Assume the following two PRN sequences: G1 = (1 1 1 0]: G2 = (1 0 1 1]: PRN sequ

Question

Assume the following two PRN sequences: G1 = (1 1 1 0]: G2 = (1 0 1 1]: PRN sequences have ideal correlation properties. The correlation between two PRN codes is calculated utilizing the formula below: R^(a, b) (n) = 1/N sigma ^i = n - 1_i = 0 x^(a) (i) middot x^(b) (i + n) where, denotes the PRN sequence for a and b, N is the length of PRN codes x^(a), x^(b), R^(a, b) (n) is the correlation function between PRN codes x^(a), x^(b) for shift n, n is the circular shift of bits for PRN code x^(b). Calculate the following: (a) Calculate R^(a, b) (n) when x = G1 and x^(b) = G1 for n = 0 to N - 1. (b) Calculate R^(a, b) (n) when x^(a) = G1 and x^(b) = G2 for n = 0 to N - 1.

Explanation / Answer

Solution:

a) R(a,b)(n) when x(a) = G1 and x(b) = G1 for n=0 to N-1

we will solve this problem using equation 1 , this is straight forward,just put the values in the equation and solve it

here, G1 = [1 1 1 0]

R(a,b)(0) = (1/4) [x(0)x(0) + x(1)x(1) + x(2)x(2) + x(3)x(3) ] = (1/4) [ 1x1 + 1x1 + 1x1 + 0x0 ] = 3/4

R(a,b)(1) = (1/4) [x(0)x(1) + x(1)x(2) + x(2)x(3) + x(3)x(4) ] = (1/4) [ 1x1 + 1x1 + 1x0 + 0x0 ] = 2/4

R(a,b)(2) = (1/4) [x(0)x(2) + x(1)x(3) + x(2)x(4) + x(3)x(5) ] = (1/4) [ 1x1 + 1x0 + 1x0 + 0x0 ] = 1/4

R(a,b)(3) = (1/4) [x(0)x(3) + x(1)x(4) + x(2)x(5) + x(3)x(6) ] = (1/4) [ 1x0 + 1x0 + 1x0 + 0x0 ] = 0

here, x(4), x(5) ,x(6) are 0,0,0 respectively obtained after appending zeros to sequence G1 which does not affect the sequence.

Thus, the new output sequence is [3/4 2/4 1/4 0]

B) R(a,b)(n) when x(a) = G1 and x(b) = G2 for n=0 to N-1

we will solve this problem using equation 1 , this is straight forward,just put the values in the equation and solve it

here, G1 = [ 1 1 1 0 ]

G2 = [1 0 1 1]

R(a,b)(0) = (1/4) [x1(0)x2(0) + x1(1)x2(1) + x1(2)x2(2) + x1(3)x2(3) ] = (1/4) [ 1x1 + 1x0 + 1x1 + 0x1 ] = 2/4

R(a,b)(1) = (1/4) [x1(0)x2(1) + x1(1)x2(2) + x1(2)x2(3) + x1(3)x2(4) ] = (1/4) [ 1x0 + 1x1 + 1x1 + 0x0 ] = 2/4

R(a,b)(2) = (1/4) [x1(0)x2(2) + x1(1)x2(3) + x1(2)x2(4) + x1(3)x2(5) ] = (1/4) [ 1x1 + 1x1 + 1x0 + 0x0 ] = 2/4

R(a,b)(3) = (1/4) [x1(0)x2(3) + x1(1)x2(4) + x1(2)x2(5) + x1(3)x2(6) ] = (1/4) [ 1x1 + 1x0 + 1x0 + 0x0 ] = 1/4

here, x(4), x(5) ,x(6) are 0,0,0 respectively obtained after appending zeros to sequence G1 which does not affect the sequence.

Thus the new output sequence is [2/4 2/4 2/4 1/4]

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