anyone <jaup@h/WSWLG{nR1³qHe> wrote in message <513f9ae4$0$59748$c3e8da3$66d3cc2f@news.astraweb.com>...
> On Tue, 12 Mar 2013 20:09:09 +0000, Antigoni wrote:
>
> > anyone <jaup@h/WSWLG{nR1³qHe> wrote in message
> > <513f768b$0$58186$c3e8da3$88b277c5@news.astraweb.com>...
> >> On Tue, 12 Mar 2013 17:59:16 +0000, Antigoni wrote:
> >>
> >> > anyone <jaup@h/WSWLG{nR1³qHe> wrote in message
> >> > <513f6b9a$0$38175$c3e8da3$f6268168@news.astraweb.com>...
> >> >> On Tue, 12 Mar 2013 17:37:07 +0000, Antigoni wrote:
> >> >>
> >> >> > Hi all!
> >> >> >
> >> >> > Suppose you have an equation f(w,x) which involves frequency (w)
> >> >> > and space (x) variables and it is solved for various values of
> >> >> > these variables between specific ranges. After the solution, I
> >> >> > come up with a 2D matrix whose rows represent frequency and
> >> >> > columns represent space. I apply an inverse Fourier transform to
> >> >> > this matrix using IFFT, in order to construct a matrix in time
> >> >> > (rows) and space (columns).
> >> >> > My problem starts when I'm trying to increase the number of points
> >> >> > used in frequency dimension. To be more specific, I run two
> >> >> > different cases:
> >> >> >
> >> >> > Case A: I solve f(w,x) for 256 points in w and 512 points in x.
> >> >> > The range of w is between (-2.5e6 , 2.5e6) Hz, which means that I
> >> >> > use a frequency step of df=19607.843. The solution seems ok and
> >> >> > when I do the IFFT, everything seems fine. Now I want to increase
> >> >> > the points...
> >> >> >
> >> >> > Case B: So, I increase the points in frequency and I solve f(w,x)
> >> >> > for 2048 points in w and 512 points in x (are kept constant). The
> >> >> > range of w remains the same, although the frequency step is
> >> >> > reduced to df=2442.599.
> >> >> > The solution in [frequency,space] domain is correct - the number
> >> >> > of points is increased along frequency dimension. This means that
> >> >> > when I look at a complete period of my signal I have more points
> >> >> > along the signal than I had in case A. However, when I apply the
> >> >> > IFFT, it seems llike it zooms out the result, keeping the number
> >> >> > of points constant along a complete period.
> >> >> >
> >> >> > How can I keep the range constant and increase the accuracy of the
> >> >> > output signal from ifft?
> >> >> > Any related answer is welcome.
> >> >> >
> >> >> > Thanks a lot!
> >> >>
> >> >> So, you changed FFT parameters for Case B but otherwise did not
> >> >> alter the original sampling rate, correct?
> >> >
> >> > OK, my mistake! I forgot to mention that the equation f(w,x) came
> >> > from analytic fourier transform from time to frequency (I didn't use
> >> > fft). For case B I changed only the number of points and the
> >> > frequency step.
> >>
> >> It may be simpler than I first thought -- so, frequency is along a row
> >> and 'space' is a column; you keep the number of columns constant, but
> >> change the frequency resolution along rows. Next, how are you applying
> >> IFFT?
> >
> > Exactly as you described! I apply ifft as follows:
> > u=fftshift(ifft(ifftshift(displ)));
> >
> > where displ is a (m,512) matrix representing displacements in
> > frequency-space domain with m being equal to 256 (case A) or 2048 (case
> > B). In the same way, u is the transformed matrix in time-space domain. I
> > know that ifft is applied along columns, that's why frequency is at
> > rows.
>
> That's a typo, or you mixed your inverse steps? The "*shift" function delivers or expects the transform data to be
> centered around zero frequency (whether using to- or from- transform function). The innermost inverse should be
> operating on 'shifted' FFT data, the next-outer inverse doesn't make much sense to me because by that point you're
> already at some kind of time domain data, and the outermost transform expects time domain data.
>
> So the deal is, you must consistently use "fft-/ifft-shift". If that is a typo, there is still some other issue.
>
> If that was just a typo, then "zooms out the result..." (above) seems mysterious to me. A fine-grained (in frequency)
> transform gives you better ability to resolve components in the frequency domain, but the inverse process won't
> improve the time domain resolution, that is, it can mitigate transform-induced error but won't make the recovered
> result better than the original time domain data. Were you after something else?
At first, thank you for the explanation.
It's not a typo, I found this way to perform inverse Fourier in a relevant website. Although, using fft(ifftshift) gives me exactly the same result.
