control systemnegative feedbackroot locus
So this is the system for which I have to draw a root locus if 'K' changes from 0 to infinity. The problem for me is that root locus is drawn for open loop transfer function but 'K' is in the feedback loop. I'm guessing that there is a way to get K in open loop transfer function but I've tried everything and just cant do it. I have no problems if K is in open loop tho. I appreciate any help.
OK, I got it. To obtain the necessary transfer function in Evan's form, one has to assume \$ \phi_c = 0 \$. Then, the block diagram in question can be converted into:
So \$ P(s) = \Large \frac{p}{\alpha} = \frac{G(s)H(s)}{1 + \frac{k_p}{s} G(s)H(s)}\$.
You can build the transfer function from the root locust. There are two branches leaving the origin, that means there are two poles there. Both the branches converge on a zero. The only way that can happen is if there are two zeros there. The final pole is on the left with the branch going off to negative infinity. You can recreate the transfer function from that.
Best Answer
No - the root locus is not drawn for the open-loop function.
It is the main purpose of the root locus of a system with feedback to show if resp. for which constant gain values the CLOSED-LOOP function will have poles in the "forbidden" region (right half of the s-plane). Hence, you have to find the poles of the function
H(s)=G(s)/[1+G(s)*H(s)] with G=1/(s+3)(s+4) and H=K(s(s+1).
That means, you have to find the zeroes of the denominator which gives the characteristic equation: 1+G(s)*H(s)=0