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Imagine a generic feedback system, where A is forward gain and F is feedback gain. I will now go over two cases where in i will analyse this system.

enter image description here

  • 1. Reference = 110, A=10, F=1

In this case, it can be easily thought that Error will be 10. Output will be 100. These values will not change as this condition is stable (maybe marginally stable, as we will see soon). Now, if i change Reference from 110 to 100, below things will happen as i move in time. Before that, to remind, this is negative feedback system as A and F are positive.

New error will be 100 - 100 = 0. As error is 0, output will be 0. As output is 0, output of feedback path will be 0. New error will be 100 - 0 = 100. For error of 100, output will be 1000. Feedback path output will thus be 1000. New error will be 100 - 1000 = -900. For -900 error, new output will be -9000. Hence, new feedback path output will be -9000. New error thus will be 100 - (-9000) = 9100. New output will be 91000.

Just to go over output sequence again, it is +0, 1000, -9000, +91000....

If i continue doing this, numbers are going to rise continuously. This system will not converge to any specific output value. So i believe output will diverge and if any saturation is there for let's say feed-forward gain A, then it will maximum reach that saturation value. Thus, what i have created is an oscillator.

To mention again, when reference was 110, output was stable. As soon as i changed step from 110 to 100, system starts to oscillate. This means system was marginally stable to begin with.

If i do that same analysis, but rather than A = 10, if i make A = 0.5 let's say (AF < 1), then systems becomes stable and output will always reach some steady state and it will not oscillate.

  • 1. Reference = -90, A= -10, F=1

Notice that the only difference i have made is A is -10 rather than +10, and reference is -90 rather than 110 as in prev case.

In this case, i can calculate that error will be 10. Output will be -100. This condition is stable(again, marginally stable as we will see soon) and numbers will not change with time. Now if i change reference from -90 to -100, system will do through transitions as mentioned below.

New error will be -100 - (-100) = 0. New output will be 0. New feedback path output will be 0. New error will be -100 - 0 = -100. For this new error, output will be 1000. Feedback path output will be thus 1000. For this, new error will be -100 - 1000 = -1100. For this error, output will be 11000. For this new output, feedback path output is 11000. Thus new error is -100 - 11000 = -11100. For this, new output will be 111000.

Just to go over output sequence again, it is +0, +1000, +11000, +111000....

As noticed, in previous case output was diverging when |AF| > 1 and signs of each output sample were opposite. In this case, |AF| > 1 and signs of each output sample are same. This means in this case, output will go on rising without changing signs. If feedforward gain has saturation, it will saturate there and output will stay there. This is not an oscillator. Similarly, there can be another case where output goes to negative saturation limit and stays there.

In the above analysis, conclusions are:

1. If AF > 1

  • If A*F is positive, one stability point can be easily found numerically. The system will be marginally stable. If i perturb the reference, output diverges and signs oscillates with consequent sample. So this is an oscillator.
  • If A*F is negative, one stability point can be easily found numerically. The system will be marginally stable. If i perturb the reference, output diverges and signs don't change with consequent sample.

1. If AF < 1

  • If A*F is positive, one stability point can be easily found numerically. If i perturb the reference, output converges to some output. So system is stable.
  • If A*F is positive, one stability point can be easily found numerically. If i perturb the reference, output converges to some output. So system is stable.

Questions:

  1. Is this analysis valid for continuous time system as well ( I believe analysis i did is valid for discrete-time system).
  2. I think above mentioned conditions, if are somehow analysed in continuous time domain, will be stable(not marginally stable) and output will always converge to some output. It will never diverge like our analysis shows. Is this correct?
  3. If point 2 is correct, how can we think of continuous time system, by intuition?
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  • \$\begingroup\$ In the very first example, it don't follow the order of your conclusions in this case, it can be easily calculated that Error voltage will be 10. Output will be 100. You first calculate the error voltage and next the output. However, to know the error voltage you need to know the previous output. You're not providing this \$\endgroup\$
    – Huisman
    Commented Dec 6, 2019 at 9:02
  • \$\begingroup\$ You are eroneously proving that negative feedback doesn't work. \$\endgroup\$
    – Andy aka
    Commented Dec 6, 2019 at 9:03
  • \$\begingroup\$ What is the signal you are feeding forward? That information is missing as well I think. \$\endgroup\$
    – Huisman
    Commented Dec 6, 2019 at 9:03
  • \$\begingroup\$ @Huisman Does the added figure is making my question clear? \$\endgroup\$
    – Omibuddyy
    Commented Dec 6, 2019 at 10:55
  • \$\begingroup\$ @Huisman When i mean "in this case, it can be easily calculated that Error voltage will be 10. Output will be 100", it can be easily seen that error of value 10 is a stable value of error for the system. \$\endgroup\$
    – Omibuddyy
    Commented Dec 6, 2019 at 10:57

