Question

Question 2 Consider the signal or system xét) with Laplace transform X(s). Which of the following is/are true, select only the correct answers. (Assume any initial conditions of the output to be 0). If x(t) = 9sin(20)u(t), then X(s) - 180/(s2+400) If x(t) = 5u(t) - 2(t-6) - Scos(5t)u(t) then X(s) = 5/s - 2e-6s - 85/(52+25). If x(t) =-6tu(t) then X(s) = -6/s. If x(t) = 6sin( 101-3)u(t-0.3), then X(s) = (604-0.3s )(2+100), If x(t) =7tuit) then X(s) = 7/32

Answer 1

The above question can be solved by using standard formulae of Laplace transform. Please refer to the below image for detailed explanation:

y2 + 400 Solution : 13. If X(t) = 9 sin (20)ult) then x(3) = from the standard formulae. g Laplace transform we know that - [ sin fé jutt)] 22 - @ using the abone formula we get :- [xct)] = [ 9 sin(20t Jult)] => 9.(20) (8² +202) => 180 = X(Y) g? + 400. 1. Statement I is true . 2> If y(t) = Sult) - 2816-6) - 8 Cay(5t)ult) then x(9) = 5 - 26-68.8%. (y2 +25) from the Standard formulae of Laplace transform We know L[ult)] = = L[s(t-a)] = was L [ Coslat Juct) ] 2 using all the above formulae in X(t) we [xlt)]- L[5u(t)-2814-6)-86(SHack] = 5-2002- get

o Statement 2 is True / 2) If y() = -6tult) then x(9),= -5 room standard formula of Laplace transform we know that L[tult)] = - (b) in using above formula in X ) we get :- L[x (U] = [-6tult) ] - Co. Statement 3 is wrong (False ) 4 Is X(t) = 68in (lot-3) ult-0.3) then . X(Y) = (Goce - 0.3%) (82+ 100 ) We can write X(t) = 6 Sin lo(t-3 ult-3). using Formula @ from above we get & L{xH) => 1 [6 in 10lt-Jul-2.1] => 6.(10). 2-0.34 - 600 -0.3% also.:- (y2 +10²) 7 (y2 +10o) Lult-a)] = eas? he Uue e = x(4) ja } & statement 4> is True.

5> If X(t) = 7tult) then 403) = 2 I using the formula 1 from above we get L[7tult)] => 2 = x(). % Statement s is true. Hence statements 1, 2, 4, 5 are true and Statement 3 is false, Answer