首页Consider a 5-bit linear-feedback shift register (LFSR) with the following recursive rule: (n + 1)-th bit = (n − 2)-th bit ⊕(n − 4)-th bit, for n ≥ 5. i. Demonstrate one step of this LFSR, with the aid of a schematic, starting from seed 11010.

Consider a 5-bit linear-feedback shift register (LFSR) with the following recursive rule: (n + 1)-th bit = (n − 2)-th bit ⊕(n − 4)-th bit, for n ≥ 5. i. Demonstrate one step of this LFSR, with the aid of a schematic, starting from seed 11010.

时间: 2023-12-31 22:03:20 浏览: 75

一个n-LFSR给定结构常数具有“由-序列密码（大全）

一个n-LFSR（给定结构常数）具有“由一个短的种子产生一个长的序列”的功能：以短的种子作为初态，产生的输出序列可以任意长！上述表明，任一n-LFSR都初步具有作为一个KG的资格；但从作为KG的效用来讲，自然更希望所使用的n-LFSR进一步是m-序列生成器。 m序列的特性

Sure! Here's a schematic of the 5-bit LFSR with the given recursive rule: ``` +--->(+)---(n-2)------+ | | | v 11010 ---| (+) | | +--->(+)<---(n-4)<---+ ``` Starting from the seed 11010, we first shift all bits to the right: ``` 11010 -> 01101 ``` Then, we compute the next bit using the recursive rule: ``` (n + 1)-th bit = (n − 2)-th bit ⊕(n − 4)-th bit = 1 ⊕ 1 = 0 ``` So the next bit of the LFSR is 0. We add this bit to the left of the shifted register: ``` 01101 -> 00110 ``` And that's one step of the LFSR starting from the seed 11010.

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