Hi all, I have the following question related from sampling. Suppose I have the original band pass signal with spectrum in [-250, -150] and [150, 250]Hz. The negative and positive spectrum are antisymmetrical themselves, which means the spectrum is not locally symmetrical with respect to the axis -200 Hz and +200Hz... For simplicity, you can take two non-equallateral triangles at -200Hz and +200Hz as example. Thus by sampling, I cannot do Fs=200Hz, since the two spectrum are not locally symmetrical, so I the Fs=200Hz, these two part will overlap at 0Hz. I can do Fs=250Hz, but then [-250, -150] will be moved to [0, 100], and [+150, +250] will be moved to [-100, 0], thus the original negative non-equallateral triangle at -200Hz is now moved to +50Hz, the original positive spectrum is moved to -50Hz... I can apply a low pass filter [-100, 100] to get the inverted spectrum, but how to invert them back? What are the best approaches to invert the spectrum back? I have thought of using multiplication a 2*cosine(2*pi*100*t) , then the spectrum in [-100, 100] will be moved to [0, 200], and also [-200, 0], then I apply a low pass filter at [-100, 100] again to get the original signal back. So after sampling, the reconstruction process has three steps: r(t) =((s(t)* (200sinc(200t))) . (2cosine(2.pi.100t)))* (200sinc(200t)) The first bracket corresponds to the first low pass filtering, the middle one is the multiplication with cosine, the third bracket is the low pass filter again. "*" denotes convolution. "." denotes multiplication. s(t) is the sampled signal, r(t) is the reconstructed signal. Any comments on this method? Are there any other method of doing inverting spectrum for the sake of low sampling rate? Thanks a lot
How to invert spectrum after sampling?
Started by ●November 28, 2004
Reply by ●November 29, 20042004-11-29
"kiki" <lunaliu3@yahoo.com> wrote in message news:cod1sc$47c$1@news.Stanford.EDU...> Hi all, > > I have the following question related from sampling. > > Suppose I have the original band pass signal with spectrum in [-250, -150] > and [150, 250]Hz. The negative and positive spectrum are antisymmetrical > themselves, which means the spectrum is not locally symmetrical with > respect to the axis -200 Hz and +200Hz... For simplicity, you can take two > non-equallateral triangles at -200Hz and +200Hz as example. > > Thus by sampling, I cannot do Fs=200Hz, since the two spectrum are not > locally symmetrical, so I the Fs=200Hz, these two part will overlap at > 0Hz. >It's not clear to me what you mean yet.... It's not the first one but is it perhaps the second or the third? | /| | |/ // | | | // // | | | // / | | | / | | | | | | | | | | | | | | | ----------|-----|------------|------------|-----|------------------ -f0 0 f0 Fig 1. Spectra are mirror images. Spectrum is even. | | | |/ | | // | | // | | / | | | | | | | | | ----------|-----|------------|------------|-----|------------------ | | | -f0 | 0 f0 | | / | // | // | /| Fig 2. Spectra are mirror images and of opposite sign. Spectrum is odd. | | /| | /| // | | // | // | | // | / | | / | | | | | | | | | | | | | | | | ----------|-----|------------|------------|-----|------------------ -f0 0 f0 Fig 3. Spectra are not mirror images In the first case you should be able to sample at f0. The result will be a complex time series - the "complex envelope". I think the second case, if the the spectrum is purely real then, being odd, the temporal result is purely imaginary. In the third case, the temporal sequence is complex to begin with. Fred
Reply by ●November 30, 20042004-11-30
"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:Bc-dnbnu5PtUyzbcRVn-rw@centurytel.net...> > "kiki" <lunaliu3@yahoo.com> wrote in message > news:cod1sc$47c$1@news.Stanford.EDU... >> Hi all, >> >> I have the following question related from sampling. >> >> Suppose I have the original band pass signal with spectrum in >> [-250, -150] and [150, 250]Hz. The negative and positive spectrum are >> antisymmetrical themselves, which means the spectrum is not locally >> symmetrical with respect to the axis -200 Hz and +200Hz... For >> simplicity, you can take two non-equallateral triangles at -200Hz and >> +200Hz as example. >> >> Thus by sampling, I cannot do Fs=200Hz, since the two spectrum are not >> locally symmetrical, so I the Fs=200Hz, these two part will overlap at >> 0Hz. >>Hi Fred, Great drawing! Thanks a lot!> > It's not the first one but is it perhaps the second or the third? > > > | > /| | |/ > // | | | // > // | | | // > / | | | / > | | | | | > | | | | | > | | | | | > ----------|-----|------------|------------|-----|------------------ > > -f0 0 f0 > Fig 1. Spectra are mirror images. Spectrum is > even.> > | > | > | |/ > | | // > | | // > | | / > | | | > | | | > | | | > ----------|-----|------------|------------|-----|------------------ > | | > | -f0 | 0 f0 > | | > / | > // | > // | > /| > Fig 2. Spectra are mirror images and of opposite > sign. Spectrum is odd. > > > | > | > /| | /| > // | | // | > // | | // | > / | | / | > | | | | | > | | | | | > | | | | | > ----------|-----|------------|------------|-----|------------------ > > -f0 0 f0 > Fig 3. Spectra are not mirror images > > In the first case you should be able to sample at f0. The result will be > a complex time series - the "complex envelope". > >I meant the first one originally... you can still sample at f0 even when the the signal is not locally symmetrical? Then the positive and negative spectrum will overlap at f=0 part and kill each other,... can you verify your claim that f0 sampling rate can still be achieved? Figure 3 is actuall easy in terms of sampling: you just need fs=f0, the positive and negative spectrums will overlap at f=0 and since they are the same, they don't kill each other and f0 is the lowest sampling rate we can achieve. Figure 2 and Figure 1 are the same in terms of sampling: my intuition is that they can not do fs=f0... since when they overlap at f=0 they mess up with each other... Using the data provided in my original post, fs=250Hz, instead of 200Hz. But this will lead to the sampled and low pass filtered spectrum inverted, negative spectrum moves to [0, 100], positive spectrum moves to [-100, 0]... then how to invert them back?> > I think the second case, if the the spectrum is purely real then, being > odd, the temporal result is purely imaginary. > > In the third case, the temporal sequence is complex to begin with. > > FredAny more thoughts are highly appeciated... Thanks alot!
