What is the link between transient and frequency instability?

My question stems from the following 2 points:

1. A LHP pole is stable, a RHP pole is unstable. This is because the transient response of a homogenous system is e^pt and so a +ve p gives an unbounded
exponential whereas a -ve p gives a decaying exponetial.

2. Two poles contribute 180deg phase shift. In a -ve feedback system, if this 180deg occurs when gain > 1, instability occurs.

How do I link 1 and 2 above?

Eg. If I have 2 LHP poles they will give a 180degC phase shift. If the gain > 1, does this suddenly make these RHP or something?

Im clearly missing something here. Any insights?

Thanks,

Diarmuid

The first statement is describing the closed loop transfer function (the overall response of the feedback system), the second referring to the loop gain.

diarmuid said:

2. Two poles contribute 180deg phase shift. In a -ve feedback system, if this 180deg occurs when gain > 1, instability occurs.

Eg. If I have 2 LHP poles they will give a 180degC phase shift. If the gain > 1, does this suddenly make these RHP or something?

Im clearly missing something here. Any insights?

Click to expand...

I think, you can answer the question by yourself.
At which frequency will a system with two poles reach a phase shift of -180 deg?
And what will be the gain magnitude at this frequency?

Hi diarmuid,
When taking about stability you have to differentiate between the poles and zeros of the open loop and the closed loop transfer function. If the open loop tf has a pole at the RHP that means that it is unstable because of the reason you said, but that does not mean that the closed loop tf is unstable. It could be the case that the open loop tf is unstable but the closed loop stabilizes the tf . To find out the stability of the closed loop tf (whether it has poles on the RHP or not) you have two choices: either you do the math find the closed loop tf and then find the poles or you draw the bode plot of the open loop tf and find out the phase margin. If the phase margin ( the phase @ 0db) is negative that means that the closed loop tf is unstable which means that it has poles on the RHP. Now why the gain at 180 degrees should be less than 1 ? well thats because the negative feedback becomes positive feedback when the phase becomes 180 degrees between the input and output. A positive feedback loop is unstable if the loop gain is larger than 1. I hope that this explains your question.
Greetings

Hello OTA_LDO:

You identified my exact confusion with the following line:

"draw the bode plot of the open loop tf and find out the phase margin. If the phase margin ( the phase @ 0db) is negative that means that the closed loop tf is unstable which means that it has poles on the RHP."

If the PM of an open loop plot is negative, why does this mean the closed loop is unstable? Could the closed loop not stabilise things?

Thanks,

Diarmuid

diarmuid said:

If the PM of an open loop plot is negative, why does this mean the closed loop is unstable? Could the closed loop not stabilise things?

Click to expand...

If the PM of the open loop is negative you have a loop gain>0 dB for a loop phase of -180deg.
Together with the phase inversion at the inverting opamp terminal (also -180deg) you have a total loop phase of -360 deg (0 deg).Thus, you have positive signal feedback at this frequency .
An open loop never can be unstable (if there are no other hidden loops). Thus, closing the loop cannot stabilize anything - in contrary, only a closed-loop can be unstable.

LvW - very helpful reply.

"An open loop never can be unstable" - makes sense given that there is no feedback. Does that mean an open loop transfer function will never
have a RHP pole?

Also, a lecturer once told me that Bode plots can only be used to test for stability if the system is open loop stable. Given that an open loop
can never be unstable was he not just stating the obvious?

Thanks,

Diarmuid

diarmuid said:

"An open loop never can be unstable" - makes sense given that there is no feedback. Does that mean an open loop transfer function will never
have a RHP pole?

Click to expand...

Yes - in case there are no "hidden" or unwanted feedback effects

diarmuid said:

Also, a lecturer once told me that Bode plots can only be used to test for stability if the system is open loop stable. Given that an open loop
can never be unstable was he not just stating the obvious?

Click to expand...

There are some circuits with more than only one single feedback loops.
In this case, it is possible that by opening one loop the remaining circuit is unstable (for my opinion, a rather rare situation).

Thanks LvW. You certainly know your loops!

Final question - why do we do open loop analysis instead of closed loop analysis?

Is it that it is just easier to do open loop rather than closed loop (easier to derive the transfer function, etc)?

Thanks,

- - - Updated - - -

Correction to :

"why do we do open loop analysis instead of closed loop analysis?"

More correctly:

"why is open loop analysis so common when determining stability of a closed loop? Why not just perform the
entire analysis on the closed loop"

diarmuid said:

"why is open loop analysis so common when determining stability of a closed loop? Why not just perform the
entire analysis on the closed loop"

Click to expand...

* A closed-loop analysis must be performed in the time domain (measurement and/or simulation) because measurements/simulations in the frequency domain cannot be performed for unstable circuits.
Thus, you only can see if the circuit with feedback is stable or not (yes/no decision). On the other hand, if the system is stable you can perform a frequency analysis of the closed loop - and you can derive some rough information about stability properties from the results (amplitude peaking near the pole frequency).

