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 f 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 f cations, and the design procedure is applied correctly, then veri f cation that the speci f cations are actually satis f 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 f 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 . Problem #B-7-12
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hwsln-421-s06-10b - ECE 421, Spring 2006 Solutions to...

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