# Positive feedback with constant transfer functions

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?

• 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? – Roger Rowland Jan 9 '17 at 21:06
• 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 – diegobatt Jan 9 '17 at 21:11

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.

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

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.

• This is exactly what is confusing the OP, as it is counter-intuitive. – Eugene Sh. Jan 9 '17 at 21:21
• 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)? – diegobatt Jan 9 '17 at 21:22
• 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. – Eugene Sh. Jan 9 '17 at 21:24
• 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. – Eugene Sh. Jan 9 '17 at 21:31
• Because posting it as an answer will require more precise terminology and less handwaving. And I am lazy for that. – Eugene Sh. Jan 9 '17 at 22:30