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Assume that we have a postive feedback block diagram like this. You can derive the closed loop gain as in the image.

If you replace the numbers with loop gain larger than one, you get a wrong result and does not make sense. The ouput should keep increasing to infinity in theory or to the rail voltages in real life.
I understand that GH = 1 is the stability limit.
You may add some delay in the feedback network but the block diagram makes sense without delay.

My question is just about the theory not about how it happens in real life.
Assume that you're given that block diagram. If you look at the block diagram, due to the loop with loop gain GH >1, the output should keep increasing.

Why can't the equation include that information?

I don't see anything wrong with the derivation and block diagram.

EDIT: There are some answers and comments saying that it is wrong because the system is time varying so you should apply a different method. Then I wondered why negative feedback works while it's also time-varying system. And Eugene Sh. in the comment said "negative feedback works because x(t) will get closer to x(t-delta) over time." However, there is no formal proof for that. Is there a model (maybe with delay) so from that model you can obtain the correct result for both negative and positive feedbacks?

enter image description here

enter image description here

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    \$\begingroup\$ @anhnha It's due to an assumption made in the equation you use to derive your result. Can you see what that assumption is and what it states? \$\endgroup\$ – jonk Jun 9 at 19:32
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    \$\begingroup\$ @anhnha No. That's a requirement made by the equation result. What's the equation setup that gets you there? You don't show it above, yet it exists. What is it? \$\endgroup\$ – jonk Jun 9 at 19:44
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    \$\begingroup\$ That is, to correct my mistake: your calculation is based on the assumption: y(t) = G*(x(t)+H*y(t)) while in reality it is y(t) = G*(x(t)+H*y(t-delta)). The latter equation can be seen as differential equation, introducing dynamics. \$\endgroup\$ – Eugene Sh. Jun 9 at 19:58
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    \$\begingroup\$ @anhnha There's a much simpler answer than is being offered so far. I don't want to steal from you the chance to find it on your own. It's one of those things you will really want to find on your own. And it's not about "differentials" or "dynamic behavior." As a hint, do yourself a favor and just draw out a nice table relating the sign of Y and the sign of X and what that means for the sign of G. Really, this is just basic algebra. Almost grade school stuff. Make two tables. One for GH>1 and one for GH<1. It will pop out like magic to you. \$\endgroup\$ – jonk Jun 9 at 20:09
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    \$\begingroup\$ Here is an answer i wrote to a similar question involving positive feedback. It doesn't exactly match the question asked here, but does involve loop gain of positive feedback systems. \$\endgroup\$ – AJN Jun 10 at 1:06
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I believe that this system, where the loop has net gain and the response functions G and H are instantaneous, is a physically-impossible and mathematically ‘pathological’ case. I am speaking a bit colloquially with this term, but will explain what I mean by it later. The upshot is that all we need to do is add an infinitesimally-small amount of memory to get the result we expect.

First, just for completeness, let me just point out the mathematical result that since G and H functions are constants, their associated time-domain response functions are \$\delta\$ functions. The outputs depend only on the instantaneous inputs, with no use whatsoever of past information….as has been pointed out already by others. Your mathematical result is correct for this case, but we can’t apply our usual intuition of a signal propagating around a loop and being summed with anything that depends upon past history. This is where our intuition misleads us for this special case of instantaneous response.

To gain more understanding, let’s consider instead a simple, but more realistic case with some memory, and then take the limit as it becomes an increasingly ‘instantaneous’ system, and see what happens. Let G be a first-order system with a pole at \$-a\$, and a gain of \$K\$, and let’s make it a unity-feedback system. Thus \$G = \frac{Ka}{s+a}\$ and \$H=1\$.

For this example, then, the closed-loop transfer function becomes

$$A_{cl} = \frac{\frac{Ka}{s+a}}{1-\frac{Ka}{s+a}} = \frac{Ka}{s+a(1-K)}$$

We can see that this system has a first-order response, with a pole at \$-a(1-K)\$. So it is stable as long as the pole is in the left-hand plane, which means \$K<1\$. What is interesting, in the context of the problem you stated (the instantaneous response system), is that even as we speed this system up \$(a\rightarrow \infty)\$ until it looks more and more like an instantaneous system, it does not approach the stable result that your instantaneous system did, as long as the gain \$K\$ is greater than 1. This is why I say that the instantaneous system is not only physically impossible, but mathematically pathological. It is not the limiting case of a system with memory in the limit that that memory duration goes to zero (the pole going to negative infinity).

