# Validity of applying Heaviside function for signal processing applications

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.

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.

• 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. Commented Apr 29, 2020 at 16:13
• @RodrigodeAzevedo Ahh, yes, of course. The graph confused me as to the obvious. Commented Apr 30, 2020 at 6:29