2-D Dirac Delta in Polar Coordinate

422 Views Asked by At

In Section 7.3.2 of Fourier Methods in Imaging (by Dr. Roger Easton Jr), it is stated that for 2-D Dirac delta located on the x-axis at a distance $\alpha > 0$,

$x = r$ and $y = r_0 \theta$

Thus the polar representation would be: $\delta[x-\alpha]\delta[y] = \delta[r-r_0]\delta[r_0 \theta]$

I do not understand why $y = r_0 \theta$ instead of $y=r_0 tan\theta$?

It also states that the radial coordinate is directed along the x-axis and the azimuthal displacement due to the angle (what angle?) is parallel to the y-axis. What angle are we talking about? Why is there a displacement parallel to the y-axis if the delta is located on x-axis?

1

There are 1 best solutions below

3
On BEST ANSWER

Since it's on the $x$ axis any $\theta$ must be small. For small $\theta$, $\tan\theta\approx\theta$