# Improving stability systems

1. ## Improving stability systems

Hi all,

I have a question related to the stability of feedback systems. I read some theory of systems compensation and there is a method which consist in adding a pole into the system.

For example if I have a pole near the imaginary axis in the s-plane, the time response of the system will be longer. What I don´t understand of compensation of adding a pole to the system is that if I add a pole with the objective to increase the phase margin of my system, this pole will be closer to the origin, so this makes mi confuse, I want to improve my stability (phase margin) but this pole is closer to the origin so the time response will be longer.

Can you help me to understand this please ?

Regards !

Joaquin

•

2. ## Re: stability systems !!!

How will a pole close to the UGB improve the stability? It can only degrade. I guess you want to introduce a zero close to the UGB, which will mean in closed loop you will have peaking eventhough you have proper phase margins. So, your settling(99%) will be bad but atleast the time to reach 90% would be quiet significant. So, it really means what you want, 90 or 99% settling.

1 members found this post helpful.

3. ## Re: stability systems !!!

Joaquin,

I am afraid, you are mixing poles of open and closed-loop systems.
You are right that a pole of the closed-loop system which is close to the imag. axis (small damping factor sigma) will cause a small phase margin.
This can be improved by adding another real dominating pole to the open-loop system which is known as "lag compensation.
This new pole will cause the magnitude to drop at lower frequencies thereby improving the phase margin. However, this method reduces the bandwidth of the closed-loop system.

1 members found this post helpful.

4. ## Re: stability systems !!!

Originally Posted by LvW
Joaquin,

I am afraid, you are mixing poles of open and closed-loop systems.
You are right that a pole of the closed-loop system which is close to the imag. axis (small damping factor sigma) will cause a small phase margin.
This can be improved by adding another real dominating pole to the open-loop system which is known as "lag compensation.
This new pole will cause the magnitude to drop at lower frequencies thereby improving the phase margin. However, this method reduces the bandwidth of the closed-loop system.

Hi LYw, thank you for your response !. For example if I have a close loop system with a negative feedback equal to 1, if I have poles closer to the imaginary axis, I understand that the system will have a slow response cause the dominants poles of the system but what make feel confuse is that you told me that a pole near the imaginary. axis will cause a small phase margin, so if I add another pole closer to the axis imaginary I think that this will make smaller the phase margin. Why did you tell me that this will increase the phase margin ?

I read some theory of compensation and in the explanation of adding a dominant pole, If I add a smaller pole in the system, then using bode and phase margin, I can see that the system increase it´s phase margin. For this case I have the same question. I improved my phase margin making a dominant pole closer to the imaginary axis. For me this is a little contradictory. Can you help me ?

Best regards !
Joaquín

5. ## Re: stability systems !!!

Originally Posted by biolycans
For example if I have a close loop system with a negative feedback equal to 1, if I have poles closer to the imaginary axis, I understand that the system will have a slow response cause the dominants poles of the system ...
Why do you expect a "slow response"? Pole close to the imag. axis have a large pole-Q (quality factor) resulting in a quick response (however, the step response has overshoot and ringing).

Originally Posted by biolycans
.., so if I add another pole closer to the axis imaginary I think that this will make smaller the phase margin. Why did you tell me that this will increase the phase margin ?
Did I? No, I have mentioned "lag compensation" by adding a REAL pole to the open-loop response.
A good example for this kind of stabilization is the open-loop gain compensation for opamps. As aresult - the bandidth is drastically reduced (typically down to 10 Hz or so).

Originally Posted by biolycans
I read some theory of compensation and in the explanation of adding a dominant pole, If I add a smaller pole in the system, then using bode and phase margin, I can see that the system increase it´s phase margin. For this case I have the same question. I improved my phase margin making a dominant pole closer to the imaginary axis. For me this is a little contradictory. Can you help me ?
You are right. This sounds contradictory. Something must be wrong - either in your description or in your test (..I improved my phase margin...)

1 members found this post helpful.

•

6. ## Re: stability systems !!!

Originally Posted by LvW
Why do you expect a "slow response"? Pole close to the imag. axis have a large pole-Q (quality factor) resulting in a quick response (however, the step response has overshoot and ringing).

