input -->

***(+-)***---> 1/(s-1) ---

***--> output**

The feedback is simply multiplied by 2.

to *(-)*

The feedback is simply multiplied by 2.

to *

- posted
18 years ago

A simple feedback control loop:

input -->***(+-)*** ---> 1/(s-1) ---***--> output**

The feedback is simply multiplied by 2.

to *(-)*

input -->

The feedback is simply multiplied by 2.

to *

Loading thread data ...

- posted
18 years ago

----- 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.

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.

- posted
18 years ago

Harry,

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).

James

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).

James

- posted
18 years ago

I get your point.... That transfer function is 1/(1+H) = 1/(1+(2/(s-1)))
= (s-1)/(s+1)

Thanks a lot!!

Harry

Thanks a lot!!

Harry

- posted
18 years ago

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? )

Harry

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? )

Harry

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