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For convolution of two functions (one is known analytically: $\omega = \frac{\delta(R-\lvert \mathbf{r}\rvert)}{4\pi R}$, the other only on a spatial grid: $f$) I want to use the convolution theorem: $f(r)\otimes\omega(r) = \mathcal{F}^{-1}\Big\{ \mathcal{F}\{f(r)\} \mathcal{F}\{\omega(r)\} \Big\} = \mathcal{F}^{-1}\Big\{ \hat{f}(k) \hat{\omega}(k) \Big\}$. The functions $f(r)$ and $\omega(r)$ go to zero at finite values of $r$. Both functions are even with respect to $r=0$.

Using spherical symmetry of $f$ and $\omega$ I end up with $\hat{f}(\mathbf{k}) = \frac{2}{\lvert \mathbf{k} \rvert} \int\limits_0^\infty r f(r) \sin(2\pi \lvert \mathbf{k} \rvert r) \mathbf{d}r $ and $\hat{\omega}(k)=\frac{\sin(2\pi R\lvert \mathbf{k} \rvert)}{2\pi\lvert \mathbf{k} \rvert}$ with parameter $R$.

L'Hôpital's rule for $k \rightarrow 0$ gives me: $\lim\limits_{\mathbf{k} \rightarrow 0} \hat{f}(\mathbf{k})= \lim\limits_{\mathbf{k} \rightarrow 0} 4 \pi \int\limits_0^\infty r^2 f(r) \cos(2\pi \lvert \mathbf{k} \rvert r) \mathbf{d}r = 4 \pi \int\limits_0^\infty r^2 f(r) \mathbf{d}r$ and $\lim\limits_{\mathbf{k}\rightarrow 0}\hat{\omega}(\mathbf{k}) = R$.

Because I don't have $f(r)$ analytically I need to use the discrete sine transform. I would like to use DST-II: $X_k = \sum\limits_{n=0}^{N-1} x_n \sin\Big[ \frac{\pi}{N} (n+\frac{1}{2}) (k+1) \Big]$ for $k=0, ..., (N-1)$.

Therefore, I end up with something like: $F_k = \frac{2}{k} \sum\limits_{n=0}^{N-1} r_n f_n \sin\Big[ \frac{\pi}{N} (n+\frac{1}{2}) (k+1) \Big]$ for $k=0, ..., (N-1)$. I think the $k$ in front of the sum and the $k$ as argument of the $\sin$ don't match (the sine does not go to zero for $k=0$.

Where is the indexing wrong? What do I have to do to convolute the two functions correctly?

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  • $\begingroup$ I partially solved the problem: The values for k=0 do not matter as long as they're finite (which is the case here). Because the inverse sine transform multiplies the functions with k instead of r. $\endgroup$ – Rlf Oct 23 '16 at 11:25

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