I used to have the transfer function given but here the root locus instead ...
Can one help with the start...
\$\begingroup\$
\$\endgroup\$
1
-
\$\begingroup\$ if you can remember how to go from the transfer function to the root locus diagram, work backwards to get the TF again \$\endgroup\$– KyranFCommented Dec 18, 2015 at 19:19
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
4
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.
-
\$\begingroup\$ this wii give me the root locus equitation 1 +k GL(s) ... the zero of the transfer function will not be found from the diagram \$\endgroup\$ Commented Dec 19, 2015 at 4:29
-
\$\begingroup\$ But I don't think i need the TF in this question. Becuse I can see from the RLoucs that the desired zeta and wn intersect so that there is a k that satisfy the need \$\endgroup\$ Commented Dec 19, 2015 at 4:39
-
\$\begingroup\$ the only problem with the p controller is i don't know how to test the steady state error so that to see if p controller work or not !! \$\endgroup\$ Commented Dec 19, 2015 at 4:41
-
\$\begingroup\$ @user3136052 Ess = 1/[1 +k GL(0)] The root locust also always starts at the open loop poles and ends at the closed loop zeros. These values combined give you the transfer function. \$\endgroup\$– vini_iCommented Dec 19, 2015 at 12:58