Floating-point to Fixed-point in (Sharc) DSP
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From: Tim Wescott on 6 Jun 2010 14:05 On 06/03/2010 02:07 AM, m26k9 wrote: > Hello, > > I am sure this is a pretty simple question, but I am having some ambiguity. > I need to confirm what is happening exactly. Any information is really > appreciated. > > So this is a Sharc DSP I am working with. Question is pretty simple. > > Rn = FIX Fx; > > FIX description from the manual is as follows: > "Converts the floating-point operand in Fx to a twos-complement 32-bit > fixed-point integer result. If the MODE1 register TRUNC bit=1, the FIX > operation truncates the mantissa towards –Infinity. If the TRUNC bit=0, > the FIX operation rounds the mantissa towards the nearest integer. The > TRUNC operation always truncates toward 0. Note that the TRUNC bit > does not influence operation of the TRUNC instruction." > > > Now the problem is, what is the "floating point OPERAND"? > The mantissa is always a fractional value with one digit left to the > decimal point. For example, the floating point will be something like > -2.345*(10-5) or 5.678*(10+8), and the mantissa will be 2.345 or 5.678. Not quite. It's binary floating point, so the exponent is in powers of 2. If it's IEEE-like, the mantissa will be unsigned with a sign bit somewhere in the word. I _think_ IEEE exponents are signed 2-s compliment, but they may be sign-magnitude as well. Learning a bit about the floating point representation of the machine would be a good thing. > Now, when it says the mantissa towards -inifinity or nearest integer, that > means the fixed-point value will be -2 or 5 (according to the above example > values)? Which definitely cannot be the case since we have 32-bits. > > Could somebody please explain. > > Cheers. Do you have a way to check this on a real machine, preferably with a debugger? I can't imagine that's what they mean to do. It would make far more sense that the conversion is to first calculate where the radix point* is in the mantissa, make it into an integer + fractional bits, then truncate or round. * It's base two, so it's not a decimal point. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
From: Raymond Toy on 6 Jun 2010 14:25 On 6/6/10 2:05 PM, Tim Wescott wrote: > On 06/03/2010 02:07 AM, m26k9 wrote: >> Hello, >> >> I am sure this is a pretty simple question, but I am having some >> ambiguity. >> I need to confirm what is happening exactly. Any information is really >> appreciated. >> >> So this is a Sharc DSP I am working with. Question is pretty simple. >> >> Rn = FIX Fx; >> >> FIX description from the manual is as follows: >> "Converts the floating-point operand in Fx to a twos-complement 32-bit >> fixed-point integer result. If the MODE1 register TRUNC bit=1, the FIX >> operation truncates the mantissa towards –Infinity. If the TRUNC bit=0, >> the FIX operation rounds the mantissa towards the nearest integer. The >> TRUNC operation always truncates toward 0. Note that the TRUNC bit >> does not influence operation of the TRUNC instruction." >> >> >> Now the problem is, what is the "floating point OPERAND"? >> The mantissa is always a fractional value with one digit left to the >> decimal point. For example, the floating point will be something like >> -2.345*(10-5) or 5.678*(10+8), and the mantissa will be 2.345 or 5.678. > > Not quite. It's binary floating point, so the exponent is in powers of > 2. If it's IEEE-like, the mantissa will be unsigned with a sign bit > somewhere in the word. I _think_ IEEE exponents are signed 2-s > compliment, but they may be sign-magnitude as well. IEEE exponents are unsigned, but they have a bias so the an exponent value of the bias represents a true exponent of 0. The mantissa might also have a hidden 1 bit, as someone else mentioned. > > Learning a bit about the floating point representation of the machine > would be a good thing. > >> Now, when it says the mantissa towards -inifinity or nearest integer, >> that >> means the fixed-point value will be -2 or 5 (according to the above >> example >> values)? Which definitely cannot be the case since we have 32-bits. >> >> Could somebody please explain. >> >> Cheers. > > Do you have a way to check this on a real machine, preferably with a > debugger? I can't imagine that's what they mean to do. It would make I agree with that. If they really meant the mantissa and it's binary floating point, the only possible values of that instruction are 0, +/-1 and +/- 2 (I think). Hardly useful. The conversion surely takes into account the exponent and probably does what you would think it would do, so -2.345e-5 becomes 0 and 5.678e8 is 567800000. (Assuming base 10 for simplicity). Ray
