Homework 02 Posted by D. Dane Quinn on Thu 04 of Sep, 2008 16:58 EDT Get it here...get it now ...and use this topic for any questions and/or answers
Homework 02 Posted by D. Dane Quinn on Thu 04 of Sep, 2008 16:58 EDT Get it here...get it now ...and use this topic for any questions and/or answers
Posted by ZachMiday on Sat 06 of Sep, 2008 14:25 EDT Problem #2: (Ogata, B-3-3) Simplify the block diagram shown in the figure and obtain the closed loop transfer function C(s)/R(s). No block diagram is shown in this paper and the bookstore is out of books. Anyone have this file they could scan in? Thanks
Posted by D. Dane Quinn on Sat 06 of Sep, 2008 19:20 EDT I have updated the problem assignment to include the figure from B-3-3. At least this is what I think the block diagram looks like...I don;t have the text at home. If anyone finds any errors in the figure, please respond within this topic. Thanks.
Posted by ZachMiday on Sun 07 of Sep, 2008 13:17 EDT On Problem #2: (Ogata, B-3-3) Looking at the area around H1, G2, and H2 and the junctions. Just G2 and H2 = G2/(1+G2H2). Just H1 and G2 = H1+G2. But due to placement of junctions/branches I keep looping around and my equations get larger and larger.
Posted by D. Dane Quinn on Sun 07 of Sep, 2008 13:30 EDT We talked about this one in class Friday. You can’t exactly loop around, because as you found out you end up with a ever increasing set of equations. Instead, whenever you eliminate a variable, like you are trying to do by looping around, you have to eliminate it in all of the equations. You are no doubt trying to eliminate some variable by looping, but it also appears in some other equation...get rid of it there too...
Posted by AndrewShinko on Sun 07 of Sep, 2008 19:22 EDT For the last problem (number four), when it is asked for the steady state Error function, isn’t E(s) simply R(s)-1, or is this a different steady-state error, possibly the one after D(s) enters into the system?
Posted by D. Dane Quinn on Sun 07 of Sep, 2008 21:48 EDT The error is E(s)=R(s)-C(s), where C(s) can be written in terms of R(s) through the closed-loop transfer function. Then the steady-state error can be determined by using the Laplace final value theorem.