Designer’s Guide Community

‹ Previous topic | Next topic ›

Pages: 1 2

Bode Diagram of Conditionally Stable System (Read 46082 times)

ndnger

Junior Member

Offline

Posts: 11
Los Angeles

Re: Bode Diagram of Conditionally Stable System
Reply #15 - May 10^th, 2009, 1:04am

I agree, my logic is flawed. I guess because that kind of treatment gives correct result for some cases only but is not fundamentally valid for all cases.

But what is fundamentally true is that the closed loop response is given by A/(1+A*beta). At the red dot frequency A is given by -k where k is positive and mod(k) >> 1. Hence the output becomes -k/(1-k) times Vin(s), for beta = 1. Since k is positive and k>1, both num and denom. of this expression are negative, hence Vo(s) = positive qty times Vin (s). Also mod(-k/(1-k)) > 1, since mod(1-k) < mod(k). Hence the error term which is input to the amplifier is a negative qty given by 1/(1-k) times Vin(s). All these quantities, i.e., input, error term, and output are bounded and the system has well defined, stable ouputs at this frequency and hence it wouldn't oscillate or latch.

Coming back to your question, of why does positive feedback not result in oscillation. So, at this particular frequency, error term becomes negative which going through the amplifier and the negation stays negative because of overall positive f/b. However input is positive and so it wouldn't add to the i/p, rather subtracts and since output is larger in magnitude, the error term stays negative.

Here's a simple example case: Suppose amp gain = -10 at the red dot frequency. Then Vo = -10/(1-10) = 10/9 of Vin.

Verr = Vin - Vo = Vin - (10/9)Vin = (-1/9)Vin, which after going through the amp becomes (10/9)Vin as above.

This analysis is general and will always be correct. Nyquist criteria is a shortcut for stability analysis and when the open loop transfer fn is not readily available.

IP Logged

pancho_hideboo

Senior Fellow

Offline

Posts: 1424
Real Homeless

Re: Bode Diagram of Conditionally Stable System
Reply #16 - May 10^th, 2009, 2:07am

ndnger wrote on May 10^th, 2009, 1:04am:

But what is fundamentally true is that the closed loop response is given by A/(1+A*beta).
At the red dot frequency A is given by -k where k is positive and mod(k) >> 1.
Hence the output becomes -k/(1-k) times Vin(s), for beta = 1.
Since k is positive and k>1, both num and denom of this expression are negative, hence Vo(s) = positive qty times Vin (s).
Also mod(-k/(1-k)) > 1, since mod(1-k) < mod(k).

Correct.

However what do you mean by "mod()" ?
modulo or mode ?

ndnger wrote on May 10^th, 2009, 1:04am:

Hence the error term which is input to the amplifier is a negative qty given by 1/(1-k) times Vin(s).
All these quantities, i.e., input, error term, and output are bounded
and the system has well defined, stable ouputs at this frequency and hence it wouldn't oscillate or latch.
Coming back to your question, of why does positive feedback not result in oscillation.

Your argument assume nonlinear effect.

I don't argue nonlinear effect.
http://www.designers-guide.org/Forum/YaBB.pl?num=1234428781/7#7

My question is why the conditionally stable system does not create poles in RHP(Right Half Plane) regardless of positive feedback.
Here I argue completely linear system.

ndnger wrote on May 10^th, 2009, 1:04am:

So, at this particular frequency, error term becomes negative which going through the amplifier
and the negation stays negative because of overall positive f/b.
However input is positive and so it wouldn't add to the i/p, rather subtracts
and since output is larger in magnitude, the error term stays negative.

Here's a simple example case: Suppose amp gain = -10 at the red dot frequency.
Then Vo = -10/(1-10) = 10/9 of Vin.
Verr = Vin - Vo = Vin - (10/9)Vin = (-1/9)Vin, which after going through the amp becomes (10/9)Vin as above.

This analysis is general and will always be correct.

What on earth is correct ?

ndnger wrote on May 10^th, 2009, 1:04am:

Nyquist criteria is a shortcut for stability analysis and when the open loop transfer fn is not readily available.

My example of nyquist diagram is for open loop transfer function.

Your story is completely contradictory to Nyquist Criteria.

If your story is correct, even a simple system which is judged as unstable by Nyquist Criteria will be stable.
http://www.engin.umich.edu/group/ctm/freq/nyq.html

In my example which is a very typical example for conditional stable system, if yellow point moves on Real-Axis inside magenda unit circle or blue point moves on Real-Axis outside magenda unit circle in Nyquist Diagram, system will be unstable, although red point remains in same point.
http://www.designers-guide.org/Forum/YaBB.pl?num=1190272820/0#0

Here some poles in RHP(Right Half Plane) will be created by this yellow point or blue point movement.

ndnger wrote on May 9^th, 2009, 10:18pm:

I'm not an expert in control theory
so I'd really like if someone can simulate this kind of system in cadence or matlab and see what actually happens.

