質問: π となりましたので √π * √πでπとなりました
回答: 相互相関係数の定義はお分かりですか? 違っていたら計算は意味ないですよ? 定義を良く調べて見てください。 r_fg(τ)=e{f(t)g(t+τ)}/[{√e(f^2(t))}{√e(g^2(t))} ここで、 e{x(t)y(t+τ)} ={1/(2π)}∫[0,2π] f(t)g(t+τ)dt ={1/(2π)}∫[0,2π] sin(t)cos(t+τ)dt ={1/(2π)}∫[0,2π] sin(t){cos(t)cos(τ)-sin(t)sin(τ)}dt ={1/(2π)}∫[0,2π] {sin(t)cos(t)cos(τ)-sin^2(t)sin(τ)}dt ={1/(2π)}∫[0,2π] {-sin^2(t)sin(τ)}dt ={-sin(τ)/(2π)}∫[0,2π] {sin^2(t)}dt ={-sin(τ)/(4π)}∫[0,2π] {1-cos(2t)}dt =-sin(τ)/2 e{f^2(t)}={1/(2π)}∫[0,2π] sin^2(t)dt ={1/(4π)}∫[0,2π] {1-cos(2t)}dt=1/2 e{g^2(t)}={1/(2π)}∫[0,2π] cos^2(t)dt ={1/(4π)}∫[0,2π] {1+cos(2t)}dt=1/2 故に r_fg(τ)=-{sin(τ)/2}/{(1/√2)*(1/√2)}=-sin(τ) ←(答え) 参考url:http://www.mech.tohoku-gakuin.ac.jp/nken/java/autocorr/theory.pdf