10.2 Fourierova analýza}\vspace{0.1cm
Příklady
Určete konvoluci funkcí $(f\ast g)(x)$:
$f(x)=\left\{ \begin{array}{r c l} 0 & &x\in(-\infty,0) \\ 2 & &x\in\langle 0,\infty) \end{array} \right.$
$g(x)=\left\{ \begin{array}{r c l} 0 & &x\in(-\infty,0)\\ 1 & &x\in\langle 0,\infty) \end{array} \right.$
$f(x)=\left\{ \begin{array}{r c l} 0 & & x\in(-\infty,0) \\ \e^{-2x} & & x\in\langle 0,\infty)\end{array}\right.$
$g(x)=\left\{ \begin{array}{r c l} 0 & & x\in(-\infty,0) \\ \e^{-x} & & x\in\langle 0,\infty) \end{array} \right.$
$f(x)=\left\{\begin{array}{r c l} 0 & & x\in(-\infty,0) \\ \e^{-x} & & x\in\langle 0,\infty) \end{array} \right.$
$g(x)=\left\{ \begin{array}{r c l} 0 & & x\in(-\infty,0) \\ \sin x & & x\in\langle 0,\infty) \end{array} \right.$
$f(x)=\left\{\begin{array}{r c l} 0 & & x\in(-\infty,0) \\ \sin x & & x\in\langle 0,\infty) \end{array} \right.$
$g(x)=1$
$f(x)=\left\{ \begin{array}{r c l} 0 & & x\in(-\infty,0) \\ \sin x & & x\in\langle 0,\infty) \end{array} \right.$
$g(x)=\left\{\begin{array}{r c l} 0 & & x\in(-\infty,0) \\ 1 & & x\in\left\langle 0,\dfrac{\pi}{2}\right\rangle \\ 0 & & x\in\left(\dfrac{\pi}{2},\infty\right) \end{array}\right.$
$f(x)=\left\{ \begin{array}{r c l} 0 & & x\in(-\infty,0) \\ \sin x & & x\in\langle 0,\infty) \end{array} \right.$
$g(x)=\sin(a x),\,a>0$
$f(x)=\left\{ \begin{array}{r c l} 0 & & x\in(-\infty,0) \\ \sin x & & x\in\langle 0,\infty) \end{array} \right.$
$g(x)=\left\{ \begin{array}{r c l} 0 & & x\in\left(-\infty,0\right) \\ 1-\dfrac{2}{\pi}x & & x\in\left\langle 0,\dfrac{\pi}{2}\right\rangle \\ 0 & & x\in\left(\dfrac{\pi}{2},\infty\right) \end{array}\right.$
Určete Fourierův obraz funkce:
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$f(x)=1$ pro $x\in\left\langle-\dfrac{1}{2},\dfrac{1}{2}\right\rangle$, $f(x)=0$ pro $x\notin\left\langle -\dfrac{1}{2},\dfrac{1}{2}\right\rangle$,$\widehat{f}(\omega)=\dfrac{2}{\omega}\sin\left(\dfrac{\omega}{2}\right)$
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$f(x)=\e^{-b x}$ pro $x\in\left\langle 0,\infty\right)$, $f(x)=0$ pro $x\in\left(-\infty,0\right)$, $b>0=\text{konst.}$,$\widehat{f}(\omega)=\dfrac{1}{b+\ii\omega}$
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$f(x)=\e^{-b|x|}$ pro $x\in\left(-\infty,\infty\right)$,$\widehat{f}(\omega)=\dfrac{2b}{b^2+\omega^2}$
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$f(x)=\e^{-b x^2}$ pro $x\in\left(-\infty,\infty\right)$,$\widehat{f}(\omega)=\sqrt{\dfrac{\pi}{b}}\ee^{-\expover{\omega^2}{4b}}$
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$f(x)=1+x$ pro $x\in\left\langle-1,0\right)$, $f(x)=1-x$ pro $x\in\left\langle 0,1\right)$, $f(x)=0$ pro $|x|>1$,$\widehat{f}(\omega)=\dfrac{2}{\omega^2}\left(1-\cos\omega\right)$
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$f(x)=x^2$ pro $x\in\left\langle-1,1\right\rangle$, $f(x)=0$ pro $|x|>1$.$\widehat{f}(\omega)=\dfrac{2}{\omega^3}\left[(\omega^2-2)\sin\omega+2\omega\cos\omega\right]$
Určete Fourierův obraz funkce:
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$f(x)=\sin x$ pro $x\in\left\langle 0,2\pi\right\rangle$, $f(x)=0$ pro $x\notin\left\langle 0,2\pi\right\rangle$,$\widehat{f}(\omega)=\dfrac{\ii\omega\left(1-\e^{-2\pi\ii\omega}\right)}{1-\omega^2}$
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$f(x)=\cos x$ pro $x\in\left\langle 0,2\pi\right\rangle$, $f(x)=0$ pro $x\notin\left\langle 0,2\pi\right\rangle$,$\widehat{f}(\omega)=-\dfrac{1-\e^{-2\pi\ii\omega}}{\omega^2-1}$
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$f(x)=\sin x\cos x$ pro $x\in\left\langle 0,2\pi\right\rangle$, $f(x)=0$ pro $x\notin\left\langle 0,2\pi\right\rangle$,$\widehat{f}(\omega)=-\dfrac{1-\e^{-2\pi\ii\omega}}{\omega^2-4}$
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$f(x)=\sin^2x\cos^2x$ pro $x\in\left\langle 0,2\pi\right\rangle$, $f(x)=0$ pro $x\notin\left\langle 0,2\pi\right\rangle$,$\widehat{f}(\omega)=-\dfrac{2\ii\left(1-\e^{-2\pi\ii\omega}\right)}{\omega\left(\omega^2-16\right)}$
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$f(x)=A^{B x}$ pro $x\in\left\langle 0,1\right\rangle$, $f(x)=0$ pro $x\notin\left\langle 0,1\right\rangle$.$\widehat{f}(\omega)=\dfrac{1-\e^{-\ii\omega}A^B}{\ii\omega-B\ln A}$
Na příkladech 10.23, 10.24, 10.25 a 10.26 ověřte platnost rovnice (10.10):
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$\widehat{f}(\omega)=-\dfrac{2\ii}{\omega}$, $\widehat{g}(\omega)=-\dfrac{\ii}{\omega}$, $\widehat{(f\ast g)}(\omega)=-\dfrac{2}{\omega^2}$
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$\widehat{f}(\omega)=\dfrac{1}{2+\ii\omega}$, $\widehat{g}(\omega)=\dfrac{1}{1+\ii\omega}$, $\widehat{(f\ast g)}(\omega)=\dfrac{1}{2-\omega^2+3\ii\omega}$
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$\widehat{f}(\omega)=\dfrac{1}{1+\ii\omega}$, $\widehat{g}(\omega)=\dfrac{1}{1-\omega^2}$, $\widehat{(f\ast g)}(\omega)=\dfrac{1}{(1+\ii\omega)(1-\omega^2)}$
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$\widehat{f}(\omega)=\dfrac{1}{1-\omega^2}$, $\widehat{g}(\omega)=\dfrac{1}{\ii\omega}$, $\widehat{(f\ast g)}(\omega)=\dfrac{1}{\ii\omega(1-\omega^2)}$