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