Why Is the RMS Coefficient of a Triangle Wave \(\frac{1}{2\sqrt{3}}\)?

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SUMMARY

The RMS coefficient of a triangle wave is definitively calculated as \(\frac{V_m}{\sqrt{3}}\). This value arises from the integration of the square of the voltage over one period, leading to the conclusion that the RMS value is not \(\sqrt{ \frac{1}{T} \int_0^T {V^2 dt} }\) as initially suggested. The correct derivation confirms that the peak voltage \(V_m\) is divided by \(\sqrt{3}\) to obtain the RMS value. Miscalculations were acknowledged during the discussion, emphasizing the importance of accurate integration in waveform analysis.

PREREQUISITES
  • Understanding of RMS (Root Mean Square) calculations
  • Familiarity with triangle wave properties
  • Knowledge of voltage waveforms and their mathematical representations
  • Basic calculus for integration of functions
NEXT STEPS
  • Study the derivation of RMS values for different waveforms
  • Learn about the properties of triangle waves in electrical engineering
  • Explore integration techniques for periodic functions
  • Investigate the implications of RMS values in power calculations
USEFUL FOR

Electrical engineers, physics students, and anyone involved in waveform analysis or signal processing will benefit from this discussion.

gulsen
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I've seen some coefficient such as \frac{1}{2\sqrt 3} for triangle waves. Who ordered that? How can I derieve it? It doesn't seem to be \sqrt{ \frac{1}{T} \int_0^T {V^2 dt} } because I found something like that for triangle wave: V_m 2\sqrt 2 f (am I wrong?? or I'm calculating the wrong thing??)
 
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Whoops! Miscalculation! It's \frac{V_m}{\sqrt 3}. Gotta sleep, I guess...
 

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