My roots don't match example roots when solving routh criterion

조회 수: 4 (최근 30일)
Jhryssa
Jhryssa 2024년 9월 24일
답변: Sam Chak 2024년 9월 25일
I can't seem to figure out how to output similar roots as my teacher's. I've written the same code as he did (modified to match my problem).
The left is what I've started to input and the right is the example he gave us. When I also run my teachers code I get similar outputs to the one on the left. How do I get me roots to match his roots? This is a Routh Stability Criterion Problem with an unknown variable

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Torsten
Torsten 2024년 9월 25일
이동: Torsten 2024년 9월 25일
syms K s
p = s^3+9*s^2+12*s+15+3*K;
solve(p,s,'MaxDegree',3)
var2 = vpa(ans);
var = subs(var2,K,-1)

추가 답변 (1개)

Sam Chak
Sam Chak 2024년 9월 25일
From Part (a), you should have obtained the stable range of K: . Part (b) instructs you to verify the stability result by substituting the midpoint value of the stable range, and you should expect that the real parts of all roots are negative.
a = -5; % lower bound of stable range
b = 31; % upper bound of stable range
Kmid= (a + b)/2 % mindpoint of that range
Kmid = 13
syms K s
p3 = s^3 + 9*s^2 + 12*s + 15 + 3*K
solve(p3, s, 'MaxDegree', 3)
sol = vpa(ans);
%% Test
var = subs(sol, K, a) % the real-valued root must be zero
var = subs(sol, K, b) % the real part of the complex-valued roots must be zero
var = subs(sol, K, Kmid) % the real parts of all roots must be negative

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