What we get, instead, is that the system stability for the positive feedback loop, in this first-order unity-feedback example, depends only on the gain of the forward transfer function, \$K\$. As long as this gain is less than 1, the system is stable. If it is greater than 1, the system is unstable. This makes intuitive sense. The positive feedback would send the response to infinity unless our loop has some attenuation.

When we consider this more realistic system, all it takes is the addition of just a tiny bit of memory (\$a\$ is extremely large but not \$\infty\$), to get the result our intuition expects: if the loop gain is \$\gt 1\$, the positive-feedback loop is unstable.

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  • \$\begingroup\$ Why does the negative feedback work while it still uses the same block diagram just different in sign? \$\endgroup\$ – anhnha Jun 12 at 20:36
  • \$\begingroup\$ They both work, for this system. Your mathematical result is correct for the instantaneous response system. What is wrong is our intuition, because our intuition is based on a system with some memory. \$\endgroup\$ – rpm2718 Jun 12 at 21:07
  • \$\begingroup\$ For negative feedback, the closed loop has a pole at -a(1+K). Because (1+K) is always positive so the pole is always on the LHP --> stable. Is this right? The closed loop gain K*a/(s + a(1+K)), now if we let the a approaches infinity then the system is K/(1 +K) and if you convert this into time domain you would get δ function --> unstable but the system is stable in realilty so why is it wrong? \$\endgroup\$ – anhnha Jun 14 at 12:07
  • \$\begingroup\$ Yes, that is correct. For negative feedback in this first-order system, the pole remains in the LHP, assuming K > -1, of course. \$\endgroup\$ – rpm2718 Jun 14 at 12:12
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    \$\begingroup\$ To address your question about letting a go to infinity in the negative feedback system....this is simply a first-order system, with a single pole on the negative real axis. As that pole moves toward negative infinity, the system does not become unstable, but it does become really fast, and approaches a 'memoryless' system. A \$delta\$ function in the time domain does not mean it is unstable. (Sorry, can't remember how to get Latex in the comment) \$\endgroup\$ – rpm2718 Jun 14 at 12:23
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There already are two answers, I'll tackle the part that confuses you.

What you have there is not a "system" -- you have a mathematical equation described, visually, as a block diagram. And because it's represented visually, you're interpreting it as a feedback system. It is not.

As I mentioned the \$G\$ and \$H\$ "blocks" are simply constants, parameters, values that describe an equation, and that equation is: \$y=Kx\$. That's because the fraction \$\frac{G}{1-GH}\$ reduces to a constant, irrespective of what values \$G\$ and \$H\$ take, or how you represent it graphically, as long as it makes up that formula. Therefore there is no loop, no block diagram, just a fancy pictogram for \$y=f(x)=Kx\$. Instead of \$G\$ and \$H\$ you could have \$p\$ and \$q\$, two random variables, and it wouldn't change a thing.

Therefore the result is not wrong, it's perfectly valid. But if you want to have feedback, then you need to account for time, and that comes through the introduction of states and an analysys done in more than one time point, as the other answers and comments explain at large.


Just for a fistful of giggles, here's another "block diagram" that has the exact same \$y=Kx\$, but with another fancy way of expressing \$K\$ as a combination of 5 elements this time, \$K=-\frac{ABC}{B(AE+CD)-1}\$:

test

As you can see there's no infinity at the end and, despite the absolute positive "feedback", almost all voltages are negative. In particular, see V(y) -- the output of the "block diagram" -- vs V(test) -- V2 with the equivalent "transfer function".

