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hwsln-421-s06-10b

hwsln-421-s06-10b - ECE 421 Spring 2006 Solutions to...

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ECE 421, Spring 2006 Solutions to ‘Textbook Problems’ for HW Assignment #10 The solutions to the additional problems appear after in a separate fi le. For these HW problems, if there is a choice of the desired dominant closed-loop pole s 1 , it 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 #B-7-11 The closed-loop poles of the given plant model with G c ( s ) = 1 are located at s = 2 ± j 2 3 . Thus, the desired dominant closed-loop pole used for design is s 1 = 2 + j 2 3 . Since that is already a closed-loop pole, | G p ( s 1 ) | = 1 and 6 G p ( s 1 ) = 180 . Therefore, no additional phase shift needs to be added, only the steady-state error needs to be considered. For the given system, the velocity error constant is K v actual = lim s 0 s · 16 s ( s + 4) ¸ = 4 (1) The speci fi ed value for K v is K v = 20 , the units are sec 1 . A special lag compensator can be designed for the system with the parameter α g being α g = z cg p cg = K v spec K v actual = e ss actual e ss spec = 20 4 = 5 (2) In order to have the compensator’s pole and zero separated by a factor of 5 and still have | G c ( s 1 ) | 1 and 6 G c ( s 1 ) 0 , they must be located must closer to s = 0 than the point s 1 . The compensator zero is normally placed to the right of the real part of s 1 by a factor of 100 , and the compensator pole is placed to the right of the zero by the factor α g . The special lag compensator is G c ( s ) = ( s + 0 . 02) ( s + 0 . 004) (3) The gain of the special lag compensator is always 1 .

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