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How to Find the Phase Difference in Degrees

  1. Sep 24, 2009 #1
    Bit of a noob question,

    I've measured the phase difference between sine waves of V(in) and V(out) using an oscilloscope to be 22.0 micro-seconds at a frequency of 10.4 kHz.

    How would I find the phase difference in degrees?

    Thanks.
     
  2. jcsd
  3. Sep 24, 2009 #2
    Found a good link:
    http://www.sengpielaudio.com/calculator-timedelayphase.htm" [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Oct 1, 2009 #3
    The simplest way would be to measure the width of one cycle (if your scope gives you frequency, simply invert that) to give you the period. In your case, the frequency is given, so the period would be 1/10400 cycles per second = 96.15 micro-seconds per cycle.

    On cycle equates to 360 degrees (that is, there are 360 degrees in one cycle). So, take your phase offset in time (22 micro-seconds), divide it by the period (96.15 micro-seconds per cycle) and multiply by 360 degrees per cycle. The result is the angular offset.

    22 micro-seconds / 96.15 micro-seconds per cycle * 360 degrees/cycle = 82.37 degrees

    Or, more simply, take the offset in time and multiply by the frequency. Then, multiply this result by 360 degrees per cycle.

    22 micro-seconds * 10400 cycles per second * 360 degrees per cycle = 82.37 degrees
     
  5. Oct 8, 2009 #4
    Just wanted to clean up my earlier post...

     
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