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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

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1 Answer 1

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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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so essentially what $$ \int^t_0 du f_T(t-u)f_s (u), t>0 $$ 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 ? –  S F May 6 '13 at 15:01
    
@SF: that is the long and short of it. –  Ron Gordon May 6 '13 at 15:13
    
do 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 (t-x) and the 2nd t with x . that does not apply in this case –  S F May 6 '13 at 15:27
    
@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. –  Ron Gordon May 6 '13 at 15:29

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