How can closed loop poles of a 2nd order system lie in the RHP?

Hello,

1.) RHP poles equate to an expanding exponential in the time domain -> unstable system.

2.) For a 2nd order system, if the 2 open loop poles result in a 180deg open loop phi shift when the loop gain > 1, oscillation occurs -> unstable system.

What I dont get is that how does pt.2 push the closed loop poles into the RHP?

Attached is the maths behind my question.

Does anyone have any insights into what I am missing here?

Thanks in advance,

Diarmuid

Diarmuid, your calculation has an error:
the denominator in your equ. (2) must be (s-po1)(s-po2).

So if both po1 and po2 are negative, then [-(po1+po2)/2] will always be positive.

Does suggest the opposite to my original question - closed loop poles always lie in the RHP?

Cant be so again Im missing something.

darmuid, see my revised answer.

But if its (s-po1)(s-po2) ... doesnt that suggest po1 and po2 are RHP? How can this be when po1 / po2 are open loop poles?

Txs

Simple example: RC lowpass:

H(s)=1/(1+sRC)

Pole: 1+sRC=0 >>> s,p=-1/RC or RC=-1/s,p

Therefore: H(s)=1/(1-s/s,p)=s,p/(s,p-s)=-s,p/(s-s,p)

Remark: Replacing the pole s,p by the so called "pole frequency" s,p=-1/RC=-wp (value of wp always positive) you can write:

H(s)=1/(1+s/wp)=wp/(s+wp).

General comment to your attachement:
A system with a stable second-order loop gain function will - after closing the loop - always result in a stable sysytem (loop phase -180 deg will be reached for infinite frequencies only).

Thanks LvW,

So Ive gone over things again with the substitution of (s-po1)(s-po2).

In fact Ive gone over this a few times (hence the delay in response) but keep coming back
to the same question - how can 2nd closed loop poles get pushed to the RHP?

Ive updated the attached with the new substitution of (s-po1)(s-po2). Ive highlighted
the sign changes for clarity. Ive also highlighted what brings me back to the original question.

Again, Im missing something here.

So in your document you are defining the open loop transfer function as being only second order. In this case yes the system will never be unstable, as LvW stated above. But be careful that you don't confuse this with a second order system (being just G) always being closed-loop stable (since often H will have more poles, meaning the loop TF is higher than second order).

Thanks for the response mtwieg. Yes I agree. I have seen 3rd order systems whose root locus shows that increasing gain pushes the closed loop poles into the RHP. In
fact this is what got me thinking about the 2nd order system.

A 2 stage opamp typically has 2 dominant poles and so can be approximated via a 2nd order system. If both those poles contributed 90deg where the loop gain > 1 then we would have +ve feedback i.e. instability.

What Im looking for is to make the connection with this instability (for my 2 stage opamp) and what it does to the closed loop poles.

Hi diarmuid,

I find the terminology a bit confusing; looking at the plot in your rev2 doc:
- G is the open-loop gain
- GH is the loop gain
- G/(1+GH) is the closed loop gain
while you seem to call GH an open-loop gain... is there really a textbook using that definition?

A second order system can be defined differently on different textbooks (only if H is real the different definitions are equivalent) you seem to have chosen a definition pointing to GH being second order, which is fine.

In your situation (GH has two LHS poles) the only ways to make the system unstable are
1) with positive feedback: A(0) would have to be negative (in your doc you imply that your 'sigma' block substracts v_in from the output of H, so this would be positive feedback) and large enough for p1 p2 + A(0) to become negative as well;
2) with an RHP zero in H canceling an RHP pole in G, the pole would disappear in GH but still be there in the closed loop response

If you are looking at a simple two-stage op-amp this type of representation can create more problems than it solves as it can easily hide loading effects (very common in IC design). You are better off looking at the small signal representation and take a look at Middlebrook and Tian work on stability.

Hope this helps.

Hello dgnani,

Thank you for your feedback. I like the way you are approaching this.

So I re-did my calculations for a closed loop with +ve feedback (i.e. unstable closed loop) and am starting to get somewhere.

