hw6soln_F2009 - MAE170 Homework 6 Solution Fall 2009 P10.5...

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MAE170 Homework 6 Solution Fall 2009 P10.5 We desire PO % < 10%, T s < 1 . 5 and zero steady-state error to a step command. From these specifications ζ > 1 1+ π 2 / ln 2 ( PO % / 100) = 0 . 5912, and ζω n > 8 3 = 2 . 667. To achieve zero state-error a PI controller is used. The architecture of the controller is a follows: G c ( s ) = k p + k I s = k p ( s + k I k p ) s (1) To design the controller we’ll first find the k p that satisfies both the PO % and the settling time requirement by settling k I = 0 for now. The characteristic equation is, 1 + k a G c G = 1 + 25 k a k p ( s + 0 . 15)( s + 6 . 667) = 0 (2) Equivalently, s 2 + 6 . 817 s + (1 + 25 k a k p ) = 0 (3) We then select a desired set of roots that meet the specifications (see figure below) s d = - ζω n ± n p 1 - ζ 2 = - 3 . 4085 ± 4 j (4) For the desired roots ζ = 0 . 6486 and ω n = 5 . 2552 1
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s 2 + 6 . 817 s + (1 + 25 k a k p ) = s 2 + 2 ζω n s + ω 2 n = s 2 + 6 . 817 s + 27 . 6171 (5) therefore, k a k p = 27 . 6171 - 1 25 = 1 . 0647 (6) So, we choose k a = 1 . 0647 and k p = 1 The response is given below with P.O = 6 . 87%, T s = 1 . 14, and e ss = 0 . 0360 Step Response Time (sec) Amplitude 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 0 0.2 0.4 0.6 0.8 1 1.2 1.4 System: Closed Loop r to y I/O: r to y Peak amplitude: 1.03 Overshoot (%): 6.87 At time (sec): 0.777 System: Closed Loop r to y I/O: r to y Settling Time (sec): 1.14 System: Closed Loop r to y I/O: r to y Final Value: 0.964 The transient response specifications are met so now we need to design the integrator part of the controller to remove the steady-state error. G c ( s ) = 1 + k I s = ( s + k I ) s (7) We will place the zero of the controller z = k I close to the controller pole at the origin so that the root locus is not significantly affected. More specifically, we will make the difference of the angles of the pole and the zero at the desired root be approximately 1 (i.e θ p - θ z = 1 ). The characteristic equation is 1 + k a G c G = 1 + 26 . 6175( s + k I ) s ( s + 0 . 15)( s + 6 . 667) = 0 (8) therefore, θ p = ( π - arctan( 4 3 . 4085 )) 180 π = 130 . 4351 and θ z = θ p - 1 = 129 . 4351 Now we can solve for z. tan(
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This note was uploaded on 08/31/2010 for the course ENGRMAE 170 taught by Professor Staff during the Fall '08 term at UC Irvine.

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hw6soln_F2009 - MAE170 Homework 6 Solution Fall 2009 P10.5...

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