I have only seen systems with this transfer function treated in stability criteria with this general form:$$\frac{KG(s)}{1\pm KG(s)}$$ so the $$\lvert H(s)\rvert =1$$ is it possible to apply criteria with H(s) distinct of 1? or must do some transformation before? thanks.

  • \$\begingroup\$ Normally, the system TF is assumed in the form: \$\frac{KG(s)}{1+KG(s)H(s)}\$, and there is no such restriction on \$H(s)\$. Give a link to the website that you are quoting from \$\endgroup\$ – Chu Jan 7 '17 at 9:29
  • \$\begingroup\$ cranck, do you expect that we should guess what you mean with H(s)? At least, you should show us an equation not only a simple expression. \$\endgroup\$ – LvW Jan 7 '17 at 9:48
  • \$\begingroup\$ This is only valid for systems with Unity feedback path if your systems feedback path has a value that is not equal to one then the system transfer function should be $$ \frac{KG(s)}{1+H(s) *KG(s)} $$ \$\endgroup\$ – Elbehery Jan 7 '17 at 10:43
  • \$\begingroup\$ Can you tell us which "stability criteria" are you talking about? Certainly all I know are applicable to H(s) distinct from 1. \$\endgroup\$ – Deep Dec 26 '17 at 16:41

If we assume G(s) is strictly positive thus: $$ \frac{KG(s)}{1+KG(s)} $$ $$H(s) = 1$$ is a closed loop negative feedback system where it is assumed that H(s) which is the feedback component which is equal to 1.

We cannot substitute the plus sign for a minus sign which would make it a positive feedback system and therefore make it unstable.

$$ \frac{KG(s)}{1+K H(s)G(s)} $$ $$ H(s)\neq 1 $$

For a system to be BIBO stable it must satisfy the magnitude and angle criteria. Which relates to the poles in the s-plane of the closed loop system.

Basically if you look a Root Locus diagram of a system all the poles should be in the left half plane and the phase plot should be within \$ \pm\pi\$ but thats not strictly true. A system can have range for which K the system is stable.


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