[SOLVED] Phase Margin open loop

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Unless the output of Amp 4 needs to feed back to the input of Amp 1 you don't need to calculate the phase margin since you are not going to make a look out of the system. You just have to make sure each amp by itself is stable since it operates in closed loop. So you should make sure that the each Amp has enough phase margin.
For AC analysis of the whole system you can look at the phase of the output to know the input output phase relationship however you want to use it but phase margin for an open loop system doesn't mean anything.

Just to add, while you haven't given a specification for what you are trying to achieve, cascading filters like that is a poor way of designing a filters. Also, you may want to change the components to something more practical - 15.9uF capacitors are rare.

Keith

Also just to add....

What is purpose of R4?

Hi guys thanks for your time,

The circuit represents a model of a process. R4 feeds into amp 4 output. To my knowledge this is known as the feedback loop on this system (i am aware of the feedback loops in each amp).
This circuit is part of a mechtronics module in a degree course. The purpose of resistor value 15.9uf is just for educational purposes. I will state the questions I have been asked, and also the ones I am stuck with, to which I am not after the exact answer just some guidelines please, or some experienced knowledge of:

i. what is the open-loop phase margin for the model?
ii. at what amplifier gain would you expect the closed loop system to be just unstable?
iii. with the gain set to the value from part ii, obtain the frequency of oscillation of the closed loop system model.
iv. Use the gain and frequency of oscillation of parts ii, and iii, to derive the Ziegler Nichols settings as described in the closed loop online tuning technique in the edexcel notes.

engineerme said:

i. what is the open-loop phase margin for the model?
ii. at what amplifier gain would you expect the closed loop system to be just unstable?
iii. with the gain set to the value from part ii, obtain the frequency of oscillation of the closed loop system model.
iv. Use the gain and frequency of oscillation of parts ii, and iii, to derive the Ziegler Nichols settings as described in the closed loop online tuning technique in the edexcel notes.

Click to expand...

To 1) For better evaluation of the simulation results, you should use a better resolution around the unity gain bandwidth (suppress the higher frequency region) - however, my rough estimate is a small positive stability margin, perhaps 1 dB and something below 10 deg. More than that - just show the output 4 (magnitude and phase). All other curves are not important. And select: Grid on - otherwise you cannot read any values.
For a correct loop gain simulation your input must be into R4 - and set the signal source at the inverting input to zero.

To 2) As mentioned above: perhaps 1 db? (depends on better accuracy of simulation).

To 3) This can be easily derived from the unity gain frequency if the stability margin is zero.

To 4) Read the Ziegler-Nichols procedure for proper PID settings (based on oscillation frequency).

---------- Post added at 21:06 ---------- Previous post was at 20:46 ----------

I am a bit confused about your power supplies. All are positive??? (Or did you allocate negative values?)

LvW
My good i cant believe i didn't spot that. Cheers LvW. After some slight alterations since you mentioned that small error (now using virtual op amps insted of 3554am- as i cant seem to get the amps to work, i have tried running the sim with power supplied to the 3554AM amp but changing the source around has no affect what so ever, and i am more confident with virtual op amps as i have used them many times before). Anyway, i have suppressed higher freq. and pinpointed unity gain, also plugged in signal source to zero, therefore my frq. is 0.6764 Hz at 0dB, and dropping that into phase analysis gives me -149.7 degrees. I understand that at 0.6764 hz gain drops, but i also understand that at -3db gain drops by -20db at 45 degrees(do you class this as the point of instability? or where unity gain lies?) i also understand that at 0.6765 hz the phase has shifted to -149.7 degrees, Is this my phase margin, i am unsure of this. what do you mean by 'if the stability margin is zero?' do you mean unity gain?. I have attached my findings, both open and closed loop, but i am more concerned with open feedback. On question two, it refers to closed loop system point of instability, can you tell this from the open loop AC analysis?

