반파대칭함수의 푸리에 계수

반파대칭함수의 푸리에 계수

fourier coefficient of half symmetry function

정리

주기가 $2L$인 함수 $f$가 반파대칭이면 $f$의 푸리에 계수는 아래와 같다.

$$ \begin{align*} a_{0} &= 0 \\ a_{n} &= \begin{cases} \dfrac{2}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt & (n=1, 3, \cdots ) \\ 0 & (n=0, 2, \cdots )\end{cases} \\ b_{n} &= \begin{cases} \dfrac{2}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt & (n=1, 3, \cdots ) \\ 0 & (n=0, 2, \cdots )\end{cases} \\ c_{n} &= \begin{cases} \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t)e^{-i\frac{n \pi t}{L} } dt & (n=\pm 1, \pm 3, \cdots) \\ 0 &( n=\pm 2, \pm 4, \cdots )\ \end{cases} \end{align*} $$

$c_{n}$은 복소 푸리에 계수이다.

증명

$$ \begin{align*} a_{0} &= \dfrac{1}{L} {\displaystyle \int_{-L}^{L} } f(t) dt \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t) dt + \dfrac{1}{L} {\displaystyle \int_{-L}^{0} } f(t) dt \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t) dt - \dfrac{1}{L} {\displaystyle \int_{-L}^{0} } f(t+L) dt \Big( \because f(t)=-f(t+L) \Big) \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t) dt - \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t) dt \quad (\text{change of variable }t+L = t) \\ &= 0 \end{align*} $$

$$ \begin{align*} a_{n} &= \dfrac{1}{L}\int_{-L}^{L} f(t) \cos \frac{n \pi t}{L} dt \\ &= {\displaystyle \dfrac{1}{L}\int_{0}^{L} }f(t) \cos \frac{n \pi t}{L} dt + \dfrac{1}{L}{\displaystyle \int_{-L}^{0}} f(t) \cos \frac{n \pi t}{L} dt \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{-L}^{0}} f(t+L) \cos \frac{n \pi t}{L} dt \quad \Big( \because f(t)=-f(t+L) \Big) \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi (t-L)}{L} dt \quad (\text{change of variable }t+L = t) \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \left( \cos\frac{n \pi t}{L}\cos (n \pi)+\sin\frac{n \pi t}{L}\sin n\pi \right) dt \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt - (-1)^n\dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \cos\frac{n \pi t}{L} dt \\ &= \begin{cases} \dfrac{2}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt & (n=1, 3, \cdots ) \\ 0 & (n=0, 2, \cdots )\end{cases} \end{align*} $$

$$ \begin{align*} b_{n} &= \dfrac{1}{L}\int_{-L}^{L} f(t) \sin \frac{n \pi t}{L} dt \\ &= {\displaystyle \dfrac{1}{L}\int_{0}^{L} }f(t) \sin \frac{n \pi t}{L} dt + \dfrac{1}{L}{\displaystyle \int_{-L}^{0}} f(t) \sin \frac{n \pi t}{L} dt \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{-L}^{0}} f(t+L) \sin \frac{n \pi t}{L} dt \quad \Big( \because f(t)=-f(t+L) \Big) \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi (t-L)}{L} dt \quad (\text{change of variable }t+L = t) \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \left( \sin\frac{n \pi t}{L}\cos (n \pi)+\cos\frac{n \pi t}{L}\sin n\pi \right) dt \\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt - (-1)^n\dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \sin\frac{n \pi t}{L} dt \\ &= \begin{cases} \dfrac{2}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt & (n=1, 3, \cdots ) \\ 0 & (n=0, 2, \cdots )\end{cases} \end{align*} $$

$$ \begin{align*} c_{n} &= \dfrac{1}{2}(a_{n}-ib_{n}) \\ &= \dfrac{1}{2}\dfrac{2}{L}{\displaystyle \int_{0}^{L} }f(t) \left( \cos\frac{n \pi t}{L} -i\sin \frac{n \pi t}{L} \right)dt \\ &= \dfrac{1}{L}{\displaystyle \int_{0}^{L} }f(t) e^{-i\frac{n \pi t }{L}} dt \end{align*} $$

$a_{n}$과 $b_{n}$의 조건에 의해

$$ c_{n} = \begin{cases} \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t)e^{-i\frac{n \pi t}{L} } dt & (n=\pm 1, \pm 3, \cdots) \\ 0 &( n=\pm 2, \pm 4, \cdots )\ \end{cases} $$

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