I am trying to implement a function mapping shown in attached image but not working, pls correct me

%{
  my code not working ,pls correct me
question:
for i=0:1:15  output(A)
    '0000'     00
    '0001'     01
    '0010'     10
    '0011'     11
    '0100'     11
    '0101'     10
    '0110'     01
    '0111'     00
    '1000'     00
    '1001'     01
    '1010'     10
    '1011'     11
    '1100'     11
    '1101'     10
    '1110'     01
    '1111'     00
  

 if   msb-1==0;  output=i(msb-2:end);

 else  msb-1==1;   output=~i(msb-2:end);  i.e., ones comp of i(msb-2:end)
%}

N=16;                
m= 2^(log2(N)-2);    % m=4 

    A=zeros(1,m);  
    j=zeros(1,m); 
    k=zeros(1,m);
    
        for i=0:1:N-1
           msb1= bitget(i,log2(N)-1); %msb1=msb-1
          if(msb1==0)
            j(i+1)=string(dec2bin(i)).extractAfter(1);
            A(i+1)=bin2dec(j);
          else
            j(i+1)=string(dec2bin(i)).extractAfter(1);
            k(i+1)=not(j-'0');
            A(i+1)=bin2dec(k);
          end
        end

[Walter Roberson]

if   msb-1==0
else msb-1==1

msb is either 0 or 1. If msb is 1, then msb-1==0 would be true. If msb is 0 then msg-1 would be -1 which would not equal 1

Unless, that is, msb is the position of the most significant bit. If so then you would need to specify which side you are counting from. If it is a position then in the case msb-1==0 that would imply that msb is in position 1, and i(msb-2:end) would be i(1-2:end) which would be i(-1:end) which would be an error.

[VARDIREDDY]

sir ,

msb means 1st position from left .

msb-1 means 2nd position from left .

msb-2 means 3rd position from left .

let i=0:1:15

if msb-1 == 0 , output = 0,1,2,3

if msb-1 == 1 , output = 3,2,1,0

for i=0:1:15

msb-1==0 happens 2 times and

msb-1==1 happens 2 times

therefore

expected output= 0,1,2,3,3,2,1,0,0,1,2,3,3,2,1,0

i am getting following errors:

Error using bin2dec

Input argument must be a character vector, string, or cell array of character vectors.

Error in untitled7 (line 13)

A(i+1)=bin2dec(j);

pls suggest corrections in code

[Torsten]

j is an array of doubles. It is not suited as input to "dec2bin".

Answer 1

[ag]

Hi VARDIREDDY,

I understand that you need to implement the following logic:

if the 2nd MSB is 0, output= 3rd MSB to end, else output = One’s complement of 3rd MSB to end.

I’ve made a few changes to your code, as shown below:

N=16; 
m= 2^(log2(N)-2); %m = 4 
A = ['']; %empty character array for result 
for i = 0:N-1 
    binStr = dec2bin(i, m); % Convert number to binary string 
    remStr = extractAfter(binStr, 2); %extract the 3rd MSB to end string 
    if(binStr(2)=='0') % check if the 2nd MSB is 0 
        A = [A; remStr]; % append the result to answer array 
    else 
        A = [A; char(not(remStr-'0') + '0')]; %append the one’s complement of the result to the answer array 
    end 
end 
A 

A = 16×2 char array
    '00'
    '01'
    '10'
    '11'
    '11'
    '10'
    '01'
    '00'
    '00'
    '01'
    '10'
    '11'
    '11'
    '10'
    '01'
    '00'

For more details, please refer to the following documentations:

• https://www.mathworks.com/help/symbolic/sym.dec2bin.html
• https://www.mathworks.com/help/matlab/ref/extractafter.html#:~:text=extractAfter(str%2Cpat)-,newStr%20%3D%20extractAfter(str%2Cpos),-Description

Hope this helps!

Best Regards,

Aryan Gupta