Understanding stability

Thread Starter

Track99

Joined Jun 30, 2022
58
This question is about understanding stability in a control system.
The 2 slides below are taken from the lecture slides that are available at:
https://control.asu.edu/Classes/MAE318/318Lecture10.pdf


1699555426659.png

Long story short, the slides are telling us that if we close the initial loop with a gain of 1000, the entire system will become unstable.

I disagree with what is being said. I think that the system will still be stable.

I believe so, because when you convert the Closed-loop Transfer Function from the Laplace domain to the time domain, we see that the system reaches steady state nicely and neatly within 2 seconds as shown in the blue graph below.

1699555965812.png

1699556075755.png

I do not see anything unstable in the graph above. ( I am assuming that a system is unstable if the graph looks like a sine wave but if it "cools down" and flattens out, then it means that it is stable).

Can someone please tell me where am I wrong, if I am wrong?
Ty
 

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LowQCab

Joined Nov 6, 2012
3,937
Are You just studying the Math, or do You have a real-World Process application ???

Is this a Homework assignment ?
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anhnha

Joined Apr 19, 2012
905
You should use the closed-loop transfer function instead of the open loop. In this case, the open-loop plant is stable since there are no poles in the right-half plane. The closed-loop transfer function is G(s)/(1+G(s)). If you examine this, you'll see there are two poles on the right-half plane (RHP), indicating instability.
 

Thread Starter

Track99

Joined Jun 30, 2022
58
You should use the closed-loop transfer function instead of the open loop. In this case, the open-loop plant is stable since there are no poles in the right-half plane. The closed-loop transfer function is G(s)/(1+G(s)). If you examine this, you'll see there are two poles on the right-half plane (RHP), indicating instability.
Yes, you are correct, but my question is, why is it that when I graph the closed-loop transfer function ( as shown in one of the pics ), the graph is showing a stable output?
 

anhnha

Joined Apr 19, 2012
905
As mentioned before, you're working on the inverse Laplace transform of the open loop, which is stable due to the absence of right-half-plane (RHP) poles. Try the inverse Laplace transform of 1/s * G(s)/(1+G(s)) and plot the results over time for comparison.
This is the closed loop transfer function:

1699679021552.png
 
Last edited:

Ajith-N

Joined Sep 14, 2020
31
Yes, you are correct, but my question is, why is it that when I graph the closed-loop transfer function ( as shown in one of the pics ), the graph is showing a stable output?
I suggest that 'anhnha' has already said it. Let me try as well. The closed-loop transfer function is G(s)/( 1 + G(s) H(s)) = 1000/(s^3 + 10s^2 + 31s +1030). You should be graphing the Laplace inverse of that to see if it is stable after closing the loop. What you graphed is the inverse of just G(s) -- the plant in open loop.
 

LvW

Joined Jun 13, 2013
1,749
Just for clarification:
* We have some stability criteria which apply to the OPEN loop (frequency domain) - better known as "loop gain" (Nyquist, Barkhausen,).
* When such a criterion is fulfilled, we can be sure that the CLOSED loop will have no poles in the right half of the s-plane.
 

MrAl

Joined Jun 17, 2014
11,263
This question is about understanding stability in a control system.
The 2 slides below are taken from the lecture slides that are available at:
https://control.asu.edu/Classes/MAE318/318Lecture10.pdf


View attachment 307115

Long story short, the slides are telling us that if we close the initial loop with a gain of 1000, the entire system will become unstable.

I disagree with what is being said. I think that the system will still be stable.

I believe so, because when you convert the Closed-loop Transfer Function from the Laplace domain to the time domain, we see that the system reaches steady state nicely and neatly within 2 seconds as shown in the blue graph below.

View attachment 307116

View attachment 307118

I do not see anything unstable in the graph above. ( I am assuming that a system is unstable if the graph looks like a sine wave but if it "cools down" and flattens out, then it means that it is stable).

Can someone please tell me where am I wrong, if I am wrong?
Ty
Hello there,

As others have said now, you have to analyze the closed loop transfer function and that will be much different than your present choice.
There are a couple points here you might want to consider.

First, your inverse transform does not look correct even for your choice of transfer function. The coefficients look too large.
Second, when you see the laplace function already factored you have very little work to do to figure out if it is stable, stable in the strictest sense that is.
When you have (s+a) in the denominator that means you have a root of -a in the solution, and -a is in the left half plane therefore stable.
When you have (s+a)*(s+b) in the denominator, it's already factored, and that has roots -a and -b, both in the left half plane.
This goes for any number of linear factors like that as long as the constant is positive.
If any of the constants are negative, that means that one root is in the right half plane so you have more work to do, and that is usually unstable.

You can do a root locus which is very interesting. That can tell you what gain will cause the root locus plot to cross the jw axis, which makes it either an oscillator or just completely unstable.

Since the roots to examine are in the denominator, you can solve for 's' in the closed loop function denominator and look at the real parts of the results. If any real parts are zero or greater, that puts them in the right half plane or right on the jw axis. If they are on the jw axis that means it's an oscillator and that may be acceptable if you are building an oscillator, but not acceptable in the more general case like this one. If they are even just a tiny bit to the right of the jw axis that makes it completely unstable. If they are close to the jw axis that means you have to choose components very carefully, but hopefully change the gain a little so the roots are farther to the left of the jw axis (in a real application).

This problem is interesting because although as others have said this one is surely unstable, but with the right gain value(s) it can be made stable. You can actually solve for the gain that puts two of the roots right on the jw axis, and then reduce it to get it to be completely stable.

One thing to note here is if you factor the denominator you have to know how to factor using complex numbers because any of the second degree factors could be complex.

If you do not come back soon I'll post the root locus plot minus the actual values just so you can get an idea how this works.
 

LvW

Joined Jun 13, 2013
1,749
Yes, you are correct, but my question is, why is it that when I graph the closed-loop transfer function ( as shown in one of the pics ), the graph is showing a stable output?
The closed-loop transfer function is unstable.This can be shown as follows:
* The phase for open-loop function (loop gain) crosses the 360 deg-line (0 deg) at a frquency where the magnitude is above 0 dB. This is a violation of the stability criterion.
* The step response for the closed-loop transfer function goes high and approaches infinity.
 
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