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To circumvent the size difference between $S$ and $v$, one can use JPL codes. These are a XOR (which is modulo $2$ addition) of two or more maximal length sequences of coprime sizes.

After 2nd thought this is not the same as I advised in my comments. For two bit sequences of coprime sizes $2^n - 1$ and $2^m - 1$, make the size the period of each sequence as periodic infinite bit sequence. Adding them with XOR gives a new infinite bit sequence with the product of both periods as period (since they are coprime). Take one period of this sequence as the resulting PRBS $x$. This can be used as circulant matrix $K$, but on a permuted $v$: $v_i$ must then be the output of the ($i$ mod $2^n - 1$)th sampler on timeframe $i$ mod $2^m - 1$.

You can either reorder $v$ or use instead the matrix $M=(S_{ij} + T_{kl})_{nm}$ (mod 2) with $i=n, j=m, k=n, l=m$ modulo the other sequence length. The autocorrelation of $K$ is the same as that of $M$ since it's merely permuted. Edit: no that does not follow. But the autocorrelation of the samplers is that of the maximal length sequence, and the autocorrelation over the timeframes the other sequence.

I'm not sure what to do if you can't exactly control the amount of timeframes, but I think it's fine to add zeroes at the end of $v$ to get the right dimension.

To circumvent the size difference between $S$ and $v$, one can use JPL codes. These are a XOR (which is modulo $2$ addition) of two or more maximal length sequences of coprime sizes.

After 2nd thought this is not the same as I advised in my comments. For two bit sequences of coprime sizes $2^n - 1$ and $2^m - 1$, make the size the period of each sequence as periodic infinite bit sequence. Adding them with XOR gives a new infinite bit sequence with the product of both periods as period (since they are coprime). Take one period of this sequence as the resulting PRBS $x$. This can be used as circulant matrix $K$, but on a permuted $v$: $v_i$ must then be the output of the ($i$ mod $2^n - 1$)th sampler on timeframe $i$ mod $2^m - 1$.

You can either reorder $v$ for JPL codes, or use instead the approach from my comments and find the matrix $M=(S_{ij} + T_{kl})_{nm}$ (mod 2) with $i=n, j=m, k=n, l=m$ modulo the other sequence length. It might be that it depends on the context which one is better. The autocorrelation of $K$ is the same as that of $M$ since it's merely permuted. Edit: no that does not follow. But the autocorrelation of the samplers is that of the maximal length sequence, and the autocorrelation over the timeframes the other sequence.

I'm not sure what to do if you can't exactly control the amount of timeframes, but I think it's fine to add zeroes at the end of $v$ to get the right dimension.

To circumvent the size difference between $S$ and $v$, one can use JPL codes. These are a XOR (which is modulo $2$ addition) of two or more maximal length sequences of coprime sizes.

This appears to be the exact but permuted same as I advised in my comments. For two bit sequences of coprime sizes $2^n - 1$ and $2^m - 1$, make the size the period of each sequence as periodic infinite bit sequence. Adding them with XOR gives a new infinite bit sequence with the product of both periods as period (since they are coprime). Take one period of this sequence as the resulting PRBS $x$. This can be used as circulant matrix $K$, but on a permuted $v$: $v_i$ must then be the output of the ($i$ mod $2^n - 1$)th sampler on timeframe $i$ mod $2^m - 1$.

By writing the circulant matrix as a sum of powers of another cyclic permutation $P$ rather than $(1 \space 2 \space ... \space 5080)$, "the tensor XOR addition" matrix $M=(S_{ij}+T_{kl})_{(i,k)(j,l)}$ (mod 2) with index order first all samplers on the first timeframe, then the second timeframe etc. can be retrieved. This matrix retains the autocorrelation of $x$. However $40$ timeframes is a bad example, in order to use JPL codes you need an amount of timeframes that is a product of coprime factors $2^{n_i} -1$ such as $31$ or $63$ or $7×15=105$. It must also be coprime to the amount of samplers.

You can either reorder $v$ or find $P$, it is defined as the cycle with on the $i$th zero-based position the number $q+iL$ (modulo the size) (where $q$ is between $0$ and $L-1$ and equals $i$ modulo $L$). Reordering $v$ means applying the inverse of $P$ to it, but this means also the results after deconvolution are permuted. In formulas, $M=PKP^T$ and reordered $v$ is $P^Tv$ so that $KP^Tv=P^TMv$. The autocorrelation of $K$ is the same as that of $M$ since it's merely permuted. Edit: no that does not follow. But the autocorrelation of the samplers is that of the maximal length sequence, and the autocorrelation over the timeframes the other sequence.

I'm not sure what to do if you can't exactly control the amount of timeframes, but I think it's fine to add zeroes at the end of $v$ to get the right dimension.

To circumvent the size difference between $S$ and $v$, one can use JPL codes. These are a XOR (which is modulo $2$ addition) of two or more maximal length sequences of coprime sizes.

