what is the relationship between R and R1 and R3 for R.cos(wt+β​)=R1.cos(w​t+β)+R3.co​s(3wt+3β)

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YOUSSEF El MOUSSATI
YOUSSEF El MOUSSATI 2024 年 1 月 8 日
コメント済み: Sam Chak 2024 年 3 月 29 日
With R =squrt(R1^2+R3^2) for R.cos(wt+β)=R1.cos(wt+β)+R3.cos(wt+β)
  2 件のコメント
Torsten
Torsten 2024 年 1 月 8 日
I don't understand your question.
Sam Chak
Sam Chak 2024 年 3 月 29 日
@YOUSSEF El MOUSSATI, Has this question been resolved? If not, could you please provide constructive feedback and comments to guide further follow-up and discussion?

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回答 (1 件)

Walter Roberson
Walter Roberson 2024 年 1 月 8 日
syms R R1 R3 w t beta
eqn1 = R == sqrt(R1^2 + R3^2);
eqn2 = R .* cos(w * t + beta) == R1 .* cos(w*t + beta) + R3 .* cos(w*t + beta);
sol = solve([eqn1, eqn2], [R1, R3])
Warning: Possibly spurious solutions.
sol = struct with fields:
R1: [2×1 sym] R3: [2×1 sym]
sol.R1
ans = 
sol.R3
ans = 
So the relationship is that R1 = R and R3 = 0, OR R1 = 0 and R3 = R

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