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LabView 6.0 fft specturm vi error

hai, i have a problem with the fft vi. a fft of a sine with 0° phase created with the sine vi produces a phase offset of -pi. why? cu andré
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This is the correct answer. A pure cosine wave has a phase of 0 while a pure sine wave has a phase of 90. All other phases are a combination of the two.

When you are dealing with sinusoidal signals (i.e. a signal whose frequency content is a single frequency), the signal can be represented as y = cos(x)+j*sin(x) to encompass all possible phases of the signal.

The magnitude is, of course, mag = sqrt(cos^2(x) + sin^2(x)).

The phase is phase = arctan(sin(x)/cos(x)).

In your case, since you are using a sine wave, the cosine component is zero and the sine component is 1, so phase = arctan(1/0) which is equal to pi/2.

It is a little odd that the sine wave is the standard signal even though zero degrees phase is measured from the cosine wave, but that is the way it is. Try changing the phase to see how it changes the results and remember to add 90 degrees to the number to get the actual phase of the signal.
Randall Pursley
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