Implementing sin function in fpga

I am attempting to implement a sine function using cordic method.  I
can't figure out how to reduce an arbitrary angle to -90 and 90 deg.
For example, if an input is 390 deg, it should be reduced to 30 deg
before passing to the cordic method.  What algorithm can i use to
efficiently reduce a arbitrary angle down to -90 and 90 deg.

Any help would be much appreciated

Thanks
pvn

pvnguyen@mail.ucf.edu wrote:

>I am attempting to implement a sine function using cordic method. 

Forget about CORDIC, computing 17-bit accurate sin(x) using linear
interpolation needs much smaller hardware and is conceptually simpler. 
It could be as fast as your RAM is -- my implementation of this method
at a speed grade 6 Cyclone generates up to 256*10^6 sin(x) values
per second.

    Best regards
    Piotr Wyderski

Rather than running the CORDIC off degrees, run it off a phase value from
0-1 corresponding to 0-360 degrees.
Just multiply an 18-bit input by 2^18/360 and use only the bottom 18 bits
from the multiply, effectively truncating the integer portion of a
fixed-decimal number.  The 18 bits are (phase/360 - int(phase/360)) * 2^18
or using a less popular notation frac(phase/360)*2^18.


<pvnguyen@mail.ucf.edu> wrote in message
news:1117566811.817537.138790@g14g2000cwa.googlegroups.com...
> I am attempting to implement a sine function using cordic method.  I
> can't figure out how to reduce an arbitrary angle to -90 and 90 deg.
> For example, if an input is 390 deg, it should be reduced to 30 deg
> before passing to the cordic method.  What algorithm can i use to
> efficiently reduce a arbitrary angle down to -90 and 90 deg.
>
> Any help would be much appreciated
>
> Thanks
> pvn

Depend on the accuracy you need, but if it is not "too bad" you can
simple use memories as LUT which you preload with the sin result.
The amount of logic is minimal and the speed is as fast as you can run
the memory.

Have fun.

Berty wrote:

> Depend on the accuracy you need, but if it is not "too bad" you can
> simple use memories as LUT which you preload with the sin result.

If you use two memories with 256 entries each (the first one for sin(k),
18-bit wide and the second one, 11-bit wide for sin(k+1)-sin(k)) and 
an unsigned 11x8 multiplier, then you can compute sin(x) with 17
significant binary digits. It will consume about 450 LEs.

    Best regards
    Piotr Wyderski

I remember that every high-school student used to learn that
 sin(90 + x) = sin(90 -x),    and sin(180+x) = -sin(180-x),
and sin-x = -sinx
Peter Alfke

I am very comfortable with using Cordic method, however i will
definitely look into linear interpolation.

No matter what algorithm I use, I would still need to determine the
offset of an angle in order to compute its sine result.  The most
obvious way to calculate the offset is to devide by 360 and extract the
remainder.  However, i was hoping not having to implement a divide
block due to speed and area.   Are there shifting combinations that can
result in an exact division of 360? What kind of algorithm does a
calculator used?

Thanks
pvnguyen

I would just perform a repetitive subtraction of 360, until the result
is smaller than 360. You are only interested in the remainder...
Peter Alfke

pvnguyen@mail.ucf.edu wrote:

> No matter what algorithm I use, I would still need to determine the
> offset of an angle in order to compute its sine result.  The most
> obvious way to calculate the offset is to devide by 360 and extract the
> remainder.  However, i was hoping not having to implement a divide
> block due to speed and area.   Are there shifting combinations that can
> result in an exact division of 360?

IMHO the best way is to scale a given angle so that 360 degrees
is represented by 2^N for some appropriately large N and then
simply throw away everything except the N least significant bits. 
Just multiply your angle by 2^N/360 and that's all.

> What kind of algorithm does a calculator used?

If you really need to know true modulo, then the Russian
peasants division algorithm would be a good idea.

    Best regards
    Piotr Wyderski

> Just multiply your angle by 2^N/360 and that's all.

I see!

John_H mention that earlier but i wasn't fully understood.  I know what
I need to do now. 

Thanks for your help

pvnguyen