What tool of Cadence do you mean ?
I don't think there are some tools suitable for this topic in Cadence. There is only common Spice which Cadence call as Spectre.

I have "Control System Toolbox" of MATLAB.
I can easily plot "Bode Diagram", "Nyquist Diagram", "Root Locus" and "Pole-Zero Plot".
Other than these classical method, we can use modern control theory such as State Space Formulation to judge stability.
Any judgememt for my system results in stable.

« Last Edit: May 11^th, 2009, 1:45am by pancho_hideboo »

Kita━━━━━━(ﾟ∀ﾟ)━━━━━━ !!!!!
http://www7.plala.or.jp/ungeromeppa/flash/kita.html
http://www.youtube.com/watch?v=mjIxGh55bMM&feature=related

IP Logged

ndnger

Junior Member

Offline

Posts: 11
Los Angeles

Re: Bode Diagram of Conditionally Stable System
Reply #17 - May 10^th, 2009, 11:37pm

mod is "the absolute value function". Sorry about he confusion. I haven't assumed any non-linearity in the system. I used the statement that "All these quantities, i.e., input, error term, and output are bounded" to convey that the transfer function and all the node voltages are not unbounded at this frequency. I have used boundedness as the condition for stability. It doesn't mean that the system is nonlinear. However, I must say that I haven't said anything about the step response of the system, which will be determined by the poles of the closed loop transfer function.

I haven't used Nyquist criteria to prove stability and as you've stated that any analysis technique gives that the system is stable. So my result is not contradicting any of these other analysis methods.

Also the michigan website link says that open loop transfer function becomes confusing when the open loop transfer function has poles on the jw axis. But your open loop system doesn't have any pole on the jw axis, so bode plot (and A/(1+A)) can be used to make judgements on stability.

Now my question to you is why should the red dot frequency create a RHP pole. In other words, why should this loop have RHP, given that the loop gain has 360 deg phase shift, but remember we are looking at the input to output transfer function, ie, A/(1+A)?

BTW, I was referring to spectre, when I said cadence.

IP Logged

pancho_hideboo

Senior Fellow

Offline

Posts: 1424
Real Homeless

Re: Bode Diagram of Conditionally Stable System
Reply #18 - May 11^th, 2009, 1:02am

ndnger wrote on May 10^th, 2009, 11:37pm:

I haven't used Nyquist criteria to prove stability and as you've stated that any analysis technique gives that the system is stable.
So my result is not contradicting any of these other analysis methods.

No. Your story is contradicting.

Again, if your story is correct, even a simple system which is judged as unstable by Nyquist Criteria will be stable.

ndnger wrote on May 10^th, 2009, 11:37pm:

Now my question to you is why should the red dot frequency create a RHP pole.
In other words, why should this loop have RHP, given that the loop gain has 360 deg phase shift,

Again see Nyquist Diagram of http://www.designers-guide.org/Forum/YaBB.pl?num=1190272820/0#0

In this situation, there is no pole for A/(1+A) in RHP.

However if yellow point moves on Real-Axis inside magenta unit circle or blue point moves on Real-Axis outside magenta unit circle in Nyquist Diagram, system will be unstable,
although red point remains in same point.
Here some poles for A/(1+A) in RHP will be created by this yellow point or blue point movement.

ndnger wrote on May 10^th, 2009, 11:37pm:

but remember we are looking at the input to output transfer function, ie, A/(1+A)?

This is your basic misunderstanding point about stability for system.

In studying stability of system, an important factor which we have to focus on is
"Characteristics Polynomial" not "Input to Output Transfer Function".
http://en.wikipedia.org/wiki/Nyquist_stability_criterion

And you have to learn "Internal Stability" which is more extended issue about Stability, here we can't argue it's stability by "Input to Output Transfer Function".
State Space Formulation is required
http://www.designers-guide.org/Forum/YaBB.pl?num=1241534678/4#4

But your misunderstanding is not related to this "Internal Stability".
Your misunderstanding is more basic.

ndnger wrote on May 10^th, 2009, 11:37pm:

Also the michigan website link says that open loop transfer function becomes confusing when the open loop transfer function has poles on the jw axis.
But your open loop system doesn't have any pole on the jw axis,
so bode plot (and A/(1+A)) can be used to make judgements on stability.

I can't understand what you want to claim.

If open loop function is completely lossless, Nyquist Diagram is never closed in finite region of complex plane.
Maybe the michigan website explains this issue in Nyquist Diagram using MATLAB, although this is not problem at all.

I just wanted to show very simple Nyquist Diagram for unstable system which has only one crossing point in negative Real-Axis
to show "If your story is correct, even a simple system which is judged as unstable by Nyquist Criteria will be stable".

Maybe the michigan website is not good example for this purpose.
And this michigan website doesn't argue "Conditionally Stable System" at all.

The followings are more proper as example.
See page.11 of http://www-control.eng.cam.ac.uk/gv/p6/Handout6.pdf
See page.3 of http://www.staff.ncl.ac.uk/j.w.finch/CS_notes_2.pdf

ndnger wrote on May 10^th, 2009, 11:37pm:

BTW, I was referring to spectre, when I said cadence.

