Thread: Simple Manchester decoder

Hi,

First post so please be gentle . I am trying to write a very very simple manchester decoder type thingy.

I have a 16-bit string that i want to decode. I understand to do this properly there is the requirement for taking into account the clock etc. However, to start with all i want to do is actually decode a 16-bit sample.

so basically i have a 16-bit binary string, and using the rule of (IEEE802.3) when encoded 1's become 10 and 0's become 01 because of the transition. so my bit string is:

1001101001011001

and i want the end result to be decoded from this.

Does this make sense?

Thanks

Weez

I think basically you just wait for a transition up or down.
I don't think you would normally do this in a program it would be more hardware attached to a transmission line.

If you were in a program say maybe you just read one bit, see if it a 1 or zero, read the next bit, if it is a 1-0 transition out put a 1,
if it is a 0-1 out put a 0. If there is no transition then out put what you previously output. If there is no transition stay as you were?

Thanks for the replies. i shall look into the decoding chips when i get further along. I shall be attacking the actual recieving of data from an external source through GPIO a bit further along. for now i want to get the interpretation of the data written up, this takes place after the transmission is just a stream of 1's and 0's... I have a simple c program that ive quickly written up to hopefully demonstrate this step in my program... however i cannot get it toi work:

Code:

#include <stdio.h>





    int Data[] = {1,0,0,1};  //Data array
        
    int decode[] = {0,0};    //decoded data array








int main()
{


/******************************************************************************
    Initialis/Print Data array
******************************************************************************/    


    int a=0;


    printf("\nBinary Sample is:\n");


    for(a=0; a<4; a++)
    {
        
        printf("%d", Data[a]);
    }    
    printf("\n");


/*******************************************************************************
    Initialise/Print Decode array
*******************************************************************************/
    
    int b=0;


    printf("\nDecoded data array set to 0:\n");


    for(b=0; b<2; b++)
    {
        printf("%d", decode[b]);
    }    
    printf("\n\n");




/*******************************************************************************
    Test for bit patterns
*******************************************************************************/


    int c;
    int d;
    a=0;    
    b=0;


    printf("Testing bit pattern.....\n");


    for(c=0;c<4;c++)
    {
        printf("\nProcessing instance bits %d & %d... \n",c++,c);        
        printf("\nData will be put into decode @ %d \n",b);


        if(Data[c] == 1 && Data[c++] == 0)
        {
        printf("\ntest 1 PASS\n");
        decode[b] = 0 ;
        printf("\nData bits %d & %d converted\n", c++,c);
        printf("\nConverted bit is %d \n",decode[b]);
        printf("\nNext TEST\n");    
        b+1;


        }
        else if(Data[c] == 0 && Data[c++] == 1)
        {


        printf("stage 2 PASS\n");
        decode[b] = 1;
        printf("\nData bits %d & %d converted\n", c,c++);
        printf("\nConverted bit is %d \n",decode[b]);
        b+1;
        printf("\nNext TEST\n");
        }
        else (printf("Error"));
        
    }


    printf("\nDecoded data is %d",decode[b]);
    
    for(b=0; b<2; b++)
    {
        
        printf("%d", decode[b]);
    }    
    printf("\n\n");


    return 0;
}

P.s sorry for the length of time between replies

cheers

Originally Posted by weez

however i cannot get it toi work:

Rule #1: "It doesn't work" is not a useful problem description. Tell us what exactly went wrong (compiler errors, wrong output, ...)

Rule #2: Always compile with warnings turned on. Then you should get something like:

Code:

$ make foo
cc -Wall -Wextra -ggdb3   -c -o foo.o foo.c
foo.c: In function ‘main’:
foo.c:76:60: warning: operation on ‘c’ may be undefined [-Wsequence-point]
foo.c:84:52: warning: operation on ‘c’ may be undefined [-Wsequence-point]
foo.c:87:9: warning: statement with no effect [-Wunused-value]
foo.c:97:54: warning: operation on ‘c’ may be undefined [-Wsequence-point]
foo.c:99:9: warning: statement with no effect [-Wunused-value]
foo.c:66:9: warning: unused variable ‘d’ [-Wunused-variable]
cc   foo.o   -o foo

Tim told you already about lines 87 and 99.

Lines 76, 84 and 97 are about your usage of "c++".

You can't use both "c++" and "c" as arguments in your printf()-calls because the order in which the arguments are evaluated isn't defined.
If the compiler evaluates "c++" first, then printf() gets the values "c" and "c + 1".
But if the compiler evaluates "c" first, then printf() gets the values "c" and "c".

Generally, I think you have misunderstood how "c++" works and in which situations you can use it.

Bye, Andreas