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From: Budding MATLAB Jockey on 31 Mar 2010 17:29
> That's because the Fourier transform of a real-valued signal is conjugate symmetric. The Fourier transform at a "positive" frequency is the complex conjugate of the Fourier transform at the corresponding negative frequency. Accordingly, if the phase is pi/4 radians at a positive frequency, it is -pi/4 at the negative frequency. If you are using it to plot a phase response, the convention is to report (show) just the positive frequencies.
>
> Of course, you should keep in mind that there is inherent uncertainty in the phase. That's why most phase response plots use unwrap().
>
> Wayne

so the real(FT) is +6 in both frequency domains
the imag(FT) us +4.5 in positive frequency domain and -4.5 in negative frequency

Therefore the phase is (based on positive freq) inverse tan(4.5/6)?

Then to get -180 deg would I expect Real(FT) = +6 and imag(FT) = -6 ?
From: Matt J on 31 Mar 2010 17:52
"Budding MATLAB Jockey " <chico1st(a)hotmail.com> wrote in message <hp0er0$mf3$1(a)fred.mathworks.com>...

> so the real(FT) is +6 in both frequency domains
> the imag(FT) us +4.5 in positive frequency domain and -4.5 in negative frequency
>
> Therefore the phase is (based on positive freq) inverse tan(4.5/6)?
================

Remember that a real sine wave is the sum of two complex sinusoids, each with its own phase, so it makes little sense to speak of THE phase, as if it were unique.

In this case, the phase of the complex sinusoidal component with positive frequency is tan(4.5/6).
The phase of the complex sinusoid with negative frequency is -tan(4.5/6)


> Then to get -180 deg would I expect Real(FT) = +6 and imag(FT) = -6 ?

No, a phase of -180 deg would require a negative real part and an infinitessimally small but negative imaginary part. Note that there is a discontinuous jump in angle across the negative real axis, as the following examples show, so there is little point in distinguishing between +180 and -180 degrees there

>> angle(-1)*180/pi

ans =

180

>> angle(-1-eps*i)*180/pi

ans =

-180




From: Greg Heath on 31 Mar 2010 22:45
On Mar 31, 12:58 am, "Budding MATLAB Jockey " <chico...(a)hotmail.com>
wrote:
> Okay this was spurred by the
> "Understanding phase in the FFT"
> thread
>
> If you have an FFT of a sin wave how do you know the phase?
> You can get your ratio of imaginary to real parts of the FFT conjugate output but how do you know which quadrant the angle is in?
>
> Thanks!

doc angle
help angle
type angle

Hope this helps.

Greg
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