$x(t)=u(t),\lambda (t)=e^{-2t} u(t)$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : DEC 2015

$$x(t) = u(t)$$

Taking Laplace transform of above equation, we get

$∴X(s) = \dfrac 1s ------------- 1 \\ h(t) = e^{-2t} u(t) $

Taking Laplace transform of above equation, we get

$∴H(s) = \dfrac1{s + 2} --------------- 2$

By convolution theorem,

$Y(s) = H(s).X(s)$

Putting the values of X(s) and H(s) from equations 1 and 2 in the above equation, we get

$∴Y(s) = \dfrac 1{s + 2}\dfrac 1s \\ ∴Y(s) = \dfrac1{s (s + 2)} ------------- 3$

Expanding above equation in partial fractions

$Y(s) = \dfrac{k_O}s + \dfrac{k_1}{(s + 2)} ------------ 4 \\ \text {Here} k_O = s Y(s) |_{s = 0} \\ = s \dfrac1{s (s + 2) } |_{s = 0} \\ ∴k_O = \dfrac12 \\ k_1 = (s + 2) Y(s) |_{s = -2}\\ = (s + 2)\dfrac 1{s (s + 2)} |_{s = -2} \\ ∴k_1 = \dfrac{-1}2 $

Substituting the values of kO, and k1 in equation 4, we get

$Y(s) = \dfrac12 \dfrac 1s - \dfrac12 \dfrac 1 {(s + 2)} ------------- 5$

Taking inverse Laplace transform of above equation

$∴y(t) = \dfrac12 u(t) - \dfrac12 e^{-2t} u(t) $

This is the response of the system.