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The dominator is $10^6$ means that 1 ppm is added. Assuming the length of the symbols is $N = 1024$ as I insert the SFO for each symbol alone, so the maximum added SFO will be 1024.

On the other hand, if we can normalize the added SFO by $N$ to have the maximum added SFO 1ppm, the above equation will be:

The dominator is $10^6$ means that 1 ppm is added. Assuming the length of the symbols is $N = 1024$ as I insert the SFO for each symbol alone, so the maximum added SFO will be $1024$ multiplied by the difference between $x[N]$ and $x[N-1]$.

On the other hand, if we can normalize the added SFO by $N$ to have the maximum added SFO 1ppm multiplied by the difference between $x[N]$ and $x[N-1]$, the above equation will be:

The dominator is $10^6$ means that 1 ppm is added. Assuming the length of the symbols is $N = 1024$, so the maximum added SFO will be 1024 too. On the other hand, if we can normalize the added SFO by $N$ to have the maximum added SFO 1ppm, the above equation will be:

The dominator is $10^6$ means that 1 ppm is added. Assuming the length of the symbols is $N = 1024$ as I insert the SFO for each symbol alone, so the maximum added SFO will be 1024.

On the other hand, if we can normalize the added SFO by $N$ to have the maximum added SFO 1ppm, the above equation will be:

Update

Normally, the added SFO effect into the signal $x[n]$ as following:

$y\left [ n \right ] = x\left [ n \right ] + n \times \frac{x\left [ n + 1 \right ] - x\left [ n \right ]}{10^6}$

The dominator is $10^6$ means that 1 ppm is added. Assuming the length of the symbols is $N = 1024$, so the maximum added SFO will be 1024 too. On the other hand, if we can normalize the added SFO by $N$ to have the maximum added SFO 1ppm, the above equation will be:

$y\left [ n \right ] = x\left [ n \right ] + n \times \frac{x\left [ n + 1 \right ] - x\left [ n \right ]}{N \times 10^6}$

Unfortunately, the rotation is still very big and the signal is completely deteriorated.