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)+i∑nsin⁡(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=eix∑neiφn=eix∑n(cos⁡(φn)+isin⁡(φn))=eix(∑ncos⁡(φn)+∑nisin⁡(φn))=eix(∑ncos⁡(φn)+i∑nsin⁡(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+B2≈1.59641058674\sqrt{A^2+B^2}\approx1.59641058674 arg⁡(A+iB)=tan⁡−1(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⁡(2πt−15.5∘)\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)}

#Algebra

Note by Gandoff Tan
12 months ago

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