EE102 Systems and Signals (Discussion)

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Laplace Transforms of signals that begin at a negative time

In some problems (for example Problem 4.10 from the text), the periodic signal of interest is defined naturally over $-\frac {T}{2} < t < \frac {T}{2}$ instead of $0 < t < T$ as we saw in the examples above. In that case, it would be more convenient to define our $f_1(t)$ as

$$f_1(t) = \begin{cases} f(t) & \text{if $-\frac{T}{2} \leq t \leq \frac{T}{2}$.} \\ 0 & \text{otherwise} \end{cases}$$

$$= f(t)[u(t+\frac{T}{2}) - u(t-\frac{T}{2})]$$

But in Chapter 3, the Laplace Transform was defined as:

Definition 1:  $$F(s) = \int_0^\infty f(t) e^{-s t} dt$$

So we need to modify the definition to take into account the fact that our signal $f_1(t)$ begins at $-T/2$ instead of 0.

What we do is define our Laplace Transform as:

Definition 2:  $$F(s) = \int_{t_0}^\infty f(t) e^{-s t} dt$$

where the lower limit is now $t_0$ instead of 0. In this way, we can work with a different initial time (for example $t_0 = -\frac{T}{2}$)

For illustration, consider $t_0 = -1$.

Laplace Transforms of signals which begin at 0 are the same for both definitions. For example, the Laplace Transform of $u(t)$ is $1/s$ according to both definitions.

But with definition (2), we can also analyze signals which begin at time $-1$ instead of 0. For example, if we use definition (1), $u(t+1)$ would have the same Laplace Transform $1/s$ as $u(t)$, since the lower limit 0 in definition (1) ignores the signal $u(t+1)$ between $-1$ and 0. But if we use definition (2), the Laplace Transform of $u(t+1)$ is $e^s/s$.

In fact, the the delay property

$${\cal L}[f(t-t_d)u(t-t_d)] = e^{-s t_d} F(s)$$

also holds for negative time-shifts of $t_d$ with $0 \geq t_d \geq -1$ when we use definition (2).

Our choice of $-1$ in the lower limit of the definition of the Laplace Transform was used just to illustrate the case when the signals of interest began at $-1$ instead of 0. In any problem, if we encounter signals which can be expressed as $f(t+t_0)u(t+t_0)$, the Laplace Transform can be taken to be

$${\cal L}[f(t+t_0)u(t+t_0)] = e^{+s t_0} F(s)$$

where ${\cal L}[f(t)u(t)] = F(s)$.

The only case where the Laplace Transform cannot be directly used is when the signals go all the way back to $-\infty$. In that case, there is no delay $t_0$ that can make the signals 0 for $t \leq 0$.

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