 # Intro to the Laplace Transform (example problem)

### Determine the Laplace transform for the following:

$$u_c(t) = \left\{ \begin{array}{ll} 0 & t \lt c \\ 1 & t \ge c \end{array} \right. \qquad (Definition \; 1)$$

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: $$F(s) = \int_{0^-}^{\infty} 1e^{-st} dt$$ $$\qquad = \Big( \frac{-1}{s}e^{-st} \Big) \Big|_0^{\infty}$$ $$\qquad = \Big( \frac{-1}{se^{st}} \Big) \Big|_0^{\infty}$$ $$\qquad = \Big( \frac{-1}{se^{s\infty}} - \frac{-1}{se^{s(0)}} \Big)$$ $$\qquad = \Big( 0 - \frac{-1}{s} \Big)$$

$$F(s) = \frac{1}{s}$$