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

enter image description here

and the input is:

enter image description here

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:

enter image description here

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?

enter image description here

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  • \$\begingroup\$ 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)? \$\endgroup\$
    – TimWescott
    Mar 19, 2019 at 0:35
  • \$\begingroup\$ I took the laplace transform of r(t), I forgot to mention it \$\endgroup\$
    – Pedro
    Mar 19, 2019 at 0:40
  • \$\begingroup\$ Please show more of your work; that may help to pinpoint where the problem lies. \$\endgroup\$
    – TimWescott
    Mar 19, 2019 at 0:47
  • \$\begingroup\$ 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. \$\endgroup\$
    – TimWescott
    Mar 19, 2019 at 0:52
  • \$\begingroup\$ I included my calculations, I hope you can understand my writing \$\endgroup\$
    – Pedro
    Mar 19, 2019 at 1:00

1 Answer 1

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

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