Hello everyone, just a little question about the good old Hilbert Transform. - Goal: obtain analytic signal (a(n)+j�(n)) from an in-phase signal (a(n)); - Tool: Blackfin BF537. I get a sampled signal, and i compute it's hilbert transform using a FIR filter with coefficients obtained using Matlab/Scilab's "Hilb" function. -Problem: I have to create the analytic signal in a numeric fashion... The signal �(n) coming out of the filter is longer than the initial signal a(n), by the amount of taps/coeffs in the filter. So, when i reassemble both signals in a .re and .im structure, the imaginary part is longer... The question is, can i just truncate the imaginary signal so that i have equivalent length signal components? Or should i use a smaller length signal applied to the input of the filter? This way, i would have: - initial signal a[n] of length N - filter h length M - yields: �[N]=a[N-M]*h[M] Although i have not noticed anyproblems with this method, I am not sure if this is a correct way of proceeding, or if some better method exists... Thanks for your time DSPers! ----------------------------------- NickelCadmium - NiCad - keeps on going, and going, and going..

# Good old Hilbert Transform question

Started by ●June 24, 2006

Reply by ●June 24, 20062006-06-24

NiCad wrote:> Hello everyone, > > just a little question about the good old Hilbert Transform. > > - Goal: obtain analytic signal (a(n)+j=E2(n)) from an in-phase signal > (a(n)); > - Tool: Blackfin BF537. I get a sampled signal, and i compute it's hilbert > transform using a FIR filter with coefficients obtained using > Matlab/Scilab's "Hilb" function. > -Problem: I have to create the analytic signal in a numeric fashion... The > signal =E2(n) coming out of the filter is longer than the initial signal > a(n), by the amount of taps/coeffs in the filter. So, when i reassemble > both signals in a .re and .im structure, the imaginary part is longer... > > The question is, can i just truncate the imaginary signal so that i have > equivalent length signal components? Or should i use a smaller length > signal applied to the input of the filter? This way, i would have: > - initial signal a[n] of length N > - filter h length M > - yields: =E2[N]=3Da[N-M]*h[M] > > Although i have not noticed anyproblems with this method, I am not sure if > this is a correct way of proceeding, or if some better method exists... > > Thanks for your time DSPers! > > ----------------------------------- > NickelCadmium - NiCad - keeps on going, and going, and going..Since you are doing a Hilbert transform via an FIR filter, then simply delay your in-phase part by the delay of the Hilbert and then pair the two together to make your analytic signal. Linear phase FIRs have a delay of (N+1)/2 samples where N is the number of taps in the filter. Clay

Reply by ●June 24, 20062006-06-24

Clay wrote:> NiCad wrote: >> Hello everyone, >> >> just a little question about the good old Hilbert Transform. >> >> - Goal: obtain analytic signal (a(n)+j�(n)) from an in-phase signal >> (a(n)); >> - Tool: Blackfin BF537. I get a sampled signal, and i compute it's hilbert >> transform using a FIR filter with coefficients obtained using >> Matlab/Scilab's "Hilb" function. >> -Problem: I have to create the analytic signal in a numeric fashion... The >> signal �(n) coming out of the filter is longer than the initial signal >> a(n), by the amount of taps/coeffs in the filter. So, when i reassemble >> both signals in a .re and .im structure, the imaginary part is longer... >> >> The question is, can i just truncate the imaginary signal so that i have >> equivalent length signal components? Or should i use a smaller length >> signal applied to the input of the filter? This way, i would have: >> - initial signal a[n] of length N >> - filter h length M >> - yields: �[N]=a[N-M]*h[M] >> >> Although i have not noticed anyproblems with this method, I am not sure if >> this is a correct way of proceeding, or if some better method exists... >> >> Thanks for your time DSPers! >> >> ----------------------------------- >> NickelCadmium - NiCad - keeps on going, and going, and going.. > > > Since you are doing a Hilbert transform via an FIR filter, then simply > delay your in-phase part by the delay of the Hilbert and then pair the > two together to make your analytic signal. Linear phase FIRs have a > delay of (N+1)/2 samples where N is the number of taps in the filter.If your HT has an even number of taps, the necessary delay will not be a whole number of sample times. If your HT has an odd number of taps, the properly delayed I signal is available at the middle tap of the HT delay structure. Both ends of a filtered signal are corrupted with transients. While the whole output is longer than the input by twice the filter length, the uncorrupted output is shorter by the same amount. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●June 28, 20062006-06-28

>Clay wrote: >> NiCad wrote: >>> Hello everyone, >>> >>> just a little question about the good old Hilbert Transform. >>> >>> - Goal: obtain analytic signal (a(n)+j�(n)) from an in-phase signal >>> (a(n)); >>> - Tool: Blackfin BF537. I get a sampled signal, and i compute it'shilbert>>> transform using a FIR filter with coefficients obtained using >>> Matlab/Scilab's "Hilb" function. >>> -Problem: I have to create the analytic signal in a numeric fashion...The>>> signal �(n) coming out of the filter is longer than the initialsignal>>> a(n), by the amount of taps/coeffs in the filter. So, when ireassemble>>> both signals in a .re and .im structure, the imaginary part islonger...>>> >>> The question is, can i just truncate the imaginary signal so that ihave>>> equivalent length signal components? Or should i use a smaller length >>> signal applied to the input of the filter? This way, i would have: >>> - initial signal a[n] of length N >>> - filter h length M >>> - yields: �[N]=a[N-M]*h[M] >>> >>> Although i have not noticed anyproblems with this method, I am notsure if>>> this is a correct way of proceeding, or if some better methodexists...>>> >>> Thanks for your time DSPers! >>> >>> ----------------------------------- >>> NickelCadmium - NiCad - keeps on going, and going, and going.. >> >> >> Since you are doing a Hilbert transform via an FIR filter, then simply >> delay your in-phase part by the delay of the Hilbert and then pair the >> two together to make your analytic signal. Linear phase FIRs have a >> delay of (N+1)/2 samples where N is the number of taps in the filter. > >If your HT has an even number of taps, the necessary delay will not be a>whole number of sample times. If your HT has an odd number of taps, the >properly delayed I signal is available at the middle tap of the HT delay>structure. Both ends of a filtered signal are corrupted with transients.> While the whole output is longer than the input by twice the filter >length, the uncorrupted output is shorter by the same amount. > >Jerry >-- >Engineering is the art of making what you want from things you can get. >����������������������������������������������������������������������� >Thanks alot for the good info Nic ----------------------------------- NickelCadmium