What will be the two-dimension Discrete Fourier Transform of sine?

Question

Suppose I have a two-dimensional sine signal $\sin⁡(2 \pi u_0 m/M + 2 \pi v_0 n/N)$, let's say $M=N=1001$ and $u_0=100$, $v_0=200$? What will the 2-dimensional Discrete Fourier Transform of this signal look like? Will it have 2 peaks or 4?

I know 1-Dimensional Discrete Fourier Transform of a sine signal will have 2 peaks, at $f_0$ and $-f_0$. Now, for the above function, if I take DFT,

1. Will it have 2 peaks, one at $(100,200)$ and the other at $(M-100, N-200)$ ${or}$
   
2. Will it have 4 peaks, 2 higher magnitude peaks at $(100,200)$, $(M-100,N-200)$ and 2 lower magnitude peaks at $(100,N-200)$, $(M-100,200)$?

Note: This was an assignment question to be implemented in MatLab/Python and the due date is over. Hence you can freely post the complete answer.

Why this doubt?: I got 2 peaks as described in part 1 and few of my classmates said they got 4 peaks as described in part 2. Hence I got confused. If required, I can post my code as well.

Thanks!

Answer 1

Well, let's take a look at the Fourier Transform of something similar:

$$\begin{align*}\mathscr{F}\left\{\sin\left(\pi ax+\pi by\right)\right\} &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \sin\left(\pi ax+\pi by\right)\space e^{-i2\pi\left(ux+vy\right)}\space dx \space dy\\ \\ &= \int_{-\infty}^{\infty} \left[\int_{-\infty}^{\infty} \sin\left(\pi ax+\pi by\right)\space e^{-i2\pi ux}\space dx \right]e^{-i2\pi vy}\space dy\\ \\ &= \int_{-\infty}^{\infty} e^{-i2\pi^2 by u/a}\dfrac{1}{|a|}\dfrac{1}{2i}\left[\delta\left(\dfrac{u}{a}-\dfrac{1}{2}\right)-\delta\left(\dfrac{u}{a}+\dfrac{1}{2}\right)\right] \space e^{-i2\pi vy}\space dy\\ \\ &= \int_{-\infty}^{\infty} e^{-i2\pi^2 by u/a}\dfrac{1}{2i}\left[\delta\left(u-\dfrac{a}{2}\right)-\delta\left(u+\dfrac{a}{2}\right)\right] \space e^{-i2\pi vy}\space dy\\ \\ &= \dfrac{1}{2i}\left[\delta\left(u-\dfrac{a}{2}\right)-\delta\left(u+\dfrac{a}{2}\right)\right]\int_{-\infty}^{\infty} e^{-i2\pi \left(\pi bu/a +v\right)y} \space dy\\ \\ &= \dfrac{1}{2i}\left[\delta\left(u-\dfrac{a}{2}\right)-\delta\left(u+\dfrac{a}{2}\right)\right] \delta\left(\dfrac{\pi b}{a}u +v\right)\\ \\ \end{align*}$$

So that is two, parallel, infinitely long impulse lines at $u = \pm \dfrac{a}{2}$ multiplied by another infinitely long impulse line that intersects those two.

It looks like you should only expect 2 peaks.