Let u(t) be a function defined on 0 \le t < \infty. The Laplace transform of u(t) is U(s) defined as:
\mathcal{L}[u] = U(s) = \int_0^\infty u(t) e^{-st} dt
Assuming the improper integral exists.
| u(t) | U(t) |
|---|---|
| 1 | \frac{1}{s} for s > 0 |
| t | \frac{1}{s^2} for s > 0 |
| e^{at} | \frac{1}{s - a} for s > a |
| h_a(t) | \frac{1}{s} e^{-as} for s > 0 |
| h_a(t)u(t-a) | U(s)e^{-as} |
| \delta_a(t) | e^{-as} |
| | | |
| | |
Useful for defining functions of several formulae as a single formula
h_a(t) = \begin{cases} 0 & \text{if} & t < a \\ 1 & \text{if} & t \ge a \\ \end{cases}
Example:
f(t) = \begin{cases} t^2 & \text{if} & 0 \le t < 1 \\ 1 & \text{if} & 1 \le t < 2 \\ t - 1 & \text{if} & 2 \le t \\ \end{cases}
Can be expressed as f(t) = t^2h_0(t) + (1-t^2)h_1(t) + (t-2)h_2(t)
If the function u(t) is piecewise continuous and there exists a constant M > 0 and \alpha such that |u(t)| \le Me^{\alpha t} then u(t) has a Laplace transform.
(u * v)(t) = \int_0^t u(\tau)v(t - \tau) d\tau
When u and v are continuous on [0, \infty) then (u * v)(t) = (v * u)(t)
Let a > 0, \epsilon > 0
b_{a,\epsilon}(t) = \begin{cases} \frac{1}{\epsilon} & \text{for } a - \frac{\epsilon}{2} \le t < a + \frac{\epsilon}{2} \\ 0 & \text{otherwise} \end{cases} = \frac{1}{\epsilon}(h_{a - \frac{\epsilon}{2}}(t) - h_{a+\frac{\epsilon}{2}}(t))
Delta function is a generalized function:
\delta_a(t) = \lim_{\epsilon \to 0} b_{a, \epsilon}(t)