# How to get general projection of rect function?

To be more clear I want to get a general expression for

PƟ(t)= ∫∫rect(x)rect(y)δ(xcosƟ+ysinƟ-t)dxdy

I also want to find particular projection for Ɵ=0 and Ɵ=45 and also Fourier Transform of the projections at 0 and 45. I kinda new at image processing and projections, thus I could not figure out a way.

• Please learn to use MathJax before posting complicated questions involving formulas. – Dilip Sarwate Nov 6 '18 at 20:15

Since your expression \begin{align*}P(\theta,t) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \Pi(y)\Pi(x)\delta\left(x\cos\theta + y\sin\theta -t\right)dxdy\\ \end{align*}

uses a delta function with only 1 dimensional argument, in the 2 dimensional integration I assume you mean an infinite impulse line (an impulse "wall") and not a point impulse (an impulse "spike").

Let's just look at the general solution first, and then handle the degenrate case of $$\theta = 0$$.

I'll be using the following properties of the delta function:

\begin{align*}\delta(ax+b) &= \dfrac{1}{|a|}\delta\left(x + \dfrac{b}{a}\right)\\ \\ \int_{-\infty}^{\infty} f(x)\delta(x-a)dx &= f(a)\\ \end{align*}

So, I believe the solution is this:

\begin{align*}P(\theta,t) &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \Pi(y)\Pi(x)\delta\left(x\cos\theta + y\sin\theta -t\right)dxdy\\ \\ &= \int_{-\infty}^{\infty} \Pi(y) \left[\int_{-\infty}^{\infty} \Pi(x)\dfrac{1}{|\cos\theta|}\delta\left(x - \left[-\dfrac{\sin\theta}{\cos\theta}y +\dfrac{t}{\cos\theta}\right]\right)dx\right]dy\\ \\ &= \dfrac{1}{|\cos\theta|}\int_{-\infty}^{\infty} \Pi(y) \Pi\left(-\dfrac{\sin\theta}{\cos\theta}y +\dfrac{t}{\cos\theta}\right)dy\\ \\ &\mbox{and if the rectangles have any overlap for the given t and \theta}\\ \\ &= \dfrac{1}{|\cos\theta|}\int_{\max\left(-\frac{1}{2},-\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)}^{\min\left(\frac{1}{2},\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)} dy\\ \\ &= \dfrac{y}{|\cos\theta|}\biggr\rvert_{\max\left(-\frac{1}{2},-\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)}^{\min\left(\frac{1}{2},\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)}\\ \\ P(\theta,t) &= \dfrac{\min\left(\frac{1}{2},\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)-\max\left(-\frac{1}{2},-\frac{1}{2}\frac{\cos\theta}{\sin\theta}+\frac{t}{\sin\theta}\right)}{|\cos\theta|} \end{align*}

So for the degenerate case of $$\theta = 0$$, the above expression runs into problems, but the integral for this special case is a bit easier:

\begin{align*}P(0,t) &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \Pi(y)\Pi(x)\delta\left(x\cos{0} + y\sin{0} -t\right)dxdy\\ \\ &= \int_{-\infty}^{\infty} \Pi(y) \left[\int_{-\infty}^{\infty} \Pi(x)\delta\left(x-t\right)dx \right]dy\\ \\ &= \Pi(t)\int_{-\infty}^{\infty} \Pi(y) dy\\ \\ &= \Pi(t) \end{align*}

The one dimensional Fourier Transform of $$P(0,t)$$ is a simple table look-up to find that it is a $$\mathrm{sinc}()$$ function.

The one dimensional Fourier Transform of the more general $$P(\theta, t)$$ is obviously more complicated. There are 4 cases there.