91影视

Given functions are $f\left(x\right)=蠄\left(x\right)$and $g\left(伪\right)=\sqrt{\frac{h}{2蟺}}蠒\left(p\right)$,and substitution $伪=\frac{2蟺p}{h}$.

Here p is the new variable, h is a constant.

The Fourier transform of a function is a complex-valued function that symbolises the original function's complex sinusoids. A Fourier transform is a mathematical operation that breaks down geographically or temporally dependent functions into functions with those same dependencies.

Solve $f\left(x\right)$and $g\left(伪\right)$by Fourier transformation from the problem 34 and substitute $伪=\frac{2蟺p}{h}$in $f\left(x\right)$and $g\left(伪\right)$.

Solve the equation$f\left(x\right)$,

$\begin{array}{c}f\left(x\right)=蠄\left(x\right)\\ f\left(x\right)=\frac{1}{\sqrt{2蟺}}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}g\left(伪\right)e^{i伪x}d伪\\ 蠄\left(x\right)=\frac{1}{\sqrt{2蟺}}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}\left(\sqrt{\frac{h}{2蟺}}蠒\left(p\right)\right)e^{i\left(\frac{2蟺p}{h}\right)x}\left(2蟺\frac{1}{h}dp\right)\\ 蠄\left(x\right)=\left(\frac{1}{\sqrt{2蟺}}\right)\left(\sqrt{\frac{h}{2蟺}}\right)\left(2蟺\frac{1}{h}\right){鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}蠒\left(p\right)e^{\frac{i2蟺px}{h}}dp\end{array}$

Solve further,

$蠄\left(x\right)=\frac{1}{\sqrt{h}}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}蠒\left(p\right)e^{\frac{i2蟺px}{h}}dp$ 鈥︹�� (1)

Solve the equation$g\left(伪\right)$,

$\begin{array}{c}g\left(伪\right)=\sqrt{\frac{h}{2蟺}}蠒\left(p\right)\\ g\left(伪\right)=\frac{1}{\sqrt{2蟺}}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}f\left(x\right)e^{鈭�i伪x}dx\\ \sqrt{\frac{h}{2蟺}}蠒\left(p\right)=\frac{1}{\sqrt{2蟺}}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}蠄\left(x\right)e^{鈭�i\left(\frac{2蟺p}{h}\right)x}dx\\ 蠒\left(p\right)=\sqrt{\frac{2蟺}{h}}\left(\frac{1}{\sqrt{2蟺}}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}蠄\left(x\right)e^{鈭�i\left(\frac{2蟺p}{h}\right)x}dx\right)\end{array}$

Solve further,

$蠒\left(p\right)=\frac{1}{\sqrt{h}}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}蠄\left(x\right)e^{鈭�i\left(\frac{2蟺p}{h}\right)x}dx$ 鈥︹�� (2)

Use the normalization factor $\frac{1}{\sqrt{2蟺}}$in front of equations (1) and (2) and also the substitution$伪=\frac{2蟺p}{h}$.

$\begin{array}{c}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\left|\frac{1}{\sqrt{2蟺}}蠄\left(x\right)\right|}^{2}dx=\frac{1}{2蟺}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\left|蠄\left(x\right)\right|}^{2}dx\\ {鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\left|\frac{1}{\sqrt{2蟺}}蠒\left(p\right)\right|}^{2}dp=\frac{1}{2蟺}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\left|蠒\left(p\right)\right|}^{2}dp\\ \frac{1}{2蟺}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\left|蠄\left(x\right)\right|}^{2}dx=\frac{1}{2蟺}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\left|蠒\left(p\right)\right|}^{2}dp\\ {鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\left|蠄\left(x\right)\right|}^{2}dx={鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\left|蠒\left(p\right)\right|}^{2}dp\end{array}$

Therefore, it is proved that if $f\left(x\right)=蠄\left(x\right)$and $g\left(伪\right)=\sqrt{\frac{h}{2蟺}}蠒\left(p\right)$then

$蠄\left(x\right)=\frac{1}{\sqrt{h}}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}蠒\left(p\right)e^{\frac{2蟺ipx}{h}}dp$,

$蠒\left(p\right)=\frac{1}{\sqrt{h}}{鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}蠄\left(x\right)e^{鈭�\frac{2蟺ipx}{h}}dx$

, and

${鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\left|蠄\left(x\right)\right|}^{2}dx={鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{\left|蠒\left(p\right)\right|}^{2}dp$.