The Block Diagram For Case 2 From Challenge Assessment 4 And Copied Into The Appendix Is Shown In Figure 3 The Object 1 (81.05 KiB) Viewed 52 times
The Block Diagram For Case 2 From Challenge Assessment 4 And Copied Into The Appendix Is Shown In Figure 3 The Object 2 (46 KiB) Viewed 52 times
The block diagram for Case 2 (from Challenge Assessment 4 and copied into the appendix) is shown in Figure 3. The objective of this project is to improve the transient response peak time to be 75% of the Case 2 peak time by designing a compensator (active or passive). Case 2 is the uncompensated system where the compensator transfer function was simply Ge(s) = K and Gp(s) = $(5+1) Use the gain K found in CA 4 to analyze the uncompensated system OR use a gain of K = 2 instead. R(S) U(s) A(S) X(s) G.(5) G.(5) $26,(s) 1/82 1/52 H(S) Figure 3. Case 2 Block Diagram for Accelerator Feedback Model Objective: Use the root locus method to design a compensator (active or passive) to reduce the peak time to 75%. Since the uncompensated system has a fairly desirable transient response and no steady-state error, maintain or decrease the overshoot of the uncompensated system in the compensated system and maintain zero steady-state error. Use MATLAB to meet the objective(s). The root locus method requires plotting the root locus for both the compensated and uncompensated system. For the compensator, the transfer function should be determined as well as the gain K. To compare the uncompensated and compensated systems, plot the lane change step response for each system on the same graph. As a designer of the automated steering system, you must determine the compensator transfer function to meet the design objective. Discuss the following: • Explain the reasoning behind your compensator design choices (at least discuss and explain the choice of compensator pole and zero placement, gain needed, and second order approximation since transient response performance characteristics depend on using a second order approximation). Since you have a choice between active and passive compensation, what are the advantages and disadvantages of your choice? • Compare and contrast the uncompensated and compensated system transient responses (performance characteristics). • Compare and contrast the expected compensated system lane change step response to the actual MATLAB simulated response (performance characteristics). What would explain any differences?
0.1 Gp Just as in Case 1: A proportional gain controller is used to control a steering actuator (output steering angle). The steering actuator controls the vehicle and thus its lateral position, X(s), which is feed back to the controller as negative feedback. The desired lateral position, R(s), is the input to the controller. In case 2, an accelerometer is used to measure and compute the lateral position. Thus, feedback of the system must be the acceleration, A(s), not the lateral position, X(s). But the lateral position must still be the output of the entire system. Thus, to go from acceleration to position as an output the vehicle transfer function becomes = 32 s(s + 1) This s2 must be taken out before the final output of lateral position. Note that the Gp(s) in Figure 3 should not have the s2 term in the numerator twice; thus, the block should have the transfer function: 0.152 s?G s(s+1) The acceleration signal, A(s), is feed back to the accelerometer. The accelerometer is modeled as a mass- spring-damper system whose transfer function is 100 H(s) $2 + 20s + 100 However, the output signal of the accelerometer needs to also be changed back to position to compare to the desired input position (just like the output of the vehicle needs to remove the sa). Just as in Case 1, the actuator dynamic in Case 2 are neglected and the actuator transfer function is Ga(s) = 2 a
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