Sirs, Generally, when fft is imimplemented in hw, due to prohibitive silicon area, the intermediate stages are allowed only 2 bit (for eg) growth. If I simply see this from a mathematical perspective, we do lose a very significant amount of bits due to this. How come this is acceptable and widely used in many hw fft implementation s? Thanks, manish
FFT in hw
Started by ●June 16, 2013
Reply by ●June 17, 20132013-06-17
On Sun, 16 Jun 2013 11:29:02 -0500, manishp wrote:> Sirs, > > Generally, when fft is imimplemented in hw, due to prohibitive silicon > area, the intermediate stages are allowed only 2 bit (for eg) growth. If > I simply see this from a mathematical perspective, we do lose a very > significant amount of bits due to this. How come this is acceptable and > widely used in many hw fft implementation s? > > Thanks, manishNo answers yet?!?! Try again, but look at it from a signal to noise perspective, quantization taken as noise. See if only adding two bits per stage looks like it makes more sense. -- My liberal friends think I'm a conservative kook. My conservative friends think I'm a liberal kook. Why am I not happy that they have found common ground? Tim Wescott, Communications, Control, Circuits & Software http://www.wescottdesign.com