Not understanding this integral of unit step

Question

Singular functions are extensively used in signals for EE, I'm trying to not get lost in the initial parts of the book I'm following. However, I cant understand how/why the author uses $\mu$ as a function of $\tau$ initially. This often appears in singular functions defined through integrals. Can someone help me please?

Answer 1

Unit step function is defined as: $$ u(t) = \cases{1 & if $t\ge0$ \\ 0 & if $t<0$} $$

Using this definition helps when solving integrals because allows to split domain of integration in a simple way and your example is one of those; $u(\tau - t_0)$ means that the value is $0$ when $\tau < t_0$ and $1$ when $\tau \ge t_0$ so when you solve the integral you write this: $$ \int_0^t u(\tau - t_0) d\tau = \int_0^{t_0} u(\tau-t_0) d\tau + \int_{t_0}^t u(\tau-t_0) d\tau = \int_0^{t_0} 0 \cdot d\tau + \int_{t_0}^t 1 \cdot d\tau $$ Even changing the variable of integration $\tau \rightarrow T = \tau - t_0$ lead to: $$ \int_0^t u(\tau - t_0) d\tau = \int_0^{t_0} u(\tau-t_0) d\tau + \int_{t_0}^t u(\tau-t_0) d\tau = \int_{-t_0}^{0} u(T) \cdot dT + \int_{0}^{t-t0} u(T) \cdot dT $$

Answer 2

If I'm understanding your confusion correctly, $\tau$ is what's called a "dummy variable" in this integration. You're probably expecting something a bit more like $$ f(t) = \int_{t_0}^{\color{red} t} u({\color{blue} t} - t_0) \, d{\color{blue} t}. $$ However, this notation is problematic, since the symbol $t$ is playing two roles in this expression: one as the variable we're integrating over ("inside" the integral, colored blue) and one as the limit of the integration (colored in red.) This can lead to confusion; for example, if I want to take the derivative of $f(t)$ $$ \frac{df}{dt} = \frac{d}{dt} \left[ \int_{t_0}^{\color{red} t} u({\color{blue} t} - t_0) \, d{\color{blue} t} \right], $$ how does this derivative with respect to $t$ act on the "blue" $t$'s in the expression in brackets?

The answer is that it doesn't, since the variable of integration is a "dummy" — it doesn't appear in $f(t)$ once you've evaluated the integral. So to remind ourselves that these $t$'s are playing different roles, we use a different symbol for the "dummy" variable: $$ f(t) = \int_{t_0}^{t} u(\tau - t_0) \, d\tau. $$ If we then try to take the derivative of this function, we have $$ \frac{df}{dt} = \frac{d}{dt} \left[ \int_{t_0}^{t} u(\tau - t_0) \, d\tau \right] $$ which is equal to $u(t - t_0)$ by the first fundamental theorem of calculus.