Hello Forum, I found in a book the Fourier transform interpreted as a filtering operation: Given a signal f(t), The complex coefficient at frequency w0 given by the Fourier transform is X(w0). Say that the signal spectrum is X(w+w0) as if it was the result of a modulation. X(w+w0) is centered at w0. The Fourier integral has {f(t)*exp(-j*w0*t)} as integrand. This step can be seen as a demodulation which shifts the spectrum of f(t) back to the origin. The {integral of f(t)*exp(-j*w0*t)} can be seen as a convolution integral of f(t)*exp(-j*w0*t) with the signal g(t)=1. This is equivalent to multiplication in the frequency domain by a delta(w) at the origin, which sifts the value of the spectrum of X(w+w0) at w0, i.e. X(w0) which is a constant, and therefore does not vary with time ,and exists for all time.... Does this make any sense? If we put limits other than infinity in the integral the convolution would not be with the constant 1 but with rectangular function..... That would make X(w0) not a constant anymore, but a function that changes with time..... thanks for any correction, or better explanation.... I think this is an interesting way to view the Fourier integral... thanks fisico32

# Filtering view of the Fourier transform

Started by ●April 30, 2010

Reply by ●April 30, 20102010-04-30

On Apr 30, 11:42�am, "fisico32" <marcoscipioni1@n_o_s_p_a_m.gmail.com> wrote:> Hello Forum, > > I found in a book the Fourier transform interpreted as a filtering > operation: > > Given a signal f(t), The complex coefficient at frequency w0 given by the > Fourier transform is X(w0). > Say that the signal spectrum is X(w+w0) as if it was the result of a > modulation. X(w+w0) is centered at w0. > > The Fourier integral has {f(t)*exp(-j*w0*t)} as integrand. This step can be > seen as a demodulation which shifts the spectrum of f(t) back to the > origin. > > The {integral of f(t)*exp(-j*w0*t)} can be seen as a convolution integral > of f(t)*exp(-j*w0*t) with the signal g(t)=1. This is equivalent to > multiplication in the frequency domain by a delta(w) at the origin, which > sifts the value of the spectrum of X(w+w0) at w0, i.e. X(w0) which is a > constant, and therefore does not vary with time ,and exists for all > time.... > > Does this make any sense? > > If we put limits other than infinity in the integral the convolution would > not be with the constant 1 but with rectangular function..... > That would make X(w0) not a constant anymore, but a function that changes > with time..... > > thanks for any correction, or better explanation.... > I think this is an interesting way to view the Fourier integral... > > thanks > fisico32Well not only can the Fourier transform be viewed as a "filter." But you will find convolution is a property common to many integral transforms and not just the Fourier transform. Also since the idea of finite limits seems to intrigue you - you may also wish to look at Green's functions. With these you replace a variable limit on the integral with infinite limits and multiply the integrand with a "gating function - an adjustable window." An advantage of doing such is the delta function can be expanded into a sum of basis functions which in some cases simplfies the integration. Clay

Reply by ●April 30, 20102010-04-30

>On Apr 30, 11:42=A0am, "fisico32" <marcoscipioni1@n_o_s_p_a_m.gmail.com> >wrote: >> Hello Forum, >> >> I found in a book the Fourier transform interpreted as a filtering >> operation: >> >> Given a signal f(t), The complex coefficient at frequency w0 given bythe>> Fourier transform is X(w0). >> Say that the signal spectrum is X(w+w0) as if it was the result of a >> modulation. X(w+w0) is centered at w0. >> >> The Fourier integral has {f(t)*exp(-j*w0*t)} as integrand. This step can=>be >> seen as a demodulation which shifts the spectrum of f(t) back to the >> origin. >> >> The {integral of f(t)*exp(-j*w0*t)} can be seen as a convolutionintegral>> of f(t)*exp(-j*w0*t) with the signal g(t)=3D1. This is equivalent to >> multiplication in the frequency domain by a delta(w) at the origin,which>> sifts the value of the spectrum of X(w+w0) at w0, i.e. X(w0) which is a >> constant, and therefore does not vary with time ,and exists for all >> time.... >> >> Does this make any sense? >> >> If we put limits other than infinity in the integral the convolutionwoul=>d >> not be with the constant 1 but with rectangular function..... >> That would make X(w0) not a constant anymore, but a function thatchanges>> with time..... >> >> thanks for any correction, or better explanation.... >> I think this is an interesting way to view the Fourier integral... >> >> thanks >> fisico32 > >Well not only can the Fourier transform be viewed as a "filter." But >you will find convolution is a property common to many integral >transforms and not just the Fourier transform. Also since the idea of >finite limits seems to intrigue you - you may also wish to look at >Green's functions. With these you replace a variable limit on the >integral with infinite limits and multiply the integrand with a >"gating function - an adjustable window." An advantage of doing such >is the delta function can be expanded into a sum of basis functions >which in some cases simplfies the integration. > >Clay >Hi Clay, thanks for the message. Please bear with me and lets see if I truly understand. The way the Fourier transform is viewed as a filter, is that given a signal x(t) and its spectrum X(w), the FT will filter such a signal and only allow the single frequency w0 to go through, as if it was an infinitely sharp passband filter with gain 1. This is done for all different frequencies w..... I am still, conceptually, not sure why the amplitudes X(w) of such filter cannot be time-dependent. I know its due to the infinity limits of integration. Can you give me a better explanation of why the Fourier transform works as a filter and exactly what type of filter it is? If the limits of integration were finite we would get the STFT (short time Fourier transform) Wy would the FT work well only for stationary signals and not for nonstationary? Two different signals can have the same power spectra but different phase spectra. the phase spectra seem to characterize the signals at every instant uniquely......that seems like the FT works well for stationary and nonstationary signals.... What am i missing? Thanks! fisico32