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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) , so there is only 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 + 1) 2 1 = 180 , l = 0 σ A = 0 + 0 ( 1) 2 1 = 1 (2) Break points can be determined from dK/ds = 0 K = s 2 ( s + 1) , dK ds = £ ( s + 1) · 2 s s 2 (1) ¤ ( s + 1) 2 = 0 = s 2 + 2 s = s ( s + 2) (3) so the break points are at s = 0 (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 + 1) 2 , H ( s ) = 1 , K 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 + 1) 2 1 = 180 , l = 0 σ A = 1 1 ( 4) 2 1 = 2 (5) Break points can be determined from dK/ds = 0 K = ( s + 1) 2 ( s + 4) , dK ds = h ( s + 4) · 2 ( s + 1) ( s + 1) 2 (1) i ( s + 4) 2 = 0 (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

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

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