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If I have a control loop with positive feedback, the closed loop transfer function expression is: $$\frac{Y(s)}{X(s)} = \frac{P(s)}{1-P(s)H(s)} $$ where \$P(s)\$ is the direct loop transfer function and \$H(s)\$ is the feedback one. I know that in this scenario the stability (or instability) of the closed loop depends entirely on the zeros of $$ 1 - P(s)H(s) $$ But what if both \$P\$ and \$H\$ are constants greater than \$1\$? In that scenario my intuition tells me that the system should be unstable given that every input would appear at the output scaled by a factor, then multiplied again to be added with the input to be scaled again and so on. So the output would go to infinity. But if I follow the closed loop transfer function the output has a fixed negative value for a positive input given by: $$ y(t) = \frac{p}{1-ph} x(t) $$ How is this possible? I simulated it with Simulink and it is consistent with the closed loop transfer, but I can't understand how I can get a negative output multiplying and adding positive things in a loop.

What am I missing here?

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  • \$\begingroup\$ What do you mean by "constant" transfer function? If you mean one that outputs a constant value regardless of the input then it's surely non-linear isn't it? \$\endgroup\$ – Roger Rowland Jan 9 '17 at 21:06
  • \$\begingroup\$ I am thinking, for example, in a ideal operational amplifier, where P = a and H=f a resistance feedback network, both ideally constanst for all frequencys \$\endgroup\$ – diegobatt Jan 9 '17 at 21:11
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The system you are looking at can't be realized because it has no delay.

Such a system is known as an algebraic loop. Simulink detecs such loops and treats it like an equation and solves for a valid output.

See: https://de.mathworks.com/help/simulink/ug/algebraic-loops.html

Just because Simulink can work with it does not mean that it can be realized or has a physical interpretation.

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Just look at the conditions for \$ H(s) \$ to be negative

\$ p \$ and \$ h \$ are always positive, but if \$ p*h > 1 \$ then the denominator is negative and the output will be negative if the input is positive.

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  • \$\begingroup\$ This is exactly what is confusing the OP, as it is counter-intuitive. \$\endgroup\$ – Eugene Sh. Jan 9 '17 at 21:21
  • \$\begingroup\$ But this contradicts the intuitive idea of positive things being multiplied and add all over on a loop, is that idea wrong? Wikipedia says that if ph>1 the transfer function doesn't exist. And for example isn't this the ideal Schmitt trigger transfer function (which is unstable)? \$\endgroup\$ – diegobatt Jan 9 '17 at 21:22
  • \$\begingroup\$ It doesn't exist as for a non-zero initial condition it will go to infinity at zero time. Which is breaking the definition of the term "function". But why to a negative infinity.. it's an interesting question. \$\endgroup\$ – Eugene Sh. Jan 9 '17 at 21:24
  • \$\begingroup\$ I think if you move into the discrete domain and the Z-transform instead of Laplace, it will make more sense, as t will include a notion of the unit delay. \$\endgroup\$ – Eugene Sh. Jan 9 '17 at 21:31
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    \$\begingroup\$ Because posting it as an answer will require more precise terminology and less handwaving. And I am lazy for that. \$\endgroup\$ – Eugene Sh. Jan 9 '17 at 22:30

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