Search tips

# Discussion Groups | Comp.DSP | Generating Maximum Length Sequence using Galois LFSR

There are 30 messages in this thread.

You are currently looking at messages 1 to .

Is this discussion worth a thumbs up?

0

# Generating Maximum Length Sequence using Galois LFSR - Nicholas Kinar - 2009-07-04 00:55:00

```Hello--

I am trying to generate a Maximum Length Sequence (MLS) using a Linear
Feedback Shift Register (LFSR).  Assuming that the taps have been loaded
into variable "taps," and that "lfsr" is the variable being used in the
shift register computation, I am trying to generate the sequence using
code similar to that found on Wikipedia at the page
http://en.wikipedia.org/wiki/Linear_feedback_shift_register.  Here is a
code snippet:

// generate sequence
for(int i = 0; i < L; i++)
{
lfsr = (lfsr >> 1) ^ (-(lfsr & 1u) & taps);

// if I have an output array with L elements,
// what do I set it equal to in this loop?
// output[i] = ??
}

However, how do I determine the output of the LFSR?  For m = 12, it
follows that L = (2^m) - 1 = 4095.  How would I generate an MLS sequence
with a length of 4095 using the above code?  The loop should repeat 4095
times.  What is the "output stream"? Is it simply the lowest bit in the
sequence?  Is this the bit that is fed back into the sequence after
passing through the logic gates?

Is there a way to test if my output is an MLS?  Does the Galois LFSR
generate exactly the same output as a Fibonacci LFSR?```

# STUPIDENT::Re: Generating Maximum Length Sequence using Galois LFSR - Vladimir Vassilevsky - 2009-07-04 12:12:00

```Nicholas Kinar wrote:

> Hello--
>
> I am trying to generate a Maximum Length Sequence (MLS) using a Linear
> Feedback Shift Register (LFSR).  Assuming that the taps have been loaded
> into variable "taps," and that "lfsr" is the variable being used in the
>  shift register computation, I am trying to generate the sequence using
> code similar to that found on Wikipedia at the page
> http://en.wikipedia.org/wiki/Linear_feedback_shift_register.  Here is a
> code snippet:
>
>
> // generate sequence
> for(int i = 0; i < L; i++)
> {
>     lfsr = (lfsr >> 1) ^ (-(lfsr & 1u) & taps);
>
>     // if I have an output array with L elements,
>     // what do I set it equal to in this loop?
>     // output[i] = ??
> }
>
>
> However, how do I determine the output of the LFSR?  For m = 12, it
> follows that L = (2^m) - 1 = 4095.  How would I generate an MLS sequence
> with a length of 4095 using the above code?  The loop should repeat 4095
> times.  What is the "output stream"? Is it simply the lowest bit in the
> sequence?  Is this the bit that is fed back into the sequence after
> passing through the logic gates?
>
> Is there a way to test if my output is an MLS?  Does the Galois LFSR
> generate exactly the same output as a Fibonacci LFSR?
>
>
>```

# Re: Generating Maximum Length Sequence using Galois LFSR - Steve Pope - 2009-07-05 02:55:00

```robert bristow-johnson  <r...@audioimagination.com> wrote:

>you need to select your taps word correctly.  first of all, for m,
>then taps > 2048, but there is more to it.  somewhere on the web,
>there is a resource of "primitive polynomials".  find that and use one
>of them (ignore the least significant bit, which is always 1, the
>other m bits of the primitive polynomial are the m bits of your taps
>word).  for m, you could write a program that guesses at a 12-bit
>word, try it out (see if it goes through all 2^12 - 1 non-zero
>states), and if it fails, try another word for taps.

Here is the simplest one for m:

0x05eb

It is probably bit reversed from the OP's set-up,
The leadign coefficient of one is deleted from
this constant.

(Unlike the OP, I always put the LS term of the polynomial
in the lsb of a binary word, but after looking at OP's
code it seems more compact to do it his way.)

Steve```

# Re: Generating Maximum Length Sequence using Galois LFSR - Nicholas Kinar - 2009-07-05 10:26:00

```Hi Robert--

Thank you for your reply!  The code that I found on Wikipedia was
different enough to make me wonder exactly what was happening.  There's
additional reference code for MLS sequence generation that can be found
here:

http://libmls0.sourceforge.net/

> your L "elements" are single bits.  the "S" of MLS is a sequence of
> *bits*, not of words.
>

>
> pick any bit of the shift register (your "lfsr") and use it.  i
> usually pick the same bit that falls off the edge
>

>
> you need to select your taps word correctly.  first of all, for m,
> then taps > 2048, but there is more to it.  somewhere on the web,
> there is a resource of "primitive polynomials".  find that and use one
> of them (ignore the least significant bit, which is always 1, the
> other m bits of the primitive polynomial are the m bits of your taps
> word).  for m, you could write a program that guesses at a 12-bit
> word, try it out (see if it goes through all 2^12 - 1 non-zero
> states), and if it fails, try another word for taps.
>

There are a few resources on "primitive polynomials" currently on the
web, but perhaps you are thinking about the following paper?

