What is the Laplace transform of this function What is the Laplace transform of the time domain signal below:

The signal V(t) starts from zero and increases with a slope m and after t1 seconds reaches to constant value a.
How can this function be written in s domain?
 A: Just take the derivative of $V(t)$, its expression is very simple and so is its transform. Then all you need is to know  how the transform of the integral is expressed in terms of the transform of the integrand.
A: We want to calculate the Laplace transform of $V(t)$. It is easy to calculate it directly from the definition:
$$\int_0^\infty V(t)e^{-st}\text{d}t=\int_0^{t_1}mt e^{-st}\text{d}t+\int_0^\infty ae^{-st}\text{d}t=m\int_0^{t_1}t e^{-st}\text{d}t+a\int_0^\infty e^{-st}\text{d}t$$
The first integral can be solved using integration by parts and the other one can be computed directly. 
A: Laplace transform definition: 
$f(s) = \int_0^\infty f(t)e^{-st} dt$
Your function is $f(t) = mt$ from 0 to $t_1$ and $f(t) = a$ from $t_1$ to $\infty$, so:
$f(s) = \int_0^{t1} mt*e^{-st} $ $dt + \int_{t1}^\infty a*e^{-st}$ dt.
A: Observe that $V(t) = mt$ for $0\leq t \leq t_1$ and $V(t) = a$ for $t > t_1$.
The Laplace transform can be computed via its definition 
$$\mathcal{L}[V(t)](s) = \int_{0}^{\infty}V(t)e^{-st}dt = \int_{0}^{t_1}mte^{-st}dt + \int_{t_1}^{\infty}ae^{-st}dt$$ $$=-\frac{mt_1e^{-st_1}}{s}+m\frac{1-e^{-st_1}}{s^2}+ \frac{ae^{-st_1}}{s}$$
