I'm doing some coursework on control theory and I'm unsure whether I'm getting the right answer for one of the questions:enter image description here

When I go through all the analysis, I get the response $$y(t) = 3U(t)(4-2e^{6t}(\cos4t+6\sin4t))$$

which is clearly unstable. This is not what I expected since the poles of the characteristic equation (\$-6\pm4j\$) lie in the left-hand plane, and the system should should therefore be stable. Should the system be stable or not?

  • \$\begingroup\$ The exponential index must be negative: the real part of your roots are -6. A 2nd order TF cannot be unstable unless it has a negative coefficient. \$\endgroup\$ – Chu Mar 25 '17 at 19:59
  • \$\begingroup\$ @Chu Thanks for the help. Can I update my question to ask about checking my method, or should I ask a new question? I really can't see where I've gone wrong with this \$\endgroup\$ – imulsion Mar 25 '17 at 20:09
  • \$\begingroup\$ @Chu: What about \$p^2\$ it does not have negative coefficients but is still unstable. Sorry for beeing picky :). \$\endgroup\$ – MrYouMath May 29 '17 at 11:19
  • \$\begingroup\$ @imulsion: As long as there are no answers you can edit your question. It is not good practice to change your question after someone invested his time to answer your first question. Btw what is your exact problem? You simply need to determine the inverse laplace transform of to obtain \$y(t)=L^{-1}\left{\frac{24s+208}{s^2+12s+52}\frac{1}{s}\right}\$. You can achieve this by partial fractions (easy) or the Bromwich integral (difficult) for the inverse Laplace transform. \$\endgroup\$ – MrYouMath May 29 '17 at 11:21
  • \$\begingroup\$ @MrYouMath, \$ p^2\$ is oscillatory ('critically stable'). \$\endgroup\$ – Chu May 29 '17 at 11:51

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