Laplace Transform: Convolution Theorem?

Question

Laplace Transform: Convolution Theorem?

Answer 1

The Laplace transform of a convolution is the product of the convolutions of the two functions. This is often used in reverse. If you know the inverse transforms of F(s) and G(s) and they are functions f(x) and g(x), then the inverse transform of F(s)G(s) is the convolution f*g(x)=int_0^x f(t)g(x-t)dt.

Answer 2

My maths is a bit rusty (I studied maths several years ago), but I found the Wikipedia article to be extremely informative.



Rather than replicate the article in part here, I'll point you to it as the source. Hope this helps!

Answer 3

your on the right track



now let the inverse transform of 1/s^2 = f(t)=t (if i remember correctly)



and let L^-1[1/((s-1)^+11)] = g(t)



plugging f(t) and g(t) in to the integral gives

h(t)=int(1/(t-r)^2*re^(r)sinrdr)



and solve

Answer 4

i found a site that may be helpful to u..but u have to use MATLAB..its easier...in that website it will show u how to solve the convolution theorem using MATLAB...i hope this will help u...

Answer 5

You are going the right way, but the integrals are a bit tricky and tedious. Define:



f(t) = t

g(t) = exp(t)sin(t)



Thus h(t) = int(0->t): f(t-r)g(r)dr = int(0->t): (t-r)*exp(r)sin(r)dr =



int(0->t): t*exp(r)sin(r)dr - int(0->t): r*exp(r)sin(r)dr



Both these integrals must be done using integration by parts, and there is a "loop trick" you need to use to solve.



For the first integral, we can treat 't' as a constant because we are intergrating with respect to 'r'. Thus, we can pull it outside of the integral:



t*int(0->t): exp(r)sin(r)dr [let's also denote the integral as @ for later use]



Now let:

u = exp(r) ---------> du = exp(r)dr

dv = sin(r)dr ---------> v = -cos(r)



Thus,



t*int(0->t): exp(r)sin(r)dr = -exp(r)cos(r) + int(exp(r)cos(r)dr)



Again, use integration by parts on the 2nd integral:



u = exp(r) -----------------> du = exp(r)dr

dv = cos(r)dr ----------------> v = sin(r)



Thus, int(exp(r)cos(r)dr) = exp(r)sin(r) - int(exp(r)sin(r))



Now let's combine the entire integral:

-------

t*int(0->t): exp(r)sin(r)dr =



-exp(r)cos(r) + exp(r)sin(r) - int(exp(r)sin(r))

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The third part of the answer is the same as the integral itself. Now let's use the @ substitution above to do the "loop trick". Making this substitution, you have:



@ = -exp(r)cos(r) + exp(r)sin(r) - @. Put all the @ on one side and solve for @.



2@ = -exp(r)cos(r) + exp(r)sin(r)



@ = int(0->t): exp(r)sin(r)dr = (1/2)[e^r*sin(r)-e^r*cos(r)]



Don't forget the 't'



t*int(0->t): exp(r)sin(r)dr = (t/2)[e^r*sin(r)-e^r*cos(r)]



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Now the 2nd integral from the top is a little trickier:

---------------int(0->t): r*exp(r)sin(r)dr------------



Use integration by parts again, this time:



u = r -------> du = dr

dv = e^r*sin(r) ------> v = look at the result above



Using the same algorithm as above and using the "loop trick", you get the following answer:



(e^r/2)[cos(r) + r*sin(r) - r*cos(r)



-------



Combine the two answers together to get h(t)



h(t) = (t/2)[e^r*sin(r)-e^r*cos(r)] - (e^r/2)[cos(r) + r*sin(r) - r*cos(r)

evaluated from [r: 0->t]



Evaluating h(t) at these limits, you get:



h(t) = (t/2) - [e^t*cos(t)]/2 + (1/2), which is the desired answer.



Use the following website to check your answers



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Hope this helps and was easy to follow