In previous questions I have used Laplace transform to find the inverse Laplace transform.
I have worked through this work booklet (http://www3.ul.ie/~mlc/support/Loughborough%20website/chap20/20_6.pdf ) and understand convolution theorem.However, this question threw me off perhaps the particular layout is the problem.
Part A which I understand how to was
$$ f(t) = \begin{cases} 2 & \text{ } 0 \leq t < T \\ 1 & t \geq T \end{cases} $$
Part B however I dont understand
I know the convolution theorem is $$ \int^t_0 f(t-x)g(x)\ dx =$$
But how to apply that to
Use convolution theorem to find the Laplace transform of
$$ \int^t_0 du f_T(t-u)f_s (u), t>0 $$
where fs(t) is the function in (1) replaced with T replaced by S
The LT of a convolution of two functions is the product of the individual LTs. If you recall, we found that
$$\hat{f_T}(p) = \frac{2 - e^{-p T}}{p}$$
Thus, the LT of the above convolution is simply
$$\hat{h_{S*T}}(p) = \left ( \frac{2 - e^{-p S}}{p} \right )\left ( \frac{2 - e^{-p T}}{p} \right )$$