Question

Can someone solve (d) using the proper equation.. not the easy way please. using the equation X(s)= (integral from minus infinity to infinity)(x(t)e^(-st)dt

6.1 Using the definition in Eq. (6.5), calculate the bilateral Laplace transform and the associated ROC for the following CT functions: (а) x() —D е-5 и() + e*и(-1); (b) x(t)e-31 (d) x(1) — е-3і cos(5t); (е) x(1) — е" соs(91)u(-1); e (f) x(t) (с) x(1) — 1? сos(10г)u(—t); otherwise

Answer 1

Laplace Transform, d) x(t) = e^-3(t) cos 56 x(t) = {e^-3t cos 5t t greaterthanorequalto 0 e^3t cos 5t t < 0 cos 5t = e^j 5t + e^-j 5t/2 rightarrow using Euler theorem x(t) = {e^-3t/2 [e^-j 5t + e^-j 5t] t greaterthanorequalto 0 e^+ 3t/2 [e^j 5t + e^-j 5t] t < 0 X(s) = integral_-infinity^infinity x(t) e^-st dt X(s) = integral_-infinity^- e^3t/2 [e^j 5t + e^-j 5t]e^-st dt + integral_0^infinity e^-3t/2 [e^-j 5t + e^-j 5t] e^-st dt \r\n X(s) = 1/2 integral_-infinity^0 e^(3 + 5j - s)t + e^(3 - 5j - s)t dt + 1/2 integral_0^infinity e^(-3 + 5j - s)t + e^-3 -5j -s)t dt X(s) = 1/2 [e^(3 + 5j - s)t/(3 + 5j - s) + e^(3 - 5j - s)t/(3 - 5j - s)]_-infinity^0 + 1/2 [e^-(3 - 5j + s)t/-(3 - 5j + s) + e^-(3 + 5j + s)t/-(3 + 5j + s)] \r\n X(s) = 1/2 (1/3 + 5j - s) + 1/3 - 5j - s) + 1/2(3 + 5j + s + 1/3 - 5j + s) X(s) = 1/2 (6 - 2s/(3 - s)^2 + 5^2) + 1/2 (6 + 2s/(3 + s)^2 - (5j)2)