It is no more than common "SPICE" although Cadence call it as "Spectre".

If I have to choose one from many SPICE type simulators, I use HSPICE which is Golden Standard Simulator.
http://www.designers-guide.org/Forum/YaBB.pl?num=1238150066/5#5

« Last Edit: May 11^th, 2009, 9:28am by pancho_hideboo »

Kita━━━━━━(ﾟ∀ﾟ)━━━━━━ !!!!!
http://www7.plala.or.jp/ungeromeppa/flash/kita.html
http://www.youtube.com/watch?v=mjIxGh55bMM&feature=related

IP Logged

ndnger

Junior Member

Offline

Posts: 11
Los Angeles

Re: Bode Diagram of Conditionally Stable System
Reply #19 - May 11^th, 2009, 1:58am

If the yellow point is on the magenta unit circle, then A = -1 and the output will be unbounded.

But if the yellow point moves inside the magenta unit circle, and the input is a single tone then the output will also be a single tone based on my earlier analysis. This is what I predict. Can you provide an alternate analysis to prove this wrong? Consider only ideal linear systems.

The step response which is given by the location of poles and zeros will however be exponentially increasing because of the creation of RHP and I concur with you on this.

IP Logged

pancho_hideboo

Senior Fellow

Offline

Posts: 1424
Real Homeless

Re: Bode Diagram of Conditionally Stable System
Reply #20 - May 11^th, 2009, 2:14am

Back to very beginning of this topic,
do you understand "Conditionally Stable System" which is sometimes called as "Nyquist Stable System" ?

ndnger wrote on May 11^th, 2009, 1:58am:

If the yellow point is on the magenta unit circle, then A = -1 and the output will be unbounded.

I didn't argue singular point at all.

ndnger wrote on May 10^th, 2009, 1:04am:

This analysis is general and will always be correct.

ndnger wrote on May 11^th, 2009, 1:58am:

But if the yellow point moves inside the magenta unit circle,
and the input is a single tone then the output will also be a single tone based on my earlier analysis.
This is what I predict.

I can't understand what you want to claim.

What on earth is your analysis useful for ?

In this case, system is unstable, here some poles for A/(1+A) exist in RHP.
This means existence of unstable natural(or eigen) modes in system.

Again, in studying stability of system, an important factor which we have to focus on is
"Characteristics Polynomial" not "Input to Output Transfer Function".

« Last Edit: May 11^th, 2009, 6:42am by pancho_hideboo »

Kita━━━━━━(ﾟ∀ﾟ)━━━━━━ !!!!!
http://www7.plala.or.jp/ungeromeppa/flash/kita.html
http://www.youtube.com/watch?v=mjIxGh55bMM&feature=related

IP Logged

vivkr

Community Fellow

Offline

Posts: 780

Re: Bode Diagram of Conditionally Stable System
Reply #21 - May 14^th, 2009, 3:17am

Hi pancho,

It is unclear to me what you really want to know by making this post. You seem to already know all the answers which you mention in the very first post.

As you point out yourself, this is a conditionally stable system. Looking at gain/phase margin in this case is not very useful. And you seem to be making incorrect deductions based on the Barkhausen criterion.

If the gain is > 1 at the point where the phase is -180 deg, this does not automatically imply instability. Instability (and oscillation) is guaranteed if the gain is exactly equal to 1 when the phase is -180 degrees, but you cannot stretch this to the case where the gain is >1.

That is the very reason to use the Nyquist test for conditionally stable systems, because the conventional way of looking the the gain and phase plots is confusing and misleading.

Best regards,

Vivek

IP Logged

zychang

Junior Member

Offline

Posts: 13
Ireland

Re: Bode Diagram of Conditionally Stable System
Reply #22 - Mar 8^th, 2012, 10:36am

Hi guys,

Just came across this post when I needed it. Great.
It seems to me that everyone has agreed that the system shown on the very first post is stable, i.e. no oscillation by itself.

But what about ringing? One nice thing about Phase Margin (180-phase@UGBW) in a two-pole system is that, it gives indication of overshoot ratio (%) for a step response. Can we apply the same relationship here?

Thanks,
Michael

IP Logged

Frank Wiedmann

Community Fellow

Offline

Posts: 677
Munich, Germany

Re: Bode Diagram of Conditionally Stable System
Reply #23 - Mar 8^th, 2012, 2:08pm

zychang wrote on Mar 8^th, 2012, 10:36am:

But what about ringing? One nice thing about Phase Margin (180-phase@UGBW) in a two-pole system is that, it gives indication of overshoot ratio (%) for a step response. Can we apply the same relationship here?

No (see for example http://www.designers-guide.org/Forum/YaBB.pl?num=1182388268/15#15). And be sure to consider http://www.designers-guide.org/Forum/YaBB.pl?num=1182388268/5#5.

IP Logged

Pages: 1 2

‹ Previous topic | Next topic ›