And to take it one step further, here are two more examples about what I've been talking about in my 2nd comment:

  • Just because the input is time varying it doesn't mean that the "block diagram" will turn into a feedback system -- it's still \$y=Kx\$:

one

  • Just because you throw an integrator in there it doesn't mean that the output will blow up if the analysis is at one time point, only. The voltages may change, depending on the operating conditions calculated by the SPICE engine, otherwise the ideal case will simply discard the integrator, consider the capacitor as opened (also, for comparison, your "block diagram" with an integrator):

two

  • And, finally, for the sake of completion, a system with feedback with a DC voltage and a varying voltage. The conditions are to be at least one state (here, the integrator) and the analysis to be done in at least two time points (here, 1 s) -- I decreased the time constant because they were blowing up too slow:

three



Given your comment it looks like maybe you're not convinced that what you have there is not a "feedback system" but a simple function, so here's another attempt that shows your "block diagram" versus the behavioural source version of \$y=Kx\$:

math

B1 multiplies V(x2) (the input) with K, which is a parameter made up of G's and H's values, 10 and 1/9, respectively. To the left you see them plotted and they completely overlap -- the plot above is their difference: 0.

You don't have a "system", and the only reason why you think it is one is because you represented it with a block diagram. It is not a system, the same way the first example I showed is not one, either.




Until now I've shown why you don't have a system and only mentioned what you would need in order to have one. Now I'll show why you need two ingredients: a state and an analysis for at least two time points.

First, the state is a concept about a system's dependence of time: at time point \$t_1\$ there is one value, and at \$t_2\$ there is another, not necessarily different, but who is dependent on the previous one.

As shown, having static values for \$G\$ and \$H\$ means that everything reduces to a simple \$y=Kx\$, but if either \$G\$ or \$H\$ are dependent on time or some other quantity except the previous state, it will still not be a system. Here is the proof:

Let's say \$H\$ is dependent on another voltage. \$x\$ varies with time, therefore it's implied to be \$x(t)\$, so then let \$H=H(t)=f(A(t))\$, where \$A(t)\$ is an external value. If \$A(t)\$ is simply \$A\$, a static value, it all reduces to \$\frac{G}{1-AG}\$, so no change compared to the previous case, while for \$A(t)\$ it becomes:

$$\frac{G}{1-A(t)G} \tag{1}$$

There is nothing special in this notation other than a time dependency. Nothing in this formula shows feedback, positive or negative, since the equivalent \$K\$ is now a simple \$K(t)\$, therefore for any \$x\$ the output \$y\$ will not depend, in any form, of the previous state of \$x\$, or on the previous state of \$A(t)\$ -- it simply follows \$A(t)\$'s value as if it was just a constant, itself. Here's a proof:

still no system

V2 generates an external signal, varying from 0.3 to 0.7 in 0.5 s, and then remaining at the last value. B1 takes V(ext) and multiplies it with V(Hy), which would be the output after the usual \$H\$ block. As you can see, the output is a linear match for B2, which has the formula in (1). You can see that V(y) and V(test) are completely overlapped.

But if you replace B1 with a block that has states, e.g. an integrator, then things change, but only if the analysis involves at least two different time points, and the system is not static -- because, as I've shown before (3rd picture), simply adding a capacitor means that the operating point will be calculated by the SPICE engine and the output will be steady.

In practice there is always noise so instabilities would accumulate. This is a simulation so it needs a bit of user help. It could start by itself, due to the non-ideal nature of the numerical precision, but that, too, requires careful considerations of timestep over timespan, to match the expected changes.

So let's change the input to be an ever so slightly changing ramp, going from 0.2 V to 0.2001 V over 0.5 s, and then remaining at the last value (similar to what V(ext) was):

now

Now look at V(y)! In fact, all that would be needed would be to have the tiniest change in the input, for the tiniest moments (therefore at the minimum two time points), or anywhere in the path, and the output would accumulate. The input, V(x), is plotted in blue, and it shows that it changes only for the first half a second. The output seems to follow linearly the input, but as soon as the input has stopped, the output hasn't. This is what the presence of a state does to the system. Now it's a system.

I said "seems to follow" because it doesn't, it actually blows up (or down, depending on the value of the input), but here the time constant of the state is 1 s so the changes accumulate slowly, but even halving it to 0.5 will show significant changes.

