Fourier series of the square root of a signal
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From: nanren888 on 8 Jan 2010 04:20 Greg Heath <heath(a)alumni.brown.edu> wrote in message <0f380e68-6065-410a-9e0e-473fd5a4db1d(a)f5g2000yqh.googlegroups.com>... > > CORRECTED FOR THE HEINOUS SIN OF TOP-POSTING!!! > > On Jan 7, 6:05 am, "Stefano " <s.mangi...(a)gmail.com> wrote: > > "Abdullah " <abdullahu...(a)yahoo.com> wrote in message <hi4amh$h3...(a)fred.mathworks.com>... > > > "Stefano " <s.mangi...(a)gmail.com> wrote in message <hi45n5$f1...(a)fred.mathworks.com>... > > > > I have an array X containing the (truncated) Fourier series expansion of a real positive even continuous (C(inf)) periodic signal x(t). > > > > > > I would like to know how the Fourier series coefficients for the square root of x(t) (I mean the real signal y(t)=sqrt(x(t))) could be obtained. Could someone provide me a reference to solve this problem? > > > > > > A rude solution would be to synthesize the sampled x(t), compute the square root of the samples and compute the coefficients by projection on the Fourier basis. What I'm looking for is a more elegant (if not less computationally expensive) way to do this. > > > > > Dear Stefano, > > > Fourier series is for periodic signals. Make sure taking sqrt of a periodic signal gives you a periodic signal > > > > It would be very surprising if, being x(t) a periodic signal, sqrt(x(t)) turned out not > > to be periodic. > > > > (I know that you can multiply it with a signal taking randomly values +1 or -1 and > > still get a square root of x(t), but that is not the case I'm interested in). > > you are confused sqrt(1*x) ~ sqrt(-1*x) for if x ~= 0. > > Hope this helps. > > Greg I really don't think this is the forum for these theory questions. I suspect that you have to ignore the potential problem with each instantaneous value of the sqrt signal being either positive or negative by assuming a positive solution. That would otherwise lead to infinite many solutions. I suspect the "even" might refer to the spectrum rather than the signal. If the element-wise product of two waveforms in time transforms to the convolution of there spectra in frequency, then the signal you are looking for has a spectrum that when convolved with itself gives you the Fourier series you started with. Convolution is equivalent to polynomial multiplication, so if you prefer some sort of (sqrt) factorisation of the polyniomial specified by the Fourier series. Essentially a deconvolution, where the two parts are the same. SQRT has been suggested in the past as a spectrum compression method, hence bandwidth reduction measure, hence allowing sampling rate reduction. If you have a concocted problem, eg no noise or other nasty things, then you can likely solve it. eg start at one end & just do the division. In practice I suspect you'll need something a whole lot more sophisticated. Could consider sqrt in the time domain? :)
From: Greg Heath on 8 Jan 2010 05:22 On Jan 7, 3:15 pm, TideMan <mul...(a)gmail.com> wrote: > On Jan 8, 12:05 am, "Stefano " <s.mangi...(a)gmail.com> wrote: > > It would be very surprising if, being x(t) a periodic signal, sqrt(x(t)) turned out not to be periodic. > > > (I know that you can multiply it with a signal taking randomly values +1 or -1 and still get a square root of x(t), but that is not the case I'm interested in). > > > Stefano > > > "Abdullah " <abdullahu...(a)yahoo.com> wrote in message <hi4amh$h3...(a)fred.mathworks.com>... > > > "Stefano " <s.mangi...(a)gmail.com> wrote in message <hi45n5$f1...(a)fred..mathworks.com>... > > > > I have an array X containing the (truncated) Fourier series expansion of a real positive even continuous (C(inf)) periodic signal x(t). > > > > > I would like to know how the Fourier series coefficients for the square root of x(t) (I mean the real signal y(t)=sqrt(x(t))) could be obtained. Could someone provide me a reference to solve this problem? > > > > > A rude solution would be to synthesize the sampled x(t), compute the square root of the samples and compute the coefficients by projection on the Fourier basis. What I'm looking for is a more elegant (if not less computationally expensive) way to do this. > > > > Dear Stefano, > > > Fourier series is for periodic signals. Make sure taking sqrt of a periodic signal gives you a periodic signal. > > Yes, but if x(t) is periodic, then depending upon its mean, it may > take negative as well as positive values, so what happens when you > take the sqrt of that? > Seems to me you're not telling us the whole story The square root of a negative number will be imaginary. The resulting fft is still valid. However, i don't know of any way to predict the coefficients of the fft(sqrt(x)) given the coefficients of fft(x). except to solve for the Dk from x = sqrt(x)*sqrt(x) sum{ Cn*exp(i*2*pi*fn*t)} = sum{ Dj*exp(i*2*pi*fj*t)} *sum{ Dk*exp (i*2*pi*fk*t)} Since the freqiencies in the sums range over [-N/2:N/2+1] *f0 , the nonlinear multiplication will yield frequencies in the range [-N:N+2]*f0. Therefore when the LHS and RHS are matched by frequency, there will be N equations with nonzero LHSs and N equations with zero LHS. The solution could be more complicated than just taking the fft of sqrt(x) directly. Hope this helps. Greg
