How to Find the Phase Difference in Degrees

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Discussion Overview

The discussion revolves around calculating the phase difference in degrees between two sine waves based on a measured time delay and frequency. The context includes practical application using an oscilloscope and mathematical reasoning related to wave properties.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks how to convert a measured phase difference in time (22 micro-seconds) to degrees at a frequency of 10.4 kHz.
  • Another participant provides a link to a calculator for time delay and phase conversion.
  • A participant suggests calculating the period of the wave (96.15 micro-seconds) and then using the formula: (phase offset in time / period) * 360 degrees to find the angular offset, resulting in approximately 82.37 degrees.
  • A later reply reiterates the calculation method, emphasizing the same formula and confirming the result of approximately 82.37 degrees.

Areas of Agreement / Disagreement

Participants generally agree on the method to calculate the phase difference in degrees, but there is no explicit consensus on the correctness of the calculations presented, as they are based on the same assumptions and formulas.

Contextual Notes

The discussion does not address potential limitations or assumptions in the calculations, such as the precision of the measurements or the accuracy of the oscilloscope used.

etotheo
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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.
 
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Found a good link:
http://www.sengpielaudio.com/calculator-timedelayphase.htm"
 
Last edited by a moderator:
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
 
Just wanted to clean up my earlier post...

zgozvrm said:
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:

\frac{1}{10400}\enspace cycles\enspace per\enspace second = \frac{1\enspace cycle}{10400\enspace Sec}\approx\frac{96.15\mu{Sec}}{cycle}.

One cycle equates to 360^\circ (that is, there are 360^\circ in one cycle). So, take your phase offset in time (22 \mu{S}),

divide it by the period \left(\frac{96.15 \mu{S}}{cycle}\right) and multiply by \frac{360^\circ}{cycle}. The result is the angular offset.

22 \mu{S}\div \frac{96.15 \mu{S}}{cycle} \times \frac{360^\circ}{cycle} \approx 82.37^\circ

Or, more simply, take the offset in time and multiply by the frequency.

Then, multiply this result by \frac{360^\circ}{cycle}.

22 \mu{S} \times \frac{10400 \enspace cycles}{Sec} \times \frac{360^\circ}{cycle} \approx 82.37 ^\circ
 

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