Editing Total harmonic distortion (section)

==Examples==
For many standard signals, the above criterion may be calculated analytically in a closed form.<ref name="iaroslav_04" /> For example, a pure [[Square wave (waveform)|square wave]] has THD<sub>F</sub> equal to
:<math>
 \mathrm{THD_F} = \sqrt{\frac{\pi^2}{8} - 1} \approx 0.483 = 48.3\%.
</math>
The [[Sawtooth wave|sawtooth signal]] possesses
:<math>
 \mathrm{THD_F} = \sqrt{\frac{\pi^2}{6} - 1} \approx 0.803 = 80.3\%.
</math>
The pure symmetrical [[triangle wave]] has
:<math>
 \mathrm{THD_F} = \sqrt{\frac{\pi^4}{96} - 1} \approx 0.121 = 12.1\%.
</math>
For the rectangular [[pulse train]] with the ''[[duty cycle]]'' ''μ'' (called sometimes the ''cyclic ratio''), the THD<sub>F</sub> has the form
:<math>
 \operatorname{THD_F}(\mu) = \sqrt{\frac{\mu(1 - \mu)\pi^2}{2\sin^2\pi\mu} -1}, \quad 0 < \mu < 1,
</math>
and logically, reaches the minimum (≈0.483) when the signal becomes symmetrical ''μ''&nbsp;=&nbsp;0.5, i.e. the pure [[Square wave (waveform)|square wave]].<ref name="iaroslav_04" /> Appropriate filtering of these signals may drastically reduce the resulting THD. For instance, the pure [[Square wave (waveform)|square wave]] filtered by the [[Butterworth filter|Butterworth low-pass filter]] of the second order (with the [[cutoff frequency]] set equal to the fundamental frequency) has THD<sub>F</sub> of 5.3%, while the same signal filtered by the fourth-order filter has THD<sub>F</sub> of 0.6%.<ref name="iaroslav_04" /> However, analytic computation of the THD<sub>F</sub> for complicated waveforms and filters often represents a difficult task, and the resulting expressions may be quite laborious to obtain. For example, the closed-form expression for the THD<sub>F</sub> of the [[sawtooth wave]] filtered by the first-order [[Butterworth filter|Butterworth low-pass filter]] is simply
:<math>
 \mathrm{THD_F} = \sqrt{\frac{\pi^2}{3} - \pi\coth\pi} \approx 0.370 = 37.0\%,
</math>
while that for the same signal filtered by the second-order [[Butterworth filter]] is given by a rather cumbersome formula<ref name="iaroslav_04" />
: <math>
 \mathrm{THD_F} =
 \sqrt{\pi \frac{\cot\dfrac{\pi}{\sqrt{2}} \cdot \coth^2 \dfrac{\pi}{\sqrt{2}} -
                 \cot^2 \dfrac{\pi}{\sqrt{2}} \cdot \coth\dfrac{\pi}{\sqrt{2}} -
                 \cot\dfrac{\pi}{\sqrt{2}} - \coth\dfrac{\pi}{\sqrt{2}}}
                {\sqrt{2} \left(\cot^2 \dfrac{\pi}{\sqrt{2}} +
                                \coth^2 \dfrac{\pi}{\sqrt{2}}\right)} +
       \frac{\pi^2}{3} - 1}
 \approx 0.181 =  18.1\%.
</math>
Yet, the closed-form expression for the THD<sub>F</sub> of the [[pulse train]] filtered by the ''p''th-order [[Butterworth filter|Butterworth low-pass filter]] is even more complicated and has the following form:<ref name="iaroslav_04" />
: <math>
 \operatorname{THD_F}(\mu, p) =
 \csc\pi\mu \cdot
 \sqrt{\mu(1 - \mu)\pi^2 - \sin^2 \pi\mu -
       \frac{\pi}{2} \sum_{s=1}^{2p} \frac{\cot \pi z_s}{z_s^2}
        \prod\limits_{\scriptstyle l=1\atop\scriptstyle l\neq s}^{2p} \frac{1}{z_s - z_l} +
       \frac{\pi}{2} \operatorname{Re} \sum_{s=1}^{2p} \frac{e^{i\pi z_s(2\mu - 1)}}{z_s^2 \sin \pi z_s}
        \prod\limits_{\scriptstyle l=1\atop\scriptstyle l\neq s}^{2p} \frac{1}{z_s - z_l}},
</math>
where ''μ'' is the [[duty cycle]], 0 < ''μ'' < 1, and
: <math>
 z_l \equiv \exp{\frac{i\pi(2l - 1)}{2p}}, \quad l = 1, 2, \ldots, 2p.
</math>