Playing With Shifts

July 9, 2007

This post is related to a very specific problem that I was solving. (Which one isn’t…?) It will hopefully find its use in low resolution SAR and sonar systems. I hope to publish it soon and this is the first try. I expect some trouble with mathematics. So, if you are good at Math, and you are reading this, let me know what you think.

Intro

Suppose we have a following function:

$s(x,y)=\sum_{i=1}^{I}\sigma_i\exp\left[-\jmath\, a_i(y/2-x)+\jmath\, b_i x\right],$

where $i\in \mathbb{N}, \sigma\in \mathbb{C}$ and $a,b\in \mathbb{R}$ . The objective is to divide this function into two functions $s_a$ and $s_b$ , i.e.

$s(x,y)=s_a(x,y)+s_b(x,y),$

such that

$s_a(x,y)=\sum_{k=1}^{K}\sigma_k\exp\left[-\jmath\, a_k(y/2-x)\right]$

$s_b(x,y)=\sum_{l=1}^{L}\sigma_l\exp\left[-\jmath\, a_l(y/2-x)+\jmath\, b_l x\right]$

$I=K+L \qquad k,l\in \mathbb{N}.$

Analysis

We do not know values of $a$ and $b$ . What we do know, however, is that in 2D spectral domain $(k_x,k_y)$ , function $s_a$ will produce peaks located on a straight line $k_y=-k_x/2$ .

To show this, suppose that $\sigma$ is a continuous amplitude function instead of a discrete one. Then,

$s_a(x,y)=\int\limits_{-\infty}^\infty\sigma (a)\exp\left[-\jmath\, a(y/2-x)\right] da.$

Fourier transform of $s_a(x,y)$ with respect to $x$ will be:

$\hat{s}_a(k_x,y)=\int\limits_{-\infty}^\infty s_a(x,y)\exp (-\jmath\, k_x x)\, dx,$

$\hat{s}_a(k_x,y)=\int\limits_{-\infty}^\infty \sigma (a)\exp(-\jmath\, ay/2)\int\limits_{-\infty}^\infty\exp(\jmath\, ax)\exp (-\jmath\,k_x x)\, dx\, da,$

$\hat{s}_a(k_x,y)=2\pi\int\limits_{-\infty}^\infty \sigma (a)\exp (-\jmath\, ay/2)\delta (a-k_x)\, da,$

$\hat{s}_a(k_x,y)=2\pi\sigma (k_x)\exp (-\jmath\, k_xy/2).$

Fourier transform of $\hat{s}_a(x,y)$ with respect to $y$ will be:

$S_a(k_x,k_y)=2\pi\int\limits_{-\infty}^\infty\hat{s}_a(k_x,y)\exp(-\jmath\, k_yy)\, dy,$

$S_a(k_x,k_y)=2\pi\int\limits_{-\infty}^\infty\sigma (k_x)\exp(-\jmath\, k_xy/2)\exp(-\jmath\, k_yy)\, dy$

$S_a(k_x,k_y)=4\pi^2\sigma(k_x)\delta(k_y+k_x/2).$

Since Dirac delta $\delta (x)$ has a value other than zero only at $x=0$ , $S_a(k_x,k_y)$ will exhibit peaks at a line $k_y=-k_x/2$ .

Solution

If we could shift the points on this line in such a way that $k_y=0$ , techniques described previously could be used for the separation. The solution is to multiply $\hat{s}_a (k_x,y)$ by $\exp(\jmath\, k_xy/2)$ .