[Scilab-users] Problems arising from truncation of %pi

[Federico Miyara]

Dear all,

In order to test the FFT on a periodic signal whose period is an exact 
submultiple of the FFT length I found a frequency fo satisfying this for 
the chosen sample rate and window length, and generated the following 
signal:

x = sin(2*%pi*fo*t);

where t is a time vector. This should be a perfectly periodic discrete 
signal but it isn't because the sin() function has a (virtually) exact 
period of pi, while %pi is exact to 16 digits only.

For instance, we have (23 digits shown)

--> sin(%pi)
  ans  =
    0.0000000000000001224647

--> sin(1e10*%pi)
  ans  =
   -0.0000022393627619559233

--> sin(1e15*%pi)
  ans  =
   -0.2362090532517409080526

The Wolfram Alpha site yields the correct value 0 in all cases (using 
their own pi).

Questions:

1) How is the sin() function extended to very large values of the 
argument? My first guess would be to compute a quarter cycle using 
Taylor and then extend it by symmetry and periodicity, but with which 
period? The best approximation available is 2*%pi. Or it is possible to 
use extended precision internally?

2) Is there a way to get a periodic discrete sine other than extending 
it periodically with the desired period?
Wouldn't this create a slight glitch at the frontier between cycles?

Regards,

Federico Miyara


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