Binary value conversion.

I have a binary sequence like this :

01011001

01010110

01010010

01001111

01001100

01010010

01011001

How can I convert all the 0 values to -1..? 1 values will be 1 as well.

[John D'Errico]

Please. There was no reason to ask the same question twice.

[Noman Abir]

The network issue. That's why it submitted twice. sorry.

Answer 1

[Image Analyst]

mask = yourSequence == 0; % Find all 0s
yourSequence(mask) = -1; % Convert 0s to -1s.

[Noman Abir]

The code is not working. This is a part of a binary sequence i found from an audio signal by using dec2bin command (with sampling, quantization and encoding).

[Image Analyst]

Attach your binary sequence variable in a .mat file.

[Noman Abir]

Here, g is my binary sequence and I want to convert all the 0 values to -1.

[Image Analyst]

Try this:

clc;
clear all;
close all;
bit=8;
levels=256;
Fs=8000;
t=0:1/Fs:(10-1)/Fs;
x = [-0.52, 1.2, 1.65, -1.99, -0.8, 1.96, 1.32, 0.76, 0, 0.01];
% stairs(t,x);
% title('Sample values stairs version figure');
max_x = 2;
min_x = -2;
step=(max_x-min_x)/levels;
lowest_level=min_x+(step/2);
highest_level=max_x-(step/2);
level=lowest_level:step:highest_level;
level_number=ceil((x-min_x)/step);
quantized_value=level(level_number);
encoded = x/quantized_value*level_number;
g = dec2bin(encoded);
g
% Convert from character array to string.
g = string(g)
[rows, columns] = size(g)
% Scan down row by row replacing 0 with -1
for row = 1 : rows
	thisRow = g(row)
	newRow = strrep(thisRow, "0", "-1")
	g(row) = newRow;
end
g

You'll get

g =

10×1 string array

"-11-11111-1"

"11-1-111-1-1"

"111-11-1-11"

"-1-1-1-1-1-1-1-1"

"-11-1-111-1-1"

"111111-11"

[Noman Abir]

It's showing error.

[Noman Abir]

Can you do this with if-else command..? (in for loop).

I have tried this but it's giving me first row valuse from the binary sequence.

I have added the full code, you can check it.

clc;
clear all;
close all;
bit=8;
levels=256;
Fs=8000;
t=0:1/Fs:(10-1)/Fs;
x = [-0.52, 1.2, 1.65, -1.99, -0.8, 1.96, 1.32, 0.76, 0, 0.01];
% stairs(t,x);
title('Sample values stairs version figure');
max_x = 2;
min_x = -2;
step=(max_x-min_x)/levels;
lowest_level=min_x+(step/2);
highest_level=max_x-(step/2);
level=lowest_level:step:highest_level;
level_number=ceil((x-min_x)/step);
quantized_value=level(level_number);
encoded = x/quantized_value*level_number;
z = dec2bin(encoded);

j = 1
for i=1:length(t)
      if z(i)=='0';
        num(i) = -1
    else
        num(i) = 1
        number = num(i)
         j = j+1
    end
end
number

[Image Analyst]

Try this if you want a number out instead of a string:

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format short g;
format compact;
fprintf('Beginning to run %s.m ...\n', mfilename);

bit=8;
levels=256;
Fs=8000;
t=0:1/Fs:(10-1)/Fs;
x = [-0.52, 1.2, 1.65, -1.99, -0.8, 1.96, 1.32, 0.76, 0, 0.01];
% stairs(t,x);
title('Sample values stairs version figure');
max_x = 2;
min_x = -2;
step=(max_x-min_x)/levels;
lowest_level=min_x+(step/2);
highest_level=max_x-(step/2);
level=lowest_level:step:highest_level;
level_number=ceil((x-min_x)/step);
quantized_value=level(level_number);
encoded = x/quantized_value*level_number;
z = dec2bin(encoded)
[rows, columns] = size(z);
for row = 1 : rows
	for col = 1 : columns
		if z(row, col) == '1'
			num(row, col) = 1;
		else
			num(row, col) = -1;
		end
	end
end
num

You'll get:

z =

10×8 char array

'01011110'

'11001100'

'11101001'

'00000000'

'01001100'

'11111101'

'11010100'

'10110000'

'01111111'

'10000000'

num =

-1 1 -1 1 1 1 1 -1

1 1 -1 -1 1 1 -1 -1

1 1 1 -1 1 -1 -1 1

-1 -1 -1 -1 -1 -1 -1 -1

-1 1 -1 -1 1 1 -1 -1

1 1 1 1 1 1 -1 1

1 1 -1 1 -1 1 -1 -1

1 -1 1 1 -1 -1 -1 -1

-1 1 1 1 1 1 1 1

1 -1 -1 -1 -1 -1 -1 -1

[Noman Abir]

I am getting all binary values stays in a String. So, I want the output as a String also where all 0 will be replaced by -1. The string code you provided before is not working.

The code :

clc;
clear all;
close all;
bit=8;
levels=256;
Fs=8000;
t=0:1/Fs:(10-1)/Fs;
x = [-0.52, 1.2, 1.65, -1.99, -0.8, 1.96, 1.32, 0.76, 0, 0.01];
% stairs(t,x);
title('Sample values stairs version figure');
max_x = 2;
min_x = -2;
step=(max_x-min_x)/levels;
lowest_level=min_x+(step/2);
highest_level=max_x-(step/2);
level=lowest_level:step:highest_level;
level_number=ceil((x-min_x)/step);
quantized_value=level(level_number);
encoded = x/quantized_value*level_number;
z = dec2bin(encoded);

Binary values output :

01011110

11001100

11101001

00000000

01001100

11111101

11010100

10110000

01111111

10000000

My needed output :

-11-11111-1

11-1-111-1-1

111-11-1-11

-1-1-1-1-1-1-1-1

-11-1-111-1-1

111111-11

11-11-11-1-1

1-111-1-1-1-1

-11111111

1-1-1-1-1-1-1-1

Can you help me with this..?