Did I? No, I have mentioned "lag compensation" by adding a REAL pole to the open-loop response.
A good example for this kind of stabilization is the open-loop gain compensation for opamps. As aresult - the bandidth is drastically reduced (typically down to 10 Hz or so).

You are right. This sounds contradictory. Something must be wrong - either in your description or in your test (..I improved my phase margin...)

Hello again, sorry to bother you again. I did this:

1) I simulated in Matlab an open loop systems where the transfer function is a(s) = 9.5 / ( (1+s/3) * (1+s/5) * (1+s/8) ) and the feedback is f=1.

Using the open loop system (a(s) * f), the phase margin and gain margin for this open loop system was approximately 0 grades and 0 dB. So I add a new polo in this system much smaller than the original dominant pole.

I used again the open loop system but in this case with the compensation (new dominant polo in the lower frequencies) and the phase margin and gain margin were positive so the system was stable.

Then I compared the step response in the time domain for the uncompensated and compensated system with the objective to see the differences between both. The step response related to the compensated system has a faster response but the overshoot bigger than the original system without compensation.

So to conclude if my phase margin is bigger the response of the close loop system in the time domain will be faster. Am I correct ?

So what make me feel confuse is that in one of your post, you told me this:

"You are right that a pole of the closed-loop system which is close to the imag. axis (small damping factor sigma) will cause a small phase margin."

This make feel confuse because, in the compensation I added a new pole closer than the other poles to the axis imaginary so in the simulation the phase margin of the open loop system increase and then the system in closed loop was stable with a response in time faster. Can you help me please ?

Regards,

Joaquin

•

7. ## Re: stability systems !!!

Originally Posted by biolycans
1) I simulated in Matlab an open loop systems where the transfer function is a(s) = 9.5 / ( (1+s/3) * (1+s/5) * (1+s/8) ) and the feedback is f=1.
Using the open loop system (a(s) * f), the phase margin and gain margin for this open loop system was approximately 0 grades and 0 dB.
Yes - I agree.

Originally Posted by biolycans
Using the open loop system (a(s) * f), the phase margin and gain margin for this open loop system was approximately 0 grades and 0 dB. So I add a new polo in this system much smaller than the original dominant pole.
I used again the open loop system but in this case with the compensation (new dominant polo in the lower frequencies) and the phase margin and gain margin were positive so the system was stable.
What means "much smaller"? What was the value you have used?
For example, the loop gain starts to decrease at wp1=3 rad/s (smallest open-loop pole); the new pole (compensation) should be, for example, at wp4=0.1 rad/s or smaller (I simulated with wp4=0.05 rad/s with excellent results)

Originally Posted by biolycans
Then I compared the step response in the time domain for the uncompensated and compensated system with the objective to see the differences between both. The step response related to the compensated system has a faster response but the overshoot bigger than the original system without compensation.
So to conclude if my phase margin is bigger the response of the close loop system in the time domain will be faster. Am I correct ?
I wonder how the step response of the closed-loop system without compensation in your simulation looks like. It is unstable!
With compensation (wp4) the step response is very good without any overshoot or ringing.

Originally Posted by biolycans
So what make me feel confuse is that in one of your post, you told me this:
"You are right that a pole of the closed-loop system which is close to the imag. axis (small damping factor sigma) will cause a small phase margin."
This make feel confuse because, in the compensation I added a new pole closer than the other poles to the axis imaginary so in the simulation the phase margin of the open loop system increase and then the system in closed loop was stable with a response in time faster. Can you help me please ?
The compensating pole (wp4) is a pole of the OPEN-loop system! I spoke about the closed-loop poles. You must not confuse both sysytems.

1 members found this post helpful.

8. ## Re: stability systems !!!

Originally Posted by LvW
Yes - I agree.

What means "much smaller"? What was the value you have used?
For example, the loop gain starts to decrease at wp1=3 rad/s (smallest open-loop pole); the new pole (compensation) should be, for example, at wp4=0.1 rad/s or smaller (I simulated with wp4=0.05 rad/s with excellent results)

I wonder how the step response of the closed-loop system without compensation in your simulation looks like. It is unstable!
With compensation (wp4) the step response is very good without any overshoot or ringing.

