Question

Given x(t) = e-tu(t), find inverse Laplace transform of
e-3s X(2s)​

Answer 1

Answer:

mujhe nahi aati soooooory

Answer 2

Answer:

The inverse Laplace transform of $L^{-1}[e^{-3s}X(2s)]=\frac{1}{2} e^{-(t-3)/2}u(t-3)$

Step-by-step explanation:

Given $x(t) = e^{-t}u(t)$

The inverse Laplace transform of a function $F(s)$ is the piecewise-continuous and exponentially-restricted real function $f(t)$ which satisfies the property:

${L}}\{f(t)\}(s)=F(s)$

where $L$ denotes the Laplace transform.

Doing the Laplace transform piecewise,

first consider the Laplace transform for $X(s)$

$L[X(s)]=L[e^{-t}u(t)]$

$=\frac{1}{s-(-1)}=\frac{1}{s+1}$

Similarly the Laplace transform for $X(2s)$ is

$L(X(2s))=\frac{1}{2s+1}$

$=\frac{1}{2s-(-1)}=\frac{1}{2} \frac{1}{s-(-1/2)}$

Therefore, $L(X(2s))= \frac{1}{2} \frac{1}{s-(-1/2)}$

Now the inverse Laplace transform for $X(2s)$ is

$L^{-1}[X(2s)]=L^{-1}\Big[\frac{1}{2} \frac{1}{s-(-1/2)}]=\frac{1}{2} e^{-t/2}u(t)$

But given to find the inverse Laplace transform of

$F(s)=e^{-3s}X(2s)$

Thus,

$L^{-1}[e^{-3s}X(2s)]=L^{-1}[X(2s)]_{t=t-3}$

$=\frac{1}{2} e^{-(t-3)/2}u(t-3)$

Therefore, the inverse Laplace transform of  

$L^{-1}[e^{-3s}X(2s)]=\frac{1}{2} e^{-(t-3)/2}u(t-3)$.