I have a problem in DFT. It was one of my past-year exam papers questions.


Let $F(u,v)$ be the 2-D Fourier transform of a 2-D continuous function $f(x,y)$. Derive in terms of $F(:,:)$ the 2-D Fourier transform of each of the following functions

1) $f(x,-2y)$

2) $f(x+2y,y)$

I know how to do 1-D Fourier transforms but not 2-D. I'm not sure how to start on it and need some guidance.

For the second part, this was my approach. Please let me know if it's right or correct me if it's wrong.

Let $τ= x + 2y$ hence $x = τ-2y$ and $dx = dτ$ $$ \begin{align} \mathfrak{F}\{f(x+2y,y)\}&=∬ f(τ,y)e^{−j2π(u(τ-2y) +vy )} dx\ dv\\ \mathfrak{F}\{f(x+2y,y)\}&= ∬ f(τ,y)e^{−j2π(uτ+(-2u+v)y )} dx\ dτ \\ \mathfrak{F}\{f(x+2y,y)\}&= F(u,-2u+v) \end{align} $$


1 Answer 1


Here is the first one:

By definition,

$\mathfrak{F}\{ f(x,-2y)\} = \iint_{- \infty}^{\infty} f(x,-2y)e^{-j2\pi (ux+vy)}dxdy$

let $\tau = -2y$ and conversely $y=\frac{-\tau}{2}$

$\mathfrak{F}\{ f(x,-2y)\} = \iint_{- \infty}^{\infty} f(x,\tau)e^{-j2\pi (ux-\frac{v\tau}{2})}dxd(-\frac{\tau}{2})$

$\mathfrak{F}\{ f(x,-2y)\} = -\frac{1}{2}\iint_{- \infty}^{\infty} f(x,\tau)e^{-j2\pi (ux-\frac{v\tau}{2})}dxd\tau$

$\mathfrak{F}\{ f(x,-2y)\} = -\frac{1}{2} F(u,-\frac{v}{2})$

The other one will be harder, but I will leave it to you.

  • $\begingroup$ it would be great if you did the other too... $\endgroup$ Commented Nov 11, 2011 at 16:40
  • 3
    $\begingroup$ @cnnlakshmen It would be really great if you tried doing the other yourself :) Jason has shown you the general way to approach the problem, so I see no reason to not give the other an honest try. Please don't treat this as a place for quick answers, but rather a place to learn. $\endgroup$ Commented Nov 11, 2011 at 16:45
  • $\begingroup$ i tried and got stuck.... thats why... asked him again... $\endgroup$ Commented Nov 11, 2011 at 16:50
  • $\begingroup$ maybe you'd like to show us your steps for the second question? You can just edit that into the question here (no need for a new one) and we can see where you're stuck. Knowing what you're doing wrong or where you're stuck will help us to point you the right way, rather than simply giving the answer, in which case you will never know what you didn't know and everything will remain hocus pocus. $\endgroup$ Commented Nov 11, 2011 at 16:56
  • $\begingroup$ I've updated the question with my answer, and I have a feeling it is wrong... please do help me check... $\endgroup$ Commented Nov 11, 2011 at 16:59

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