Subject: Re: HILBERT TRANSFORM From: Ramdas Kumaresan <kumar@xxxxxxxx> Date: Mon, 27 Nov 2006 15:32:59 -0500 List-Archive:<http://lists.mcgill.ca/scripts/wa.exe?LIST=AUDITORY>I assume that you missed a 't' on the right side of your expression for s(t). An easy way to compute the HT of any sequence is to take the fft, set the frequency bins from the Nyquist frequency to the sampling frequency (\pi to two \pi) to zero and then inverse transform the sequence. Then the imaginary part of the resulting sequence has the HT of the original sequence within a scale factor of 2. Ramdas Kumaresan Quoting Fred Herzfeld <herzfeld@xxxxxxxx>: > Hello list, > > Suppose I have a finite sampled set of data of length 2^n of a function > s(t) = cos(2*pi*f-phi) where f and phi are constant with time and s(t) > is sampled at time intervals so that s(t) falls exactly into the second > FFT bin. If I perform the FFT, modify the output by 90 degrees and > perform an inverse FFT, the output of the FFT will be exactly the > Hilbert Transform of the original series s(t) namely HT[s(t)]. > > Now I repeat the same process except that I change the frequency (still > constant) to 7*f/8 and call the time series S1(t). The output of the FFT > will now contain many non zero bins. In theory I should still be able to > modify the FFT output and do the inverse FFT to get the Hilbert > Transform of S1(t). > > (1) Can I really compute HT[S1(t)] correctly ? > (2) If I can, how should the output of the FFT of S1(t) be modified > > Any thoughts on this will be very much appreciated. > > Fred > > -- > Fred Herzfeld, MIT '54 > 78 Glynn Marsh Drive #59 > Brunswick, Ga.31525 > USA > >