Talk:Gold code

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This page was revised 10 November 2006 by 59.182.20.199.

The revision includes the new statement:

This is not true. Look at Table 3-I and Figure 3-10 in IS-GPS-200D, a 974 kB file available at:

http://www.navcen.uscg.gov/gps/geninfo/default.htm

http://www.navcen.uscg.gov/gps/geninfo/IS-GPS-200D.pdf

This shows a way to form the C/A codes (Gold codes) by tapping 2 bit locations in the G2 register and modulo 2 summing the result. There are "10 choose 2", or 45, ways to do this. However, Table 3-I shows only 36 codes. The reason is that the other 9 are unbalanced (typically 480 bits of one value and 543 of the other). This is why the pseudorandom noise (PRN) codes 34 and 37 are identical.

Recommend removing the last sentence from this Wikipedia entry.

[ Being new to Wikipedia participation, I take to heart the admonition to avoid arbitrarily making a change, and the recommendation to first discuss it with the previous editor. ]

Banchang (talk) 22:37, 18 November 2007 (UTC)[reply]

The link to the Xilinx app note XAPP217 is dead and I am unable to find the note anywhere on xilinx.com.

MJ (talk) 17:49, 23 August 2010 (UTC)[reply]

I don't think i understand this enough to do it myself; but IMO it would be good if the article had an example of how to construct Gold codes and how they are used, using short enough lengths that people can easily keep up with. --TiagoTiago (talk) 01:16, 24 May 2013 (UTC)[reply]

Good idea. Let's try to make a set of small Gold codes. According to this there are two MLFSRs of length 4, generating sequences of length 15, so if we XOR different phases of these we should get length-15 Gold codes. The two shift register polynomials are 9 and C, which I take to mean 8 + 1 and 8 + 4, indicating XOR of the last of four bits (the 8) with either the first (the 1) or the third (the 4). Start with a 1 in the register, and repeatedly left shift it, adding in the new XOR-generated bit on the right, to get these sequences of register contents:

      9    C
 1 0001 0001
 2 0011 0010
 3 0111 0100
 4 1111 1001
 5 1110 0011
 6 1101 0110
 7 1010 1101
 8 0101 1010
 9 1011 0101
10 0110 1011
11 1100 0111
12 1001 1111
13 0010 1110
14 0100 1100
15 1000 1000
16 0001 0001

That looks good, since we got a period of 15 for both.

So taking the right-most bits from 1 to 15 as the base sequences, they are 1111010110010001 and 1001101011110001. Now let's XOR those together, shifting the second one leftward to every possible phase; and see how balanced they are by counting 1s of the 15 bits:

  1111010110010001
+ 1001101011110001
= 0110111101100000 7 ones

  1111010110010001
+ 0011010111100011
= 1100000001110010 6 ones

  1111010110010001
+ 0110101111000110
= 1001111001010111 10 ones

  1111010110010001
+ 1101011110001100
= 0010001000011101 6 ones

  1111010110010001
+ 1010111100011001
= 0101101010001000 6 ones

Etc. But before I do all 15 phases, someone should check me on this, and maybe decide if it's worth completing... We expect to get 7 or 8 ones on about half of them, so maybe I'm off some place? Or maybe I just need to keep going? Or maybe at this short length there are no sequences that satisfy "such that their absolute cross-correlation is less than or equal to ..."? Dicklyon (talk) 03:25, 24 May 2013 (UTC)[reply]

I'm just "parking" here a Gold code sequence found in an off-line source for later analysis and possible integration:

0x6, 0x9, 0x0, 0xA, 0xE, 0xC, 0x7, 0xC, 0xD, 0x2, 0x1, 0x5, 0xD, 0x8, 0xF, 0x9, 0xA, 0x4, 0x2, 0xB, 0xB, 0x1, 0xF, 0x3, 0x4, 0x8, 0x5, 0x7, 0x6, 0x3, 0xE

--Matthiaspaul (talk) 11:27, 3 March 2018 (UTC)[reply]

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