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Say I have 2 signals:

  • A = 1kHz Sine
  • B = 1kHz Sine, 30° lagging the first one.

If I plot the FFT-Phase of these 2 waveforms, and get the Phase values at 1kHz, I would get:

  • A = -90°
  • B = -120°

Questions:

  1. Should the phase difference be (B - A) or (A - B)?
  2. What is the conventional meaning for a negative phase difference? Lagging or leading?
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I believe the terms “leading” and “lagging” can be misunderstood, so I don’t recommend using that terminology.

Due to the traditional definition of the discrete Fourier transform (DFT), a measured DFT sample phase of zero degrees means the input sinusoid was a cosine wave with an initial phase of zero degrees at time t = 0. So when you measure a DFT sample’s phase to be –90 degrees (your ‘A’ signal) that means the input sinusoid was a cosine wave with an initial phase of –90 degrees which is equal to a sine wave with an initial phase of zero degrees at time t = 0.

Your signal ‘B’ phase of –120 degrees means that input signal was a sine wave with an initial phase of -30 degrees at time t = 0.

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  • $\begingroup$ Thanks! I understand that the FFT phase is wrt to a cosine, but the question is mainly on the "phase difference". I've rephrased my post for clarity. $\endgroup$ – Ryuu Jun 15 '16 at 11:09
  • $\begingroup$ I'm not sure why this answer is getting an up-vote. I am asking for the meaning of the sign when 2 phases are subtracted against each other, not what the sign in the FFT-phase results represents. The FFT-Phase method used is just an example. $\endgroup$ – Ryuu Jun 16 '16 at 5:31
  • $\begingroup$ Forgive me for what may appear to be an incomplete answer. Phase values are defined as “x degrees relative to …” So (A-B) = 30 degrees means: “The phase of A is +30 degrees relative to B at time t = 0.” (B-A) = -30 means: “The phase of B is -30 degrees relative to A at time t = 0.” $\endgroup$ – Richard Lyons Jun 16 '16 at 16:04
  • $\begingroup$ Thank you, sir :) One final clarification: "-30° relative to A at time t = 0" --> is that to the left or right of t = 0? $\endgroup$ – Ryuu Jun 17 '16 at 1:50
  • $\begingroup$ Time t = 0 is a single instant in time. If you plot the phases of ‘A’ and ‘B’, over time, their instantaneous phases will change (fluctuate) as time advances. And the difference between their phases will fluctuate as time advances. However, your FFT results tells you that at the single time instant of t = 0 the phase of ‘B’ is exactly 30 degrees less than the phase of ‘A’. $\endgroup$ – Richard Lyons Jun 18 '16 at 9:31

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