Therefore in order to be a system it needs two things: a state (something that makes the system dependent on its previous value), and an analysis of at least two time points (during which, if the analysis is ideal, there needs to be a slight perturbation). Your "block diagram" has none of these.

And wherever I used quotes it was because whatever was in the quotes did not represent what was described -- i.e. "block diagram" is certainly not what you have there, not a "system", there is no "feedback".





why negative feedback that uses the same block diagram just the sign different gives the correct result with just block diagram?

Despite everyone telling you otherwise you're still looking at it from a system's point of view. There is no feedback in there, so just because you get -18 it doesn't mean it's the wong result. It isn't. It's very correct, because there is no system, only a mathematical equation in disguise. Look at your drawing, again:

pic

Extracting the "transfer function" is done just like in a block diagram (except there is no system involved, here):

$$\begin{align} e&=x+Hy \tag{2} \\ y&=Ge \tag{3} \\ &=G(x+Hy) \\ &=Gx+HGy \\ y-HGy&=Gx \\ y(1-HG)&=Gx \\ &\Rightarrow \\ y&=\dfrac{G}{1-GH}x \tag{4} \end{align}$$

If what I've shown (and you say you read everything, several times) in my 2nd picture is not enough, then try what jonk suggested in the comments. Impose \$G=1\$ and step through various values of \$H\$ such that you have three cases: \$GH<1\$, \$GH=1\$, and \$GH>1\$:

   | GH=0.5 | GH=2.0
 x |    y   |    y
___________________
 0 |    0   |    0
 1 |    2   |   -1
 2 |    4   |   -2
 3 |    6   |   -3
 4 |    8   |   -4
 5 |   10   |   -5
 6 |   12   |   -6
 7 |   14   |   -7
 8 |   16   |   -8
 9 |   18   |   -9

There is no column for \$GH=1\$ because it results in a division by zero.

I won't bother repeating the same conclusions everyone here agrees with, except you, because at this point it looks like you're not expecting an answer to a question, but a confirmation of what you've made up your mind to. If I am wrong, my apologies, but given your comments just now (after a week) it seems you still cling to the idea that you have feedback in there.

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  • \$\begingroup\$ Why does the negative feedback work while it still uses the same block diagram just different in sign? \$\endgroup\$ – anhnha Jun 12 at 20:36
  • \$\begingroup\$ @anhnha It seems you haven't read the answer. It may be a bit long, I agree, but it addresses the fact that you don't have a "system", why you don't have one, and what should be done in order to have one. \$\endgroup\$ – a concerned citizen Jun 12 at 21:34
  • \$\begingroup\$ @anhnha I've added some more examples about the part where I said what is needed in order to have a system. \$\endgroup\$ – a concerned citizen Jun 13 at 7:01
  • \$\begingroup\$ Do you have any textbook references for the concepts of "system" and "feedback" requiring time dependence, disqualifying OP's diagram? \$\endgroup\$ – knzy Jun 13 at 21:52
  • \$\begingroup\$ @knzy Here's a free book from MIT. See the definition in 1.1, right in the beginning. All systems behave in time. A feedback cannot be instantaneous because any change at the input would mean an instantaneous change at the output and, through the feedback, at the input, which means a clash in values. Fun fact: Scilab's Xcos doesn't even allow OP's "system" to be simulated because of this. So, as I said, repeatedly, OP's drawing is not a system, it only looks like one because of the drawing. \$\endgroup\$ – a concerned citizen Jun 14 at 6:24
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Why can't the equation include that information? I don't see anything wrong with the derivation and block diagram.

It won't and it never will, because for you to know the output of any time varying system, you must know two things:

  1. The initial condition of the system (t=0) OR the current state and current time
  2. the transfer function of how the system changes inputs over time

The problem is for block diagrams, the system states are not usually given in the diagram, they are given in a problem statement with the diagram.

For the diagram above, there are no system states, the system does not change over time and will give only a gain or multiplier of the input. You put X in you'll get Y out, instantly. There is no feedback, because the transfer function can be reduced to a constant. Therefore it wont' keep increasing forever, it will simply be a multiplier of the input.

