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chaaru datta
chaaru datta el 9 de Nov. de 2023
Movida: Dyuman Joshi el 9 de Nov. de 2023
Hello all, I am working on research work related to GSSK (generalized space shift keying) in wireless communication.
In the GSSK scheme, out of total number of antennas at transmitter (Nt), only nt transmit antennas remain active during a single transmission.
My aim is to design a vector for transmitted signals (N).
The relationship between N, Nt and nt is given as:
Suppose the Nt = 8 and nt =3 then number of transmitted signals i.e., N is 32.
I am able to make only 28 unique vectors but not 32. The 28 unique vectors are shown below:
x_1 = al*[1 1 0 0 0 0 0 0 ]'; x_2 = al*[1 0 1 0 0 0 0 0 ]'; x_3 = al*[1 0 0 1 0 0 0 0 ]'; x_4 = al*[1 0 0 0 1 0 0 0 ]';
x_5 = al*[1 0 0 0 0 1 0 0 ]'; x_6 = al*[1 0 0 0 0 0 1 0 ]'; x_7 = al*[1 0 0 0 0 0 0 1 ]';
x_8 = al*[0 1 1 0 0 0 0 0 ]'; x_9 = al*[0 1 0 1 0 0 0 0 ]'; x_10 = al*[0 1 0 0 1 0 0 0 ]'; x_11 = al*[0 1 0 0 0 1 0 0 ]'; x_12 = al*[0 1 0 0 0 0 1 0 ]';
x_13 = al*[0 1 0 0 0 0 0 1]';
x_14 = al*[0 0 1 1 0 0 0 0]'; x_15 = al*[0 0 1 0 1 0 0 0]'; x_16 = al*[0 0 1 0 0 1 0 0]'; x_17 = al*[0 0 1 0 0 0 1 0]'; x_18 = al*[0 0 1 0 0 0 0 1]';
x_19 = al*[0 0 0 1 1 0 0 0]'; x_20 = al*[0 0 0 1 0 1 0 0]'; x_21 = al*[0 0 0 1 0 0 1 0]'; x_22 = al*[0 0 0 1 0 0 0 1]';
x_23 = al*[0 0 0 0 1 1 0 0]'; x_24 = al*[0 0 0 0 1 0 1 0]'; x_25 = al*[0 0 0 0 1 0 0 1]';
x_26 = al*[0 0 0 0 0 1 1 0]'; x_27 = al*[0 0 0 0 0 1 0 1]';
x_28 = al*[0 0 0 0 0 0 1 1]';
where 1 indicates the corresponding antenna being activated out of total Nt antennas.
My query is that if N = 32 for this example, then how can we achieve 32 unique vectors.
Any help in this regard will be highly appreciated.
  2 comentarios
Dyuman Joshi
Dyuman Joshi el 9 de Nov. de 2023
Editada: Dyuman Joshi el 9 de Nov. de 2023
You can not.
There are only 28 unique possible combinations when numbers from [1 8] are taken two at a time, without repetitions -
Though I do not know why you have taken combinations with two at a time.
n=8;
y = nchoosek(1:n,2);
size(y)
ans = 1×2
28 2
disp(y)
1 2 1 3 1 4 1 5 1 6 1 7 1 8 2 3 2 4 2 5 2 6 2 7 2 8 3 4 3 5 3 6 3 7 3 8 4 5 4 6 4 7 4 8 5 6 5 7 5 8 6 7 6 8 7 8
chaaru datta
chaaru datta el 9 de Nov. de 2023
Movida: Dyuman Joshi el 9 de Nov. de 2023
@Dyuman Joshi Thank u sir for your response.

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Walter Roberson
Walter Roberson el 9 de Nov. de 2023
Movida: Walter Roberson el 9 de Nov. de 2023
(8)
(3)
is 8!/(3! 5!) = 56. log2 of that is between 5 and 6. floor of that is 5. So your formula calculates N = 5, not N = 32
Taking 8 objects 3 at a time without respecting order can be represented as an ordered list of bits of length 8 in which exactly 3 bits are set. But your code uses bit lists in which only 2 bits are set instead of 3.
  1 comentario
chaaru datta
chaaru datta el 9 de Nov. de 2023
@Walter Roberson Thank u sir for your response....Now I had edited my mistake

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