From: Jaime Andrés Aranguren Cardona on 6 Jun 2010 15:18 On 3 jun, 11:07, "m26k9" <maduranga.liyanage(a)n_o_s_p_a_m.gmail.com> wrote: > Hello, > > I am sure this is a pretty simple question, but I am having some ambiguity. > I need to confirm what is happening exactly. Any information is really > appreciated. > > So this is a Sharc DSP I am working with. Question is pretty simple. > > Rn = FIX Fx; > > FIX description from the manual is as follows: > "Converts the floating-point operand in Fx to a twos-complement 32-bit > fixed-point integer result. If the MODE1 register TRUNC bit=1, the FIX > operation truncates the mantissa towards Infinity. If the TRUNC bit=0, > the FIX operation rounds the mantissa towards the nearest integer. The > TRUNC operation always truncates toward 0. Note that the TRUNC bit > does not influence operation of the TRUNC instruction." > > Now the problem is, what is the "floating point OPERAND"? > The mantissa is always a fractional value with one digit left to the > decimal point. For example, the floating point will be something like > -2.345*(10-5) or 5.678*(10+8), and the mantissa will be 2.345 or 5.678. > Now, when it says the mantissa towards -inifinity or nearest integer, that > means the fixed-point value will be -2 or 5 (according to the above example > values)? Which definitely cannot be the case since we have 32-bits. > > Could somebody please explain. > > Cheers. Hi, The operand is Fx. For instance: R1 = FIX F0; Converts the floating point value stored in F0 to a fixed point value stored in R1. Say F0 stores the value 123.456; aplying the above instruction will store the value 123 in R1. Could not be easier, thanks to SHARCs! JaaC
From: Albert van der Horst on 7 Jun 2010 16:16 In article <g7CdnVzpH6pQ7ZrRnZ2dnUVZ_qednZ2d(a)giganews.com>, m26k9 <maduranga.liyanage(a)n_o_s_p_a_m.gmail.com> wrote: >Hello, > >I am sure this is a pretty simple question, but I am having some ambiguity. >I need to confirm what is happening exactly. Any information is really >appreciated. > >So this is a Sharc DSP I am working with. Question is pretty simple. > >Rn = FIX Fx; > >FIX description from the manual is as follows: >"Converts the floating-point operand in Fx to a twos-complement 32-bit >fixed-point integer result. If the MODE1 register TRUNC bit=1, the FIX >operation truncates the mantissa towards -Infinity. If the TRUNC bit=0, >the FIX operation rounds the mantissa towards the nearest integer. The >TRUNC operation always truncates toward 0. Note that the TRUNC bit >does not influence operation of the TRUNC instruction." > > >Now the problem is, what is the "floating point OPERAND"? >The mantissa is always a fractional value with one digit left to the >decimal point. For example, the floating point will be something like >-2.345*(10-5) or 5.678*(10+8), and the mantissa will be 2.345 or 5.678. Start writing the fp numbers as fixed point. Write those numbers as - 0.00002345 and 57800000000.0 Truncating towards 0 means leaving off all behind the decimal point. Towards -infinity means leaving off a .000 or else the highest number that is still smaller, in those cases: -1 57800000000 >Now, when it says the mantissa towards -inifinity or nearest integer, that >means the fixed-point value will be -2 or 5 (according to the above example >values)? Which definitely cannot be the case since we have 32-bits. The second number is too large for 32 bits. This probably results in the largest 32 bit number. Write a small test program if the manual leaves that unclear. > >Could somebody please explain. > >Cheers. Groetjes Albert -- -- Albert van der Horst, UTRECHT,THE NETHERLANDS Economic growth -- being exponential -- ultimately falters. albert(a)spe&ar&c.xs4all.nl &=n http://home.hccnet.nl/a.w.m.van.der.horst
From: m26k9 on 7 Jun 2010 21:40
Dear Tim, Ray, Jaac and Albert. Thank you so much. I think I got it now. The SHARC uses IEEE single-precision format. So what I need to do is to write is as a decimal without an exponent, and round it to the possible integer? For example, the value is 2.345e-5, I write it as 0.00002345, which converted to fixed-point will give zero. If I had 2.345e+5, then I write 234500, so my fixed-point value is 234500. Thank you very much for the clarification. Cheers. |