Question

A signal x(t) having unilateral Laplace transform as,

$X\left(s\right)=\frac{1}{e^{s+3}}.\frac{s^2}{(s+1)(s+2)}$

A function G(s) having relation with X(s) is defined as X(s) = $\frac{1}{e^{s+3}}G\left(s\right)$.

Select the correct statement(s).

• A. $x\left(t\right)=\frac{1}{e^3}\delta (t-1)-e^{-t}+4e^{-2t}$
• B. $x\left(t\right)=\frac{1}{e^3}\left[\delta \right(t-1)+e^{-t}u(t-1)-4e^{-2t}u(t-1\left)\right]$
• C. $g\left(t\right)=\delta \left(t\right)-e^{-t}+4e^{-2t}$
• D. $g\left(t\right)=\delta \left(t\right)+[e^{-t}-4e^{-2t}]u\left(t\right)$

Answer 1

Answer: (B), (D)

$g\left(t\right)=\delta \left(t\right)+[e^{-t}-4e^{-2t}]u\left(t\right)$

$X\left(s\right)=\frac{1}{e^{s+3}}.\frac{s^2}{(s+1)(s+2)}$

$X\left(s\right)=\frac{1}{e^{s+3}}G\left(s\right)$

$G\left(s\right)=\frac{s^2}{(s+1)(s+2)}=\frac{s^2}{s^2+3s+2}$

$G\left(s\right)=1-\frac{3s+2}{s^2+3s+2}=1-\frac{3s+2}{(s+1)(s+2)}$

$=1-\frac{A}{s+1}-\frac{B}{s+2}$

$A={lim}_{s\to -1}\frac{3s+2}{s+2}|=-1$

$B={lim}_{s\to -2}\frac{3s+2}{s+1}=4$

$G\left(s\right)=1+\frac{1}{s+1}-\frac{4}{s+2}$

$g\left(t\right)=\delta \left(t\right)+[e^{-t}-4e^{-2t}]u\left(t\right)$

$Now,X\left(s\right)=\frac{1}{e^{s+3}}G\left(s\right)=\frac{1}{e^3}{e}^{-s}G\left(s\right)$

$x\left(t\right)\leftrightarrow \frac{1}{e^3}g(t-1)$

$x\left(t\right)=\frac{1}{e^3}\left[\delta \right(t-1)+(e^{-(t-1)}-4e^{-2(t-1)}\left)u\right(t-1\left)\right]$