# How to represent impulse function in 2D?

To be more specific I want to show that impulse function in 2D can be represented as $$β(r)=δ(r)/πr$$.

Also I want to show that each projection of a two dimensional impulse function at the origin is a Delta function.

I could not find any useful information about these two problems, if anyone could help I will be appreciated.

• Hi! In 2D (or higher dimensions) the are point impulses like $\delta(x,y)$ or line impuleses like $\delta_2(x)$; the latter representing set of point impulses placed continuously on the $y$ axis. So in 2D polar coordinate system which kind of impulse representation are you up to? – Fat32 Oct 28 '18 at 17:14
• @Fat32 There are also ring impulses: ${}^2\delta(r - r_0)$ :) – Andy Walls Oct 28 '18 at 17:17
• @AndyWalls The circle impulse you mean $\delta_2(x^2+y^2-r_0^2)$ :-)). And what about $\delta_2(r +\cos(\theta) -1 )$ :-) – Fat32 Oct 28 '18 at 17:48

So dealing with generalized functions like the Dirac delta requires some care, and when dealing with N-dimensional versions you need to be very explicit with your notation to keep things straight.

I'll denote the 2 dimensional delta function in polar coordinates at the origin as $${}^2\delta(r, \theta) = {}^2\delta(r)$$, since for the special case of the origin, $$\theta$$ doesn't matter.

For this derivation, $${}^2\delta(r)$$ represents the following limiting sequence of functions (that are asymmetrical about 0):

$${}^2\delta(r) = \lim_{\epsilon \rightarrow 0^+} \dfrac{1}{\pi\epsilon^2} \quad 0 < r < \epsilon$$

Using this limiting sequence of functions, the 2-D delta function at the origin in polar coordinates has the following property of "integrating to 1" for the 2-D integration:

$$\int_0^{2\pi} \int_0^{\infty} {}^2\delta(r) \space r \space \mathrm{d}r \space \mathrm{d}\theta = 1$$

To separate the $$r$$ and $$\theta$$ portions of $${}^2\delta(r)$$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:

$${}^1\delta_a(r) = \lim_{\epsilon \rightarrow 0^+} \dfrac{1}{\epsilon} \quad 0 < r < \epsilon$$

Using this limiting sequence of functions, the 1-D delta function at the origin has the following property of "integrating to 1" for the 1-D integration: $$\int_0^{\infty} {}^1\delta_a(r) \space \mathrm{d}r = 1$$

We can then equate the two above integrals to get the relationship between $${}^2\delta(r)$$ and $${}^1\delta_a(r)$$

\begin{align}\\ \int_0^{\infty} {}^1\delta_a(r) \space \mathrm{d}r &= 1\\ &= \int_0^{2\pi} \int_0^{\infty} {}^2\delta(r) \space r \space \mathrm{d}r \space \mathrm{d}\theta\\ &= \int_0^{\infty} {}^2\delta(r) \space 2\pi r \space \mathrm{d}r\\ \end{align}

so we have

$${}^2\delta(r) = \dfrac{{}^1\delta_a(r)}{2\pi r}$$

This differs from your desired result by a factor of $$2$$, because I did not allow the 1 dimensional delta function to be symmertic around $$0$$, which wouldn't make sense for a polar origin where $$r$$ cannot be less than $$0$$ and $$\theta$$ ranges in $$[0, 2\pi)$$.

If one allows $$r$$ to go negative and $$\theta$$ to range in $$[0, \pi)$$, then one can show for a 1 dimensional delta function that is symmetric around $$0$$ $${}^2\delta(r) = \dfrac{{}^1\delta(r)}{\pi |r|}$$

(note the absolute value bars!)

This is a good example of why delta functions require careful definitions and considerations when making statements.

To show the projections of a 2-D delta function are themselves delta functions, start with a suitable limiting sequence of functions for $${}^2\delta(x,y)$$, like an infinitely thin and tall rectangle function both in x and in y, such that

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} {}^2\delta(x,y) \space \mathrm{d}x \space \mathrm{d}y = 1$$

and the answer should fall out from the limiting sequence of functions you use.

• Hi thank you so much now I have better understanding of delta function I completely understand the first part, but not the second part I did not get what you mean by the answer should fall out from the limiting sequence of functions you use. Maybe I was not so clear, I want to show that each projection of a two dimensional impulse function at the origin is a δ(t). I could not figured out how to convert 2δ(x,y) dx dy to δ(t). I hope that this is not too much to ask I m kinda new at this subject one more thanks in advance. – truckdriver Oct 29 '18 at 17:55