descrabmle

This commit is contained in:
Jinghao Shi 2017-04-17 12:53:44 -04:00
parent 556794ae2e
commit 25c5d67904

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@ -208,16 +208,83 @@ using the same logic.
Scrambler/Descrambler Logic
Suppose the current input bit is :math:`B`, the output bit :math:`B'` and the
Suppose the current input bit is :math:`B_n`, the scrambled bit :math:`B^s_n` and the
internal state of the scrambler is updated as follows:
.. math::
B' &\leftarrow X^1 \oplus B\\
X^1 &\leftarrow X^7 \oplus X^4\\
X^i &\leftarrow X^{i-1}, i = 2, 3,\ldots, 7
B^s_n &\leftarrow X_n^1 \oplus B_n\\
X_{n+1}^1 &\leftarrow X_n^7 \oplus X_n^4\\
X_{n+1}^i &\leftarrow X_n^{i-1}, i = 2, 3,\ldots, 7
At the transmitter side, for each packet, the scrambler is loaded with pseudo
random value. The very first 7 bits of the data bits is preset to zero before
scrambling, so that the receiver can estimate the value using the scrambled
bits.
where :math:`X^i_n` is the scrambler state before the nth input bit, :math:`n=0,
1, 2,\ldots`.
At the transmitter side, for each packet, the scrambler is initialized with
pseudo random value. The very first 7 bits of the data bits is preset to zero
before scrambling, so that the receiver can estimate the value using the
scrambled bits.
Now let's see how the receiver recovers the initial state of the transmitter's
scrambler. There are two ways to interpret this.
First, we can *calculate* the initial state. Since the first 7 un-scrambled bits
(:math:`B_0` to :math:`B_6`) are all zeros, the scrambled bits can be obtained
by:
.. math::
B^s_0 &= X_0^7 \oplus X_0^4\\
B^s_1 &= X_1^7 \oplus X_1^4 = X_0^6 \oplus X_0^3\\
B^s_2 &= X_2^7 \oplus X_2^4 = X_0^5 \oplus X_0^2\\
B^s_3 &= X_3^7 \oplus X_3^4 = X_0^4 \oplus X_0^1\\
B^s_4 &= X_4^7 \oplus X_4^4 = X_0^3 \oplus B^s_0\\
B^s_5 &= X_5^7 \oplus X_5^4 = X_0^2 \oplus B^s_1\\
B^s_6 &= X_6^7 \oplus X_6^4 = X_0^1 \oplus B^s_2\\
From which we can reverse calculating the value of :math:`X` as follows:
.. math::
X_0^1 &= B^s_6 \oplus B^s_2\\
X_0^2 &= B^s_5 \oplus B^s_1\\
X_0^3 &= B^s_4 \oplus B^s_0\\
X_0^4 &= B^s_3 \oplus X_0^1 = B^s_3 \oplus B^s_6 \oplus B^s_2\\
X_0^5 &= B^s_2 \oplus X_0^2 = B^s_2 \oplus B^s_5 \oplus B^s_1\\
X_0^6 &= B^s_1 \oplus X_0^3 = B^s_1 \oplus B^s_4 \oplus B^s_0\\
X_0^7 &= B^s_0 \oplus X_0^4 = B^s_0 \oplus B^s_3 \oplus B^s_6 \oplus B^s_2\\
This interpretation does not lead to efficient Verilog implementation since we
need to first buffer the first 7 bits, calculate the initial state and then
descramble from the first 7 bits again.
The second interpretation is that: **the first 7 scrambled bits are the state
after scrambling the 7 bits**. In other words, we have:
.. math::
X_7^7 &= B^s_0\\
X_7^6 &= B^s_1\\
X_7^5 &= B^s_2\\
X_7^4 &= B^s_3\\
X_7^3 &= B^s_4\\
X_7^2 &= B^s_5\\
X_7^1 &= B^s_6\\
For instance, take a look at :math:`X_7^7`,
.. math::
X_7^7 = X_6^6 = \ldots = X_1^1 = X_0^7 \oplus X_0^4
We also know that:
.. math::
B^s_0 &= X_0^7 \oplus X_0^4 \oplus B_0\\
&= X_0^7 \oplus X_0^4 \oplus 0 \\
&= X_0^7 \oplus X_0^4
Therefore :math:`X_7^7 = B^s_0`. This way we directly get the state to
descramble the next bit :math:`B^s_7`, resulting a very simple Verilog
implementation.