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Unformatted text preview: Example on Root Locus Sketching and Control Design MCE441 - Spring 05 Dr. Richter April 25, 2005 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We wish to: K ( s 2 +6 . 5 s +12) s 1 ( s +1)( s +2) U ( s ) Y ( s ) R ( s ) +- Figure 1: Control System 1. Provide a detailed hand-sketch of the root locus. 2. Find the gain K so that the system responds with an overshoot of 1 percent. 3. Find the gain K so that the system responds with an overshoot of less than 1 percent and a settling time as fast as possible. Root Locus Sketching The poles of the open-loop transfer function G ( s ) K ( s ) are given by s = 0 , s =- 1 , s =- 2 and there is a pair of complex zeros at s =- 3 . 25 ± 1 . 199 i The order of the system (number of poles) is n = 3 and the number of zeros is m = 2. According to this, m = 2 poles go to the zeros as K → ∞ and the remaining pole goes to infinity following an asymptote. The angle of the asymptote is obviously 180 ◦ , but in general, we find asymptote angles with the formula φ A = 180 n- m (2 k + 1) where k = 0 , 1 , ..n- m- 1. In our case, n- m- 1 = 0, so φ A = 180 ◦ Again, in this case it is unnecessary to find the center of the asymptotes, but in the general case it is obtained by subtracting the sum of zeros from the sum of poles and dividing by n- m . In this case σ A = (- 1- 2)- (- 3 . 25 + 1 . 199 i- 3 . 25- 1 . 199 i ) 1 = 3 . 5 With this information, we still have two possibilities for the shape of the root locus that comply with the basic requirements, as shown in the figure below. In order to determine which one of the possibilities is 1 Possibility 1 Possibility 2 Figure 2: Two possibilities for the root locus...
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This note was uploaded on 01/11/2011 for the course MECH ENG AE361 taught by Professor Waltereversman during the Fall '10 term at Missouri S&T.
- Fall '10