How to Find the Phase Difference in Degrees

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SUMMARY

This discussion focuses on calculating the phase difference in degrees between two sine waves using time measurements. The user measured a phase difference of 22.0 microseconds at a frequency of 10.4 kHz. The calculation involves determining the period of the wave, which is 96.15 microseconds per cycle, and using the formula: (phase offset in time / period) * 360 degrees. The final result is an angular offset of 82.37 degrees.

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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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