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Definitions and Some Properties

  Consider a signal f(t), a function of time t (e.g. a current or voltage waveform), for positive times t. The Laplace Transform of f(t) is a function F(s), where $s = \sigma + j \omega$ is a complex variable, defined by

$$F(s) = {\cal L}[f](s) = \int_{0^-}^\infty e^{-st} f(t) dt$$ (19)

We usually use capital letters to denote Laplace transforms. The complex variable $s = \sigma + \omega$ is called complex frequency.   (Note that for AC steady state analysis, we use $s=j\omega$ and don't use $\sigma$; see §4.1.)

In general there is a correspondance

$$f(t) \longleftrightarrow F(s)$$

and it is possible to find f(t) from F(s). These are called Laplace Transform Pairs; an important example is

$$e^{-at} \longleftrightarrow \frac{1}{s+a} .$$

So if F(s) = 2/(s+3), then f(t)= 2e^-3t for $t \geq 0$.

The Laplace transform is linear, which means that

$$\begin{array}{rl} {\cal L}[af(t) + bg(t) ] & = a {\cal L}[ f(... ...[ g(t)] \\ \\ {\rm or} & \\ & = a F(s) + b G(s) . \end{array}$$

So the Laplace transform of a superposition is the superposition of the individual Laplace transforms.

We have seen that Ohm's Law for capacitors and inductors involves derivatives. Let's see how derivatives transform:

$${\cal L}[\frac{df}{dt}](s) = s {\cal L}[f(t)](s) - f(0^-) .$$ (20)

The Laplace transform of a derivative is s times the Laplace transform of the function plus the initial condition f(0^-). So if f(t)= 2e^-3t for $t \geq 0$, then ${\cal L}[f^\prime(t)](s)=-6/(s+3)$ (check this both ways!).

Electric circuits often have switches, and we can model switches using a special function called the unit step function u(t) defined by (see Figure 22)

$$u(t) = \left\{ \begin{array}{rl} 0 & \ \ {\rm if} \ \ t \leq 0 \\ 1 & \ \ {\rm if} \ \ t > 0 . \end{array} \right.$$ (21)

  

Figure 22: Unit step function.

Exercise. Show that

$${\cal L}[u(t)](s) = U(s) = \frac{1}{s} .$$

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