Solutions_F11_ECH157_HW5

Solutions_F11_ECH157_HW5 - 12.2 Problem 2: G = GvG p Gm =...

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1.6(1 0.5 ) (3 1) vpm s GGGG ss == + The process transfer function contains a zero at s = +2. Because the controller in the Direct Synthesis method contains the inverse of the process model, the controller will contain an unstable pole. Thus, Eqs. (12-4) and (12-5) give: ( ) () 31 11 τ 2 τ 10 . 5 c cc s G Gs s + Modeling errors and the unstable controller pole at s = +2 would render the closed-loop system unstable. Modify the specification of Y/Y sp such that G c will not contain the offending (1-0.5 s ) factor in the denominator. The obvious choice is . 5 τ 1 sp c d Ys  =  +  Then using Eq.(12-3b), 2 τ 1 c c s G + =− + which is not physically realizable because it requires ideal derivative action. Modify Y/Y sp , 2 . 5 ( τ = + sp c d Y Then Eq.(12-3b) gives 2 2 τ 4 τ 1 c s G s + + + which is physically realizable.
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Solution: Cohen and Coon Me/hod: From the solution of Exercise II.l4, we have ( ) 2.86 - 038s g s = e 1.9s + I We already have the system response in the FOPDT form, thus directly from Table 12.2, we can
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This note was uploaded on 01/28/2012 for the course ECH 157 taught by Professor Palagozu during the Fall '08 term at UC Davis.

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Solutions_F11_ECH157_HW5 - 12.2 Problem 2: G = GvG p Gm =...

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