HW8 Solution - Michigan State UniversityHW8 Solution.pdf Author aspelun1 Created Date 4/14/2016...
Transcript of HW8 Solution - Michigan State UniversityHW8 Solution.pdf Author aspelun1 Created Date 4/14/2016...
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E7.10 (a) The characteristic equation is
1 +K(s+ 2)
s(s+ 1)= 0 .
Therefore,
K = −(s2 + s)
(s+ 2),
and
dK
ds= −
s2 + 4s + 2
(s+ 2)2= 0 .
Solving s2+4s+2 = 0 yields s = −0.586 and −3.414. Thus, the system
breakaway and entry points are at s = −0.586 and s = −3.414.
(b) The desired characteristic polynomial is
(s+ 2 + aj)(s + 2− aj) = s2 + 4s+ 4 + a2 = 0 ,
where a is not specified. The actual characteristic polynomial is
s2 + (1 +K)s+ 2K = 0 .
Equating coefficients and solving for K yields K = 3 and a =√
2.Thus, when K = 3, the roots are s1,2 = −2±
√
2j.
(c) The root locus is shown in Figure E7.10.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
xxo
Real Axis
Imag Axis
*
*
K=3, s=-2+1.414j
s=-0.58s=-3.41
FIGURE E7.10
Root locus for 1 +Ks+2
s(s+1)= 0.