1
\$\begingroup\$

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?

\$\endgroup\$
2
  • \$\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
    Commented Jan 9, 2017 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
    Commented Jan 9, 2017 at 21:11

2 Answers 2

Reset to default
1
\$\begingroup\$

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.

\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$
6
  • \$\begingroup\$ This is exactly what is confusing the OP, as it is counter-intuitive. \$\endgroup\$
    – Eugene Sh.
    Commented Jan 9, 2017 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
    Commented Jan 9, 2017 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.
    Commented Jan 9, 2017 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.
    Commented Jan 9, 2017 at 21:31
  • 1
    \$\begingroup\$ Because posting it as an answer will require more precise terminology and less handwaving. And I am lazy for that. \$\endgroup\$
    – Eugene Sh.
    Commented Jan 9, 2017 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.