How is $u(at+b) = u(t+b/a)$ for $a>0$ and $u(-t-b/a)$ for $a<0$?

Question

I've come across this equality that I don't know how to prove. I've tried using the delta function due to its relationship to the step function, but I can't reach any results.

Answer 1

• General definition of the unit-step function: $$u(t-t_0) = \begin{cases} 1 &t>t_0\\ 0 &t<t_0 \end{cases} $$
• Let's write it for $u(at+b)$: $$u(at+b) = \begin{cases} 1 &at>-b \\ 0 &at<-b \end{cases} \tag{1} $$
• Now recall from basic algebra that:

$$at > -b \implies \begin{cases} t>-\frac{b}{a} & a > 0\\ t<-\frac{b}{a} & a < 0\\ \end{cases} $$ and similarly, that: $$at < -b \implies \begin{cases} t<-\frac{b}{a} & a > 0\\ t>-\frac{b}{a} & a < 0\\ \end{cases} $$

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So for $a>0$ we can write $(1)$ as: $$u(at+b) = \begin{cases} 1 &t>-\frac{b}{a} \\ 0 &t<-\frac{b}{a} \end{cases} \quad= u(t+b/a) $$ and for $a < 0$ we can write $(1)$ as: $$u(at+b) = \begin{cases} 1 &t<-\frac{b}{a} \\ 0 &t>-\frac{b}{a} \end{cases} \quad= 1 - u(t+b/a) = u(-t-b/a) $$

the last step involves the time inversion property of the unit step function, which states: $$u(-t) = 1 - u(t)$$