Post by thread:[PIC]: fixed point binary representation

Hi there, Please light me, I have headache trying to understand why: 3.578(dec) = 11.1001 0011 1111 0111 1100 ...(bin) sorry for bothering you with such stick, but searching at piclist I found only theory which I can't match to above example. Thank's again, Vasile -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

The binary number after the decimal point is a always represents a fraction of 1. For a 20bit accurate representation of 0.578 just multiply 0.578 * 2^20 (that's 2 raised to the power of 20) = 606076.928 which truncated and converted to binary gives you 1001 0011 1111 0111 1100. For a 10bit accurate result 0.578 * 2^10 = 591.872 which truncated in binary is 1001 0011 11 So hopefully you see that ANY fraction can be represented by a number of bits (N) in a binary word so long as you (and you code) understand that 2^N = 1 Does that help? Pete > {Original Message removed}

Pete, it helps more than you imagine! Finaly I've undestand this brilliant but un-exemplified theory: A number in binary form is represented by a sum of powers of two: 31 30 0 N = I * 2 + I * 2 + ... I * 2 + Q32.32 31 30 0 -1 -2 -32 + F * 2 + F * 2 + ... F * 2 31 30 0 But what you have written this correct? > For a 10bit accurate result 0.578 * 2^10 = 591.872 which truncated in binary > is 1001 0011 11 seems it depends of justifying, so will be: 0010 0100 1111 , isn't it ? Thank's, Vasile On Thu, 15 Mar 2001, Peter Betts wrote: {Quote hidden}> The binary number after the decimal point is a always represents a fraction > of 1. > > For a 20bit accurate representation of 0.578 just multiply 0.578 * 2^20 > (that's 2 raised to the power of 20) = 606076.928 which truncated and > converted to binary gives you 1001 0011 1111 0111 1100. > > For a 10bit accurate result 0.578 * 2^10 = 591.872 which truncated in binary > is 1001 0011 11 > > So hopefully you see that ANY fraction can be represented by a number of > bits (N) in a binary word so long as you (and you code) understand that 2^N > = 1 > > Does that help? > > Pete > > > > > {Original Message removed}

> > For a 10bit accurate result 0.578 * 2^10 = 591.872 which > > truncated in binary is 1001 0011 11 > seems it depends of justifying, so will be: 0010 0100 1111 , > isn't it ? Err? No. I don't think so it is still XX.1001 0011 11 .1001 0011 11 can be expressed as (2^-1) + (2^-4) + (2^-7) + (2^-8) + (2^-9) + (2^-10) {as negative index bits -1,-4,-7,-8,-9 and -10 are set} If you calculate (2^-1) + (2^-4) + (2^-7) + (2^-8) + (2^-9) + (2^-10) = 0.5771 Which for a 10bit result is as close to 0.578 as you can get. Using more bits in the fraction will give you greater accuracy and you will tend towards your ideal of 0.578 with decreasing error. Pete -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

Thank's very much ! Vasile On Thu, 15 Mar 2001, Peter Betts wrote: {Quote hidden}> > > For a 10bit accurate result 0.578 * 2^10 = 591.872 which > > > truncated in binary is 1001 0011 11 > > > seems it depends of justifying, so will be: 0010 0100 1111 , > > isn't it ? > > > Err? No. I don't think so it is still XX.1001 0011 11 > > .1001 0011 11 can be expressed as > > (2^-1) + (2^-4) + (2^-7) + (2^-8) + (2^-9) + (2^-10) > > {as negative index bits -1,-4,-7,-8,-9 and -10 are set} > > If you calculate > (2^-1) + (2^-4) + (2^-7) + (2^-8) + (2^-9) + (2^-10) = 0.5771 > > Which for a 10bit result is as close to 0.578 as you can get. > > Using more bits in the fraction will give you greater accuracy and you will > tend towards your ideal of 0.578 with decreasing error. > > Pete > > -- > http://www.piclist.com hint: The PICList is archived three different > ways. See http://www.piclist.com/#archives for details. > > -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

Perhaps the following example will help you understand how fractions are expressed in binary. I will begin with something you are hopefully already familiar with, which is straight binary integer representation. I will then extend the example to include fractional representation as well. 10000000 = 128 01000000 = 64 00100000 = 32 00010000 = 16 00001000 = 8 00000100 = 4 00000010 = 2 00000001 = 1 .10000000 = 1/2 (128/256) .5 .01000000 = 1/4 (64/256) .25 .00100000 = 1/8 (32/256) .125 .00010000 = 1/16 (16/256) .0625 .00001000 = 1/32 (8/256) .03125 .00000100 = 1/64 (4/256) .015625 .00000010 = 1/128 (2/256) .0078125 .00000001 = 1/256 (1/256) .00390625 Now let's decode 11.10010011 Since 11. = 3 we know the number is 3.something To determine the "something" fractional part, consider the fraction to be (128+16+2+1)/256 = 147/256 = .57421875 maximum error is 1/256 = .00390625 so the truncated 8 bit fraction will have an actual value somewhere between . 57421875 and .578125 which is an error of .7% ********* Here is a shortcut way to convert 3.578 to binary: first write the 3 as 11b multiply .578*256 and you get 147.968 convert 147 to binary and you get 10010011b combine the whole number and fraction parts to get 11.10010011b If you need more digits, convert the .968 remainder the same way: .988*256=247.808 247 converts to 128+64+32+16+4+2+1 = 11110111 11.10010011 111101111 Need even more digits? Convert the .808 remainder the same way: .808*256=206.848 206 converts to 128+64+8+4+2 = 11001110 11.10010011 11110111 11001110 etc. I hope this helps. Fr. Tom McGahee {Original Message removed}