フーリエ変換の問題

■問題

$3$\displaystyle{ f\left({x}\right)=\left{{ \begin{array} \frac{\hspace{2}k\hspace{2}}{\hspace{2}L\hspace{2}} & - \frac{\hspace{2}L\hspace{2}}{\hspace{2}2\hspace{2}}< x< \frac{\hspace{2}L\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ 0& \left|{ \frac{\hspace{2}L\hspace{2}}{\hspace{2}2\hspace{2}} }\right|\geq x & \vspace{6}\\ \end{array} }$

のフーリエ余弦変換 $3$\displaystyle{ \hspace{1}{F}_{c}\left({{\omega}}\right) }$ を求めよ．また， $3$\displaystyle{ { \lim }\limits_{ L\rightarrow 0 }\hspace{1}{F}_{c}\left({{\omega}}\right) }$ を求めよ．

■解答

$3$\displaystyle{ \hspace{1}{F}_{c} ${\omega}$ = \frac{\hspace{2}k\hspace{2}}{\hspace{2} L{\omega} \hspace{2}}\sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}{\pi}\hspace{2}} }\sin \frac{\hspace{2} L{\omega} \hspace{2}}{\hspace{2}2\hspace{2}} }$ ， $3$\displaystyle{ { \lim }\limits_{ L\rightarrow 0 }\hspace{1}{F}_{c}\left({{\omega}}\right)=\frac{\hspace{2}k\hspace{2}}{\hspace{2} \sqrt{ 2{\pi} } \hspace{2}} }$

■解き方

$3$\displaystyle{ \hspace{1}{F}_{c}\left({{\omega}}\right)=\sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}{\pi}\hspace{2}} }\displaystyle{ \int _{0}^{\infty} f\left({x}\right)\cos {\omega}xdx } }$

$3$\displaystyle{ =\sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}{\pi}\hspace{2}} }\displaystyle{ \int _{0}^{ \frac{\hspace{2}L\hspace{2}}{\hspace{2}2\hspace{2}} } \frac{\hspace{2}k\hspace{2}}{\hspace{2}L\hspace{2}}\cos {\omega}xdx } }$

$3$\displaystyle{ =\sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}{\pi}\hspace{2}} }\cdot \frac{\hspace{2}k\hspace{2}}{\hspace{2}L\hspace{2}}\cdot \left[{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}{\omega}\hspace{2}}\sin {\omega}x }\right] _{0}^{ \frac{\hspace{2}L\hspace{2}}{\hspace{2}2\hspace{2}} } }$

$3$\displaystyle{ =\sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}{\pi}\hspace{2}} }\cdot \frac{\hspace{2}k\hspace{2}}{\hspace{2}L\hspace{2}}\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2}{\omega}\hspace{2}}\sin \frac{\hspace{2} L{\omega} \hspace{2}}{\hspace{2}2\hspace{2}} }$

$3$\displaystyle{ =\frac{\hspace{2}k\hspace{2}}{\hspace{2} L{\omega} \hspace{2}}\sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}{\pi}\hspace{2}} }\sin \frac{\hspace{2} L{\omega} \hspace{2}}{\hspace{2}2\hspace{2}} }$

$3$\displaystyle{ { \lim }\limits_{ L\rightarrow 0 }\hspace{1}{F}_{c}\left({{\omega}}\right)={ \lim }\limits_{ L\rightarrow 0 }\frac{\hspace{2}k\hspace{2}}{\hspace{2} L{\omega} \hspace{2}}\sqrt{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}{\pi}\hspace{2}} }\sin \frac{\hspace{2} L{\omega} \hspace{2}}{\hspace{2}2\hspace{2}} }$

$3$\displaystyle{ ={ \lim }\limits_{ L\rightarrow 0 }\frac{\hspace{2}k\hspace{2}}{\hspace{2} \sqrt{ 2{\pi} } \hspace{2}}\cdot \frac{\hspace{2} \sin \frac{\hspace{2} L{\omega} \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} \frac{\hspace{2} L{\omega} \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}} }$

$3$\displaystyle{ =\frac{\hspace{2}k\hspace{2}}{\hspace{2} \sqrt{ 2{\pi} } \hspace{2}} }$

（∵ここを参照）

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最終更新日： 2023年7月7日