help-octave
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## RE: I Need Help..

 From: Damian Harty Subject: RE: I Need Help.. Date: Mon, 25 Nov 2013 10:35:14 +0000

```> > 1)      Why only use the real part of the results?

> Assuming the original signal is real, taking the real part of the inverse FFT
> and multiplying by 2
> is essentially restoring the spectral symmetry of the FFT. Basically because
> the low-pass
> filtering should have preserved the 8 conjugate-symmetric bins at the end of
> the array.

But a "real" signal (I'm unaware of any real world processes that generate
imaginary numbers in the time domain, if I'm honest) generates imaginary
components of the FFT depending on the phasing of the sine component? So
sin(wt) gives an imaginary coefficient and cost (wt) gives a real coefficient,
sin (wt+phi) gives a complex coefficient depending on phi.

So it's still not really clear to me why slice off the imaginary part of the
signal. Multiplying by 2 satisfies Parseval's theorem and is all good, from my
point of view - but taking only the real component (as distinct from the
magnitude of the complex number) manifestly doesn't satisfy Parseval and
represents some kind of "polarizing filter" on the data, letting through
components with a phase angle close to pi/2 unhindered but completely blocking
components with a phase angle of zero Without more context it's difficult to
understand if it's deliberate and very clever or just a bit muddled.

> > 2)      Why use the DC part from another signal?

> I'm also not sure, without context of what the ec signal is, but it's
> essentially subtracting the mean of ec from the low-pass filtered
> original_data, right?

Yes, that's how I read it. I have no idea why.

> Back to the original question, the cutoff frequency of your low-pass
> filtering depends on
> the size of original_data. If L = length(original_data), then the frequency
> axis is in units of
> Fs/L and the cutoff frequency is 9*Fs/L.

Agreed.

Damian

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