I am struggling with a systems theory problem, the task is as follows:
u(t) -> H(s) -> y(t)
H(s) being the transfer function
$$ H(s) = H(s) = \frac{s+1}{s(s+2)^{2}} $$
$$ u(t) = e^{-5t} $$
So using Laplace transformation i get
$$ u(s) = \frac{1}{s+5} $$ From here I try to calculate Y(s) $$ Y(s) = \frac{1}{s+5} * \frac{s+1}{s(s+2)^{2}} $$ And this is were I am stuck, I have used partial fraction decompostion like so: $$ Y(s) = \frac{1}{s+5} * \frac{s+1}{s(s+2)^{2}} = \frac{k_{1}}{s} + \frac{k_{2}}{s+2} + \frac{k_{3}}{(s+2)^{2}} + \frac{k_{4}}{s+5} $$ However that has not worked, so could anyone please help me out? Give some guidance on how to move forward and get the y(t).
Thank You!
We have $$Y(s) = \left(\frac1{5+s}\right)\left(\frac{1+s}{s(2+s)^2}\right) = \frac{1+s}{s(2+s)^2(5+s)}. $$ The partial fraction decomposition is of the form $$\frac{1+s}{s(2+s)^2(5+s)} = \frac As + \frac B{2+s} + \frac C{(2+s)^2} + \frac D{5+s}. $$ This implies $$A(2+s)^2(5+s) + B(s(2+s)(5+s)) + C(s(5+s)) + D(s(2+s)^2) = 1+s $$ for all $s$. Taking $s=0$ yields $$20A=1\implies A=\frac1{20}, $$ taking $s=-2$ yields $$-6C = -1\implies C=\frac16, $$ taking $s=-5$ yields $$-45D = -4 \implies D = \frac4{45}, $$ and finally, taking $s=-1$ yields $$4A - 4B - 4C -D = 0\implies B = \frac1{4}\left(\frac4{20}-\frac46-\frac4{45}\right)=-\frac5{36}. $$ It follows that $$Y(s) = \frac{1}{20 s}-\frac{5}{36 (s+2)}+\frac{1}{6 (s+2)^2}+\frac{4}{45 (s+5)} ,$$ and so $$y(t) = \mathcal L^{-1}\{Y(s)\}=\frac1{20}e^{-t} -\frac5{36}e^{-2t} + \frac16 te^{-2t} +\frac4{45}e^{-5t}. $$