There is a timely unchanged continuous function :
$$H(s)=\frac{s-1}{s+1}$$
At the entry of the system exists a $x(t)$ which Laplace's transformation is:
$$X(s)=\frac{(5s^2 - 15s + 7)}{(s-2)^3(s-1)}$$
Which is the impulse response?and which is the exit signal of the system?
Could someone show me the answers and elaborate me,on how this goes cause I have some clues but I am lacking i some aspects,Thanks in advance.
Here is how you advance. Use partial fraction to get
$$H(s)=1-\frac{2}{1+s}$$
Taking the inverse Laplace gives
$$h(t)=\delta(t)-2e^{-t},$$
where $\delta(t)$ is the Dirac function. Note that, the Laplace transform of the functions $\delta(t)$ and $e^{-t}$ are $1$ and $\frac{1}{s+1}$. For the second one use partial fraction
$$X(s)= \frac{17}{\left( s-2 \right)^2} - \frac{17}{ \left( s-2 \right)} +\frac{17}{ \left( s-1 \right)}-\frac{27}{ \left( s-2 \right)^3} .$$
Now, try to finish the problem.