Sum of nn Out-of-phase Sinusoids with the Same Frequency

nei(x+φn)=n(cos(x+φn)+isin(x+φn))=ncos(x+φn)+nisin(x+φn)=ncos(x+φn)+insin(x+φn)\begin{aligned} \sum\limits_{n}e^{i(x+\varphi_n)}&=\sum\limits_{n}(\cos(x+\varphi_n)+i\sin(x+\varphi_n))\\ &=\sum\limits_{n}\cos(x+\varphi_n)+\sum\limits_{n}i\sin(x+\varphi_n)\\ &=\sum\limits_{n}\cos(x+\varphi_n)+i\sum\limits_{n}\sin(x+\varphi_n)\\ \end{aligned}

nei(x+φn)=neix+iφn=neixeiφn=eixneiφn=eixn(cos(φn)+isin(φn))=eix(ncos(φn)+nisin(φn))=eix(ncos(φn)+insin(φn))=eix(A+iB)=eixA2+B2eiarg(A+iB)=A2+B2eix+iarg(A+iB)=A2+B2ei(x+arg(A+iB))=A2+B2(cos(x+arg(A+iB))+isin(x+arg(A+iB)))=A2+B2cos(x+arg(A+iB))+iA2+B2sin(x+arg(A+iB))\begin{aligned} \sum\limits_{n}e^{i(x+\varphi_n)}&=\sum\limits_{n}e^{ix+i\varphi_n}\\ &=\sum\limits_{n}e^{ix}e^{i\varphi_n}\\ &=e^{ix}\sum\limits_{n}e^{i\varphi_n}\\ &=e^{ix}\sum\limits_{n}(\cos(\varphi_n)+i\sin(\varphi_n))\\ &=e^{ix}\left(\sum\limits_{n}\cos(\varphi_n)+\sum\limits_{n}i\sin(\varphi_n)\right)\\ &=e^{ix}\left(\sum\limits_{n}\cos(\varphi_n)+i\sum\limits_{n}\sin(\varphi_n)\right)\\ &=e^{ix}(A+iB)\\ &=e^{ix}\sqrt{A^2+B^2}e^{i\arg(A+iB)}\\ &=\sqrt{A^2+B^2}e^{ix+i\arg(A+iB)}\\ &=\sqrt{A^2+B^2}e^{i(x+\arg(A+iB))}\\ &=\sqrt{A^2+B^2}(\cos(x+\arg(A+iB))+i\sin(x+\arg(A+iB)))\\ &=\sqrt{A^2+B^2}\cos(x+\arg(A+iB))+i\sqrt{A^2+B^2}\sin(x+\arg(A+iB)) \end{aligned}

{nsin(x+φn)=A2+B2sin(x+arg(A+iB))ncos(x+φn)=A2+B2cos(x+arg(A+iB))A=ncos(φn)B=nsin(φn)\boxed{ \begin{cases} \sum\limits_{n}\sin(x+\varphi_n)=\sqrt{A^2+B^2}\sin(x+\arg(A+iB))\\ \sum\limits_{n}\cos(x+\varphi_n)=\sqrt{A^2+B^2}\cos(x+\arg(A+iB))\\ A=\sum\limits_{n}\cos(\varphi_n)\\ B=\sum\limits_{n}\sin(\varphi_n) \end{cases}}

Example:

=sin(2π(t+e))+sin(2πt+69)+sin(2πt+42)=sin(2πt+2πe)+sin(2πt+69)+sin(2πt+42)=A2+B2sin(2πt+arg(A+iB))\begin{aligned} &\phantom{=}\sin(2\pi(t+e))+\sin(2\pi t+69)+\sin(2\pi t+42^\circ)\\ &=\sin(2\pi t+2\pi e)+\sin(2\pi t+69)+\sin(2\pi t+42^\circ)\\ &=\sqrt{A^2+B^2}\sin(2\pi t+\arg(A+iB)) \end{aligned} {A=cos(2πe)+cos(69)+cos(42)1.53856064504B=sin(2πe)+sin(69)+sin(42)0.425861365902\begin{cases} A=\cos(2\pi e)+\cos(69)+\cos(42^\circ)\approx1.53856064504\\ B=\sin(2\pi e)+\sin(69)+\sin(42^\circ)\approx-0.425861365902 \end{cases} A2+B21.59641058674\sqrt{A^2+B^2}\approx1.59641058674 arg(A+iB)=tan1(BA)15.4716657207\arg(A+iB)=\tan^{-1}\left(\frac BA\right)\approx-15.4716657207^\circ sin(2π(t+e))+sin(2πt+69)+sin(2πt+42)1.60sin(2πt15.5)\sin(2\pi(t+e))+\sin(2\pi t+69)+\sin(2\pi t+42^\circ)\approx\boxed{1.60\sin(2\pi t-15.5^\circ)}

Note by Gandoff Tan
2 weeks, 2 days ago

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