problem in code for 16 qam

Hi i wrote a code to seperate a binary signal to groups of 4 bits and maping them to a complex number,but didnt work and i cant find why. Please help!

clear all
close all
clc

a = [1 1 1 1 0 0 0 0 1 1 1 1]

k=1
for i=1:4:16
  if a(i:i+3)=[0000]
    b(k)=1-3j
  else a(i:i+3)=[1111]
    b(k)=3+3j
  end
  k=k+1
end

Answer 1

[Vieniava]

add spaces in binary vectors:

a(i:i+3)=[0 0 0 0]
...
a(i:i+3)=[1 1 1 1]

[Vieniava]

secondly, your a vector is too short.

Look: when i=16 yout try to acces to (i+3)=19th element of a.

[Vieniava]

next, after "if" you should use isequal(a(i:i+3), [0 0 0 0])

Answer 2

[jordi10]

that was helpfull thanks!!!!

[Vieniava]

the best "thanks" for hint-giver here is to single click on "Accept Answer"

Answer 3

[jordi10]

can you please help me with the demodulation? here is my code,the problem is with the equality of matrix elements

clear all
close all
clc

a=[1 1 1 1 0 0 0 0 1 1 1 1]

%------ 16QAM MODULATION ------%
k=1
for  i=1:4:length(a)
  if a(i:i+3)==[0 0 0 0]
    b(k)= -3-3j
  elseif a(i:i+3)==[0 0 0 1]
    b(k)= -3-1j
  elseif a(i:i+3)==[0 0 1 0]
    b(k) = -3+3j
  elseif a(i:i+3)==[0 0 1 1]
    b(k) = -3+1j
  elseif a(i:i+3)==[0 1 0 0]
    b(k) = -1-3j    
  elseif a(i:i+3)==[0 1 0 1]
    b(k) = -1-1j
  elseif a(i:i+3)==[0 1 1 0]
    b(k) = -1+3j
  elseif a(i:i+3)==[0 1 1 1]
    b(k) = -1+1j    
  elseif a(i:i+3)==[1 0 0 0]
    b(k) = 3-3j
  elseif a(i:i+3)==[1 0 0 1]
    b(k) = 3-1j
  elseif a(i:i+3)==[1 0 1 0]
    b(k) = 3+3j    
  elseif a(i:i+3)==[1 0 1 1]
    b(k) = 3+1j
  elseif a(i:i+3)==[1 1 0 0]
    b(k) = 1-3j
  elseif a(i:i+3)==[1 1 0 1]
    b(k) = 1-1j
  elseif a(i:i+3)==[1 1 1 0]
    b(k) = 1+3j
  elseif a(i:i+3)==[1 1 1 1]
    b(k) = 1+1j
  end
  k = k+1
end

%------ 16QAM DEMODULATION ------%

v = 1
k = 1
for z=1:1:length(b)
  if b(k) == [-3-3j]
    x(v) = [0 0 0 0]
  elseif b(k) == [-3-1j]
    x(v) = [0 0 0 1]
  elseif b(k) == [-3+3j]
    x(v) = [0 0 1 0]
  elseif b(k) == [-3+1j]
    x(v) = [0 0 1 1]
  elseif b(k) == [-1-3j]
    x(v) = [0 1 0 0]    
  elseif b(k) == [-1-1j]
    x(v) = [0 1 0 1]
  elseif b(k) == [-1+3j]
    x(v) = [0 1 1 0]
  elseif b(k) == [-1+1j]
    x(v) = [0 1 1 1]    
  elseif b(k) == [3-3j]
    x(v) = [1 0 0 0]
  elseif b(k) == [3-1j]
    x(v) = [1 0 0 1]
  elseif b(k) == [3+3j]
    x(v) = [1 0 1 0]    
  elseif b(k) == [3+1j]
    x(v) = [1 0 1 1]
  elseif b(k) == [1-3j]
    x(v) = [1 1 0 0]
  elseif b(k) == [1-1j]
    x(v) = [1 1 0 1]
  elseif b(k) == [1+3j]
    x(v) = [1 1 1 0]
  elseif b(k) == [1+1j]
    x(v) = [1 1 1 1]
  end
  k = k + 1
  v = v + 1
end

