Test 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010

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1 Test 2 SOLUTIONS ENGI 5821: Control Systems I March 15, 2010 Total marks: 20 Name: Student #: Answer each question in the space provided or on the back of a page with an indication of where to find the answer. There are 5 questions on this exam, which consists of 6 pages. A separate formula sheet is available for this test. 1

2 Q1 [3] (a) [1] Consider a unity feedback system with forward transfer function G(s) = 1 (s+1)(s+2). Briefly describe how you would modify this system to yield a steady-state error of 0 for a ramp input. Add a compensator in cascade with G(s). The compensator should have the transfer function 1 s 2. (b) [2] Transform the following block-diagram into a signal flow graph. Make sure you label all systems and all signals. 2

3 Q2 [4] Find the closed-loop transfer function for the following system. It is not necessary to simplify the algebra in your answer. You are encouraged to draw the intermediate steps. You can begin by noticing that both of the upper feedback paths have pick-off points at C(s). Both of these can be combined in parallel to yield H(s) + 1. A parallel form in the forward path can also be recognized. Incorporating both of these reductions yields the following: It can now be observed that G(s) is within a unity feedback loop. We can collapse this loop to G(s)/(1 + G(s)) and then combine it in cascade with the transfer function 2. Finally we reduce the positive feedback loop above to yield the overall closed-loop transfer function T (s): T (s) = = 2G(s) 1+G(s) 1 2G(s) 1+G(s) 2G(s) 1 2G(s)H(s) G(s) 3

4 Q3 [4] Consider the following closed-loop transfer function: T (s) = 1 s(s + 1) Determine the steady-state error for an impulse input to this system (r(t) = δ(t)). Recall the definition of error: E(s) = R(s) C(s) In this case, R(s) = 1. Since we already have the closed-loop transfer function, C(s) is easy to calculate: C(s) = R(s)T (s) = T (s) Therefore, E(s) = 1 T (s). We apply the final value theorem to get the steady-state error: e( ) = lim se(s) s 0 ( ) s = lim s s 0 s(s + 1) ( = lim 1 ) s 0 s + 1 = 1 4

5 Q4 [4] Determine the range of values of K for which the following closed-loop transfer function is stable. Assume K 0. T (s) = K s 4 + 3s 3 + 3s 2 + 2s + K We apply the Routh-Hurwitz criterion and consider sign changes in the first column. s K s s K s 1 s K 7 K The last row will always be positive as long as K > 0. Therefore we have a stable system as long as K > 0 and the sign of the second-last row remains positive: The system is stable for K 14 9K > 0 K < 14 9 ( ) 0,

6 Q5 [5] Consider a unity-feedback system with the following forward transfer function: G(s) = s2 2s + 2 (s + 1)(s + 2) For each of the labelled points below (A-G), state whether or not each point lies on the root locus. You must provide a justification for each point, referencing the definition of the root locus or the rules for sketching the root locus. Some points will require calculations to judge, while others can be ruled in or out by one or more of the sketching rules. Give your final answers in the form of a table. 2j -3 A C D E B j F G -j -2j The open-loop system has poles at -1 and -2 and zeroes at 1±j. The rules for sketching the root locus (RL) will enable us to judge whether points C, D, and E lie on the RL. point on the RL justification C No Lies to the left of an even number (2) of finite real-axis poles or zeroes D Yes Lies to the left of an odd number (1) of finite real-axis poles or zeroes E No Lies to the left of an even number (0) of finite real-axis poles or zeroes Now consider points A and B. The sketching rules say nothing about such points (they lie off the real-axis and are not themselves poles or zeros of the open-loop system). So we have to test whether they satisfy the definition of the root locus. Consider the sum of angles from zeros and poles to A: -3 A 2j j -j j KG(A) = = 180 o which is clearly an odd multiple of 180 o. The magnitude property can be satisfied by choosing an appropriate K. Therefore, point A is on the RL. Now consider point B: 6

7 B 2j j j j KG(B) = = 225 o which is clearly not an odd multiple of 180 o. Therefore point B is not on the RL. Point F is symmetrical about the real-axis from point A. One of the sketching rules states that the RL is symmetric above and below the real-axis. Therefore point F lies on the RL. Similarly, G is symmetric with B. Since B is not on the RL, neither is G. point on the RL justification A Yes KG(A) is an odd multiple of 180 o (see above) B No KG(B) is not an odd multiple of 180 o (see above) F Yes Symmetry with A G No Symmetry with B 7

ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27

ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27 1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system