If you include 's' or a laplace in the transfer function, that means it has memory, and it can remember it's state and will modify it's future output based on it's previous state, usually s is indicated in the diagrams or transfer functions, and then you must also know it's previous state at some point to know what the future output will be.

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  • \$\begingroup\$ Could you explain why the negative feedback works? The formula is kind of same just different in sign and input and output are also functions of time. \$\endgroup\$ – anhnha Jun 9 at 20:42
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    \$\begingroup\$ In the equation you gave, I see no negative feedback. I know the equation is written that way, but without laplace there can be no feedback, there has to be a delay, an integrator or derivative term. \$\endgroup\$ – Voltage Spike Jun 9 at 20:44
  • \$\begingroup\$ Right, I didn't talk about negative feedback but just to ask about your two conditions of varying time system. Does that only apply for positive feedback? \$\endgroup\$ – anhnha Jun 9 at 20:47
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    \$\begingroup\$ No, those conditions apply to any LTI system, open loop or closed loop. Negative or positive feedback \$\endgroup\$ – Voltage Spike Jun 9 at 20:49
  • \$\begingroup\$ if so can you explain in negative it works even the system is time varying? \$\endgroup\$ – anhnha Jun 9 at 20:56
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As others have stated, the block diagram you have drawn is physically unrealisable, so our physical intuition does not apply here. What you have found by \$\frac{Y}{X}=\frac{A}{1-AB}\$ is a DC equilibrium point of your "feedback" system. It's true that if we feed in an input X and compute the output Y using the aforementioned equation, we obtain a mathematically consistent input/output pair.

For example, if \$A=1\$ and \$B=2\$ and \$x(t)=1\$, then \$y(t)=-1\$. The error signal is \$e(t)=By(t)+x(t)=-2+1=-1\$. Finally, we can recompute the output \$y(t)=Ae(t)=-1\$, which is consistent with the original "transfer function" equation. So far so good. Note that I'm performing these computations in the time domain, since there is no actual time dependence involved (the results are the same in the Laplace domain). All this is saying is that your block diagram represents a system of linear equations which can be represented as an invertible matrix (as long as \$1+AB\neq0\$).

Now we get to the problem at hand: this system is physically unrealizable, but what if we perturb the input from its equilibrium state? Using the previous example, let's apply \$x(t)=1+\delta\$ for \$t>0\$ where, say, \$\delta>0\$ (the same analysis could be applied for negative \$\delta\$). If the output \$y\$ somehow found its way to \$y(t)=-1-\delta\$ for \$t>0\$, the system would once again be consistent. But does this actually happen in a real system if we introduce an infinitesimal amount of delay into the loop? If we apply the \$\delta\$ perturbation to \$x(t)\$ at \$t=0^+\$, the error signal will be \$e(0^+)=-1+\delta\$ which will push \$y\$ more positive. This is the opposite direction the output should move in order to reach the equilibrium point. Repeating this process (recalculating \$e\$ and \$y\$) should make it clear that this system is unstable.

In order to perform this perturbation analysis, we had to assume that the system does not respond instantaneously, which is the case for physically realizable systems. For the case in the question, there are no dynamics, so the only mathematically consistent \$x\$ and \$y\$ pairs are constants for all time. Instability in the usual engineering sense requires a dynamically changing output that tends towards infinity (it's not just a "DC" infinity), so this static system cannot be unstable in that sense.

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Short and simple answer: The result is not "wrong" - it is correct.

You can prove it by hand calculation or by simulation. However, the result assumes that there are absolutely no external (perturbing) influences - and this assumption is very unrealistic (noise, temperature and power supply variations).

As a mechanical analogon you can think of two balls - the smaller one is riding upon a larger one: No problem under ideal conditions (which never can occur). However, the smallest mechanical disturbance (e.g. air movement) will destroy this stable system.

Remark: Regarding simulation, only a TRAN simulation (time domain) or a variation of the DC power supply will reveal the unstability of such a system. Note that any ac simulation cannot reveal such instability.

Special case: Note that a realistic system with pos. feedback is not always unstable. For a pos. loop gain LG between zero and unity (0<LG<1), the system is stable (reduced bandwidth). The oscillation condition is at LG=1 and for LG>1 we have saturation.

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