How would i write this sum of step functions as a singular function?

Question

I would like to know, how i could express $$H(t)-H(t-b)$$ as a single function, I know that $H(t)=1,t\ge 0$ and $H(t)=0,t<0$ and that $H(t-b)=1,t\ge b$ and $H(t-b)=0,t<b$. I've tried drawing these but my sketch doesnt seem right, The unit step function starts at $t=0$ and goes straight up to $1$ and then stays constant and the $H(t-b)$ goes to $1$ when $t=b$ and then stays constant, however i'm struggling to see what $H(t)-H(t-b)$ would look like, and how i would write it as a pointwise function. Note that $b>0$

If anyone could help me with this it would be appreciated.

Answer 1

It may help to rewrite each of the two component functions so that they are defined on exactly the same "pieces" of the domain. Since $b>0$, we have

$$H(t)=\begin{cases}0, & t<0 \\ 1, & 0\leq t < b \\ 1, & t \geq b \end{cases}$$

$$H(t-b)=\begin{cases}0, & t<0 \\ 0, & 0\leq t < b \\ 1, & t \geq b \end{cases}$$ Putting these together by subtracting, you have

$$\boxed{H(t)-H(t-b)=\begin{cases}0, & t<0 \\ 1, & 0\leq t < b \\ 0, & t \geq b \end{cases}}$$

Answer 2

$h(t)-h(t-b)$ will be zero for all $t<0$ and then jump to $1$ at $t=0$. It will stay at $1$ until $t=b$, where it'll be zero again for all $t\geq b$.

Generally, step functions are used to represent piecewise functions without brackets, but to go back the other way

$$ h(t)-h(t-b) = \left\{ \begin{array}{ll} 0 & \quad t < 0\\ 1 & \quad 0 \leq t < b \\ 0 & \quad t \geq b \end{array} \right. $$