Electronic – Why is total harmonic distortion defined in the following manner

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So the total harmonic distortion is defined as

$$THD = \sqrt{\frac{V_{2}^2 + V_{3}^2 + V_{4}^2…+V_{n}^2}{V_{1}^2}}$$

which means Total harmonic distortion is defined as the square root of the ratio of the average power delivered by harmonic components divide power delivered by non-harmonic component. So why include the square root when it would make more intuitive sense not to include it in the first place?

Best Answer

Because the voltages are in rms. Remember that the rms voltage is the equivalent DC voltage source that delivers the same amount of power to a resistive load. The rms is given by

$$v_\mathrm{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2 \ \mathrm{d}t}.$$

In our case \$v\$ is a sum of pure sine waves. Hence it is easier to work in the frequency domain, if you do not have a burning desire of evaluating integrals containing products of sine waves.

The rms in the frequency domain is given by

$$V_\mathrm{rms} = \sqrt{\frac{1}{N^2} \sum_{i = 1}^N |V_i |^2},$$

where \$V_i\$ are the frequency components and \$N\$ is the number of samples. Now remember that a sine wave satisfies

$$|V| = \frac{A}{2} (\delta(f - f_0) + \delta(f + f_0)).$$

Hence a single harmonic has two components with amplitude \$A / 2\$. Thus

$$V_\mathrm{rms}^2 = \frac{1}{N^2} \frac{A^2}{2}.$$

Consider what happens when we take the rms of a waveform comprising two distinct harmonics with amplitudes \$A_1\$ and \$A_2\$. Then you have

$$V^2_\mathrm{rms} = \frac{1}{N^2} \left( \frac{A_1^2}{2} + \frac{A_2^2}{2} \right).$$

Substituting the rms of the first harmonic \$V_1\$ and the rms of the second harmonic \$V_2\$ yields

$$V_\mathrm{rms} = \sqrt{V_1^2 + V_2^2}.$$

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