+ Post New Thread
Results 1 to 13 of 13

8th January 2014, 02:23 #1
 Join Date
 Aug 2013
 Posts
 50
 Helped
 0 / 0
 Points
 406
 Level
 4
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 splane, 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

8th January 2014, 09:22 #2
 Join Date
 Apr 2012
 Posts
 72
 Helped
 10 / 10
 Points
 1,050
 Level
 7
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.

8th January 2014, 09:22

8th January 2014, 09:51 #3
 Join Date
 May 2008
 Location
 Germany
 Posts
 5,691
 Helped
 1706 / 1706
 Points
 39,714
 Level
 48
Re: stability systems !!!
Joaquin,
I am afraid, you are mixing poles of open and closedloop systems.
You are right that a pole of the closedloop 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 openloop 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 closedloop system.Last edited by LvW; 8th January 2014 at 10:14.
1 members found this post helpful.

8th January 2014, 13:15 #4
 Join Date
 Aug 2013
 Posts
 50
 Helped
 0 / 0
 Points
 406
 Level
 4
Re: stability systems !!!
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

8th January 2014, 15:51 #5
 Join Date
 May 2008
 Location
 Germany
 Posts
 5,691
 Helped
 1706 / 1706
 Points
 39,714
 Level
 48
Re: stability systems !!!
Why do you expect a "slow response"? Pole close to the imag. axis have a large poleQ (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 openloop response.
A good example for this kind of stabilization is the openloop 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...)
1 members found this post helpful.

9th January 2014, 04:19 #6
 Join Date
 Aug 2013
 Posts
 50
 Helped
 0 / 0
 Points
 406
 Level
 4
Re: stability systems !!!
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 closedloop 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

9th January 2014, 04:19

9th January 2014, 09:47 #7
 Join Date
 May 2008
 Location
 Germany
 Posts
 5,691
 Helped
 1706 / 1706
 Points
 39,714
 Level
 48
Re: stability systems !!!
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 openloop 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 closedloop 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 OPENloop system! I spoke about the closedloop poles. You must not confuse both sysytems.
1 members found this post helpful.

9th January 2014, 20:04 #8
 Join Date
 Aug 2013
 Posts
 50
 Helped
 0 / 0
 Points
 406
 Level
 4
Re: stability systems !!!
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

9th January 2014, 20:50 #9
 Join Date
 May 2008
 Location
 Germany
 Posts
 5,691
 Helped
 1706 / 1706
 Points
 39,714
 Level
 48
Re: stability systems !!!
Joaquin, I don`t know if it is necessary to mention the following: Openloop poles move to another position after closing the loop.
Simple example: First order openloop 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.

9th January 2014, 22:31 #10
 Join Date
 Aug 2013
 Posts
 50
 Helped
 0 / 0
 Points
 406
 Level
 4
Re: stability systems !!!
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   
Sorry I forgot to said that in a book (is in spanish) here is the link: http://books.google.com.ar/books?id=QK148EPC_m0C&pg=PA420&lpg=PA420&dq=adicion+polo+compensacion&source=bl&ots=209sn97Xlg&sig=rwTRWwlIM8vpu32qNbSSalB0p6c&hl=es&sa=X&ei=GtnMUrC4JobLkAfOxoDwDg&ved=0CDQQ6AEwAQ#v=onepage&q=adicion%20polo%20compensacion&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
Joaquin

10th January 2014, 09:49 #11
 Join Date
 May 2008
 Location
 Germany
 Posts
 5,691
 Helped
 1706 / 1706
 Points
 39,714
 Level
 48
Re: stability systems !!!
1 members found this post helpful.

10th January 2014, 12:53 #12
 Join Date
 Aug 2011
 Location
 Dublin, Ireland
 Posts
 143
 Helped
 10 / 10
 Points
 1,708
 Level
 9
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
would help your understanding.
1 members found this post helpful.

10th January 2014, 12:53

10th January 2014, 21:26 #13
 Join Date
 Aug 2013
 Posts
 50
 Helped
 0 / 0
 Points
 406
 Level
 4
Re: stability systems !!!
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 !
+ Post New Thread
Please login