I wasn't sure if this question was more suitable for math.stackexchange, but I suspect it's more-so a signal processing question (albeit, a theoretical one) than a mathematical one.

I am currently studying the textbook An Introduction to Laplace Transforms and Fourier Series, second edition, by Phil Dyke. Chapter 2.1 Real Functions describes Heaviside's unit step function as follows:

Sometimes, a function $F(t)$ represents a natural or engineering process that has no obvious starting value. Statisticians call this a time series. Although we shall not be considering $F(t)$ as stochastic, it is nevertheless worth introducing a way of "switching on" a function. Let us start by finding the Laplace transform of a step function the name of which pays homage to the pioneering electrical engineer Oliver Heaviside (1850 - 1925). The formal definition runs as follows.

Definition 2.1 Heaviside's unit step function, or simply the unit step function, is defined as

$$H(t) = \begin{cases} 0 & t < 0, \\ 1 & t \ge 0. \end{cases}$$

Since $H(t)$ is precisely the same as $1$ for $t > 0$, the Laplace transform of $H(t)$ must be the same as the Laplace transform of $1$, i.e., $1/s$. The switching on of an arbitrary function is achieved simply by multiplying it by the standard function $H(t)$, so if $F(t)$ is given by the function shown in Fig. 2.1 and we multiply this function by the Heaviside unit step function $H(t)$ to obtain $H(t)F(t)$, Fig 2.2 results. Sometimes it is necessary to define what is called the two sided Laplace transform

$$\int_{-\infty}^\infty e^{-st} F(t) \ dt,$$

which makes a great deal of mathematical sense. However, the additional problems that arise by allowing negative values of $t$ are severe and limit the use of the two sided Laplace transform. For this reason, the two sided transform will not be pursued here.

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What I'm having difficulty understanding is how this procedure is valid from a signal processing perspective. Mathematically, we can see that by applying the unit step function, all values of the function for $t < 0$ become $0$. This is valid from a mathematical perspective, but it seems to eliminate all of the information associated with values $t < 0$, which leads me to wonder, since the unit step function is used for signal processing applications, how this is valid from a signal processing perspective? Couldn't values of the function for $t < 0$ contain valuable information, and aren't we deleting this information when applying the Heaviside function to this function?

I would greatly appreciate it if people would please take the time to explain this.

  • $\begingroup$ Note that the Heaviside step is very useful to write impulse responses. Assuming causality, there should be no signal for $t < 0$. Hence, no useful information is lost. $\endgroup$ Commented Apr 29, 2020 at 16:13
  • $\begingroup$ @RodrigodeAzevedo Ahh, yes, of course. The graph confused me as to the obvious. $\endgroup$ Commented Apr 30, 2020 at 6:29


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