Please see the attached for details. In particular:

- Is my conclusion that for a real closed loop pole, the larger A0 and (or) H, the further into the RHP that pole will be pushed?
- How does a complex pole for an unstable system get pushed then into the RHP?

Think we might be getting somewhere with this!

Thanks for all the help so far guys.

Diarmuid

Diarmuid, I didn´t check your calculation in detail. I did read the last sentence only.
Therefore, perhaps the following helps:
In case of positive dc feedback you are producing a REAL pole in the RHP - perhaps together with a a complex pole pair (but this is not important).
Hence, your circuit will go immediately into saturation.

For comparison with an oscillator: This circuit with feedback has (at t=0, switch-on) a pole-pair in the RHP but NOT a real pole (because the oscillator must not have positive dc feedback).
Hence, such a circuit shows a rising sinusoidal signal.

Hi Diarmuid,

Sorry did not get to review your doc in full, it will be a busy week for me...
Here a few tips:
If the sigma block in your diagram has a plus on the feedback input then the sign in the loop gain (at the denominator of the closed loop gain ) is wrong. Just change the sigma symbol so you can keep the equations as they are.
I guess you assume the second order denominator of G to have real coefficients.
In that case the poles are either real or complex conjugate.

If complex the real part is the mean of the poles so in this case it is enough for the sum of the poles to be positive to have instability (another way to say it would be that G non-dominant pole is in the RHP)

If real, the condition for having an RHP pole can be written using Descartes sign rule... I don't think your inequality covers all cases unless you are making other assumptions...

Posted via Topify using iPhone/iPad

Right so spent a lot of time plotting root locus plots of the 2nd order systems on Matlab last night.

Here are the conclusions I have:

1. A oscillatory unstable 2nd order system will have RHP complex poles (this we know). To achieve those poles, the average of the 2 open
loop poles must be -ve so that Re(s) becomes +ve via the -b formula hence residing in the RHP.

For the average of the 2 open loop poles to be -ve, (-((p1+p2)/2)), at least one open loop pole must reside in the RHP. Completely baffled how this would occur!

2. For a non-oscillatory unstable 2nd order systems, one closed loop pole looks like it will get pushed into the RHP as A(0) or H increases.

Im going to leave this topic as Im not really getting anywhere but for completion, my exact problem statements are:

- How do closed loop poles of an oscillating unstable 2nd order system end up in the RHP? If it is because at least one open loop
pole is in the RHP, then how does an open loop pole end up in the LHP? Is this physically possible?

Thanks for all the feedback guys.

- - - Updated - - -

@LvW...just seen your post there and it prompted me to go over oscillator theory.

The below are good slides on VCO design.

**broken link removed**

Page 10 gives the poles of an RLC circuit. Page 13 states that these move into the RPH for gmRp > 1.

Working through the equations I dont see how. Do you have an explanation consistent with the maths?

Txs,

diarmuid said:

The below are good slides on VCO design.

**broken link removed**

Page 10 gives the poles of an RLC circuit. Page 13 states that these move into the RPH for gmRp > 1.

Working through the equations I dont see how. Do you have an explanation consistent with the maths?

Click to expand...

Yes - but the "good" slides are somewhat confusing (on page 10):

* RL is not defined (it should be the tank parallel resistor)
* It is not noted that Rp=RL||Ro with Ro=1/Go.
* The final solution is not exact (note the ~ signs).

The correct solution for the loop gain contains the element (s/C)*(1/Rp - gm).

Hence, this term disappears for Rp=1/gm (oscillation case) and can be positive or negative, depending on the Rp, gm values (resulting in an LHP or RHP pole pair).

So I found the exact answer to my question in Thomas Lee "The Design of CMOS Radio-Frequency Integrated Circuits", pg. 472.

"stricly speaking, a two-pole negative feedback system is never unstable in the BIBO sense"

This explains why the closed loop poles always have either zero or -ve real part but never +ve real part i.e. cannot exist in the RHP.

To close out this post I have attached pg. 472 from Lee's book.

Also attached is the maths behind the 3 paragraphs on this page.

To finalise everything, I have attached 4 concluding remarks that I think sum things up pretty well.

Hopefully if anyone has this question in future they will find what they are looking for easily here.

Diarmuid