kp=10
kp=3
Closed-loop system response and characteristics, Proportional gain 10 < paste transfer function T...
PROBLEMA: (25%) A closed-loop control system is shown below Ds) T(O) U(A) C(s) (a) Show that a proportional controller (C(s)-kp) will never make the closed-loop system stable. (8%) (Hint: you need to calculate the closed-loop pole locations and make discussion for the two possible cases.) (Medim) (b) When a PD controller is used (C(s)kp+ kps), calculate the steady state tracking error when both R(s) and D(s) are unit steps. (8%) (Easy) (e) Suppose R(s) is a unit step and D(s)...
E. If you double the value of kp, what are the new closed-loop pole locations and [5 points] how much overshoot does the step response have? Hint: It is possible to determine the original value for kp. However, with the knowledge at this point, you can compute the pole locations without actually knowing kp (simply double the zero-order term in the denominator polyno- mial). Problem 2 You are confronted with a process that has the unknown transfer function G(s). It...
Q3. Consider a single loop unity feedback control system of the open loop transfer function (a) Find the range of values of the gain K and the parameter p so that: (i) The overshoot is less than 10%. (ii)The settling time is less than 4 seconds Note: , 4.6 M. = exp CO 40% (b)What are the three elements in a PID controller? Considering each in turn, explain the main ways in which varying the parameters affects the closed-loop system...
just the final answer 4. What are the four major characteristics of a closed-loop step response? a. Rise time, overshoot, settling time and steady-state b. Rise time, overflow, settling time and steady-state c. Rise start, overshoot, settling time and tracking error. error. error d. Rise start, overflow, settling time and tracking error. 5. The Ki response will decrease the: a) Steady-state error. b) Rise time ) Oversheo d) Tracking error I. The poles and zeros of a rational system function...
Assume the following closed-loop system with a PID controller. Match the step responses with the appropriate controller parameters. R(s) + PID Y(s) Controller G(s) 1. Step Response 1.5 data 0.5 0 10 40 50 20 30 Time (seconds) Kp = 2, Td = 1, Ti = 5 2. Step Response 1.5 =1, Kp = 5, Td Ti = 5 0.5 D 10 40 50 Кр = 10, Td = 1, T = 5 20 30 Time (seconds) Step Response 3....
Consider the closed loop system defined by the following block diagram. a) Compute the transfer function E(s)/R(s). b) Determine the steady state error for a unit-step 1. Controller ant Itly Ro- +- HI- 4단Toy , c) d) e) reference input signal. Determine the steady state error response for a unit-ramp reference input signal. Determine the locations of the closed loop poles of the system. Select system parameters kp and ki in terms of k so that damping coefficient V2/2 and...
Problem 5 (15 points) A unity negative feedback closed-loop system has the closed- loop transfer function given below. (s + 4) G (5) " (s +50) (s + 2)(s + 5) Compute the percent overshoot and rise time of the step response. Process D(3) GE) HX) Figure 1.
question b or the control system in Figure 1: C(s) Find the closed-loop transfer function T(s)-- R(s) a) b) Find a value of Kp that will yield less than 15% overshoot for the closed-loop system. (Note: ignore the zero dynamics to calculate Kp initially). c IIsing vour K from nart h) write a MATI AR scrint that calculates the closedloon Motor Plant R(s)+ C(s) Controller 10 Kp (s+9) s2 +6s15 12 Figure 1: Unity feedback with PD control or the...
R(s) Q1 (25p). The closed loop control system is shown in the figure. The response of the control system to a unit step reference input is shown in below. C(s) Kp 0.6 S2 + 5s + A 4 a) Find and wn. b) Find Ke and A c) If we use a proportional-integral (PI) controller instead of a proportional (P) controller, how the graph would be expected to change. For two different integral coefficients (bigger Ki_1 and smaller K_2), make...
A system having an open loop transfer function of G(S) = K10/(S+2)(3+1) has a root locus plot as shown below. The location of the roots for a system gain of K= 0.248 is show on the plot. At this location the system has a damping factor of 0.708 and a settling time of 4/1.5 = 2.67 seconds. A lead compensator is to be used to improve the transient response. (Note that nothing is plotted on the graph except for that...