problem_11_21

problem_11_21 - Problem 11.21 of the Palm textbook Plant...

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Unformatted text preview: Problem 11.21 of the Palm textbook Plant TF: GP ( s ) = 4 2 3σ + 3 Design a PID controller that gives a DOMINANT CL pole corresponding to: • a damping ratio of ζ = 0.5 . • a time constant of τ = 1 . SOLUTION PID controller: GC ( s ) = ΚΠ + ΚΙ Κ σ2 + ΚΠσ+ ΚΙ + Κ∆ σ = ∆ σ σ General PID closed-loop characteristic equation: Κ σ2 + ΚΠσ+ ΚΙ 4 0 = 1 + Γ Χ ( σ) Γ Π ( σ) = 1 + ∆ ÷ ÷ 2 σ 3σ + 3 OR DCL ( s ) = σ( 3σ2 + 3) + 4 ( Κ∆ σ2 + ΚΠσ+ ΚΙ ) = 3σ3 + 4 Κ∆ σ2 + ( 3 + 4 ΚΠ ) σ+ 4 ΚΙ = σ3 + 4 4 4 Κ∆ σ2 + 1 + ΚΠ÷ σ+ ΚΙ = 0 3 3 3 (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 ζ = 0.5 ). These dominant poles are governed by the 2 following complex factor: s 2 + 2ζω ν σ+ ω ν . Since we have a third-order CLCE, we need an additional real factor of s + β , where b is, at this point, unknown. Therefore, our total design CLCE is of the form: 2 0 = ( σ2 + 2ζω ν σ+ ω ν ) ( σ+ β) 2 2 = σ3 + ( 2ζω ν + β) σ2 + ( 2βζω ν + ω ν ) σ+ βω ν (2) Comparing coefficients for our general PID CLCE of (1) with those of the design CLCE of (2), we have the following three equations: 4 K D = 2ζω ν + β 3 4 2 1 + ΚΠ = 2βζω ν + ω ν 3 4 2 ΚΙ = βω ν 3 Since we desire ζ = 0.5 and τ = (3) 1 1 1 = 1 , this gives: ω n = = = 2 . Substitution of zwn z 0.5 this into equations (3) gives: 4 K D = 2ζω ν + β = 2 + β 3 4 2 1 + ΚΠ = 2βζω ν + ω ν = 2β + 4 3 4 2 ΚΙ = βω ν = 4 β 3 ⇒ ⇒ 3 ( 2 + β) 4 3 ΚΠ = ( 2β + 3) 4 Κ∆ = ⇒ ΚΙ = 3β 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 > ζω ν = 1 . ...
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problem_11_21 - Problem 11.21 of the Palm textbook Plant...

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