Question

$\text{Consider a signal y(t) which is related to two signals}{x}_1\left(t\right)\text{and}{x}_2\left(t\right)\text{by,}y\left(t\right)=x_1(t-2)\ast x_2(-t+3)\text{where}{x}_1\left(t\right)=e^{-2t}.u\left(t\right)\text{and}{x}_2\left(t\right)=e^{-3t}u\left(t\right).\text{Then the Laplace transform of}Y\left(s\right)\text{is}$

• A. $\frac{e^{-s}}{(s+2)(s+3)}$
• B. $\frac{e^s}{(s+2)(s+3)}$
• C. $\frac{e^{-5s}}{(s+2)(s-3)}$
• D. $\frac{e^{-5s}}{(s+2)(3-s)}$

Answer 1

The correct option is D $\frac{e^{-5s}}{(s+2)(3-s)}$

$x_1\left(t\right)\stackrel{L.T}{⟶}\frac{1}{s+2}$

$x_2\left(t\right)\stackrel{L.T.}{⟶}\frac{1}{s+3}$

$x_1(t-2)\stackrel{L.T.}{⟶}\frac{e^{-2s}}{s+2}$

$x_2(t+3)\stackrel{L.T.}{⟶}\frac{e^{3s}}{s+3}$

$x_2(-t+3)\stackrel{L.T.}{⟶}\frac{e^{-3s}}{3-s}$

$y\left(t\right)=x_1\left(t\right)\ast x_2\left(t\right)$

$Y\left(s\right)=X_1\left(s\right)\cdot X_2\left(s\right)$

$Y\left(s\right)=\frac{e^{-2s}}{(s+2)}.\frac{e^{-3s}}{(3-s)}$

$Y\left(s\right)=\frac{e^{-5s}}{(s+2)(3-s)}$