Remember, all the binary values stays in a Sring.

[Walter Roberson]

-11-11111-1

Exactly what is that? Is that a character vector ['-' '1' '1' '-' '1' '1' '1' '1' '1' '-' '1'] ? Is it the string array ["-1" "1" "-1" "1" "1" "1" "1" "-1"] ?

What is the overall output? Is it a character array? If so then is it acceptable if the short lines are padded with blanks?

[Noman Abir]

I am working with a voice signal in CDMA method. I got this binary output (using "dec2bin" command) from the voice after doing many applications.

This is a digital voice binary output and I have to transmit it through the channel.

[Image Analyst]

No. What we need to know is the class of the output. As you know, there are many different types of variables. We need to know which one you want because we keep getting conflicting messages from you. Here are some possibilities:

1. int32
2. uint8
3. double
4. character array
5. string array

Now, pick just one and give us the number from the list above. They're all different so make sure you pick the right one.

[Noman Abir]

Maybe, I got my solution. If I need any help about it, I will inform you guys. Thank you all.

[Noman Abir]

Ok, I need some more help from you guys.

I can't post the questions with some unspecified error. Tha'ts why I am asking questions here.

I am looking for a way to multiply every row of a matrix to different values? Let assume we have:

A=[1 2 3;

4 5 6;

7 8 9]

B=[a b c]

I am looking for a way to have:

C=[1*a 2*b 3*c;

4*a 5*b 6*c;

7*a 8*b 9*c]

Thanks in advance for your comments.

[Walter Roberson]

A=[1  2   3;
4   5   6;
7   8   9]

A = 3×3

1 2 3 4 5 6 7 8 9

●

B=[7 -2 5]

B = 1×3

7 -2 5

bsxfun(@times, A, B)

ans = 3×3

7 -4 15 28 -10 30 49 -16 45

●

Note: if you want symbolic values, syms a b c and you want C to be symbolic, then in your old release more work would have to be done. In releases since R2016b it is easier,

syms a b c
B = [a b c]

B = 

A .* B

ans = 

but in your old release, bsxfun does not support @times for symbolic values.

... But considering you are talking about communications, I do not think you need symbolic values, so I will not bother to code it up at the moment.

[Image Analyst]

Another option:

a = 2;
b = 4
c = 10;

A=[1  2   3;
4   5   6;
7   8   9]
B=[a  b  c]
C = A .* repmat(B, [size(A, 2), 1])

[Noman Abir]

Thanks. It's working.

[Noman Abir]

I Have some values in a array like this.

A = -4

-4

-4

4

4

4

I want to convert all negative values to 0 and all positive values to 1.

How can I do that.??

[Image Analyst]

A = double(A>0)

[Noman Abir]

Can you pleae provide me full code?

[Walter Roberson]

double(A>0)

is the full code, with the exception that it does not handle the possibility that A contains 0 exactly, which is something that you do not define the output for.

[Noman Abir]

That code is not working in my version. Can you guys provide me an if-else process (using for loop) where 0 and all negative values will be replaced by "0" and all positive values will be replaced by "1".

Assume the sequence as :

A = [ -3 2 0 4 -2 1 3 -4 4]

[Walter Roberson]

A = [ -3   2  0  4  -2  1  3  -4  4]

A = 1×9

-3 2 0 4 -2 1 3 -4 4

●

B = double(A>0)

B = 1×9

0 1 0 1 0 1 1 0 1

●

Is it possible that what you have is instead

AC = { '-3'  '2'  '0'  '4'  '-2'  '1'  '3'  '-4'  '4'}

AC = 1x9 cell array
    {'-3'}    {'2'}    {'0'}    {'4'}    {'-2'}    {'1'}    {'3'}    {'-4'}    {'4'}

B = double(str2double(AC)>0)

B = 1×9

0 1 0 1 0 1 1 0 1

●

[Walter Roberson]

AP = [ -3   2  0  4  -2  1  3  -4  4];
A = AP;
for K = 1 : numel(A)
    if A(K) <= 0
        A(K) = 0;
    else
        A(K) = 1;
    end
end
disp(A)

     0     1     0     1     0     1     1     0     1

A = AP;
A = double(A>0);
disp(A)

     0     1     0     1     0     1     1     0     1

[Noman Abir]

Thank you. I have got my solution.

[Noman Abir]

I need some more help from you @Walter Roberson & @Image Analyst.

I have added 2 different MATLAB files where "receive_cdm.m" is my main code and "y_send.mat" is a file I am getting some values from it and loaded it in main code section.

If you look at the code then I have assigned 1 parameter c1. Forget about other parameters.

The following tasks are explained in the code.

I have done it with proper MATLAB commands and equations.

But, I am getting error when I calculate it manually and matching it with MATLAB calculations. (What I should get and what I am getting from MATLAB)

Can anyone check my code, please..?

It would be very helpful for me.

Answer 2

[Walter Roberson]

num = (z-'0')*2-1;