Stahnke, W. 1973. "Primitive Binary Polynomials," Mathematics of
Computation 27(124): 977-980.

The paper gives a table of exponents of terms of primitive binary
polynomial terms up to m = 168, where the length of the sequence is L (2^m) - 1.  I've tested this table up to m = 30, and it seems to
generate an MLS.

>>  The loop should repeat [after] 4095 times.
>
> yes.  that's how you confirm it's MLS. start with a state of 0x0001
> and count how many iterations it takes to get back to 0x0001.  if it's
> 4095 and your "taps" word has zeros in the bits other than the least-
> significant 12 bits, then it's an MLS
>

You know, that's an interesting way of checking whether something is an MLS.

>>  What is the "output stream"? Is it simply the lowest bit in the
>> sequence?
>
> sure.  any bit will do.
>

>>  Is this the bit that is fed back into the sequence after
>> passing through the logic gates?
>
> it's the bit that triggers the XOR operation in your code, so in that
> way, it feeds back.
>

>> Is there a way to test if my output is an MLS?  Does the Galois LFSR
>> generate exactly the same output as a Fibonacci LFSR?
>

Perhaps a quick test of whether the sequence is an MLS could be found here:

http://www.libinst.com/mlsmeas.htm

A shifted version of the sequence is simply lined up and multiplied.
There's probably also a theoretical way of proof regarding whether
something is an MLS, but that can be left to the mathematicians.  It
just makes me wonder how you would test your own code.

Thanks, Robert.```

# Re: Generating Maximum Length Sequence using Galois LFSR - Nicholas Kinar - 2009-07-05 10:31:00

```Thanks, Steve!

> Here is the simplest one for m:
>
> 0x05eb
>
> It is probably bit reversed from the OP's set-up,
> The leadign coefficient of one is deleted from
> this constant.
>

Yes, it appears that the Galois shift register requires bit reversed
versions.

> (Unlike the OP, I always put the LS term of the polynomial
> in the lsb of a binary word, but after looking at OP's
> code it seems more compact to do it his way.)
>
> Steve

Yeah, it's kind of an interesting way of doing things.```

# Re: Generating Maximum Length Sequence using Galois LFSR - Steve Pope - 2009-07-05 16:18:00

```Nicholas Kinar  <n...@usask.ca> wrote:

>Thanks, Steve!

No problemo.

>> Here is the simplest one for m:
>>
>> 0x05eb
>>
>> It is probably bit reversed from the OP's set-up,
>> The leading coefficient of one is deleted from
>> this constant.

>Yes, it appears that the Galois shift register requires bit reversed
>versions.

The way you're doing it, yes.

>> (Unlike the OP, I always put the LS term of the polynomial
>> in the lsb of a binary word, but after looking at OP's
>> code it seems more compact to do it his way.)

>Yeah, it's kind of an interesting way of doing things.

It's good to keep track of the algebra.  For a generating
polynomial G of degree m over GF(2), you are computing the remainder

x^n mod G

for a series of values n=0, n=1,....

This remainder is a polynomial over GF(2) of (at most) degree m-1.
How you represent it in a bit field is totally up to you,
and your method is as correct as any.