So now since R4 connects to the output of amp 4 and we have ideal opamps, the question I have is how are you simulating it? Are you simulating the circuit in the image in AC analysis with the input source being V9? I assume that is what you are doing since your phase starts from 0 deg so doesn't show an inversion as it should for a negative feedback loop. If that then you don't get the loop Gain and Phase. So to answer your questions:
1. what is the open-loop phase margin for the model?
Since Phase margin has meaning for a feedback loop and I assume here we mean the feedback loop connecting R4 to the output, you have to break that loop and insert a signal to R4 and measure the output to get the gain and phase for the loop. This is easy for ideal opamps since we don't have any biasing constraints but when you use non ideal component models you will have to ensure that the DC bias points are maintained when you run the AC analysis by breaking the loop. I do this normally by adding resistors which have different values for AC and DC analysis. This simulation will give you the Phase Margin which would be the 360-Phase (or depends on how your wave plotter interprets phase values) value when the gain crosses 0 dB.
2. at what amplifier gain would you expect the closed loop system to be just unstable?
This is basically your gain at DC + your |gain margin|. Gain Margin is the gain value in dB at which your phase becomes 360 or when the loop turns from negative feedback to positive feedback.
3. with the gain set to the value from part ii, obtain the frequency of oscillation of the closed loop system model.
This is basically the frequency at which the phase becomes 360 since when you increase the gain without changing anything else everything will just shift up in the plot and you will see a phase of 360 when the gain is 0dB satisfying the instability criteria at that frequency, causing oscillation.

Hope this post is more helpful than my last one :smile:

Hi engineerme,

I only can underline what aryajur has explained above.

* The fundamental principle of loop gain analysis is to OPEN the loop and inject an ac signal (Vac=1V) into the open loop (in your case: R4).
This should be the ONLY ac source in the circuit; thus: set V9 to zero. Then, the output of the 4th opamp is identical to the loop gain (because your input is 1V).

* By doing this, I don't expect any bias problems (as mentioned by aryajur) since each opamp has local feedback.

* The frequency resolution (frequency range) as shown in your last graph is OK.

* Be aware that using ideal opamps the picture may be a bit to optimistic - however, I don't know what your task is and if ideal opamps are allowed.
(Why do you have problems to use properly connected power supplies and real opamp models?).

* It is perhaps interesting to see the behaviour of the closed loop (as shown in your last posting) - however, it does not tell anything about the amount of stabiulity margin you have. However, the phase response is a yes/no indication if the closed loop is stable or not.

The loop is closed with a -1 gain block, in so far, the open loop frequency response tells directly about feedback loop stability. The 30° phase margin is somewhat small, which results in a respective closed loop gain peaking. Time domain behaviour will show a considerable overshoot in return.

You didn't tell about the nature of the simulated process or the control requirements, thus I don't want to guess about acceptable control quality.

When dealing with control problems, the control process or "plant" response is usually given and can't be changed at will. So you need to focus on the controller part for optimization of closed loop behaviour.

Engineerme, in order to avoid confusion on your side I like to comment the above reply from FvM.

1) It is correct, your last attachement reveals a phase margin of approx. 30 deg - although you didn't simulate the loop gain in a correct way.
Let me explain: You have used V9 to feed an ac signal into an inverting opamp. And the gain of this amplifier is simply "-1" due to R4 unconnected (no error message because of this?).

2) This does NOT give the correct loop gain - however, there is only an 180 deg error in the phase response, which easily can be taken into account during evaluation of the simulation results. That's what FvM has done.
The reason is as follows: Using the correct input for loop gain simulation (R4) the gain of this amplifier will be 0.5*(1+1)=+1. Thus, there is only a differene in sign between both cases.

3) Summary: You can use the last drawings to identify the stability margins. As mentioned by FvM the phase margin is approx. 30 deg. and the gain margin (to be found at the -180 deg frequency) is approx. GM=+6 dB.
However: I am afraid this value for GM is not very exact because your phase response should show a real very steep jump from -180 to +180 deg. This is not the case - and this is an indication that you have not used a sufficient number of points during ac anmalysis. You should use at least 100 points per decade. Then, you can identify the 180 deg frequency with good accuracy and find the correct gain margin.

4) Now, if the gain of the whole loop would be larger by the amount of GM the loop gain would cross the 0 dB line exactly at the 180 deg frequency. And this indicates the frequency the closed-loop would oscillate with. This frequency is needed for determining the proper PID parameters.

5) Because of the very low operating frequencies I don't expect remarkable differences between ideal and real opamp models.

Thanks for correcting. I didn't review the circuit exactly. In fact, there's no -1 gain block rather than a differential amplifier with the loop closed through the non-inverting port. The loop gain sign is only correct due to another inversion in the "open loop" measurement.

I agree, that the open loop gain should be better measured for the correct path, although the result is identical in this case.

I also agree about suitable number of frequency sweep points.

Cheers guys. I am hoping this is what i should be getting, can you confirm please

Yes, it looks good. Thus, you can identify both margins (in fact: you have already) with good accuracy.