*However, there is another method (introduced by Nyqist), which can give more detailed and exact information about stability properties - and this is to investigate the properties under open lopp conditions (leading to the definition of loop gain).
Nyquist has found out that it is possible to derive from the loop gain all the information necessary to evaluate stability properties of the system after closing the loop - in particular if the system is "on the safe side" or not. That means: How many margin the system has until it becomes unstable. This is important, because in practice some parts or units within the feedback loop can alter (aging) or there are some other unwanted, but unavoidable, parasitic effects (phase shift!).

Very interesting. I have heard about the Nyquist stability criterion which is next for me to look into.

On the first point - why exactly can you not do ac analysis on an unstable loop?

Eg. In simulation, I configure my closed loop, apply an ac source at its input and check the ac response. If it is
stable I presume I would see something like < 1dB gain for a 360deg phase shift and if unstable > 1dB. Would
this not suffice?

Thanks,

From the above discussion I have the following comments:
first of all an open loop system can be unstable. An example for that is an inverted pendulumn which is a common control challange in system theory textbooks.
Second the root locus technique could be used for stability testing and it works for unstable open loop tf. It is a technique in which you can have a plot in the s plane that shows you how the closed loop poles move as you increase the open loop gain. It is pretty easy to plot with Matlab for example. You only have to define the open loop tf and then plot the root locus for it and see where your closed loop poles are.
Third an unstable closed loop is not only unstable for the frequency where the Phase is larger than 180degrees and the gain is larger than one but it is also unstable for a step response.

From the above comments I also have some comments:

OTA_LDO said:

From the above discussion I have the following comments:
first of all an open loop system can be unstable. An example for that is an inverted pendulumn

Click to expand...

No. If there are no hidden or unknown loops a circuit without feedback is always stable. How would you explain such an unstability? Do you know how stability is defined?
By the way: An inverted pendulum is a good example for a system with positive feedback.

OTA_LDO said:

Second the root locus technique could be used for stability testing and it works for unstable open loop tf. It is a technique in which you can have a plot in the s plane that shows you how the closed loop poles move as you increase the open loop gain. It is pretty easy to plot with Matlab for example. You only have to define the open loop tf and then plot the root locus for it and see where your closed loop poles are.

Click to expand...

Sounds a bit confusing to me. We "plot the root locus" for the open-loop TF - and "see where your closed loop poles are"?
No. Instead, we plot the locus of all roots - and these are the poles of the closed-loop TF - as a function of the gain within the loop.

OTA_LDO said:

Third an unstable closed loop is not only unstable for the frequency where the Phase is larger than 180degrees and the gain is larger than one but it is also unstable for a step response.

Click to expand...

Again, do you know how stability is defined? You are mixing frequency and time domain.
In short: If a circuit with feedback is unstable, there are only two alternatives: Either the output is latched at one of the supply voltages or it is oscillating.

- - - Updated - - -

diarmuid said:

On the first point - why exactly can you not do ac analysis on an unstable loop?
Eg. In simulation, I configure my closed loop, apply an ac source at its input and check the ac response. If it is
stable I presume I would see something like < 1dB gain for a 360deg phase shift and if unstable > 1dB. Would
this not suffice?

Click to expand...

@1: An ac analysis can give correct results only if the circuit is in its linear operating region (dc operating point) and if the output signal depends on the input signal only (that means: no oscillation in the tim domain.)
There is, indeed, the following problem: The ac analysis of an unstable system will result in an amplitude response that looks quite OK - but, nevertheless, it is false. It is only the phase response which is an indication of instability: It exhibits a rising characteristic that cannot be true.

@2: No, you did not understand the principle. It is the loop gain (the gain of the loop which is opened at one node) which is used for stability analysis. Plot the magnitude and phase of the loop gain in a BODE diagram and apply the stabiliy check. Alternatively, you can use one single drawing (magnitude and phase combined) - and that is the principle of the Nyquist diagram.
And the transfer function of the open loop tells you if the circuit will be stable (or not) after closing the loop!

LvW - that clears things up for me.

@1: AC analysis can only be performed on an LTI system. The oscillatory response of an unstable
system makes it time varying (or non-TI). Correct?

@2: As you said "Bode plot looks at the phase / magnitude of the loop gain". This I wasn't realising completely.

Thanks very much for your help. You have an excellent / intuitive understanding of stability.

Diarmuid

@1: A very good question. In general, I would say that - even in case of osciillation - the system remains LTI. However, as everybody - who is familiar with oscillators - knows, a working scillator requires some sort of amplitude-dependent non-linearity. This is necessary to keep a fixed a and stable amplitude of the oscillationg output.
And therefore the question: Still LTI?
There will be a (small) non-linearity and a (small) time dependence within the system.