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hwsln-421-s06-10a

# hwsln-421-s06-10a - ECE 421 Spring 2006 Solutions to...

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ECE 421, Spring 2006 Solutions to ‘Additional Problems’ for HW Assignment #10 The solutions to the textbook problems B-7-11, B-7-12, B-7-13, B-7-15, and B-7-16 appear after in a separate fi le. None of the problems has a unique solution–each of the problems has an in fi nite number of valid solutions. This includes the choice of the speci fi ed closed-loop pole s = s 1 (except for Problem #5 when s 1 is given) as well as the choice of compensator. The placement of the compensator pole and zero to provide the correct transient performance can be done using the procedure in the text (except for Problem #5 where it does not provide closed-loop stability), or either the pole or zero can be placed at a speci fi ed location. In either case, the poles and zeros of the compensator designed to establish s 1 as a closed-loop pole are placed to satisfy the phase angle criterion, and the gain is computed to satisfy the magnitude criterion. Once that part of the compensator is designed, steady-state error can be checked to determine whether or not a “special” lag compensator is needed to reduce that error. For these HW problems, the choice of the desired dominant closed-loop pole s 1 can be done based on the equations for the standard second-order system. If s 1 is chosen correctly based on the transient performance speci fi cations, and the design procedure is applied correctly, then veri fi cation that the speci fi cations are actually satis fi ed is not necessary. Problem #1 The speci fi ed value of settling time of T s = 2 seconds means that the absolute value of the real part of s 1 is | Re [ s 1 ] | = 4 / 2 = 2 . The speci fi cation on overshoot can be used to determine the damping ratio ζ and the corresponding angle β. Using the speci fi ed 10% overshoot, the damping ratio and angle are ζ = ¯ ¯ ln ¡ 10 100 ¢¯ ¯ q π 2 + £ ln ¡ 10 100 ¢¤ 2 = 0 . 5912 , β = cos 1 ( ζ ) = 53 . 8 (1) Therefore, the desired closed-loop pole is s 1 = μ 4 T s · [ 1 + j tan ( β )] = 2 + j 2 . 7288 (2) The phase shift of G p ( s 1 ) is 6 G p ( s 1 ) = tan 1 μ 2 . 7288 0 2 ( 5) tan 1 μ 2 . 7288 0 2 0 tan 1 μ 2 . 7288 0 2 ( 2) (3) tan 1 μ 2 . 7288 0 2 ( 8) tan 1 μ 2 . 7288 0 2 ( 10) = 42 . 3 126 . 2 90 24 . 5 18 . 8 = 217 . 2 = +142 . 8 The compensator phase angle at s 1 is 6 G c ( s 1 ) = 180 ( 217 . 2) = 180 142 . 8 = 37 . 2 . Since this angle is positive, a lead compensator is needed. The pole/zero placement procedure presented in the text will be used, which yields the following calculation. 6 ( s 1 ) = tan 1 μ 2 . 7288 0 2 0 = 126 . 2 (4) 6 ( s 1 z c ) = 126 . 2 + 37 . 2 2 = 81 . 7 , 6 ( s 1 p c ) = 126 . 2 37 . 2 2 = 44 . 5 d zc = Im ( s 1 ) tan ( 6 ( s 1 z c )) = 0 . 3962 , d pc = Im ( s 1 ) tan ( 6 ( s 1 p c )) = 2 . 777 1

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The compensator zero is placed at s = 2 0 . 3962 = 2 . 3962 , and the compensator pole is placed at s = 2 2 . 777 = 4 . 777 . The compensator gain is computed so that the series combination of the plant and lead compensator satis fi es the magnitude criterion at s 1 .
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hwsln-421-s06-10a - ECE 421 Spring 2006 Solutions to...

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