I have the following information in a problem: an input $$r(t) = 60u(t)$$, with u(t) being the unit step function, and a transfer function $$H(s) = \frac{k_1}{s^{2} + s + k_{1}}$$, where k1 is a variable constant from a controller. I want to find the output y(t) but so far the farthest I can get is $$Y(s) = H(s)r(s) = \frac{k_1}{s^{2} + s + k_{1}} * \frac{60}{s}$$ Supposedly I can break this up into a partial fraction decomposition with three terms but I have no idea how to proceed because of the k1 term. Any help would be greatly appreciated
\$\begingroup\$
\$\endgroup\$
3
-
\$\begingroup\$ I want to verify something a classmate of mine did, they tried going around the problem by doing $$y(t) = \mathcal{L}^{-1}(H(s)) * r(t)$$ but is that a correct method of doing it? \$\endgroup\$– GiornoSassakiOct 27, 2020 at 0:00
-
\$\begingroup\$ You can express in partial fractions as \$\large \frac{A}{s}+\frac{Bs+C}{s^2 +s+k_1}\$, but the form of the solution will depend on \$k_1\$. There are five possible forms. \$\endgroup\$– ChuOct 27, 2020 at 1:29
-
\$\begingroup\$ @Chu Thank you for the reminder! :) I'm getting closer to an answer with $$Y(s) = \frac{60}{s} + \frac{-60s - 60}{s^{2} + s + k_1}$$. Thankfully the inverse laplace of the first term is just 60u(t), that just leaves the other two terms \$\endgroup\$– GiornoSassakiOct 27, 2020 at 1:58
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
3
Here's a little help with the partial fraction decomposition: -
$$\dfrac{1}{s}\cdot \dfrac{k}{s^2 + s +k}\hspace{1cm} =\hspace{1cm} \dfrac{1}{s}+ \dfrac{-s-1}{s^2 + s + k}$$
Can you take it from here?
If not try this inverse laplace solver.
answered Oct 26, 2020 at 22:42
-
\$\begingroup\$ I appreciate this, but I would also like to know how you got that partial fraction decomposition. I tried using the quadratic formula on the expression with the k1 to find the roots but I think I just ended up with a mess. And once I do find that PFD, how can I get the expression into a form that resembles common laplace transforms? \$\endgroup\$ Oct 26, 2020 at 23:56
-
\$\begingroup\$ If you use the link I gave you in my answer, you can work it out from there. Sure the PFD of the 2nd fraction is trickier to follow but if you stick some real numbers in, you can figure out how the number (that replaces k) evolves into the formula. \$\endgroup\$– Andy akaOct 27, 2020 at 0:03
-