Post by thread:Doing ratio or percentage

On Tue, 28 Sep 1999 11:43:13 +0100 Michael Rigby-Jones <spam_OUTmrjonesTakeThisOuTNORTELNETWORKS.COM> writes: >These are 8 bit values right? This is pretty simple, you need to >steal a >multiply routine from somewhere that does 8bit * 8bit with a 16 bit >result. >The Microchip routine is attached (no flames please...it's tiny) The >idea >is: > >C = (A*B)/256 > >The 256 number is handy as no division is actually needed, you just >take the >most significant byte of the 16 bit result. > >Hope this is of some use. > > >Mike Rigby-Jones > I'm not sure if this is what the original poster wanted, but IF it is, it's similar to some code I wrote to have a grand master control in a lighting control system control the outputs of the A/D values from the original channel pots. There we have individual channel pots giving A/D values from 0 to 255 (0x00 to 0xff). The master pot gives the same result. We want the master pot to scale the channel pots. Turns out simple enough... Out = Chan * Master/255 Note that it's divided by 255, not 256. Otherwise, if both channel and master were 255, we'd get 254 as an output, when we want 255. Dividing by 255 is certainly not as easy as dividing by 256 (throwing away the bottom 8 bits). Perhaps with rounding that would work (divide by 256 and if bit 7 of the remainder is 1, increment the result). What I did was to check bit 7 of the master pot value. If it's 1, I added 1 to the result of the divide by 256. Works great! Harold ___________________________________________________________________ Get the Internet just the way you want it. Free software, free e-mail, and free Internet access for a month! Try Juno Web: dl.http://www.juno.com/dynoget/tagj.

On Tue, 28 Sep 1999, Harold Hallikainen wrote: {Quote hidden}> On Tue, 28 Sep 1999 11:43:13 +0100 Michael Rigby-Jones > <.....mrjonesKILLspam@spam@NORTELNETWORKS.COM> writes: > >These are 8 bit values right? This is pretty simple, you need to > >steal a > >multiply routine from somewhere that does 8bit * 8bit with a 16 bit > >result. > >The Microchip routine is attached (no flames please...it's tiny) The > >idea > >is: > > > >C = (A*B)/256 > > > >The 256 number is handy as no division is actually needed, you just > >take the > >most significant byte of the 16 bit result. > > > >Hope this is of some use. > > > > > >Mike Rigby-Jones > > > > I'm not sure if this is what the original poster wanted, but IF it is, > it's similar to some code I wrote to have a grand master control in a > lighting control system control the outputs of the A/D values from the > original channel pots. > There we have individual channel pots giving A/D values from 0 to 255 > (0x00 to 0xff). The master pot gives the same result. We want the > master pot to scale the channel pots. Turns out simple enough... > > Out = Chan * Master/255 > > Note that it's divided by 255, not 256. Otherwise, if both channel an d > master were 255, we'd get 254 as an output, when we want 255. > Dividing by 255 is certainly not as easy as dividing by 256 (throwing > away the bottom 8 bits). Perhaps with rounding that would work (divide > by 256 and if bit 7 of the remainder is 1, increment the result). What I > did was to check bit 7 of the master pot value. If it's 1, I added 1 to > the result of the divide by 256. Works great! I'm not sure what Quentin wants. Quentin? If you want a quick and easy routine to scale a number to a percentage: ;******************************************************************* ;scale_hex2dec ; The purpose of this routine is to scale a hexadecimal byte to a ;decimal byte. In other words, if 'h' is a hexadecimal byte then ;the scaled decimal equivalent 'd' is: ; d = h * 100/256. ;Note that this can be simplified: ; d = h * 25 / 64 = h * 0x19 / 0x40 ;Multiplication and division can be expressed in terms of shift lefts ;and shift rights: ; d = [ (h<<4) + (h<<3) + h ] >> 6 ;The program divides the shifting as follows so that carries are automatically ;taken care of: ; d = (h + (h + (h>>3)) >> 1) >> 2 ; ;Inputs: W - should contain 'h', the hexadecimal value to be scaled ;Outputs: W - The scaled hexadecimal value is returned in W ;Memory: temp ;Calls: none scale_hex2dec MOVWF temp ;Hex value is in W. CLRC ;Clear the Carry bit so it doesn't affect RRF RRF temp,F CLRC RRF temp,F CLRC RRF temp,F ;temp = h>>3 ADDWF temp,F ;temp = h + (h>>3) RRF temp,F ;temp = (h + (h>>3)) >> 1 ADDWF temp,F ;temp = h + ((h + (h>>3)) >> 1) RRF temp,F CLRC RRF temp,W ;d = W = (h + (h + (h>>3)) >> 1) >> 2 RETURN For Harold's case: C = A*B/255 There is another trick that you can attempt. Recall the power series for division: N N / / e \ / e \2 / e \3 \ ------- = --- * | 1 - |---| + |---| - |---| + ... | v + e v \ \ v / \ v / \ v / / In Harold's equation, N=A*B, v+e = 255. If you let v=256 and e=-1 then the series simplifies to: A*B A*B / / 1 \ / 1 \2 / 1 \3 \ -------- = --- * | 1 + |---| + |---| + |---| + ... | 256 - 1 256 \ \256/ \256/ \256/ / Or just keeping the first two terms: A*B A*B / 1 \ -------- ~= --- * | 1 + --- | 255 256 \ 256 / If A & B are 8-bit quantites, then the second term in the series would produce 0 (you'd be dividing the product A*B by 256 twice). However, as Harold notes, you may use the second term to round the result. Specifically, if you treated the multiplication (conceptually) as floating point then you'll end up with a fractional component that's greater than 0.5 if A*B is greater than 2^15. Or stated differently, if the most significant bit is set in the product, then increment the result. Scott