Talk:Control Systems/Root Locus

Rule 4
Rule 4: A point on the real axis is a part of the root-locus if it is to the left of an odd number of poles or zeros.
 * Does this mean: to the left of either an odd number of poles, or an odd number of zeroes, or both? Or does it mean: to the left of an odd number of (poles and zeroes together)? E.g. if there are 1 zero and 3 poles, then the total number is 4, which is even. If there are 1 zero and 2 poles, then the total number is 3, which is odd. What is the rule exactly? --Gerrit 14:11, 20 October 2007 (UTC)

Its both. The root locus on the real axis is to the left an odd number of roots on the real axis. --Ben f

Example 1 is incorrect
The closed-loop transfer function given in Example 1 is:


 * $$T(s) = { 1 \over 1 + 2s}$$

But earlier on the page, the closed-loop transfer function is defined as:


 * $$T(s) = { K G(s) \over 1 + KG(s)H(s) }$$

Combining these two expressions, it follows that:


 * $$KG(s) = 1$$

and


 * $$H(s) = 2s$$

Thus, the open loop transfer function is:


 * $$ KG(s)H(s) = {a(s) \over b(s)} = 2s $$

which has a single zero at


 * $$s=0$$

and a single pole for


 * $$|s| \rightarrow \infty$$

Based on the closed-loop transfer function, it follows that as K approaches 0, the location of the pole must satisfy


 * $$\angle s = -\pi $$

which means that the pole is on the negative real axis at $$s = -\infty$$.

It then follows that the root locus begins at minus infinity on the negative real-axis, and then moves forward along the negative real-axis until reaching the origin. This result is quite different from the analysis and associated graph shown in Example 1. 71.233.119.156 (talk) 02:50, 1 July 2009 (UTC)

Computer simulation shows that the example is correct. In Matlab/Octave: rlocus(tf(1,[2 1])) --129.97.120.36 (talk) 20:11, 15 April 2010 (UTC)

Example 2
Hi, i might be very wrong here(sorry if i am) but I think there's a typo in the transfer function given, i think it ought to be s^2 +4.5s +5.625. (i checked by plotting it in octave)