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Regarding Fourier transform, I read that the translation property and frequency-shift property are a duality. What does that mean and why is it true? Is there a physical implications? Thanks.

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The duality is captured well in some engineering texts when you look at the Fourier transform pair:

$f(t-T) \Leftrightarrow e^{-jwT}F(w)$ (time-shifting)

$e^{jwT}f(t) \Leftrightarrow F(w-W)$ (frequency-shifting)

The physical interpretation of $f(t-T)$ is a delay in time. The second shift theorem is, as one author puts it, "not so straightforward" to interpret physically.

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The second property can be related to AM (amplitude modulation), where you multiply the signal by a sinusoid and the spectrum is shifted in frequency. Granted, the sinusoid is real, but the concept is the same. – leonbloy Nov 14 '11 at 23:16

Let $f(x)$ be an $L^1$ function. If we translate $f$ by $y$, we get a new function:

$$g(x):= f(x-y) \,.$$

Now, what is the Fourier transform of this new function? Lets calcualte it:

$$\hat{g}(\chi) = \int_{R^d} g(x) e^{2 \pi i x \cdot \chi} dx = \int_{R^d} f(x-y) e^{2 \pi i x \cdot \chi} dx $$ $$= \int_{R^d} f(x-y) e^{2 \pi i (x-y) \cdot \chi}e^{2 \pi i y \cdot \chi} dx = e^{2 \pi i y \cdot \chi}\hat{f}(\chi) \,.$$

the same result holds in locally compact abelian groups.

A simple physical implication is the fact that the diffraction of a solid is invariant on the possition...

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