2
\$\begingroup\$

it is known that the stability of a system with negative feedback may be analized through the Nyquist stability criterion, which is based on observing the number of turns around the point (-1;0) in the complex plane.

But, what if we want to apply the same criterion for a positive feedback system? Is it correct or should be modified in some ways?

Can we say also for positive feedback that the system is unstable if and only if there are at least 1 pole with positive real part?

\$\endgroup\$
1
\$\begingroup\$

This criterium (Nyquist) can be applied to all feedback systems.

Note that positive feedback does not necessarily mean "instability". Furthermore, for real systems (with parasitics) negative feedback will always turn to positive feedback for rising frequencies. Hence, for this purpose (stability check) it makes no sense and it is not necessary to distinguish between "positive and negative" feedback.

Remember, this criterion is applied to the loop gain - that means: The feedback loop is opened!

\$\endgroup\$
  • \$\begingroup\$ So basically we look only at the open loop transfer function Nyquist plot. But it seems strange to me that the result of this analysis does not depend on if the feedback is negative or positive \$\endgroup\$ – Kinka-Byo Jan 22 at 8:52
  • \$\begingroup\$ if it depends only on the open loop gain, it means that after I have chosen the "main" open loop device, I may put any positive or negative feedback network and the result will be the same \$\endgroup\$ – Kinka-Byo Jan 22 at 8:53
  • \$\begingroup\$ Of course, the result is different in both cases. But your question was if the criterion may be applied in both cases - and the answer is "YES", \$\endgroup\$ – LvW Jan 22 at 9:23
  • \$\begingroup\$ But if the open loop gain is the same, the Nyquist plot will be the same. So I'll get the same results in both cases while applying the Nyquist criterion \$\endgroup\$ – Kinka-Byo Jan 22 at 9:25
  • \$\begingroup\$ Why do you think the loop gain would be the same? It has another sign (pos. resp. negative). \$\endgroup\$ – LvW Jan 22 at 9:40
0
\$\begingroup\$

For negative feedback you look at the number of encirclements of \$\{-1,0\}\$.

For positive feedback you look at the number of encirclements of \$\{1,0\}\$.

This is the only modification that needs to be made. This is because for negative feedback the equation is \$1+G(s)=0\$, while for positive feedback the equation is \$1-G(s)=0\$.

Here is an example. $$G(s)=\frac{2 \left(s^2-s+2.5\right)}{(s+1.5)^2+1}$$

There are no unstable open-loop poles. The solution of \$(s+1.5)^2+1=0\$ is \$\{-1.5-1. i,-1.5+1. i\}\$.

The closed-loop system with positive feedback is

$$\frac{1}{1-G(s)}=\frac{2 \left(s^2-s+2.5\right)}{(s+1.5)^2+1 -2. \left(s^2-s+2.5\right)}$$

The roots of \$(s+1.5)^2+1 -2. \left(s^2-s+2.5\right)=0\$ are \$\{0.37868, 4.62132\}\$.

The Nyquist plot confirms that with positive feedback there will be two unstable closed-loop poles because of the two encirclements of \$\{1, 0\}\$.

enter image description here

Alternatively, as is LvW's answer you can consider \$1-G(s)=0\$ as \$1+(-G(s))=0\$.

Now you are back to a standard negative feedback configuration and will look at encirclements of \$\{-1,0\}\$, but you have to consider the loop transfer function \$-G(s)\$.

For the example above you get two encirclements of \$\{-1,0\}\$ and the same conclusion as before.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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