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problem_1121

# problem_1121 - of(2 we have the following three equations 4...

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Problem 11.21 of the Palm textbook Plant TF: G P s ( ) = 4 3 s 2 + 3 Design a PID controller that gives a DOMINANT CL pole corresponding to: a damping ratio of ! = 0.5 . a time constant of ! = 1 . SOLUTION PID controller : G C s ( ) = K P + K I s + K D s = K D s 2 + K P s + K I s General PID closed-loop characteristic equation : 0 = 1 + G C s ( ) G P s ( ) = 1 + K D s 2 + K P s + K I s ! " # \$ % & 4 3 s 2 + 3 ! " # \$ % & OR D CL s ( ) = s 3 s 2 + 3 ( ) + 4 K D s 2 + K P s + K I ( ) = 3 s 3 + 4 K D s 2 + 3 + 4 K P ( ) s + 4 K I = s 3 + 4 3 K D s 2 + 1 + 4 3 K P ! " # \$ % & s + 4 3 K I = 0 (1) Design closed-loop characteristic equation : The problem asks us to find values of K P , K I and K D such that the dominant poles are complex (recall that we need ! = 0.5 ). These dominant poles are governed by the following complex factor: s 2 + 2 !" n s + " n 2 . Since we have a third-order CLCE, we need an additional real factor of s + b , where b is, at this point, unknown. Therefore, our total design CLCE is of the form: 0 = s 2 + 2 !" n s + " n 2 ( ) s + b ( ) = s 3 + 2 !" n + b ( ) s 2 + 2 b !" n + " n 2 ( ) s + b " n 2 (2) Comparing coefficients for our general PID CLCE of (1) with those of the design CLCE

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Unformatted text preview: of (2), we have the following three equations: 4 3 K D = 2 !" n + b 1 + 4 3 K P = 2 b n + " n 2 4 3 K I = b n 2 (3) Since we desire ! = 0.5 and = 1 "# n = 1 , this gives: n = 1 = 1 0.5 = 2 . Substitution of this into equations (3) gives: 4 3 K D = 2 n + b = 2 + b # K D = 3 4 2 + b ( ) 1 + 4 3 K P = 2 b n + n 2 = 2 b + 4 # K P = 3 4 2 b + 3 ( ) 4 3 K I = b n 2 = 4 b # K I = 3 b What should we use for “b”? Recall that b is the third (real) pole of the CLCE. We simply need to choose b such that we are insured that the design poles above are DOMINANT. That is, we need to choose b such that its corresponding time constant is less than our design poles. In other words, we need b > n = 1 ....
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