求教桶形移位器原理

ziyue1987

桶形移位器的原理是什么,那位大牛给解释一下！

mainhand

顺便附上tutorial
Understanding Linear Feedback Shift Registers – The Easy Way
Written by Dan Healy for UCSC’s CMPE125 class.



This tutorial will teach you how to use LFSRs, why other tutorials on the subject are so confusing, and how you can go about understanding the underlying mathematics if you really want to know.

First, a glossary.

Linear Feedback Shift Register (LFSR):

An n-bit shift register which pseudo-randomly scrolls between 2n-1 values, but does it very quickly because there is minimal combinational logic involved.
Once it reaches its final state, it will traverse the sequence exactly as before.
It has many applications you should already be familiar with if you’re reading this.

Primitive polynomial:
(very basically) A polynomial of degree n that has the form: 1 + … + xn, where (…) are zero or more terms with a coefficient of 1.
xn and 1 are always present.
For each degree, there can be many different primitive polynomials.
These polynomials also must satisfy other mathematical conditions, if you’re really interested see http://mathworld.wolfram.com/PrimitivePolynomial.html or google it.
One important property to note is that their reciprocals also form primitive polynomials (that is, they come in pairs). Example: 1 + x3 + x4 is Degree 4, its reciprocal is 1 + x + x4 (10011 and 11001), and both are primitive.

Taps:
Lines that run from the output of one register within the LFSR into XOR gates that determine input to another register within the LFSR.
These are chosen based on the primitive polynomial.

Type 1 or External LFSRs:
One way of implementing LFSRs; all XOR gates are fed sequentially into one another and end up as the input to the least (or most, either is correct) significant bit of the LFSR.
Simply put, the XORs are external from the shift register.

Type 2 or Internal LFSRs:
Another LFSR implementation; XOR gates feed into different registers within the LFSR, and are not sequential.
Simply put, the XORs are inside the shift register.



Other tutorials on this subject are confusing because they only address one type of implementation, don’t explain how they got the taps, or they show the implementation of an LFSR and then in a table show the taps corresponding to a different primitive polynomial of the same degree.
Keep all of this in mind if you look elsewhere.


Consider a simple 3-bit LFSR.
The only primitive polynomials for degree 3 are 1 + x2 + x3 and 1 + x + x3 (they are reciprocals of each other, 1011 and 1101).
Since we have two primitive polynomials, and we have two different implementation strategies, we therefore have four unique ways of implementing the LFSR.
In fact, each of these implementations can differ according to which register is the most significant bit (either way will have 2n-1 states, but with different sequences).


A and B illustrate a Type 1 / External LFSR.

C and D illustrate a Type 2 / Internal LFSR.

A and C illustrate the implementation of 1 + x2 + x3
B and D illustrate the implementation of 1 + x + x3

Since xn and 1 are always present in primitive polynomials, you can think of them as being used as the output of the shift register and the input of the shift register, respectively.


For an n-bit LFSR, you need to discover a primitive polynomial associated with it to implement it.
Tap tables on the internet will list taps as such:

N = 3, Taps at 0, 1, 3 (this corresponds to 1 + x + x3)


This tutorial uses the 0 and 3 as taps because they correspond to the powers of x in the primitive polynomial.
Sometimes Tap tables will omit the 0 and 3, since they must be present.
Other times, they may use 0 to mean the output of the least significant register (so for N = 3, their taps would be listed as "0, 2" instead of "1, 3").
They usually only list the taps associated with one primitive polynomial, but more than one exists.
All of this explains why different tap tables will sometimes show you different numbers!




Here’s another pair of examples, for n = 8 and using the primitive polynomial 1 + x2 + x3 + x4 + x8:

Internal:


External:



Warning!
The images above show the Reset line used incorrectly.
The registers should be seeded to a non-zero value; all-zeroes is called the lock-up state and will not change.
Therefore, at least one register must be preset.
Your choice in seed value determines the order of states, and ultimately the value at the 2n-1 state.
Note that it is also possible to design LFSRs to have an all-ones lock-up state instead, but this will not be discussed here for brevity.
Special thanks to all those who have pointed out this omission.

Here is a good tap table provided by Scott R. Gravenhorst! In this link, the taps listed omit "0." Again, keep in mind that other polynomials do exist.

Good luck!

Images blatantly stolen from http://www.ee.ualberta.ca/~ellio ... rivers_Ed/lfsr.html and http://www.edacafe.com/books/ASIC/Book/CH14/CH14.7.php

falloutmx

你就想象有3个水桶ABC，通过管子连接，C的水做输出用。只要C的水用掉一点，就能通过管子从B桶补充，B桶同理从A桶补充，而A桶每损失1/4（举例）的水，就从水龙头灌入A桶灌满为止。这样你不管怎么从C桶里要水都不会导致C桶的水变空（除非水源没水了）

zhyeah

这位仁兄回得真精辟

3# falloutmx

scm

学习了

dikedisk

非常需要的啊

wdtfjfe

不错要是中文就好了！