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In the book, Advanced Topics in Shannon Sampling and Interpolation Theory by Robert J. Marks II, one may find an interpolation formula for reconstructing a two dimensional signal from regular polar samples in the frequency domain:

Let $f(x,y)$ be space-limited to $2A$. Let its Fourier transform in polar coordinates $F(\rho,\phi)$ be angularly band-limited to $K$. Then $F(\rho,\phi)$ can be reconstructed from its polar samples via

$$F(\rho,\phi)=\sum_{n=-\infty}^\infty\sum_{k=0}^{N-1}\widetilde{F}\left(\frac{n}{2A},\frac{2\pi k}{N}\right)\operatorname{sinc}\left[\frac{2A(\rho-n)}{2A}\right]\frac{\sin\left[\frac{1}{2}(N-1)\left(\phi-\frac{2\pi k}{N}\right)\right]}{N\sin\left[\frac{1}{2}\left(\phi-\frac{2\pi k}{N}\right)\right]},$$ where $N$ is assumed even and $$\widetilde{F}\left(\frac{n}{2A},\frac{2\pi k}{N}\right)=\begin{cases}F\left(\frac{n}{2A},\frac{2\pi k}{N}\right),&n\ge 0, \\ F\left(-\frac{n}{2A},\frac{2\pi k}{N}+\pi\right),&n<0.\end{cases}$$

I figured that the argument of the cardinal sine function contained a typo, since the $2A$ in the numerator and denominator would appear to cancel. Thus I would guess that the author actually had

$$\operatorname{sinc}\left[\frac{\rho-2An}{2A}\right]$$

in mind. Correct me if I'm wrong?

Then the last quotient with the sine functions looks very similar to the Dirichlet kernel:

$$D_N\left(\phi-\frac{2\pi k}{N}\right)=\frac{\sin\left[\left(N+\frac{1}{2}\right)\left(\phi-\frac{2\pi k}{N}\right)\right]}{\sin\left[\frac{\phi-\frac{2\pi k}{N}}{2}\right]}$$

but is not quite the same.

Incidentally, my attempts to implement the interpolation formula in the theorem via Matlab for specific examples have failed completely with regards to reconstructing a signal. Perhaps someone can point me in the direction of a correct formula for signal reconstruction in the frequency domain from polar samples?

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I haven't worked through the algebra, but this page just before that equation:

enter image description here

suggests that $$ \operatorname{sinc}\left[2A(\rho-\frac{n}{2A})\right] $$ might be the correct form of the $\operatorname{sinc}$ term.

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  • $\begingroup$ Then this is the same as $$\operatorname{sinc}\left[\frac{\rho-n\Delta\rho}{\Delta\rho}\right],$$ with $\Delta\rho=\frac{1}{2A}$. Interestingly I also tried to implement this formula on Matlab too, but with no success. $\endgroup$ – Jason Born Jun 9 '17 at 18:08

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