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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...)

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I am unsure exactly what is being asked. What is your definition of the Fourier Transform? Presumably it is $$F(f)(\xi) = \sqrt{2 \pi} \int_{\mathbb{R}} \exp(- 2\pi i x \xi) f(x) dx.$$ Is this correct? Furthermore, are you asking how to put equations (1) and (2) together to get equation (3)? –  JavaMan Nov 15 '11 at 14:00
    
@JavaMan: Thanks, $F(f)(\xi) = \int_{\mathbb{R}} \exp(- 2\pi i x \xi) f(x) dx.$ and Yes, (1)+(2)$\to$(3). –  paul Nov 15 '11 at 14:13
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Try not to learn the formulas, instead you should try to learn the definition and to understand why (1) and (2) holds. Using the definition it is really not that hard to see why (3) holds –  AD. Nov 15 '11 at 15:06
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By the way, you might consider to use the notation $f_a(x)=f(x-a)$; $F(f(x-a))(k)$ is a bit confusing (the transform is with respect on $x$). –  AD. Nov 15 '11 at 15:08
    
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. –  Willie Wong Nov 17 '11 at 8:48
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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).

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