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Now K(s) is obviously the negative feedback loop which is ${H(s) \over 1+H(s)G(s)}$

When I substitute ${H(s) = {1 \over s-2}}$ I get K(s) = ${ 1 \over s-2 +G(s) }$

For the system to be stable I know that the poles have to be on the negative half of th imaginary plane. Not sure how to find G(s).

  • $\begingroup$ Do we want to find any one function that will make this stable? In that case, some straightforward trial and error should do the trick. Or are you supposed to solve for all such functions? $\endgroup$ – Phonon Jun 19 '14 at 16:25


Since $K(s) = \frac{1}{s-2+G(s)}$, then the roots of $s-2+G(s)$ must lie on the left side of the complex plane. Let's suppose that $G(s)$ is of the form $as + b$. Can you solve it then?

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  • 2
    $\begingroup$ I was about to write a full solution but I just realised it is #homework... +1 ! $\endgroup$ – jojek Jun 19 '14 at 16:33
  • $\begingroup$ You will get s= (b+2)/(a+1). So s must be less than that? $\endgroup$ – martinap Jun 23 '14 at 10:44
  • $\begingroup$ @martinap This about which values of $a$ and $b$ will make this system stable. $\endgroup$ – Phonon Jun 23 '14 at 17:20

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