How do I calculate this inverse Laplace Transform

I have this transfer function and input, and I have to manually calculate what the systems response will be. The transfer function is:

and the input is:

I multiplied the two functions and used fraction decomposition by partial fraction expansion and calculated the inverse laplace transform and this is what I got:

But the response I get from that function does not correspond to the one I get using matlab. This is the matlab code I used:

t=0:0.1:100;
FT6=tf(0.1,[1 1 0.1])
r=4*sin(10*t);
V=lsim(FT6,r,t);
plot(t,V,'-')


How do I calculate it properly by hand?

• You have a transfer function in the Laplace domain, and an input signal in the time domain -- did you just multiply $FT_6(s)$ to $r(t)$, or did you take the Laplace transform of r(t)? – TimWescott Mar 19 '19 at 0:35
• I took the laplace transform of r(t), I forgot to mention it – Pedro Mar 19 '19 at 0:40
• Please show more of your work; that may help to pinpoint where the problem lies. – TimWescott Mar 19 '19 at 0:47
• Whatever else is going on, you've got a 2nd-order lowpass filter that's excited by a signal that starts at zero (i.e. $r(0) = 0$). So you would expect that not only the response, but its first two derivatives would be zero at $t = 0$. Yet your result has an initial value that is not zero. – TimWescott Mar 19 '19 at 0:52
• I included my calculations, I hope you can understand my writing – Pedro Mar 19 '19 at 1:00

I understand you have $$$$R(s) = \frac{40}{s^2+10^2}$$$$ So $$$$Y(s) = FT_6(s)R(s)=\frac{4}{(s^2+10^2)(s^2+s+0.1)}$$$$ Then you proceed by taking partial fractions of Y(s) $$$$Y(s) = \frac{A+sB}{(s^2+10^2)} +\frac{C+sD}{(s^2+s+0.1)}$$$$ Then you expand the term $$$$\frac{C+sD}{s^2+s+0.1} = \frac{C+sD}{(s+\frac{1}{2})^2-0.15} = D\frac{s+\frac{1}{2}}{(s+\frac{1}{2})^2-0.15} + (C-\frac{D}{2})\frac{1}{(s+\frac{1}{2})^2-0.15}$$$$ Now you can use a table (keeping in mind the table items 7,8,21 and 22).