The compensating pole (wp4) is a pole of the OPEN-loop system! I spoke about the closed-loop poles. You must not confuse both sysytems.

Hi,

Related to the wp4 for the compensation I chose a wp4=0.1. An the results were good in comparison with the original open loop system without compensation.

For example my open loop system with the compensation (new pole in wp4=0.1), produced a better phase margin. What you told me is that if I add the wp4 in the close loop system for the original transfer function the phase margin will decrease. am I correct ?

I did this:

The phase margin and gain margin of my original open loop system (a(s) * f), where a(s)= 9.5 / ( (1+s/3) * (1+s/5) * (1+s/8) ) and f=1 were PM=0 degrees and GM=0 dB.

I calculated the close loop system and I added a new pole in wp4=0.1 to this system. the characteristic equation of my close loop system with the new pole is:

0.083 * s^4 + 1.3083 * s^3 + 6.713 * s^2 + 100.658 * s + 10.5

the numerator of the close loop system is 9.5. then in matlab I did the margin(9.5,characteristic equation); The results were that the phase margin and gain margin decrease, so the system became unstable.

Am I correct ?

Thank you very much for your help !

Joaquin

9. ## Re: stability systems !!!

Originally Posted by biolycans
Hi,
Related to the wp4 for the compensation I chose a wp4=0.1. An the results were good in comparison with the original open loop system without compensation.

OK - that is in agreement with my results.

For example my open loop system with the compensation (new pole in wp4=0.1), produced a better phase margin.

Yes - correct

What you told me is that if I add the wp4 in the close loop system for the original transfer function the phase margin will decrease. am I correct ?

No, I didn´t tell you this. Where? Read my post#3 again. A closed-loop pole near the imag. axis reduces the margin. However, I have added wp4 to the open-loop system, thereby improving the margin.

I did this:
The phase margin and gain margin of my original open loop system (a(s) * f), where a(s)= 9.5 / ( (1+s/3) * (1+s/5) * (1+s/8) ) and f=1 were PM=0 degrees and GM=0 dB.
I calculated the close loop system and I added a new pole in wp4=0.1 to this system. the characteristic equation of my close loop system with the new pole is:
0.083 * s^4 + 1.3083 * s^3 + 6.713 * s^2 + 100.658 * s + 10.5
the numerator of the close loop system is 9.5. then in matlab I did the margin(9.5,characteristic equation); The results were that the phase margin and gain margin decrease, so the system became unstable.
Am I correct ?
No, this is in contradiction to your ststement in the first line where you have mentioned that results were good.
More than that, my simulation result for the closed-loop system (wp4=0.05 rad/s) was: two conjugate-complex pole pairs.

Thank you very much for your help !

Joaquin
Joaquin, I don`t know if it is necessary to mention the following: Open-loop poles move to another position after closing the loop.
Simple example: First order open-loop system 100/(1+s/0.1). For feedback factor "1" (100% feedback) the pole will move from 0.1 rad/s to 10 rad/sec.

1 members found this post helpful.

10. ## Re: stability systems !!!

Originally Posted by LvW
Joaquin, I don`t know if it is necessary to mention the following: Open-loop poles move to another position after closing the loop.
Simple example: First order open-loop system 100/(1+s/0.1). For feedback factor "1" (100% feedback) the pole will move from 0.1 rad/s to 10 rad/sec.
Hi,

I understood what you told me now. Let me explain what I understood. First I have a system a(s) and a feedback=1. For this case the phase margin and gain margin are bad or really small, so what I want to do is to compensate the system, that means increase the value of phase and gain margin. Am I correct ?

So What I should do is to add a new pole closer to the imaginary axis in the open loops system and when I do this, the closed loop system became more stable, so the phase and gain margin are bigger.

So to conclude if I have a closed loop system (with no compensation) with their poles close to the imaginary axis, the phase margin will be small or negative. So to correct this I add a new pole (wp4) closer to the imaginary axis in the open loop system. When a simulate the response of the closed loop system, the response is better than the original system. and the roots of the characteristic equation of the closed loop system (with compensation wp4) are more far than the poles in the open loop system (including the wp4). Am I correct ?

