how to find complex polynomial solution
조회 수: 12 (최근 30일)
이전 댓글 표시
I want to find roots of CDP, T's are constant real values
CDP1 =T_6*(1i*w).^6 +T_5*(1i*w).^5 +T_4*(1i*w).^4 +T_3*(1i*w).^3 +T_2*(1i*w).^2 +T_1*(1i*w) +T_0;
CDP2 =Tp_4*(1i*w)^.4 +Tp_3*(1i*w).^3 +Tp_2*(1i*w).^2 +Tp_1*(1i*w) +Tp_0;
CDP= CDP1*(1i*w).^1.4 +CDP2;
댓글 수: 2
답변 (1개)
Alex Mcaulley
2019년 7월 30일
Then, after defining all the constant values:
syms w
CDP1 = T_6*(1i*w).^6 + T_5*(1i*w).^5 + T_4*(1i*w).^4 + T_3*(1i*w).^3 + T_2*(1i*w).^2 + T_1*(1i*w) + T_0;
CDP2 = Tp_4*(1i*w)^.4 + Tp_3*(1i*w).^3 + Tp_2*(1i*w).^2 + Tp_1*(1i*w) + Tp_0;
CDP = CDP1*(1i*w).^1.4 + CDP2;
sol = double(solve(CDP))
댓글 수: 8
Walter Roberson
2019년 7월 30일
w^10 is okay. You then multiply by i and raise the result to 0.9 or 4.9. By the power law, (A*B)^C is A^C*B^C so (i*w^10)^0.9 is i^0.9 * (w^10)^0.9 and that second part is not polynomial
In the case where w is nonnegative real if you are willing to treat 0.9 as 9/10 exactly (which it is not) then you could multiply out to get w^9. But if that is what you want then you need to code it: with the 1i in there, matlab would never compute it that way. You would be getting a different branch of 0.9 power.
참고 항목
카테고리
Help Center 및 File Exchange에서 Polynomials에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!