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How to implement this in matlab?

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tojeena

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n=1011(binary) 11 in decimal and b= 3(decimal) and finally getting answer as 011(binary).the b LSB of n to produce the subexponential code of n.How to implement it in matlab?
n=1011 b=3 z=1|011 ans=011.What will be the program for this in matlab .?
 

n=1011(binary) 11 in decimal and b= 3(decimal) and finally getting answer as 011(binary).the b LSB of n to produce the subexponential code of n.How to implement it in matlab?
n=1011 b=3 z=1|011 ans=011.What will be the program for this in matlab .?

Hi tojeena, the solution for your problem is you should convert all the number you want to compute either to binary or to decimal and then add all the number finally convert the answer based on your requirements...

to convert decimal to binary use dec2bin(input) and vice versa...

Hope this may be helpful....
 

Thank you for your reply.Actually the input is wavelet transformed coefficients bit plane coded,into binary.:!:
 

Hi tojeena , you get the output as 011 may i know on what basis you have calculated it or there are any equation

will you elaborate your problem in order to find the solution...
 

[its the coding for subexponential codes
 

Hi tojeena,
I need the logic or computation example or equation behind that subexponential, or else just send me the code that produce the output as 011...
 

Thank you so much.Expecting your reply.
 

The reference for the algorithm of subexponential encoding and decoding is from the link https://upcommons.upc.edu/pfc/bitstream/2099.1/9926/1/PFC_Marcial_Clotet.pdf page 24(encoding) and 32(decoding).

Thank you..

Hi tojeena,

I have written a code on the idea of the reference paper, and it is give below,

clear all;
clc;
h1 = input('enter the value of k= ', 's');
h2= input('enter the value of n= ', 's');
k = str2num(h1);
n = str2num(h2);
compute = power(2,k);
display(compute);
g=log2(n);
if (n>compute)
b=floor(g);
u=b-k+1;
end
display(b);
display(u);

I have just initialize the code for you
on this basis of above code you develop further on your own idea...

All the best in advance...

Hope It may helpful to you...
 
Thank you so much..I will check this.
 

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