Archemeny Posted October 28, 2014 Report Share Posted October 28, 2014 Quote Link to comment Share on other sites More sharing options...
sanctus Posted October 30, 2014 Report Share Posted October 30, 2014 Looks like homwork, but assuming your equation for y(t) is right (did not check) then it is so simple that I can give you the answer without feeling to compromise your learning curve...You have found this:[latex]y(t)=Ae^{-2t}+frac{K}{5}e^{3t}[/latex] Additionally the intial equations tell you that y(0)=0 This gives in your found equation:[latex]0=A{-2\cdot 0}+frac{A}{5}e^{3\cdot 0}[/latex] Which solving for A yields:[latex] A=\frac{-K}{5}[/latex] Quote Link to comment Share on other sites More sharing options...
sanctus Posted October 30, 2014 Report Share Posted October 30, 2014 Sorry for latex tags atm not working, but you should understand what I wrote even if you do not know latex Quote Link to comment Share on other sites More sharing options...
Archemeny Posted October 30, 2014 Author Report Share Posted October 30, 2014 I think the boundary condition shcould be changed to this : system is initially at rest Cause the boundary Condition Y(o) =o tells us only about t=o not (t=o +) , remember We solve this equation For t>o not t=o or to . . .I must say I found this problem in One of the examples of the book signals and systems by Oppenheim . . .anyway thanks a lot For your attention Sanctus . . . Quote Link to comment Share on other sites More sharing options...
sanctus Posted October 30, 2014 Report Share Posted October 30, 2014 What do you mean by 0+? t tending to zero from the right? If so, why does my approach not work? I mean[latex]lim_{t\to0+}[/latex] of y(t) is well what I wrote, no? Quote Link to comment Share on other sites More sharing options...
Archemeny Posted October 30, 2014 Author Report Share Posted October 30, 2014 maybe I didn't understand you very well but my main question is: why y(0)= y (0+) ? [Note that y(0+) is y tending Zero from right ] Quote Link to comment Share on other sites More sharing options...
sanctus Posted October 31, 2014 Report Share Posted October 31, 2014 It does not have to be, but if:[math] lim_{t\to 0+}y(t) =lim_{t\to 0-}y(t)[/math] then you have y(0)=y(0+)=y(0-) And since your y(t) is a continuos function with no singularity the above condition holds. Quote Link to comment Share on other sites More sharing options...
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