Steve```

# Re: Generating Maximum Length Sequence using Galois LFSR - robert bristow-johnson - 2009-07-05 17:16:00

```On Jul 4, 8:02 pm, spop...@speedymail.org (Steve Pope) wrote:
> robert bristow-johnson  <r...@audioimagination.com> wrote:
>
> >On Jul 4, 12:55 am, Nicholas Kinar <n.ki...@usask.ca> wrote:
> >> for(int i = 0; i < L; i++)
> >>    lfsr = (lfsr >> 1) ^ (-(lfsr & 1u) & taps);
> >i dunno what the minus sign is for.  i don't think it belongs there.
>
> Looks right to me.  Read the code again.

yeah, you're right.  the code is correct.  i almost went and changed
it at Wikipedia, but thought better.

On Jul 5, 10:26 am, Nicholas Kinar <n.ki...@usask.ca> wrote:
>
>
> > your L "elements" are single bits.  the "S" of MLS is a sequence of
> > *bits*, not of words.
>
> > pick any bit of the shift register (your "lfsr") and use it.  i
> > usually pick the same bit that falls off the edge

i forgot to mention that to get a virtually zero-mean MLS signal for
the purposes of measuring impulse response (or just for some noise
that is guaranteed to be "white" in some sense of the word), then you
take that bit, (we'll call it a[n] where n is discrete time) which is
0 or 1 and create this signed MLS signal:

x[n] = (-1)^a[n]

so, it's nearly zero mean.  there are L/2 ones and L/2-1 zeros in a[n]
(because the zero state in the shift register is the only state
missing and if it were included, there would be an equal L/2 ones and
zeros, but it's not).  then there is one more -1 than +1 in x[n] over
L samples.  the mean is -1/L

this way, when you autocorrelate, you get

Rx[k] = MEAN{ x[n] * x[n+k] }

( "*" means simple multiplication.)

which is

Rx[k] = MEAN{ (-1)^a[n] * (-1)^a[n+k] }

= MEAN{ (-1)^(a[n] + a[n+k]) }

which happens to be the same as

Rx[k] = MEAN{ (-1)^(a[n] XOR a[n+k]) }

now it turns out that if you take that periodic MLS bit sequence, copy
it, circularly rotate one of the copies by k samples, and XOR the two
MLSes together, it turns out that you can show that the same recursive
generating algorithm that produces the MLS also governs that XOR
product of two with a lag in between.  except, we know for MLS, the
zero state gets mapped back to the zero state.  for an MLS, those are
the two loops; zero goes straight to zero, but any non-zero state of
the shift register goes through the whole MLS, which are all the other
non-zero states, before it gets back to your original non-zero state.
so that MEAN above for k not a multiple of L (nor zero), is -1/L.  but
the mean for k=0 or any other integer multiple of L is the mean of a
sequence of ones.

> > you need to select your taps word correctly.  first of all, for m,
> > then taps > 2048, but there is more to it.  somewhere on the web,
> > there is a resource of "primitive polynomials".  find that and use one
> > of them (ignore the least significant bit, which is always 1, the
> > other m bits of the primitive polynomial are the m bits of your taps
> > word).  for m, you could write a program that guesses at a 12-bit
> > word, try it out (see if it goes through all 2^12 - 1 non-zero
> > states), and if it fails, try another word for taps.
>
> There are a few resources on "primitive polynomials" currently on the
> web, but perhaps you are thinking about the following paper?
>
> Stahnke, W. 1973. "Primitive Binary Polynomials," Mathematics of
> Computation 27(124): 977-980.
>
> The paper gives a table of exponents of terms of primitive binary
> polynomial terms up to m = 168, where the length of the sequence is L > (2^m) - 1.  I've tested this table up to m = 30, and it seems to
> generate an MLS.

right, and you will see m+1 "coefficients" (that are either 0 or 1) in
the primitive polynomial.  but the two end bits are always 1 and the
bit-reverse sequence can be shown to also be a primitive polynomial.
so, depending on which direction you're shifting, you leave off one of
the bits.  you were shifting right so the "1" bit in the left is kept,
and you use the m most-significant bits on your word "taps" (right
justified).

> >>  The loop should repeat [after] 4095 times.
>
> > yes.  that's how you confirm it's MLS. start with a state of 0x0001
> > and count how many iterations it takes to get back to 0x0001.  if it's
> > 4095 and your "taps" word has zeros in the bits other than the least-
> > significant 12 bits, then it's an MLS
>
> You know, that's an interesting way of checking whether something is an MLS.

it's similar to a recent experience i had where i missed that the
simplest and obvious way to display the step response of a filter (IIR
or FIR), assuming you know the coefficients, is to simple bang a step
input to a simulated filter with the same coefficients.  somebody at
music-dsp had to point that out to me, where i was thinking of doing
Heaviside partial fraction expansion of the transfer function times 1/
(z-1), inverse Z transform, and getting it from there.

> Thanks, Robert.

FWIW

BTW, if you Google "Maximum Length Sequence" and look at the link
right below the one you referred to, there is a Little MLS Tutorial
that explains the math behind most of this.  it doesn't derive
primitive polynomials, but does prove how MLS can get your time-
aliased impulse response from where you can get your frequency
response.

--

r b-j                  r...@audioimagination.com

"Imagination is more important than knowledge."```

# Re: Generating Maximum Length Sequence using Galois LFSR - Steve Pope - 2009-07-05 17:40:00

```robert bristow-johnson  <r...@audioimagination.com> wrote:

>yeah, you're right.  the code is correct.  i almost went and changed
>it at Wikipedia, but thought better.

example, that one is calculating a remainder.  I do not
know anyone who uses the terminology Fibonacci LFSR,
and the polynomial in the "Fibonacci" section is reversed
from any normal usage.  There is no mathematical difference
between the sequences described in the "Fibonacci" and