Laplace Transform of Cos at – Laplace Transform – Engineering Mathematics 3


Hello everyone today we are gonna see or we are gonna derive what is the Laplace transform of cos At now to derive the Laplace transform of cos AT formula we are gonna use the definition of Laplace transform so let me write that we want to find Laplace transform of cos AT so your L denotes the Laplace transform so as I said that we are gonnA use the definition of Laplace transform so what is the definition of Laplace transform so here I will say that by definition of Laplace transform we have Laplace of F of T is equal to integration 0 to infinity e raised to minus ST F of T DT so this is the definition of Laplace transform now let us solve the given function cos of 80 or let us find out the Laplace transform of Casa baya T so therefore Laplace of cos of 80 equal to so here we can say that this F of T in the definition is cos 80 so on the right hand side also I will substitute F of T s cos 80 so here we’ll get integration from 0 to infinity e raised to minus s T cos of a T DT now to get the Laplace of cost 80 we have to solve this integration with respect to T so to find out the value of integration of e raised to minus s T into cos 80 we are going to use one formula that is a property of integration and what is the property of integration so let us see so the property is given like this that integration of e raised to a t cos B T DT is equal to e raised to a T upon a square plus B Square in bracket a cause BT plus B sine B T so this is that formula or I will say it is a property of integration now let’s compare our term with the formula so here we can see that this ear h-2a T is equal to e raised to minus s T it means we should have a equal to minus s and if Cosby T is Casa T then we should have B equal to a so to get the value of e raised to minus s T cos a T by this formula we will substitute a as minus s and B as a in the right hand side so by doing that we will get e raised to a t is minus s to e raised to minus s T upon a square that is minus s square which is positive a square plus B Square B is a so we’ll get a square now in bracket we’ll get a a is minus s so minus s cos B T B is a so we’ll get cos of a T plus B Nobi is a so a sine B T so B is a so we’ll get a T so here we got the term now here in this term we’ll substitute the upper and lower limit to evaluate the integration now let’s substitute the upper limit first now upper limit is infinity so by substituting the upper limit here we will get a raise to minus infinity upon s square plus a square into minus s cos of infinity plus a sine of infinity so I am substituting T as infinity everywhere because I am having integration with respect to T after that negative sign and now we’ll substitute the lower limit now lower limit is 0 so this will become e raise to 0 upon s square plus a square into minus s cause of now T is 0 so 0 into a is 0 plus a sine of 0 into a again 0 now we’ll use some formulas that is e raise to minus infinity equal to 0 this is the first formula then we know that e raised to 0 equal to 1 I will use it over here then we’ll use the formula of sine 0 which is 0 and the value of cos 0 is equal to 1 so by using this formulas here we will get e raised to minus infinity 0 so 0 into anything is 0 so we will get whole term as 0 minus here I raised to 0 is 1 upon s square plus a square next in bracket will get minus s cos 0 is 1 so 1 into minus s is minus s only next a into sine 0 where sine 0 is 0 so 0 into a is 0 so I close the bracket now because of this minus sign here we’ll get minus minus plus s so it means we’ll get value as s upon s square plus a square so we got the Laplace transform of cos a t as s upon s square plus a square so here we derived the value of Laplace transform of casa t from the definition of Laplace transform thank you