Real value to binary

How can i convert my real value to binary digits?

For example; i have values within [0,25] range and 15 binary digits to represent a variable.(4 variables totally)

How can i do that and also reverse of that?

[Walter Roberson]

Fixed point or floating point? And to confirm, the input values are not necessarily integer?

25 requires 5 bits; should we assume 5 bits of integer and 10 bits of fraction?

[b]

it will be 3 places after the decimal point.

so i think that i will have 10^3*25 equal size.

that is equal to

2^14 < 25000 < 2^15. so i will have 15 bits. ??

[Walter Roberson]

3 decimal places requires 10 bits to resolve (2^(-10) = 1/1024)

You get slightly better resolution if you use 2^10 than if you use 10^3. It depends though on whether resolution is your goal or if "3 decimal places" is your goal.

Answer 1

[Sean de Wolski]

doc dec2bin

Answer 2

[Honglei Chen]

Are you asking things like dec2bin and bin2dec?

x = 20;
dec2bin(20,15)

Answer 3

[Walter Roberson]

dec2bin(round(Values * 2^10), 15) - '0'

[b]

will i enter my number to "Values"?

And within the range [0,15]

??

[Walter Roberson]

Yes, enter you number in Values.

If your range is up to "just less than" 16, then

dec2bin(round(Values * 2^11), 15) - '0'

For 2.154 the result would be

001000100111011

The first 4 bits (0010) are the 2, and the last 11 bits approximate 0.154

[b]

Superb, my friend!

Thanks a lot :)

[b]

Sorry for that but i want to ask another thing..

my range is between 0 and 25.

when i put e.g. 2.5 to this formula,this gives me wrong result

[Walter Roberson]

If you go up to 25 then you need the 2^10 not 2^11

Answer 4

[b]

Actually, if i put 2.154 as a real value within [0,25]

it should not give me 000000000000010. it represents for my range 0.001 something like that.