# Fourier transform of a translated variable

• Aug 2nd 2010, 01:41 AM
amrasa81
Fourier transform of a translated variable
Hello,

I would like to derive the Fourier transform of a translated variable. I read in some references (including wikipedia) that,

$\displaystyle \widehat{T_{\delta_{x}}v(k)} = e^{-\mathrm{i}k\delta_{x}}\widehat{v(k)}$,
where $\displaystyle T_{\delta_{x}}$ is the translation operator translating the scalar or vector field $\displaystyle v(x)$ by $\displaystyle \delta x$, $\displaystyle k$ is the wave number, $\displaystyle \mathrm{i}$ is the imaginary number and $\displaystyle ^{\widehat{}}$ denotes the fourier transform.

I have tried deriving it myself but have some doubts. Following is the derivation:
By definition, $\displaystyle \widehat{v(k)} = \int_{-\infty}^{\infty} v(x) e^{\mathrm{i}kx} dx$
$\displaystyle = \int_{-\infty}^{\infty} v(x) e^{\mathrm{i}kx} e^{\mathrm{i}k\delta x} e^{-\mathrm{i}k\delta x} dx$
$\displaystyle = e^{-\mathrm{i}k\delta x} \int_{-\infty}^{\infty} v(x) e^{\mathrm{i}k(x+\delta x)} dx$
Translating the variable $\displaystyle x$ to $\displaystyle x+\delta x$, we get,
$\displaystyle = e^{-\mathrm{i}k\delta x} \int_{-\infty}^{\infty} v(x+\delta x) e^{\mathrm{i}k(x+\delta x)} d(x+\delta x)$
Thus,
$\displaystyle \widehat{T_{\delta_{x}}v(k)} = e^{-\mathrm{i}k\delta_{x}}\widehat{v(k)}$

My questions are:
1. Is my derivation correct? If not, then how to derive it?
2. If it is correct, then while translating $\displaystyle x$ by $\displaystyle x+\delta x$, shouldn't the exponential term also be translated, which means that the exponential term will be $\displaystyle e^{\mathrm{i}k(x+2\delta x)}$ ?

Please tell me what am I doing wrong or where I am getting confused.

Thanks,