# 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 instead of as we saw in the examples above. In that case, it would be more convenient to define our as

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

Definition 1:

So we need to modify the definition to take into account the fact that our signal begins at instead of 0.

What we do is define our Laplace Transform as:

Definition 2:

where the lower limit is now instead of 0. In this way, we can work with a different initial time (for example )

For illustration, consider .

Laplace Transforms of signals which begin at 0 are the same for both definitions. For example, the Laplace Transform of is according to both definitions.

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

In fact, the the delay property

also holds for negative time-shifts of with when we use definition (2).

Our choice of 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 instead of 0. In any problem, if we encounter signals which can be expressed as , the Laplace Transform can be taken to be

where .

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

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Next: Finding Fourier Series Coefficients Up: ee102 Previous: Examples

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Sankaran Panchapagesan
2002-06-05