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In PCIe Gen 3 on-wards there is 128b/130b encoding used which scrambles incoming bit using a LFSR with polynomial: $$G(X) = X^{23} + X^ {21} + X^{16} + X^{8} + X^5 + X^2 + 1.$$ Is there any way to perform required scrambling on 2 bit in one clock using parallel scramblers, where output of incoming msb will be dependent on lsb's output.

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    $\begingroup$ I don't think so. If you were able to, you can divide and conquer attack on a single LFSR, if I've understood correctly. $\endgroup$ – kelalaka Aug 6 at 20:16
  • $\begingroup$ How can we decimate the given polynomial for getting odd and even sequence out of the sequence generated by this LFSR polynomial G(X)=X23+X21+X16+X8+X5+X2+1. I need two new LFSR which will follow only the odd sequence and even sequence of that whole sequence. $\endgroup$ – Akash Pradhan Aug 6 at 21:07
  • $\begingroup$ Please note that the PCIe uses LFSR not for security, but to reduce excessive di/dt. $\endgroup$ – forest Aug 7 at 1:32
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    $\begingroup$ @forest what is di/dt $\endgroup$ – kodlu Aug 7 at 2:25
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    $\begingroup$ Bursts of sequential 1s and 0s can cause sudden changes in power consumption, and can cause electromagnetic interference in a high-speed bus. Using an LFSR as a stream cipher makes it so that any data going over a bus appears random. DDR3/4 does it, high-speed USB does it, PCIe does it, etc. It ensures that the electronic stress and EMI is independent of the kind of data that is being sent. $\endgroup$ – forest Aug 7 at 2:43
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The online magma calculator

http://magma.maths.usyd.edu.au/calc/

tells me this LFSR corresponds to a primitive polynomial since $$G(X) = X^{23} + X^ {21} + X^{16} + X^{8} + X^5 + X^2 + 1$$ is primitive.

Thus, some phase of the output sequence is given by $$tr(\alpha^t)$$ where $tr:GF(2^{23})\rightarrow GF(2)$ is the trace map and $\alpha \in GF(2^{23})$ is primitive.

The minimal polynomial of $\alpha^2$ is also $G(X)$ and it is the polynomial generating the decimation by 2 (the minimal polynomial of $\alpha^2.$ But this is the same polynomial since an element and its square are conjugates. This is because $$ G(X)=\prod_{i=1}^{23} (X-\alpha^{2^{i-1}}), $$ from standard finite field results.

The decimation by 2 is $s(t+2^{22})$ if $s(t)$ is the original sequence in its characteristic phase, i.e., the phase that satisfies $$s(2t)=s(t)$$ for all $t \pmod{2^{23}-1}.$

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  • $\begingroup$ could you add the two polynomials? $\endgroup$ – kelalaka Aug 7 at 12:11

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