How Does Duality in Fourier Transform Help Solve F(exp(iax)f(x))(k)=X(k-a)?

[c299792458]

Homework Statement

How do I use the $(FF(f))(x)=2\pi f(-x)$ where $F$ is the Fourier transform.
and $F(f(x-a))(k)=\exp(-ika) X(k)$ where $X(k)=F(f(x))$

to get $F(\exp(iax)f(x))(k)=X(k-a)$

Homework Equations

Please see above.

The Attempt at a Solution

The variables confuse me. I can do it by brute force plugging and chugging but I really don't see how to do it using the duality property...

 

[KataKoniK]

Hi there! The duality property of the Fourier transform states that if F is the Fourier transform operator, then F(F(x))=x. This means that if we apply F to a function, and then apply F again, we should get back the original function.

In this case, we have F(f(x-a))(k)=exp(-ika) X(k), where X(k)=F(f(x)). So, if we apply F to both sides, we get F(F(f(x-a))(k))=F(exp(-ika) X(k)).

Using the duality property, we know that F(F(f(x-a))(k))=f(x-a), and F(exp(-ika) X(k))=exp(-ika) X(k-a).

Therefore, we have f(x-a)=exp(-ika) X(k-a). To get the desired result of F(exp(iax)f(x))(k)=X(k-a), we can rearrange the equation to get F(exp(iax)f(x))(k)=exp(ika) X(k-a).

I hope this helps! Let me know if you have any other questions.