hwsln-421-s06-08

hwsln-421-s06-08 - ECE 421, Spring 2006 Solutions to HW...

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ECE 421, Spring 2006 Solutions to HW Assignment #8 Problem B-6-1 G ( s )= K ( s +1) s 2 ,H ( s )=1 ,K 0 (1) On the real axis, the root locus lies to the left of s = 1 . There are two poles ( n =2) and one zero ( m =1) ,sothereison ly n m =1 asymptote. The angle and center (not really used if there is only one asymptote) of the asymptote are φ A = 180 (2 l 2 1 = 180 ,l =0 σ A = 0+0 ( 1) 2 1 (2) Break points can be determined from dK/ds K = s 2 ( s , dK ds = £ ( s · 2 s s 2 (1) ¤ ( s 2 =0= s 2 +2 s = s ( s +2) (3) so the break points are at s (where there are 2 open-loop poles) and s = 2 . Plot is attached. The complex conjugate part of the root locus for this transfer function is a perfect circle centered at the location of the zero ( s = 1) with radius equal to the geometric mean of the distance from the zero to each of the two poles. The radius is r = 1 · 1=1 , so the break points are 1 unit to the left and to the right of the zero location. Problem B-6-2 G ( s K ( s +4) ( s 2 ( s 0 (4) On the real axis, the root locus lies to the left of s = 4 . There are 2 poles and 1 zero, so there is only one asymptote. The angle and center of the asymptote are φ A = 180 (2 l 2 1 = 180 σ A = 1 1 ( 4) 2 1 =2 (5) Break points can be determined from dK/ds K = ( s 2 ( s , dK ds = h ( s · 2( s ( s 2 (1) i ( s 2 (6) 0= s 2 +8 s +7=( s +1)( s +7) (7) so the break points are at s = 1 (where there are 2 open-loop poles) and s = 7 . Plot is attached. The complex conjugate part of the root locus for this transfer function is a perfect circle centered at the location of the zero ( s =
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This note was uploaded on 10/25/2010 for the course ECE 421 taught by Professor Cook,g during the Spring '08 term at George Mason.

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hwsln-421-s06-08 - ECE 421, Spring 2006 Solutions to HW...

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