Question

Given y(t) = 3 + cos(6t) - ecos(6t) and Y(s) defined as a Y(s) = hs + s²+k + s+p s2 + qs+t s+b match variables to correct values that make Laplace transform correct. Laplace transform table handout may be used as reference, a [Choose] b [Choose h [Choose) k [Choose] р [Choose 9 [Choose [Choose)

Answer 1

given

y(t)= e3t+cos6t-e3tcos6t

Now by using the standard laplace transfrom relations

L(e^(at)) = 1/(s-a)

L(cos(at))= s/(s2+a2)

L(eatcos(bt))= (s-a)/((s-a)2+b2)

Now by applying the above relations to the present question, We get

L(y(t))=Y(s) = [1/(s-3)]+[s/(s2+36)] - [(s-3)/((s-3)2+36)]

so comparing the above equation with the equation given in the question we get

Unknown Variable	Value
a	1
b	-3
h	1
k	36
p	***
q	-6
t	45

***There is a small issue in finding value of 'p' because it cannot be found in the given form of the expression.

In the following image the correct explaination with steps is given.

It S 3Ł 3t YLt) = e + cos (67)- e cos(67) (5-3) Llylt)) S-3 8436 (5-39736. (5-3) (5-3) ($ +36) 5 65+45 1 s = t Y(S) = S (5-3) (5-3) 8436 5-65+45 Scanned with in Scanner

NOTE:

Please do comment to get clarified in any part of the solution.

The value of 'p' cannot be found out with the given expression.

There should be a -ve sign infront of the 3rd term, If it is the case then the value of 'p' is -3.

Thank you.