Consider the open-loop plant: P(S) = 10/(s2 + 2s + 5) and a controller: k(s +1) /(s) = s(s +2) structured in a unity feedback control architecture: = w r u y C(s) P(s) (f) (numerical) Suppose the maximum control effort the actuator can provide is Umar = 0.52. Run a separate iteration, this time beginning with k = 0.01, and incrementing by 0.01 to determine the largest value of k such that the actuator does not exceed its limit. Simulate for 30 seconds during each iteration. With this k, display a 1 x 2 subplot showing the control command u(t) and the output y(t). On the u(t) plot, show the maximum value Umax being less than 0.52. Also display the k value that yields this response. Note: the loop will probably take a long time to run since it's running the simulation each time.
(a) (numerical) Next, run the simulation for 60 seconds, this time introducing a unit step distur- bance w(t) beginning at 30 seconds. With the value of k found in (f), plot the output y(t) and the control command u(t). Is the system capable of rejecting a constant disturbance? E(S) (h) (by hand) As a follow-up to
(b), compute the transfer function T(s) in terms of k, and W(s) note the number of zeros at the origin. Keep this number in mind as we move into the next topic, system TYPE & steady-state error. It can tell us definitively what types of disturbances we can reject and what types of references we can track.

Explanation:

r the open-loop plant:

P(S) = 10/(s2 + 2s + 5) and a controller: k(s +1) /(s) = s(s +2) structured in a unity feedback control architecture: = w r u y C(s) P(s) (f) (numerical) Suppose the maximum control effort the actuator can provide is Umar = 0.52. Run a separate iteration, this time beginning with k = 0.01, and incrementing by 0.01 to determine the largest value of k such that the actuator does not exceed its limit. Simulate for 30 seconds during each iteration. With this k, display a 1 x 2 subplot showing the control command u(t) and the output y(t). On the u(t) plot, show the maximum value Umax being less than 0.52. Also display the k value that yields this response. Note: the loop will probably take a long time to run since it's running the simulation each time.

(a) (numerical) Next, run the simulation for 60 seconds, this time introducing a unit step distur- bance w(t) beginning at 30 seconds. With the value of k found in (f), plot the output y(t) and the control command u(t). Is the system capable of rejecting a constant disturbance? E(S) (h) (by hand) As a follow-up to

(b), compute the transfer function T(s) in terms of k, and W(s) note the number of zeros at the origin. Keep this number in mind as we move into the next topic, system TYPE & steady-state error. It can tell us definitively what types of disturbances we can reject and what types of references we can track.

answer:

i lost in getting someone back.

i learned that people are really two-faced and no matter how angelic someone appears they always have a devil side and that you cant trust anyone anymore no matter how dear they are to you.

if your wondering who it waz it waz mah still miss her tho

~batmans wife dun dun

answer:

okay

explanation: