lt t_0 \\ 1 & t \ge t_0 \end{array} \right. \qquad (Definition \; 2) $$
In our example "c" is not annotated and therefore: $$ c=0 $$ The graph of this unit step function appears as follows:
From examination of the graph we see that: $$ f(t) = 1 $$ Recall the definition of the Laplace transform (equation #1 from the previous page): $$ \mathcal{L}(f(t)) = F(s) = \int_{0^-}^{\infty} f(t)e^{-st} dt $$ Plugging our value for f(t) into this definition gives us: $$ F(s) = \int_{0^-}^{\infty} 1e^{-st} dt $$ $$ \qquad = \Big( \frac{-1}{s}e^{-st} \Big) \Big|_0^{\infty} $$ $$ \qquad = \Big( \frac{-1}{se^{st}} \Big) \Big|_0^{\infty} $$ $$ \qquad = \Big( \frac{-1}{se^{s\infty}} - \frac{-1}{se^{s(0)}} \Big) $$ $$ \qquad = \Big( 0 - \frac{-1}{s} \Big) $$
$$ F(s) = \frac{1}{s} $$