[Nouman Khan Safi]

if resolved this issue you can check it out

clear all
close all
clc
a=[1 1 1 1 0 0 0 0 0 0 1 0];
fprintf('Message Signal:\n')
disp(a)
%------ 16QAM MODULATION ------%
k=1;
for  i=1:4:length(a)
  if a(i:i+3)==[0 0 0 0]
    b(k)= -3-3j;
  elseif a(i:i+3)==[0 0 0 1]
    b(k)= -3-1j;
  elseif a(i:i+3)==[0 0 1 0]
    b(k) = -3+3j;
  elseif a(i:i+3)==[0 0 1 1]
    b(k) = -3+1j;
  elseif a(i:i+3)==[0 1 0 0]
    b(k) = -1-3j;    
  elseif a(i:i+3)==[0 1 0 1]
    b(k) = -1-1j;
  elseif a(i:i+3)==[0 1 1 0]
    b(k) = -1+3j;
  elseif a(i:i+3)==[0 1 1 1]
    b(k) = -1+1j;    
  elseif a(i:i+3)==[1 0 0 0]
    b(k) = 3-3j;
  elseif a(i:i+3)==[1 0 0 1]
    b(k) = 3-1j;
  elseif a(i:i+3)==[1 0 1 0]
    b(k) = 3+3j;    
  elseif a(i:i+3)==[1 0 1 1]
    b(k) = 3+1j;
  elseif a(i:i+3)==[1 1 0 0]
    b(k) = 1-3j;
  elseif a(i:i+3)==[1 1 0 1]
    b(k) = 1-1j;
  elseif a(i:i+3)==[1 1 1 0]
    b(k) = 1+3j;
  elseif a(i:i+3)==[1 1 1 1]
    b(k) = 1+1j;
  end
  k = k+1;
end
fprintf('Modulated Signal \n')
disp(b)

%------ 16QAM DEMODULATION ------%
v = 1;
k = 1;
r = [];
for z=1:1:length(b)
  if b(k) == [-3-3j]
    r(length(r)+1:length(r)+4) = [0 0 0 0 ];
  elseif b(k) == [-3-1j]
    r(length(r)+1:length(r)+4) = [0 0 0 1 ];
  elseif b(k) == [-3+3j]
    r(length(r)+1:length(r)+4) = [0 0 1 0 ];
  elseif b(k) == [-3+1j]
    %x(v) = [0 0 1 1]
    r(length(r)+1:length(r)+4) = [0 0 1 1 ];
  elseif b(k) == [-1-3j]
    r(length(r)+1:length(r)+4) = [0 1 0 0 ];
  elseif b(k) == [-1-1j]
    r(length(r)+1:length(r)+4) = [0 1 0 1 ];
  elseif b(k) == [-1+3j]
    r(length(r)+1:length(r)+4) = [0 1 1 0 ];
  elseif b(k) == [-1+1j]
    r(length(r)+1:length(r)+4) = [0 1 1 1 ];
  elseif b(k) == [3-3j]
    r(length(r)+1:length(r)+4) = [1 0 0 0 ];
  elseif b(k) == [3-1j]
    r(length(r)+1:length(r)+4) = [1 0 0 1 ];
  elseif b(k) == [3+3j]
    r(length(r)+1:length(r)+4) = [1 0 1 0 ];
  elseif b(k) == [3+1j]
    r(length(r)+1:length(r)+4) = [1 0 1 1 ];
  elseif b(k) == [1-3j]
    r(length(r)+1:length(r)+4) = [1 1 0 0 ];
  elseif b(k) == [1-1j]
    r(length(r)+1:length(r)+4) = [1 1 0 1 ];
  elseif b(k) == [1+3j]
    r(length(r)+1:length(r)+4) = [1 1 1 0 ];
  elseif b(k) == [1+1j]
    r(length(r)+1:length(r)+4) = [1 1 1 1];
  end
  k = k + 1;
  v = v + 1;
end
fprintf('Demodulated Singal:\n')
disp(r)

Answer 4

[Walter Roberson]

Encoding, everything at once:

LuT = [-3 -1 +3 +1];

A = reshape(A, 4, []);
b = LuT(A(1,:)*2+A(2,:)) + j*LuT(A(3,:)*2+A(4,:));

Decoding, everything at once:

RLut = [0 0; nan nan; 0 1; nan nan; 1 1; nan nan; 1 0].';

x = reshape(RLut(:,[real(b);imag(b)] + 4),4,[]).';

Answer 5

[Hrasek]

do you know modem.qammod function? Check it out