0
\$\begingroup\$

I'll make up a problem to explain what I'm having trouble with. $$\dot x=\begin{bmatrix}0&1\\-5&-6\end{bmatrix}x+\begin{bmatrix}0\\1\end{bmatrix}u=Ax+Bu\\\\y=\begin{bmatrix}0&1\end{bmatrix}x=Cx$$

Now because of the matrix form we can easily get the transfer function $$T(s)=\frac{s}{s^2+6s+5}$$

That's usally how the problems start. At some point I'm given that a value from the matrix changes, say -5 becomes κ-5, altering the transfer function. Basically, I have to draw the root locus as k changes but in order to do that I must know if what I've found is the open loop or the closed loop transfer function. Which one is it?

\$\endgroup\$
  • \$\begingroup\$ Please upload a complete 'genuine' problem. The wording may be important. \$\endgroup\$ – Chu Aug 30 '17 at 8:32
  • \$\begingroup\$ It isn't @Chu , I've talked to some other students. They all find the problems incomplete and missing info. I'll delete all the incomplete questions I've uploaded or edit them and add the full answer. \$\endgroup\$ – John Katsantas Aug 30 '17 at 9:06
  • 1
    \$\begingroup\$ Clearly, the TF in this question is a CLTF - the highest state is a function of the input and the lower states. Hence setting the denominator = 0 gives the CE. For the root locus, one of the parameters is varied from 0 to infinity. \$\endgroup\$ – Chu Aug 30 '17 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.