Next |
Prev |
Up |
Top
|
JOS Index |
JOS Pubs |
JOS Home |
Search
After IFFT, imaginary part is not quite zero due to finite
numerical precision:
sfilt = ifft(Sfilt);
rmserrpct = 100*norm(imag(sfilt))/norm(sfilt) % check
sfilt = real(sfilt);
- For a signal of length
, this was 2-3 times faster
than conv() in Matlab
- Note approximately equal amounts of ``pre-ringing'' and
``post-ringing'' due to filter being linear phase (symmetry would be
exact if signal were left-right symmetric)
Next |
Prev |
Up |
Top
|
JOS Index |
JOS Pubs |
JOS Home |
Search
[Comment on this page via email]