So, what you are implying is that I can't improve the time domain resolution by increasing the frequency steps? Is there anyway to do it?
Thanks again for the reply!![]()
> On Tue, 12 Mar 2013 20:09:09 +0000, Antigoni wrote:
>
> > anyone <jaup@h/WSWLG{nR1³qHe> wrote in message
> > <513f768b$0$58186$c3e8da3$88b277c5@news.astraweb.com>...
> >> On Tue, 12 Mar 2013 17:59:16 +0000, Antigoni wrote:
> >>
> >> > anyone <jaup@h/WSWLG{nR1³qHe> wrote in message
> >> > <513f6b9a$0$38175$c3e8da3$f6268168@news.astraweb.com>...
> >> >> On Tue, 12 Mar 2013 17:37:07 +0000, Antigoni wrote:
> >> >>
> >> >> > Hi all!
> >> >> >
> >> >> > Suppose you have an equation f(w,x) which involves frequency (w)
> >> >> > and space (x) variables and it is solved for various values of
> >> >> > these variables between specific ranges. After the solution, I
> >> >> > come up with a 2D matrix whose rows represent frequency and
> >> >> > columns represent space. I apply an inverse Fourier transform to
> >> >> > this matrix using IFFT, in order to construct a matrix in time
> >> >> > (rows) and space (columns).
> >> >> > My problem starts when I'm trying to increase the number of points
> >> >> > used in frequency dimension. To be more specific, I run two
> >> >> > different cases:
> >> >> >
> >> >> > Case A: I solve f(w,x) for 256 points in w and 512 points in x.
> >> >> > The range of w is between (-2.5e6 , 2.5e6) Hz, which means that I
> >> >> > use a frequency step of df=19607.843. The solution seems ok and
> >> >> > when I do the IFFT, everything seems fine. Now I want to increase
> >> >> > the points...
> >> >> >
> >> >> > Case B: So, I increase the points in frequency and I solve f(w,x)
> >> >> > for 2048 points in w and 512 points in x (are kept constant). The
> >> >> > range of w remains the same, although the frequency step is
> >> >> > reduced to df=2442.599.
> >> >> > The solution in [frequency,space] domain is correct - the number
> >> >> > of points is increased along frequency dimension. This means that
> >> >> > when I look at a complete period of my signal I have more points
> >> >> > along the signal than I had in case A. However, when I apply the
> >> >> > IFFT, it seems llike it zooms out the result, keeping the number
> >> >> > of points constant along a complete period.
> >> >> >
> >> >> > How can I keep the range constant and increase the accuracy of the
> >> >> > output signal from ifft?
> >> >> > Any related answer is welcome.
> >> >> >
> >> >> > Thanks a lot!
> >> >>
> >> >> So, you changed FFT parameters for Case B but otherwise did not
> >> >> alter the original sampling rate, correct?
> >> >
> >> > OK, my mistake! I forgot to mention that the equation f(w,x) came
> >> > from analytic fourier transform from time to frequency (I didn't use
> >> > fft). For case B I changed only the number of points and the
> >> > frequency step.
> >>
> >> It may be simpler than I first thought -- so, frequency is along a row
> >> and 'space' is a column; you keep the number of columns constant, but
> >> change the frequency resolution along rows. Next, how are you applying
> >> IFFT?
> >
> > Exactly as you described! I apply ifft as follows:
> > u=fftshift(ifft(ifftshift(displ)));
> >
> > where displ is a (m,512) matrix representing displacements in
> > frequency-space domain with m being equal to 256 (case A) or 2048 (case
> > B). In the same way, u is the transformed matrix in time-space domain. I
> > know that ifft is applied along columns, that's why frequency is at
> > rows.
>
> That's a typo, or you mixed your inverse steps? The "*shift" function delivers or expects the transform data to be
> centered around zero frequency (whether using to- or from- transform function). The innermost inverse should be
> operating on 'shifted' FFT data, the next-outer inverse doesn't make much sense to me because by that point you're
> already at some kind of time domain data, and the outermost transform expects time domain data.
>
> So the deal is, you must consistently use "fft-/ifft-shift". If that is a typo, there is still some other issue.
>
> If that was just a typo, then "zooms out the result..." (above) seems mysterious to me. A fine-grained (in frequency)
> transform gives you better ability to resolve components in the frequency domain, but the inverse process won't
> improve the time domain resolution, that is, it can mitigate transform-induced error but won't make the recovered
> result better than the original time domain data. Were you after something else?
At first, thank you for the explanation.
It's not a typo, I found this way to perform inverse Fourier in a relevant website. Although, using fft(ifftshift) gives me exactly the same result.
So, what you are implying is that I can't improve the time domain resolution by increasing the frequency steps? Is there anyway to do it?
Thanks again for the reply!