Page 187 - MATLAB an introduction with applications

172 ——— MATLAB: An Introduction with Applications

(b) We note that the constant ζ points (0 < ζ < 1) lie on a straight line having angle θ from the jω axis as
shown in Fig. E3.21(b).
jω
Constant ζ line
ω n
θ

σ
–ζω n
Fig. E3.21(b)
From Fig. E3.21(b), we obtain
ζω
sinθ= n =ζ
ω n
Also that ζ = 0.6 line can be defined by
s = – 0.75a + ja
where a is a variable (0 < a < ∞). To obtain the value of K such that the damping ratio ζ of the dominant
closed-loop poles is 0.6, we determine the intersection of the line s = –0.75a + ja and the root locus. The
intersection point can be obtained by solving the following simultaneous equations for a.
s = – 0.75a + ja ...(1)
s(s + 3)(s + 5) + K = 0 ...(2)
From Eqs. (1) and (2), we obtain
(– 0.75a + ja)(–0.75a + ja + 3)(–0.75a + ja +5) + K = 0
or
2
2
3
(1.8281a – 2.1875a – 3a + K) + j(0.6875a – 7.5a + 15a)= 0
3
Equating the real part and imaginary part of the above equation to zero, respectively, we obtain
2
3
1.8281a – 2.1875a – 3a + K = 0 ...(3)
3
0.6875a – 7.5a + 4a = 0 ...(4)
2
Equation (4) can be rewritten as
a = 0
or 0.6875a – 7.5a + 4 = 0
2
or
2
a – 10.90991a + 5.8182 = 0
or (a – 0.5623)(a – 10.3468) = 0
Therefore, a = 0.5323 or a = 10.3468
From Eq.(3), we obtain
K = –1.8281a + 2.1875a + 3a = 2.0535 for a = 0.5626
3
2
3
2
K = –1.8281a + 2.1875a + 3a = –1759.74 for a = 10.3468
Since the K value is positive for a = 0.5623 and negative for a = –10.3468, we select a = 0.5623. The required
gain K is 2.0535.

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