Reply by ●November 30, 20042004-11-30
I think it depneds if you have some noise in +-[250;350] that you need to get rid of first (in which case you would have to do other things before). If there is no noise in +-[50,150], why not simply sampling to 150Hz to get your signal r(t)? Seb Hi all,> >I have the following question related from sampling. > >Suppose I have the original band pass signal with spectrum in [-250,-150]>and [150, 250]Hz. The negative and positive spectrum are antisymmetrical>themselves, which means the spectrum is not locally symmetrical withrespect>to the axis -200 Hz and +200Hz... For simplicity, you can take two >non-equallateral triangles at -200Hz and +200Hz as example. > >Thus by sampling, I cannot do Fs=200Hz, since the two spectrum are not >locally symmetrical, so I the Fs=200Hz, these two part will overlap at0Hz.> >I can do Fs=250Hz, but then [-250, -150] will be moved to [0, 100], and >[+150, +250] will be moved to [-100, 0], thus the original negative >non-equallateral triangle at -200Hz is now moved to +50Hz, the original >positive spectrum is moved to -50Hz... > >I can apply a low pass filter [-100, 100] to get the inverted spectrum,but>how to invert them back? What are the best approaches to invert thespectrum>back? I have thought of using multiplication a > >2*cosine(2*pi*100*t) , > >then the spectrum in [-100, 100] will be moved to [0, 200], and also[-200,>0], then I apply a low pass filter at [-100, 100] again to get theoriginal>signal back. > >So after sampling, the reconstruction process has three steps: > >r(t) =((s(t)* (200sinc(200t))) . (2cosine(2.pi.100t)))* (200sinc(200t)) > >The first bracket corresponds to the first low pass filtering, the middle>one is the multiplication with cosine, the third bracket is the low pass>filter again. "*" denotes convolution. "." denotes multiplication. s(t)is>the sampled signal, r(t) is the reconstructed signal. > >Any comments on this method? Are there any other method of doinginverting>spectrum for the sake of low sampling rate? > >Thanks a lot > > >This message was sent from http://www.DSPRelated.com
Reply by ●December 2, 20042004-12-02
"S�b" <sebns76@yahoo.fr> wrote in message news:10qonntlfspib28@news.supernews.com...>I think it depneds if you have some noise in +-[250;350] that you need to > get rid of first (in which case you would have to do other things > before). > > If there is no noise in +-[50,150], why not simply sampling to 150Hz to > get your signal r(t)? > Seb > > > > Hi all, >> >>I have the following question related from sampling. >> >>Suppose I have the original band pass signal with spectrum in [-250, > -150] >>and [150, 250]Hz. The negative and positive spectrum are antisymmetrical > >>themselves, which means the spectrum is not locally symmetrical with > respect >>to the axis -200 Hz and +200Hz... For simplicity, you can take two >>non-equallateral triangles at -200Hz and +200Hz as example. >> >>Thus by sampling, I cannot do Fs=200Hz, since the two spectrum are not >>locally symmetrical, so I the Fs=200Hz, these two part will overlap at > 0Hz. >> >>I can do Fs=250Hz, but then [-250, -150] will be moved to [0, 100], and >>[+150, +250] will be moved to [-100, 0], thus the original negative >>non-equallateral triangle at -200Hz is now moved to +50Hz, the original >>positive spectrum is moved to -50Hz... >> >>I can apply a low pass filter [-100, 100] to get the inverted spectrum, > but >>how to invert them back? What are the best approaches to invert the > spectrum >>back? I have thought of using multiplication a >> >>2*cosine(2*pi*100*t) , >> >>then the spectrum in [-100, 100] will be moved to [0, 200], and also > [-200, >>0], then I apply a low pass filter at [-100, 100] again to get the > original >>signal back. >> >>So after sampling, the reconstruction process has three steps: >> >>r(t) =((s(t)* (200sinc(200t))) . (2cosine(2.pi.100t)))* (200sinc(200t)) >> >>The first bracket corresponds to the first low pass filtering, the middle > >>one is the multiplication with cosine, the third bracket is the low pass > >>filter again. "*" denotes convolution. "." denotes multiplication. s(t) > is >>the sampled signal, r(t) is the reconstructed signal. >> >>Any comments on this method? Are there any other method of doing > inverting >>spectrum for the sake of low sampling rate? >> >>Thanks a lot >> >> >> > >Hi, there is no noise... but can you really get Fs=150Hz? Have you plotted to see the aliasing? thanks a lot
