Fourier series of the square root of a signal
in [Matlab]
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From: Stefano on 7 Jan 2010 03:24 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.
From: Abdullah on 7 Jan 2010 04:49 "Stefano " <s.mangione(a)gmail.com> wrote in message <hi45n5$f13$1(a)fred.mathworks.com>... > I have an array X containing the (truncated) Fourier series expansion of a real positive even continuous (C(inf)) periodic signal x(t). > > I would like to know how the Fourier series coefficients for the square root of x(t) (I mean the real signal y(t)=sqrt(x(t))) could be obtained. Could someone provide me a reference to solve this problem? > > A rude solution would be to synthesize the sampled x(t), compute the square root of the samples and compute the coefficients by projection on the Fourier basis. What I'm looking for is a more elegant (if not less computationally expensive) way to do this. Dear Stefano, Fourier series is for periodic signals. Make sure taking sqrt of a periodic signal gives you a periodic signal.
From: Stefano on 7 Jan 2010 06:05 It would be very surprising if, being x(t) a periodic signal, sqrt(x(t)) turned out not to be periodic. (I know that you can multiply it with a signal taking randomly values +1 or -1 and still get a square root of x(t), but that is not the case I'm interested in). Stefano "Abdullah " <abdullahumer(a)yahoo.com> wrote in message <hi4amh$h39$1(a)fred.mathworks.com>... > "Stefano " <s.mangione(a)gmail.com> wrote in message <hi45n5$f13$1(a)fred.mathworks.com>... > > I have an array X containing the (truncated) Fourier series expansion of a real positive even continuous (C(inf)) periodic signal x(t). > > > > I would like to know how the Fourier series coefficients for the square root of x(t) (I mean the real signal y(t)=sqrt(x(t))) could be obtained. Could someone provide me a reference to solve this problem? > > > > A rude solution would be to synthesize the sampled x(t), compute the square root of the samples and compute the coefficients by projection on the Fourier basis. What I'm looking for is a more elegant (if not less computationally expensive) way to do this. > > Dear Stefano, > Fourier series is for periodic signals. Make sure taking sqrt of a periodic signal gives you a periodic signal.
From: TideMan on 7 Jan 2010 15:15 On Jan 8, 12:05 am, "Stefano " <s.mangi...(a)gmail.com> wrote: > It would be very surprising if, being x(t) a periodic signal, sqrt(x(t)) turned out not to be periodic. > > (I know that you can multiply it with a signal taking randomly values +1 or -1 and still get a square root of x(t), but that is not the case I'm interested in). > > Stefano > > "Abdullah " <abdullahu...(a)yahoo.com> wrote in message <hi4amh$h3...(a)fred.mathworks.com>... > > "Stefano " <s.mangi...(a)gmail.com> wrote in message <hi45n5$f1...(a)fred.mathworks.com>... > > > I have an array X containing the (truncated) Fourier series expansion of a real positive even continuous (C(inf)) periodic signal x(t). > > > > I would like to know how the Fourier series coefficients for the square root of x(t) (I mean the real signal y(t)=sqrt(x(t))) could be obtained. Could someone provide me a reference to solve this problem? > > > > A rude solution would be to synthesize the sampled x(t), compute the square root of the samples and compute the coefficients by projection on the Fourier basis. What I'm looking for is a more elegant (if not less computationally expensive) way to do this. > > > Dear Stefano, > > Fourier series is for periodic signals. Make sure taking sqrt of a periodic signal gives you a periodic signal. Yes, but if x(t) is periodic, then depending upon its mean, it may take negative as well as positive values, so what happens when you take the sqrt of that? Seems to me you're not telling us the whole story..............
From: Greg Heath on 8 Jan 2010 04:03
CORRECTED FOR THE HEINOUS SIN OF TOP-POSTING!!! On Jan 7, 6:05 am, "Stefano " <s.mangi...(a)gmail.com> wrote: > "Abdullah " <abdullahu...(a)yahoo.com> wrote in message <hi4amh$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 |