1
$\begingroup$

In my engineering class, I was told that

$(FF(f))(x)=2\pi f(-x) $, where $F$ is the Fourier transform ----(1)

and $F(f(x-a))(k)=\exp(-ika) X(k)$ where $X(k)=F(f(x))$ ----(2)

implies $F(\exp(iax)f(x))(k)=X(k-a)$. ----(3)

Reworded: Perhaps my question would be clearer if I said why does the the duality property of the FT ----(1) allow us to obtain the modulation property ----(2) from the translation property ----(3)?

But I don't see how that is done... I am quite happy with getting $F^{-1}X(k-a)=\exp(iax)f(x)$ by brute force calculation. I would like to see how to use duality though. (It was said that the duality should simplify stuff...)

  • 0
    The "duality property" **is** something that follows from the computations. At least at this level of discussion. They take on more abstract formulations if you study harmonic analysis on groups, but I seriously doubt whether that will give you any more insight.2011-11-17

1 Answers 1

1

Write (2) with $-a$ instead of $a$, $F(f(x+a))(k)=\exp(ika)X(k) \tag{2'}$ Apply $F$ to both sides of (2'): $F(F(f(x+a)))=F(\exp(ika)X(k)) $ then use (1) on the left: $2\pi f(-x+a)=F(\exp(ika)X(k)) $ Since the Fourier transform of $X$ is $2\pi f(-x)$, the result can be read as $ F(\exp(ika)X(k)) = F(X)(x-a)$ which is your (3).