Question

(a) find a combination of step functions that is equal to 4 when 2 ≤t ≤3 and equal to 0 otherwise. (b) write the following function in one line (i.e. without cases) using step functions: f(t)

Answer 1

(a) Recall the definition of the step function,

`\theta(t) = \begin{cases} 1 & \text{if } t \ge 0 \\ 0 & \text{if } t < 0 \end{cases}`

Then we have

•
`\theta(t-2) = 1` if
`t-2\ge0`, or
`t\ge2`

•
`\theta(t-3) = 1` if
`t-3\ge0`, or
`t\ge3`

and we can combine these to get

`\theta(t-2) - \theta(t-3) = \begin{cases} 1 & \text{if }2\le t < 3 \\ 0 & \text{if }t<2 \text{ or } t\ge3 \end{cases}`

then scale up by 4 to get the value we want over [2, 3].

`f(t) = \boxed{4\bigg(\theta(t-2) - \theta(t-3)\bigg)}`

At least that's what I get using the aforementioned definition of "step function". I don't see a way to have both
`f(2)=4` and
`f(3)=4` with this definition...

(b) No function was given...