From: Stefano on 8 Jan 2010 06:17 Greg Heath <heath(a)alumni.brown.edu> wrote in message <306666f4-2005-4ffe-9152-e131178c7473(a)h9g2000yqa.googlegroups.com>... > On Jan 7, 3:15 pm, TideMan <mul...(a)gmail.com> wrote: > > On Jan 8, 12:05 am, "Stefano " <s.mangi...(a)gmail.com> wrote: > > > It would be very surprising if, being x(t) a periodic signal, sqrt(x(t)) turned out not to be periodic. > > > > > (I know that you can multiply it with a signal taking randomly values +1 or -1 and still get a square root of x(t), but that is not the case I'm interested in). > > > > > Stefano > > > > > "Abdullah " <abdullahu...(a)yahoo.com> wrote in message <hi4amh$h3...(a)fred.mathworks.com>... > > > > "Stefano " <s.mangi...(a)gmail.com> wrote in message <hi45n5$f1...(a)fred.mathworks.com>... > > > > > I have an array X containing the (truncated) Fourier series expansion of a real positive even continuous (C(inf)) periodic signal x(t). > > > > > > > I would like to know how the Fourier series coefficients for the square root of x(t) (I mean the real signal y(t)=sqrt(x(t))) could be obtained. Could someone provide me a reference to solve this problem? > > > > > > > A rude solution would be to synthesize the sampled x(t), compute the square root of the samples and compute the coefficients by projection on the Fourier basis. What I'm looking for is a more elegant (if not less computationally expensive) way to do this. > > > > > > Dear Stefano, > > > > Fourier series is for periodic signals. Make sure taking sqrt of a periodic signal gives you a periodic signal. > > > > Yes, but if x(t) is periodic, then depending upon its mean, it may > > take negative as well as positive values, so what happens when you > > take the sqrt of that? > > Seems to me you're not telling us the whole story > > The square root of a negative number will be imaginary. The > resulting fft is still valid. However, i don't know of any way to > predict > the coefficients of the fft(sqrt(x)) given the coefficients of fft(x). > except > to solve for the Dk from > > x = sqrt(x)*sqrt(x) > > sum{ Cn*exp(i*2*pi*fn*t)} = sum{ Dj*exp(i*2*pi*fj*t)} *sum{ Dk*exp > (i*2*pi*fk*t)} > > > Since the freqiencies in the sums range over [-N/2:N/2+1] *f0 , > the nonlinear multiplication will yield frequencies in the range > [-N:N+2]*f0. Therefore when the LHS and RHS are matched > by frequency, there will be N equations with nonzero > LHSs and N equations with zero LHS. > > The solution could be more complicated than just taking the fft > of sqrt(x) directly. > > Hope this helps. > > Greg Thank you Greg and nanren888 Synthesis by means of ifft, taking the sqrt in time domain and analysis again by means of fft does a satisfying enough job. Greg: I had writted the system of nonlinear equations but couldn't find an efficient way to solve it. Nanren: no noise apart from numerical rounding error is considered in this (signal synthesis) problem, yet I supposed that a closed form expression might be known.
From: Dave Robinson on 8 Jan 2010 08:30
"Stefano " <s.mangione(a)gmail.com> wrote in message <hi45n5$f13$1(a)fred.mathworks.com>... > I have an array X containing the (truncated) Fourier series expansion of a real positive even continuous (C(inf)) periodic signal x(t). > > I would like to know how the Fourier series coefficients for the square root of x(t) (I mean the real signal y(t)=sqrt(x(t))) could be obtained. Could someone provide me a reference to solve this problem? > > A rude solution would be to synthesize the sampled x(t), compute the square root >of the samples and compute the coefficients by projection on the Fourier basis. What >I'm looking for is a more elegant (if not less computationally expensive) way to do >this. Just a thought here. The action of squaring the signal is a multiplication, and is not linear, hence you just cant take the squareroot of the spectrum. However Matlab appears to treat 'logs'/'antilogs' very gracefully thus exp(log(0)) gives you back 0 not an error condition. Likewise exp(log(-4)) gives you back -4 (more or less anyway). Could you do what you want homomorphically namely find the log of your squared signal, Fourier transform it, divide that by 2, do any LINEAR processes you want in the frequency domain, apply the inverse transform, then antilog the result. The only thing that concerns me is the ambiguity with a square root having two valid answers Regards Dave Robinson |