Convert to heaviside function: $$f(t) = \begin{cases}e^t ,& 0 \leq t \leq 1 \\0 ,& t > 1\end{cases}$$
My attempt: $f(t) = U(t) e^t - U(t-1) e^t $
I think my solution is not right because at f(t=1), it doesn't give the right value. How would I go about fixing this issue.
Thanks!
Simple.
f(t) = $e^tU(-t+1)U(t)$
EDIT
Your function f has the value of 0 on the entire x domain, in the exception of [0, 1] where it has the value of $e^t$.
You can start with f(t) = $e^t$. The problem now is that for t < 0 or t > 1 the value of f(t) is still $e^t$. We start by clearing out f(t) if t > 1. We know that the Heaviside function U(t) is 1 for t>= 0 and 0 otherwise. Consequently U(-t+1) is 0 for t < 1, 0 otherwise.
Still need to clear out f for t < 0. U(t) does the job since it is equal to 0 for t < 0.
Multiply everything and you are done!
$$ e^t U(1-t) U(t)\qquad \text{(with $1-t$, not $t-1$)}. $$ ${{{{{{{{{{{{{{{}}}}}}}}}}}}}}}$
$f(t) = U(t) e^t +U(t-1) U(1-t)e^t-U(t-1) e^t $
I keep almost the same function as you have, just modifying it's behaviour at the point $x=1$.
My attempt: $f(t)=U(t)e^t −U(t−1)e^t$
$\begin{align} U(t)-U(t-1) & = \begin{cases} 1 & : t \ge 0 \\ 0 & : t\lt 0\end{cases}-\begin{cases} 1 & : t \ge 1 \\ 0 & : t\lt 1\end{cases} \\ & = \begin{cases} 0 & : t \ge 1 \\ 1 & : 0\le t \lt 1 \\ 0 & : t\lt 0 \end{cases} \\ \color{gray}{ \operatorname{\bf 1}_{[0,\infty)}(t)-\operatorname{\bf 1}_{[1,\infty)}(t)} & = \operatorname{\bf 1}_{[0,1)}(t) & \text{half-open interval} \\[1ex] U(t)\cdot U(-t+1) & = \begin{cases} 1 & : t \ge 0 \\ 0 & : t\lt 0\end{cases}\times\begin{cases} 1 & : t \le 1 \\ 0 & : t\gt 1\end{cases} \\ & = \begin{cases} 0 & : t \gt 1 \\ 1 & : 0\le t \le 1 \\ 0 & : t\lt 0 \end{cases} \\ \color{gray}{ \operatorname{\bf 1}_{[0,\infty)}(t)\times\operatorname{\bf 1}_{(-\infty,1]}(t)} & = \operatorname{\bf 1}_{[0,1]}(t)& \text{closed interval} \end{align}$
A minute but important distinction.