After 2nd thought this is not the same as I advised in my comments. For two bit sequences of coprime sizes $2^n - 1$ and $2^m - 1$, make the size the period of each sequence as periodic infinite bit sequence. Adding them with XOR gives a new infinite bit sequence with the product of both periods as period (since they are coprime). Take one period of this sequence as the resulting PRBS $x$. This can be used as circulant matrix $K$, but on a permuted $v$: $v_i$ must then be the output of the ($i$ mod $2^n - 1$)th sampler on timeframe $i$ mod $2^m - 1$.

You can either reorder $v$ or use instead the matrix $M=(S_{ij} + T_{kl})_{nm}$ (mod 2) with $i=n, j=m, k=n, l=m$ modulo the other sequence length. The autocorrelation of $K$ is the same as that of $M$ since it's merely permuted. Edit: no that does not follow. But the autocorrelation of the samplers is that of the maximal length sequence, and the autocorrelation over the timeframes the other sequence.

I'm not sure what to do if you can't exactly control the amount of timeframes, but I think it's fine to add zeroes at the end of $v$ to get the right dimension.

To circumvent the size difference between $S$ and $v$, one can use JPL codes. These are a XOR (which is modulo $2$ addition) of two or more maximal length sequences of coprime sizes.

This appears to be the exact but permuted same as I advised in my comments. For two bit sequences of coprime sizes $2^n - 1$ and $2^m - 1$, make the size the period of each sequence as periodic infinite bit sequence. Adding them with XOR gives a new infinite bit sequence with the product of both periods as period (since they are coprime). Take one period of this sequence as the resulting PRBS $x$. This can be used as circulant matrix $K$, but on a permuted $v$: $v_i$ must then be the output of the ($i$ mod $2^n - 1$)th sampler on timeframe $i$ mod $2^m - 1$.

By writing the circulant matrix as a sum of powers of another cyclic permutation $P$ rather than $(1 \space 2 \space ... \space 5080)$, "the tensor XOR addition" matrix $M=(S_{ij}+T_{kl})_{(i,k)(j,l)}$ (mod 2) with index order first all samplers on the first timeframe, then the second timeframe etc. can be retrieved. This matrix retains the autocorrelation of $x$. However $40$ timeframes is a bad example, in order to use JPL codes you need an amount of timeframes that is a product of coprime factors $2^{n_i} -1$ such as $31$ or $63$ or $7×15=105$. It must also be coprime to the amount of samplers.

You can either reorder $v$ or find $P$, it is defined as the cycle with on the $i$th zero-based position the number $q+iL$ (modulo the size) (where $q$ is between $0$ and $L-1$ and equals $i$ modulo $L$). Reordering $v$ means applying the inverse of $P$ to it, but this means also the results after deconvolution are permuted. In formulas, $M=PKP^T$ and reordered $v$ is $P^Tv$ so that $KP^Tv=P^TMv$. The autocorrelation of $K$ is the same as that of $M$ since it's merely permuted.

I'm not sure what to do if you can't exactly control the amount of timeframes, but I think it's fine to add zeroes at the end of $v$ to get the right dimension.

To circumvent the size difference between $S$ and $v$, one can use JPL codes. These are a XOR (which is modulo $2$ addition) of two or more maximal length sequences of coprime sizes.

This appears to be the exact but permuted same as I advised in my comments. For two bit sequences of coprime sizes $2^n - 1$ and $2^m - 1$, make the size the period of each sequence as periodic infinite bit sequence. Adding them with XOR gives a new infinite bit sequence with the product of both periods as period (since they are coprime). Take one period of this sequence as the resulting PRBS $x$. This can be used as circulant matrix $K$, but on a permuted $v$: $v_i$ must then be the output of the ($i$ mod $2^n - 1$)th sampler on timeframe $i$ mod $2^m - 1$.

By writing the circulant matrix as a sum of powers of another cyclic permutation $P$ rather than $(1 \space 2 \space ... \space 5080)$, "the tensor XOR addition" matrix $M=(S_{ij}+T_{kl})_{(i,k)(j,l)}$ (mod 2) with index order first all samplers on the first timeframe, then the second timeframe etc. can be retrieved. This matrix retains the autocorrelation of $x$. However $40$ timeframes is a bad example, in order to use JPL codes you need an amount of timeframes that is a product of coprime factors $2^{n_i} -1$ such as $31$ or $63$ or $7×15=105$. It must also be coprime to the amount of samplers.

You can either reorder $v$ or find $P$, it is defined as the cycle with on the $i$th zero-based position the number $q+iL$ (modulo the size) (where $q$ is between $0$ and $L-1$ and equals $i$ modulo $L$). Reordering $v$ means applying the inverse of $P$ to it, but this means also the results after deconvolution are permuted. In formulas, $M=PKP^T$ and reordered $v$ is $P^Tv$ so that $KP^Tv=P^TMv$. The autocorrelation of $K$ is the same as that of $M$ since it's merely permuted. Edit: no that does not follow. But the autocorrelation of the samplers is that of the maximal length sequence, and the autocorrelation over the timeframes the other sequence.

I'm not sure what to do if you can't exactly control the amount of timeframes, but I think it's fine to add zeroes at the end of $v$ to get the right dimension.