Using Convolution Theorem to find the Laplace transform
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
$$\hat{h_{S*T}}(p) = \left ( \frac{2  e^{p S}}{p} \right )\left ( \frac{2  e^{p T}}{p} \right )$$
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S F
Hi Im a mathematics student, trying to learn about topics which I am struggling with at school
Updated on August 01, 2022Comments

S F over 1 year
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 $$ \int^t_0 f(tx)g(x)\ dx =$$
But how to apply that to
Use convolution theorem to find the Laplace transform of
$$ \int^t_0 du f_T(tu)f_s (u), t>0 $$
where fs(t) is the function in (1) replaced with T replaced by S

S F over 10 yearsso essentially what $$ \int^t_0 du f_T(tu)f_s (u), t>0 $$ means is to find the Laplace transform using convolution We find the laplace transform using the previous method and then we multiply the laplace transform by itself but replacing t by s for the first function ?

Ron Gordon over 10 years@SF: that is the long and short of it.

S F over 10 yearsdo I need to evaluate $$\hat{h_{S*T}}(p) = \left ( \frac{2  e^{p S}}{p} \right )\left ( \frac{2  e^{p T}}{p} \right )$$ between 0 and t ? in my workbook we would replace the first t with (tx) and the 2nd t with x . that does not apply in this case

Ron Gordon over 10 years@SF: No, this is the transform. An interesting exercise is to actually evaluate that convolution  it is not trivial so far as I can see. But you were asked to find the transform which, as I showed, is rather straightforward.