The Laplace transform.of f(t)=sin(t)

now we're going to find the laas we're going to find the laas transform of the S function the s t okay we will use the same Technique we will apply the definition so the the the plus transform of sin T is going to be the integral between zero and infinity of the sin T * e to minus St DT here we can use the uh the definition of the the improper integral so F of s is going to be the limit as X goes to Infinity of the integral between Z and X of s the S of t e to minus s t DT okay so we first focus on this integral here we see that to find this one we have to do integration by part we will do it twice because the first one we get the cosine and after that we get back back to S and we're going to see what we can get from there okay so the the technique is that we will in differentiate the sign and integrate the exponential okay here differentiate integrate we differentiate the sign we get cosine of T and minus sin T and here we have e to minus St when we integrate we get min -1 / s e to minus s and doing it one more time we get 1 / S 2 e to- s okay we have a plus minus plus okay and using this uh uh uh TT multiplication if want to stop here and integrate we multiply directly okay so we get in this case minus so this is this value here this integration here so it's going to be this first two we'll get min -1 / s e to minus s t sin T minus cine t/ s² e to minus s minus because want to stop here this is between Z and X when to stop here we integrate again between Z and x uh the value of uh s let's take the one / s Square outside we'll get sin t e to minus s t Okay so because we need to go back here and once we take the limit okay we get F of s this limit here the s t e to minus St it's always going to be uh a value that we have to determine okay but once we do that by taking uh X to Infinity we see that the exponential of the S minus s x always win the same thing for uh for the powers okay let's let's see clearly what we mean here so let's evaluate this one this is the limit as X goes to Infinity so I'm going to take this one to the front this one here so this one is going to be Min -1 / s² integral between 0 and x sin T DT e to minus s t minus the limit of one let's say 1 / s here e to minus Xs sinine X okay so I'm evaluating this one here and when I plug zero here I get zero and I move to the next one so here I have uh one minus cosine x e to - s x/ s² and once I do have once I do evaluate at zero I get minus 1 / s² because the cosine of Z is one and this one at zero is one so here the limit now for this part here this is zero this is zero okay because the exponential e to minus s x WIS over the cosine and the sign and therefore this limit here is zero and all I'm left with is this one here so here I have F of s so it's -1 / s² the integral let's let's try to get to definition the integral between 0 and Infinity of sin t e to - s t -1 / s² here I need to have a plus here because it's zero one it's plus one / s squ sorry it's 1 / s + 1 / s² because when I do evaluate at zero with this minus I get plus so here what's this one here this is the one that I'm looking for this is my laplus transform yes this is my laplus transform here this one here therefore I have F of s + 1 / s² f of s is 1 / s² okay so now using some algebra okay I have s² + one F of s okay over s² is 1 / s² and therefore for I have my f of s is just 1/ s² 1/ 1 + s² this is the laas transform of the sign okay note here so the key here is to showing that this integral is zero except when when X is zero I get one over s² and once I get here I I take the limit but but I remember that this is just the definition of the laas transform of the sign I take it to the left and I get S2 + 1 / S2 and that one is equal to 1 / s² and once I do the algebra I get S2 + 1 F of s/ s² is 1 / S2 and and I simplify I get F of s is 1 / 1 + s² and this is the L plus transform of sin T it's 1 / 1 + s² okay thank you very much

Channel: Archimedes Notes

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Id: Q_oDkRCDZNY

Length: 8min 11sec (491 seconds)

Published: Thu Jul 11 2024