CCS/TMS320C5505: Square root implementation

Hi Satyajit,

I've forwarded this to the C5000 software experts. Their feedback should be posted here.

BR
Tsvetolin Shulev

look "sqrtv" from Jeff Axelrod 1.00, A. Aboagye, 8/31/98, with N = 5 you get accuracy = 100 * 2/32767 = 0.0061% , it is very quick and
you can change them to 32bit - with a little table and only 20 ASM_lines..

HI Lars,

Thanks for reply.
Where can I find "sqrtv" from Jeff Axelrod 1.00, A. Aboagye, 8/31/98.

Hi, ok, now it's named "from Li Yuan 01/30/2002" , but I've checked this this morning. It is no more
then a chinese copy & paste of the version from Jeff Axelrod with large indexes (XAR5/XAR6) and a SI_Bug workaround.
All coments are the same! You find it in dsplib/55x_src
-make a long table until 2^31 - not only 30 and look at this, then you can make an unsigned version until 65000^2

Lars

Hi,

Actually I am very novice to dsp.
I was trying to find dsplib/55x_src folder in my PC but did not get it.
I found one src folder at location
C:\ti\ccsv6\tools\compiler\c6000_7.4.18\lib\src
but src folder it does not have "sqrtv" file

Can you please tell me where can I get dsplib/55x_src

its part of csl - actual is 307 get it from ti's site
but think the next question is "how to change to 32bit..."
you know C55 ASM?

Hi,

Thanks for reply.
Actually I recently started working on dsp around 2 months back.
I have only worked on c coding and not assembly.
I am not much familiar with c55 asm.


Below is the c code to generate the square root of sine wave. Firstly, I am converting the negative values to positive i.e. converting the range "-32767 to 32676" to "0 to 32767". Once the input is only positive, I am using square root function to get square root of input.

signed short int generate_squareroot(signed short int input)
{
signed short int SquareRoot = 0;
signed short int SquareRoot2 = 0;

input = input >> 1;
input = input + 0x3FFF;

SquareRoot = input;
sqrt_16(&SquareRoot, &SquareRoot2, 1); // take square root of sinusoid and store the squared root into SquareRoot
return SquareRoot2;
}


But as you can see, the sqrt_16 can only take 16 bit input. If input is 16 bit, it can take any values from 0 to 32767. But once square root is done, the signal can take values from 0 to 181. As such, I am losing a lot precision when I square root.

What I plan is to give input in the range 0 to 2^31 so that its square root can take values between 0 to 2^15.5

Do you think is it possible

uint32_t root = 0;
uint32_t Bit;

for ( Bit = 0x80000000L; Bit > 0; Bit >>= 1 )
{
uint32_t trial = root + Bit;
if ((uint64_t)trial * (uint64_t)trial < input)
{
root += Bit;
}
}
return root;


slow version...

Thank you very much for your timely assistance. I will try this and will get back to you.
Thanks

HI,

As per your suggestion, I tried below code. But it not seems to be working. Output is equal to input in below case.

// Square root of 32 bit number
signed long int generate_squareroot(signed long int input)
{
signed long int SquareRoot = 0;
signed long int root = 0;
signed long int Bit;
signed long int trial =0;

// convert input range from (-2^31)-(2^31) to (0)-(2^31)
input = input >> 1;
input = input + 0x3FFFFFFF;

SquareRoot = input;

for ( Bit = 0x40000000L; Bit > 0; Bit >>= 1 )
{
trial = root + Bit;
if ((unsigned long long)trial * (unsigned long long)trial < SquareRoot)
{
root += Bit;
}
}
return root;
}