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Harman [31]
3 years ago
8

Set up the following characteristic equations in the form suited to Evanss root-locus method. Give L(s), a(s), and b(s) and the

parameter, K, in terms of the original parameters in each case. Be sure to select K so that a(s) and b(s) are monic in each case and the degree of b(s) is not greater than that of a(s).
a) s + (1/τ) = 0 versus parameter τ
b) s2 + cs + c + 1 = 0 versus parameter c
c) (s + c)3 + A(Ts + 1) = 0
i. versus parameter A
ii. versus parameter T
iii. versus the parameter c, if possible. Say why you can or can not. Can a plot of the roots be drawn versus c for given constant values of A and T by any means at all.
d) 1 + (kp + k1/s + kDs/Ts + 1)G(s) = 0. Assume that G(s) = A c(s)/d(s), where c(s) and d(s) are monic polynomials with the degree of d(s) greater than that of c(s).
i. versus kp
ii. versus kI
iii. versus kD
iv. versus τ

Engineering
1 answer:
Sunny_sXe [5.5K]3 years ago
4 0

Answer:

attached below is the detailed solution and answers

Explanation:

Attached below is the detailed solution

C(iii) : versus the parameter C

The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C

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