2 Answers 2

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When the input was 110, because the output is steady, it does not imply that the system was stable to begin with. It can be said to be in metastable state and any perturbation caused the output to diverge (bounded input --> unbounded output and thus BIBO unstable).

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  • \$\begingroup\$ Edited question to accommodate thought system being marginally stable as any perturbation resulted in diverging output. Yet, the question is, is the given system, if analysed in continuous time domain, will act like what i have mentioned in the question? Essentially, our analysis shows the system is marginally stable and output will diverge. But is it really the case? \$\endgroup\$
    – Omibuddyy
    Commented Dec 9, 2019 at 5:00
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Reworking your logic (something that shouldn't have to happen in questions on SE). I see that your system is: $$ y[n+1] = A \cdot e[n], $$ $$ e[n] = r[n] - F \cdot y[n], $$ $$ A = 10,~~ F = 1.$$ Which by the way, \$A\$ is not a very generic system, it is just a static-gain. Doing the analysis, $$ y[n+1] = 10 e[n], $$ $$ e[n] = r[n] - y[n], $$ $$ y[n+1] = 10(r[n] - y[n]), $$

Applying the Z transform, $$ Y(z)z = 10(R(z) - Y(z)), $$ $$ Y(z)(z-10) = 10R(z), $$ $$ \frac{Y(z)}{R(z)} = \frac{10}{(z-10)}. $$ Notice that you have a pole at \$ z_0=10\$, and that is an unstable system, the time response is: $$ y[n] = (10 \cdot 10^i * r)_n. $$ So you could have just \$r[0]=1\$ and all other \$r[i]=0\$, and still have the response: $$ y[n] = 10 \cdot 10^n. $$

For the purpose of your analysis, I would suggest using the Z transform and reading a bit on discrete control, you will find the stability criterion for linear systems in any textbook. Also, that analysis changes for continuous systems, and is already pretty solid on any textbook.

Bonus:

use Octave to test stuff

pkg load control
sys = feedback(10/z,1,"-");
pzmap(sys);
step(sys);
r = [110 100 100 100 100 100];
t = [1 2 3 4 5 6];
lsim(sys,r,t)
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  • \$\begingroup\$ The system showed in the question is not a discrete time system (I don't have any delay in there). But my analysis itself has made that system discrete time. The real question i have is in terms of understanding how to really understand discrete and continuous time system with same line of thought. To make myself clear, system showed in question, if had even infinitesimally small delay, will act like your analysis. But if there's no delay as shown, system will be stable. Is this statement true? \$\endgroup\$
    – Omibuddyy
    Commented Dec 9, 2019 at 8:56
  • \$\begingroup\$ If there was small delay in lets say feedback path, whatever analysis you showed and i mentioned in the question holds true. But if there is no delay, our analysis doesn't hold true. However, i would like to understand why we fail to analyse in second case, like the way we analysed in first case. \$\endgroup\$
    – Omibuddyy
    Commented Dec 9, 2019 at 9:04
  • \$\begingroup\$ In the sense, the analysis you did and the one which is showed in question, are both incorrect for the system given in question. \$\endgroup\$
    – Omibuddyy
    Commented Dec 9, 2019 at 9:10
  • \$\begingroup\$ "But my analysis itself has made that system discrete time" that is why I though it was a discrete one to start with, the question is really unclear on the fact you are dealing with a continuous one. \$\endgroup\$
    – jDAQ
    Commented Dec 9, 2019 at 17:24
  • \$\begingroup\$ If you want to analyse this as a continuous system, just do so, don't confuse people with a long questions that mostly "explains" your analysis procedure and barely mention the fact the system is continuous even though your whole procedure is not. Use the Laplace transform and figure it out. \$\endgroup\$
    – jDAQ
    Commented Dec 9, 2019 at 17:27

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