In terms of force (F), acceleration (a), velocity (v), and position (y), with variables of impedance (R) and mass (m), I have a system where:
$y(t) = \frac{v(0) m}{2 R} (1-e^{\frac{-2 R t}{m}})$
$v(t) = v(0) e^\frac{-2 R t}{m}$
$f(t) = -2 R v(0) e^\frac{-2 R t}{m}$
I would like to derive the $y(t)$ equation from $F(t)$ using the Laplace domain as follows:
$F(s) = m A(s)$
$F(s) = m (s^{2} Y(s)- sy(0) - y'(0))$
$F(s) = m (s^{2} Y(s)- sy(0) - v(0))$, and if y(0) = 0, then:
$F(s) = m (s^{2} Y(s) - v(0))$
Rearranging, we get:
$Y(s) = \frac{F(s) + mv(0)}{ms^{2}}$
$Y(s) = \frac{F(s) + mv(0)}{m} * \frac{1}{s^{2}}$
But how do I now convert this back to the time domain and sub in $f(t)$? According to calculators and reference tables $\frac{1}{s^{2}} = t$, but if I just put $t$ in there for $\frac{1}{s^{2}}$ and change $F(s)$ to $f(t)$ it doesn't work.
So what is the correct operation to now do an inverse Laplace transform and get the $y(t)$ equation using $f(t)$?
It looks like your $Y(s)=v(0)/s^2 + F(s)/(ms^2)$, so you should get $$ y(t) = v(0) t + \int_0^t (t-u) f(u) du /m$$ by using linearity of the LT to handle the sum of terms, pattern matching from the a table of LT pairs to handle $1/s^2$, and using the convolution formula for the product $F(s)/s^2$.