Fourier Transform¶

The Fourier transform is:

$F[f(x)] \equiv \tilde f(\omega) = \int_{-\infty}^{\infty} f(x) e^{-i\omega x}\,\d x F^{-1}[\tilde f(\omega)] = f(x) = {1\over2\pi}\int_{-\infty}^{\infty} \tilde f(\omega) e^{+i\omega x}\,\d \omega$

To show that it works:

$F^{-1} F [f(x)] = {1\over2\pi}\int_{-\infty}^{\infty} \left[\int_{-\infty}^{\infty} f(x) e^{-i\omega x}\,\d x\right] e^{+i\omega x}\,\d \omega = {1\over2\pi}\int_{-\infty}^{\infty} \left[\int_{-\infty}^{\infty} f(x') e^{-i\omega x'}\,\d x'\right] e^{+i\omega x}\,\d \omega = = \int_{-\infty}^{\infty} f(x') \left[{1\over2\pi}\int_{-\infty}^{\infty} e^{i\omega (x- x')}\,\d \omega \right] \,\d x' = \int_{-\infty}^{\infty} f(x') \delta(x-x') \,\d x' =f(x)$

Laplace Transform¶

Laplace transform of $f(x)$ is:

$L[f(x)] = \int_0^{\infty} f(x) e^{-s x}\,\d x L^{-1}[\bar f(s)] = {1\over2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty} \bar f(s) e^{s x}\,\d s = \sum_{s_0} \res_{s=s_0} (\bar f(s) e^{s x})$

The contour integration is over the vertical line $\sigma+i\omega$ and $\sigma$ is chosen large enough so that all residues are to the left of the line (that’s because the Laplace transform $\bar f(s)$ is only defined for $s$ larger than the residues, so we have to integrate in this range as well). It can be shown that the integral over the left semicircle goes to zero:

$\left|\int_\Omega e^{sx}g(s) \d s \right| =\left|\int_{\pi\over2}^{3\pi\over2} e^{(\sigma + Re^{i\varphi})x} g(\sigma+Re^{i\varphi})iRe^{i\varphi}\d\varphi\right| \le \le R \max_\Omega |g(z)| e^{\sigma x} \int_{\pi\over2}^{3\pi\over2}\left| e^{xRe^{i\varphi}} \right|\d\varphi = = R \max_\Omega |g(z)| e^{\sigma x} \int_{\pi\over2}^{3\pi\over2}e^{xR \cos \varphi} \d\varphi = = R \max_\Omega |g(z)| e^{\sigma x} \int_0^{\pi}e^{-xR \sin \varphi} \d\varphi = < {\pi e^{\sigma x}\over x} \max_\Omega |g(z)|$

so the complex integral is equal to the sum of all residues of $\bar f(s)e^{sx}$ in the complex plane.

To show that it works:

$L^{-1} L [f(x)] = {1\over2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty} \left[\int_0^{\infty} f(x) e^{-s x}\,\d x\right] e^{s x}\,\d s = {1\over2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty} \left[\int_0^{\infty} f(x') e^{-s x'}\,\d x'\right] e^{s x}\,\d s = = \int_0^{\infty} f(x') \left[{1\over2\pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} e^{s (x- x')}\,\d s \right] \,\d x' = \int_0^{\infty} f(x') \delta(x-x') \,\d x' =f(x)$

where we used:

${1\over2\pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} e^{s (x- x')}\,\d s = {1\over2\pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} e^{s (x- x')}\,\d s = {1\over2\pi i} \int_{-\infty}^{\infty} e^{(\sigma+i\omega) (x- x')}\,i\d \omega = = {e^{\sigma (x- x')}\over2\pi} \int_{-\infty}^{\infty} e^{i\omega (x- x')}\,\d \omega = e^{\sigma (x- x')}\delta(x - x') =\delta(x - x')$

and it can be derived from the Fourier transform by transforming a function $U(x)$ :

$U(x) = \begin{cases} f(x)e^{-\sigma x} &\text{for $x\ge0$}\cr 0 &\text{for $x<0$}\cr \end{cases}$

and making a substitution $s = \sigma + i\omega$ :

$L[f(x)] \equiv \bar f(s) = F[U(x)] \equiv \tilde U(\omega) = \int_{-\infty}^{\infty} U(x) e^{-i\omega x}\,\d x = \int_0^{\infty} f(x) e^{-\sigma x} e^{-i\omega x}\,\d x = \int_0^{\infty} f(x) e^{-s x}\,\d x L^{-1}[\bar f(s)] \equiv f(x) = U(x) e^{\sigma x} = F^{-1}[\tilde U(\omega)]e^{\sigma x} = F^{-1}[\bar f(s)]e^{\sigma x} = F^{-1}[\bar f(\sigma+i\omega)e^{\sigma x}] = {1\over2\pi}\int_{-\infty}^{\infty} \bar f(\sigma + i\omega)e^{\sigma x} e^{i\omega x}\,\d \omega = {1\over2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty} \bar f(s) e^{s x}\,\d s = \sum_{s_0} \res_{s=s_0} (\bar f(s) e^{s x})$

Where the bar ( $\bar f$ ) means the Laplace transform and tilde ( $\tilde U$ ) means the Fourier transform.

Fourier Transform¶

Laplace Transform¶

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Fourier Transform¶

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