rewriting a piecewise function in terms of the Heaviside step function and then taking the derivative

Question

Okay, so I was following along this document (edit: archived document), and I had a couple of questions. So I understand the logic of turning a function off then turning it on.

Question 1: But what if we took this function: $$f(t)=\begin{Bmatrix}t^3 & 0\leq t < 2\\ 5 & 2\leq t < 20\\ e^{-t} & 20 \leq t\end{Bmatrix}$$ which the document says would be converted to: $$f(t)=t^3H(t-0)-t^3H(t-2)+5H(t-2)-5H(t-20)+e^{-t}H(t-20)$$ and rewrit it like this instead: $$g(t)=\begin{Bmatrix}t^3 & t < 2\\ 5 & 2\leq t < 20\\ e^{-t} & 20 \leq t\end{Bmatrix}$$ How would I rewrit that function, because it doesn't look like $t^3$ is ever "turned on", and only functions that have been "turned on" can be "turned off". So would that function be rewritten as: $$g(t)=5H(t-2)-5H(t-20)+e^{-t}H(t-20)$$

Question 2a: Also, for the original rewritten piecewise function, how would I take the derivative of that function? I know that $$\frac{\mathrm{d}}{\mathrm{d} x}H(x)=\delta(x)$$ so if I wanted to take $f'(x)$, would it just be: $$f'(t) = 3t^2H(t-0)+t^3H'(t-0)-3t^2H(t-2)-t^3H'(t-2)+0+5H'(t-2)-0-5H'(t-20)-e^{-t}H(t-20)+e^{-t}H'(t-20)$$ $$= 3t^2H(t-0)+t^3\delta(t-0)-3t^2H(t-2)-t^3\delta(t-2)+5\delta(t-2)-5\delta(t-20)-e^{-t}H(t-20)+e^{-t}\delta(t-20)$$ $$= 3t^2H(t)+t^3\delta(t)-3t^2H(t-2)-t^3\delta(t-2)+5\delta(t-2)-5\delta(t-20)-e^{-t}H(t-20)+e^{-t}\delta(t-20)$$

Question 2b: Is $$H(t-0) = H(t) = 1$$ and is $$\delta(t-0) = \delta(t)=0 $$?

Answer 1

to start with the end, 2b, $H(t-0) = H(t)$ is one for $t\geq 0$ and zero in $t<0$

Regarding $\delta(t)$ it is not a function but a distribution and it is definitely not zero.

for question 1:

$$f(t)=t^3-t^3H(t-2)+5H(t-2)-5H(t-20)+e^{-t}H(t-20) $$

for 2a

$$f'(t)=3t^2-3t^2H(t-2) + t^3 \delta(t-2)+5\delta(t-2)-5\delta(t-20)-e^{-t}H(t-20)+ e^{-t} \delta(t-20) $$

we can simplify since $\delta(t)$ is zero away from $t$ to get

$$f(t)=3t^2-3t^2H(t-2) + 8 \delta(t-2)+5\delta(t-2)-5\delta(t-20)-e^{-t}H(t-20)+ e^{-20} \delta(t-20) $$