Verifying that translation by $h$ in time is the same as modulating by $-h$ in frequency (Fourier Analysis)

by user1770201   Last Updated January 14, 2018 13:20 PM

According to a time-frequency equivalency table in an undergraduate book on Harmonic Analysis, we have that:

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or, in equation form:

$$ \widehat{\tau_h f}(\xi) = M_{-h} \widehat{f}(\xi). $$

Problem. This doesn't seem true to me.

Attempt. Recall that in the context of Fourier Analysis, the Fourier Transform $\widehat{f}(\xi)$ satisfies

$$ \widehat{f}(\xi) = \int_{- \infty}^{\infty} f(x) e^{-2 \pi i \xi x} dx $$


$$ M_{-h} \widehat{f}(\xi) = e^{2 \pi i (-h) \xi} \left( \int_{- \infty}^{\infty} f(x) e^{-2 \pi i x \xi} dx \right) = \int_{- \infty}^\infty f(x) e ^{-2 \pi i \xi (x + h)} dx $$


$$ \widehat{\tau_h f}(\xi) = \widehat{f(\xi - h)} = \int_{- \infty}^{\infty} f(x) e^{- 2 \pi i x( \xi - h)} dx $$

so that

$$ \widehat{\tau_h f}(\xi) = \int_{- \infty}^{\infty} f(x) e^{- 2 \pi i x( \xi - h)} dx \ne \int_{- \infty}^\infty f(x) e ^{-2 \pi i \xi (x + h)} dx = M_{-h} \widehat{f}(\xi). $$

What am I missing?

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Updated October 09, 2017 05:20 AM