Is this very simple system stable?

A simple feedback control loop:
input --> *(+-)* ---> 1/(s-1) ---*--> output
The feedback is simply multiplied by 2.
to *
Reply to
Harry Lin
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----- Original Message ----- From: "Harry Lin" Newsgroups: comp.robotics.misc Sent: Monday, August 23, 2004 8:40 PM Subject: Is this very simple system stable?
H/(1+G*H)=(1/(s-1))/(1+2*(1/(s-1)))=1/((s-1)+2)=1/(s+1) Okay, I agree with you here.
This doesn't mean it is unstable. If the loop gain was one then you would know it is metastable. Actually a non inverting first or second order system with poles in the left half plain will be unstable for any negative feedback. Your example seems to show that an inverting unstable first order system will be unstable for any positive feedback or equivalently a no inverting unstable first order system will be unstable for any negative feedback. By inverting I mean a DC gain less then one.
Positive feedback is not necessarily unstable. Notice though that H(s) is unstable. Maybe this is why the result is less then intuitive.
Bad intuition about loop gains and stability. Interesting tough.
Reply to
John Creighton
The system is stable. I believe the misstep is in the assumption that constant input to the system will produce the same constant input to the 1/(s-1) block. After transients have settled, the input to the 1/(s-1) block will actually be opposite in sign from the system input.
The transfer function from the system input to the input of the 1/(s-1) block is (s-1)/(s+1).
Reply to
James English
I get your point.... That transfer function is 1/(1+H) = 1/(1+(2/(s-1))) = (s-1)/(s+1)
Thanks a lot!!
Reply to
Harry Lin
Hi James,
Here's another question:
The bode plot of the open loop transfer function 2/(s-1) shows that
At w=0, phase = -180 deg and gain = (20*log(2)), a negative margin
Is that an indication of "unstable"?
If not, what are the correction conditions for a stable system according to its bode plot?
(positive gain margin, positive phase margin, am I right? )
Reply to
Harry Lin

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