This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ECE 147A FEEDBACK CONTROL SYSTEMS  THEORY AND DESIGN F08 Final Exam Closed book and notes. No calculators or cell phones. Show all work. Name: Solution Problem 1: /15 Problem 2: /15 Problem 3: /15 Problem 4: /15 Problem 5: /15 Problem 6: /15 Total: /90 Miscellaneous: tan − 1 (1 / 3) ≈ 18 degrees, tan − 1 (1 / 2) ≈ 26 degrees, tan − 1 (1) = 45 degrees. tan − 1 (2) ≈ 63 degrees, tan − 1 (3) ≈ 72 degrees, tan − 1 (5) ≈ 78 degrees, tan − 1 (10) ≈ 84 degrees, tan − 1 (20) ≈ 87 degrees, sin(45 ◦ ) = √ 2 / 2 ≈ . 7, sin(50 ◦ ) ≈ . 77, sin(55 ◦ ) ≈ . 82, sin(60 ◦ ) = √ 3 / 2 ≈ . 87. A lead compensator λ ( s + z ) s + λz ( λ > 1) introduces a maximum possible positive phase φ max satisfying λ = 1+sin( φ max ) 1 − sin( φ max ) at the frequency √ λ · z where the magnitude is √ λ . Name: Problem 1 1. Find the five roots of the equation s 5 3 s 4 5 s 3 + 15 s 2 + 4 s 12 = 0 . Hint: use the Routh array and explain how the Routh array agrees with your answer. → s 5 15 4 → s 43 1512 s 3 → (1 / 6) of derivative of auxiliary polynomial2 5 → (2 / 3) * s 2 58 → 5 * s 1 9 → s8 The Routh array for the polynomial is given above. The polynomial has three roots with positive real part since there are three sign changes in the first column for the rows indicated by → ’s. The roots of the auxiliary polynomial are roots of the original polynomial. The auxiliary polynomial can be factored as 3 s 4 + 15 s 2 12 = 3( s 4 5 s 2 + 4) = 3( s 2 4)( s 2 1) . Thus, we see that four of the roots are at s = 2 , 2 , 1 , 1 . The last root can be determined by solving for r in the equation ( s r )( s 4 5 s 2 + 4) = s 5 3 s 4 5 s 3 + 15 s 2 + 4 s 12 , which is satisfied with r = 3 . 2. A Nyquist plot for loop gain for a feedback loop in the negative unity feedback configuration is shown below. The loop has two poles at the origin. All other poles have negative real part. Is the closedloop feedback system stable? If not, explain how the loop gain should be modified to guarantee closedloop stability. If so, estimate the gain margin, phase margin, vector margin, and the amount of time delay the loop can withstand and still result in closedloop stability.1.510.5 0.510.80.60.40.2 0.2 0.4 0.6 0.8 1 Nyquist Diagram Real Axis Imaginary Axis Name: (Blank page formerly) Yes, the closedloop feedback system is stable. The gain margin can be estimated by inverting the magnitude of the point where the Nyquist plot crosses the negative real axis. Thus, the gain margin is about 10 . The phase margin can be estimated by finding the angle between the negative real axis and the point in the third quadrant where the Nyquist plot crosses the circle of radius one centered at the origin....
View
Full
Document
This note was uploaded on 08/03/2010 for the course ECE PROF. VOLK taught by Professor Volkanrodoplu during the Spring '10 term at UCSB.
 Spring '10
 VolkanRodoplu

Click to edit the document details