0
$\begingroup$

I am currently doing a control systems course and I was lead to believe that if a pole lies on the imaginary axis in the s-plane (i.e s=0), then the system was marginally/critically stable.

However, I come across this transfer function and plotted it in Simulink.

enter image description here

Analysis shows that there are 3 poles at s=-1, s=-3 and s=0. So because there is a pole at s=0, the system should be marginally stable right? But the output of the transfer function shows this:

enter image description here

To me, that is showing the system is unstable because the output is getting bigger and bigger as time goes on. Is this correct? Or is my understanding still flawed?

If the system is unstable, then why aren't any of the poles in the right half of the s-plane?

Thanks

$\endgroup$
2
$\begingroup$

Your system is open loop stable as the poles are at $s=-1$, $s=-3$ and $s=0$. Note, that if the order of the pole at $s=0$ is greater then 1, then the open loop system is also unstable.

But closing the loop changes the poles of the system. If $F(s)$ is your transfer function of the open loop system, then the transfer function of the closed loop system is:

$$G(s)=\frac{F(s)}{1+F(s)}=\frac{21}{s^3+4s^2+3s+21}.$$

The poles of $G(s)$ are $s \approx -4.4022$, $s \approx 0.2011 - 2.1748 i$ and $s \approx 0.2011 + 2.1748 i$. Hence, the closed loop system is unstable.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.