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If we have a periodic signal (square wave) with amplitude of 100V. Then what would the RMS value of the fundamental harmonic be compared to the RMS value of the third harmonic?

After some suggestion in the comments my solution was this:

enter image description here

Am I correct?

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    \$\begingroup\$ It depends on the shape (instantaneous amplitude profile) of the waveform. \$\endgroup\$ – Andy aka Sep 19 '19 at 11:28
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    \$\begingroup\$ The third harmonic of a triangle wave has another RMS value than the third harmonic of a square wave. The shape is extremely important. \$\endgroup\$ – Harry Svensson Sep 19 '19 at 12:40
  • \$\begingroup\$ @Andyaka I edited my question in order to be more precise. Thanks for pointing it out \$\endgroup\$ – maverick98 Sep 19 '19 at 13:42
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    \$\begingroup\$ 100 Volt per period? So after more periods have passed, the amplitude gets larger? The amplitude is just 100V, period. Not per period. \$\endgroup\$ – Bart Sep 19 '19 at 13:47
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Depends on what kind of periodic signal it is. As @Andy Aka mentioned: "Amplitude Profile" or How its amplitude changes with time (Triangular, Square or others ?).

Whatever be the complex periodic wave you have (say amplitude =\$ V_o \$ , Period = \$T = 2\pi/\omega\$), it can be represented as sum of fundamental frequency and its harmonics.

$$V(t) = V_f(t) + V_2(t)+V_3(t).....V_n(t) $$ where fundemental frequency component is -

$$V_f(t) = V_osin(\omega t)$$

Harmonics are - $$V_2(t) = V_2sin(2\omega t)$$ $$V_3(t) = V_3sin(3\omega t)$$ $$.$$ $$V_n(t) = V_nsin(n\omega t)$$ So, the nth harmonic's RMS value would be: $$V_{n(rms)} = \sqrt{\frac{1}{T}\int_0^TV_n^2sin^2(n\omega t)dt}$$

The amplitude of the wave \$V_0\$ is not enough information to determine what would be the amplitude of the nth harmonic ie., \$V_n\$ . So you cannot formulate a direct relation between rms value of the fundamental wave and its nth harmonic just with that information.

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    \$\begingroup\$ What happened to the odd harmonics? \$\endgroup\$ – Andy aka Sep 19 '19 at 12:12
  • \$\begingroup\$ oh ya good catch :D \$\endgroup\$ – Mitu Raj Sep 19 '19 at 12:15
  • \$\begingroup\$ @Μitu Raj if it is a square wave, can we know then? \$\endgroup\$ – maverick98 Sep 19 '19 at 12:19
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    \$\begingroup\$ You can expand its fourier series and find it. \$\endgroup\$ – Mitu Raj Sep 19 '19 at 12:24
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    \$\begingroup\$ check out fourier analysis section here: en.wikipedia.org/wiki/Square_wave @maverick \$\endgroup\$ – Mitu Raj Sep 19 '19 at 12:30

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