Timeline for Reshaping data vector into a matrix for deconvolution using a circulant matrix
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2023 at 10:58 | answer | added | Maarten Havinga | timeline score: 1 | |
| Sep 15, 2023 at 12:48 | comment | added | ACR | I think the key question still is how do go about circumventing the $S$ and the data vector $v$ size mismatch. The solution would still be acting on segment by segment. I would appreciate if you post it as an answer with little bit of more details because comments get deleted. | |
| Sep 15, 2023 at 10:15 | comment | added | Maarten Havinga | An easier formula for $S'$ might be $(S_{ij}+T_{kl})_{(i,k)(j,l)} \space (mod \space 2)$ where the index space is now the cartesian product of index spaces of $S$ and $T$ (so $i,j$ indicate the sampler while $k,l$ indicate the timeframe. | |
| Sep 15, 2023 at 10:08 | comment | added | Maarten Havinga | A 40 element maximal length sequence does not exist, better would be 63 or 31 timeframes. Define $(e(A))_{ij} = (-1)^{A_{ij}}$ and find a PRBS of the required length (amount of timeframes) with circulant matrix $T$. Then use $S'=e^{-1}(e(S) \otimes e(T))$ as new $S$. An easier formula for $S'$ might be $(S+T)_{(i,k)(j,l)} mod 2$ where the index space is now the cartesian product of index spaces of $S$ and $T$. | |
| Sep 14, 2023 at 19:39 | comment | added | ACR | Maarten Havinga, Your approach seems interesting, can you add a little bit more about the approach, starting with 127x127 circulant matrix $S$, and you have a $v$ which is 5080 elements. How do you suggest going about from there? | |
| Sep 14, 2023 at 18:20 | comment | added | Maarten Havinga | The tensor product method deconvolves the time series for each sampler independently from the deconvolution of all $127$ samplers at one time. | |
| Sep 14, 2023 at 17:59 | comment | added | Maarten Havinga | So the best noise reduction goes for any Hadamard matrix with the constant $1$ row and column deleted and mapping $1$ to $0$ and $-1$ to $1$. However I'm not sure if there's a reason for the circulant condition on $S$, if not just take such a 0-1 representation of a Hadamard matrix of the same size as $v$. To preserve this circulant property on $L$ inputs use the tensor product method I mentioned above. | |
| Sep 14, 2023 at 17:39 | comment | added | Maarten Havinga | I believe you need a maximal determinant binary matrix of the same size as $v$. A tensor product of two circulant matrices from 2 PRBS in $\pm 1$ form (then translated back to binary) should reduce noise a lot. From wikipedia, Let A be an n × n matrix and let B be an m × m matrix. Then $\left| \mathbf{A} \otimes \mathbf{B} \right| = \left| \mathbf{A} \right| ^m \left| \mathbf{B} \right| ^n$. It's the inverse of the determinant (which is tiny) that reduces noise. | |
| Sep 13, 2023 at 12:04 | answer | added | kodlu | timeline score: 1 | |
| Sep 12, 2023 at 15:06 | comment | added | ACR | This question is related to mathematical application to chemistry research where Hadamard Transforms are employed to signal to noise ratio enhancement. I wrote a very detailed query before (now deleted) with the background but it hid the main question which I summarized here. Any pointer or reference would be helpful. | |
| Sep 12, 2023 at 13:11 | review | Close votes | |||
| Oct 6, 2023 at 3:08 | |||||
| Sep 12, 2023 at 12:22 | history | asked | ACR | CC BY-SA 4.0 |