Another and I hope final question.

If I have an open loop system (a(s) * f), where f=1 and the phase and gain margin are GM=5 dB and PM=23 degrees for example, if I increase the gain of the open loop system with a value of 5 dB, then the GM and PM of this system are approximately 0 (in the limit of stability). So what I can see here is that the GM and PM are really related each other. So my question is: Can I improve the gain margin, with no changes in the phase margin and viceversa ?

Thank you again. Regards.

Joaquin

- - - Updated - - -

There is a part in the book that says: the effect of adding a pole in a open loop system has this consequencies:

1) fast response of the system
2) put the roots locus to the right so it reduce the relative stability. Is this correct ? If I add a pole in the open loop system, then the close loop system becomes more stable

Joaquin

•

11. ## Re: stability systems !!!

Originally Posted by biolycans
Hi,
I understood what you told me now. Let me explain what I understood. First I have a system a(s) and a feedback=1. For this case the phase margin and gain margin are bad or really small, so what I want to do is to compensate the system, that means increase the value of phase and gain margin. Am I correct ?

Yes - of course.

So What I should do is to add a new pole closer to the imaginary axis in the open loops system and when I do this, the closed loop system became more stable, so the phase and gain margin are bigger.
So to conclude if I have a closed loop system (with no compensation) with their poles close to the imaginary axis, the phase margin will be small or negative. So to correct this I add a new pole (wp4) closer to the imaginary axis in the open loop system. When a simulate the response of the closed loop system, the response is better than the original system. and the roots of the characteristic equation of the closed loop system (with compensation wp4) are more far than the poles in the open loop system (including the wp4). Am I correct ?

The last sentence sounds too simple (...more far). The closed-loop poles are not only shifted (if compared with open-loop conditions), but are completely modified. In most cases, they become conjugate complex (as in your example: 4 real poles are tansferred to 2 complex pole pairs)

Another and I hope final question.
If I have an open loop system (a(s) * f), where f=1 and the phase and gain margin are GM=5 dB and PM=23 degrees for example, if I increase the gain of the open loop system with a value of 5 dB, then the GM and PM of this system are approximately 0 (in the limit of stability). So what I can see here is that the GM and PM are really related each other. So my question is: Can I improve the gain margin, with no changes in the phase margin and viceversa ?

No, in your example only the GM is reduced to zero. The PM is not necessarily zero (can be positive or negative). Both margins are "related" - however, in a manner that depends on the particular system. Answer to the last question: No (see my answer in the other thread which you have opened.)

Thank you again. Regards.

Joaquin

- - - Updated - - -

Sorry I forgot to said that in a book (is in spanish) here is the link: http://books.google.com.ar/books?id=...sacion&f=false

There is a part in the book that says: the effect of adding a pole in a open loop system has this consequencies:

1) fast response of the system
2) put the roots locus to the right so it reduce the relative stability. Is this correct ? If I add a pole in the open loop system, then the close loop system becomes more stable

No, cannot be stated in general. Depends on the particular system and on the ploe location.

Joaquin
General hint: Sometime it helps to visualize the effect of an additional pole using the BODE plot. Try to find a book which explains the different methods how the open-loop response of an opamp is compensated for universal applications

1 members found this post helpful.

12. ## Re: stability systems !!!

biolycans,

I would recommend you read through the post:

"What is the link between transient and frequency instability?"

In this post LvW answered a lot of my fundamental questions regarding stability. I think the info

1 members found this post helpful.

13. ## Re: stability systems !!!

Originally Posted by LvW
General hint: No, in your example only the GM is reduced to zero. The PM is not necessarily zero (can be positive or negative). Both margins are "related" - however, in a manner that depends on the particular system. Answer to the last question: No (see my answer in the other thread which you have opened.)
Hi,

When I increment the gain of the open loop system, the phase margin and gain margin became zero again. I don´t understand why you told me that the PM is not necessarily zero. For this case I increment the gain that makes my close system being in the limit of stability. This means that I reduced my gain margin.

How can I modify only the phase margin of the open loop system so my system goes to the limit of instability ?

Regards !

--[[ ]]--