Sarahisme Posted May 20, 2006 Share Posted May 20, 2006 hey this is not so much a problem that i can't do, its problem that seems to have 2 answers, yet i thought laplace transforms are unique.... if i have [math] F(s) = \frac{2s-1}{(s-1)^2s^2} [/math] then if i take the inverse transform it straight from here (either by using maple, or relising that [math] \frac{2s-1}{(s-1)^2s^2} = \frac{s^2 - (s-1)^2}{(s-1)^2s^2} [/math] i get [math] f(t) = -t + te^t [/math] however if i split up F(s) to [math] F(s) = \frac{2s-1}{(s-1)^2s^2} = \frac{2}{(s-1)^2s} - \frac{1}{(s-1)^2s^2} [/math] and then take the inverse transform (either by using maple or convolution integrals), i get [math] f(t) = 2te^t-4e^t+2t+t^2 +4 [/math] can anybody give an insight as to why this is happening? (as even maple seems to get 2 quite different answers....) -Sarah Link to comment Share on other sites More sharing options...
Yggdrasil Posted May 20, 2006 Share Posted May 20, 2006 Are you sure you're evaluating the last inverse laplace transform correctly? I get that the inverse laplace transform gives: [math]f(x) = 2 (xe^x * 1) - (xe^x * x)[/math] where * represents convolution. In other words: [math]f(x) = 2\int_0^xte^tdt - \int_0^xte^t(x-t)dt[/math] Which gives: [math]f(x) = 2\int_0^xte^tdt - x\int_0^xte^tdt + \int_0^xt^2e^tdt[/math] [math]f(x) = (2-x)\int_0^xte^tdt + \int_0^xt^2e^tdt[/math] Apply differentiation by parts to the second term: [math]f(x) = (2-x)\int_0^xte^tdt + \int_0^xt^2d(e^t)[/math] [math]f(x) = (2-x)\int_0^xte^tdt + (t^2e^t|_0^x - \int_0^xe^td(t^2))[/math] [math]f(x) = (2-x)\int_0^xte^tdt + (t^2e^t|_0^x - 2\int_0^xte^tdt)[/math] [math]f(x) = -x\int_0^xte^tdt + x^2e^x[/math] Apply differentiation by parts again: [math]f(x) = -x\int_0^xtd(e^t) + x^2e^x[/math] [math]f(x) = -x(te^t|_0^x - \int_0^xe^tdt) + x^2e^x[/math] [math]f(x) = -x(xe^x- e^t|_0^x) + x^2e^x[/math] [math]f(x) = -x(xe^x- (e^x - 1) + x^2e^x[/math] [math]f(x) = -x^2e^x + xe^x - x + x^2e^x[/math] [math]f(x) = xe^x - x[/math] So, I see no contradiction. Link to comment Share on other sites More sharing options...
Sarahisme Posted May 20, 2006 Author Share Posted May 20, 2006 for this bit, when i used convolution integrals i used t^2 instead of t, i.e. [math] f(x) = 2\int_0^xte^tdt - \int_0^xt^2e^{x-t}dt [/math] but i did put it into maple, and it gives the answer in my original post.... so i can see what you did is correct, i just want to know why if you used t^2 instead of t(x-t) you get an almost completely different answer....?? Link to comment Share on other sites More sharing options...
Yggdrasil Posted May 20, 2006 Share Posted May 20, 2006 for this bit' date=' when i used convolution integrals i used t^2 instead of t, i.e. [math'] f(x) = 2\int_0^xte^tdt - \int_0^xt^2e^{x-t}dt [/math] but i did put it into maple, and it gives the answer in my original post.... so i can see what you did is correct, i just want to know why if you used t^2 instead of t(x-t) you get an almost completely different answer....?? That's an incorrect usage of the convolution. Here are the correct forms of the convolution: [math](xe^x * x) = \int_0^xte^t(x-t)dt = \int_0^xt(x-t)e^{x-t}dt[/math] What you wrote is a completely different convolution: [math]\int_0^xt^2e^{x-t}dt = (e^x * x^2)[/math] Link to comment Share on other sites More sharing options...
Sarahisme Posted May 20, 2006 Author Share Posted May 20, 2006 ok , fair enough but what about the fact that maple gives 2 different answers? Link to comment Share on other sites More sharing options...
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