Mathematical Derivations of Fourier Coefficients ($b_n$)
1. Standard Square Wave (2-Level)
For an ideal square wave of amplitude $V$, evaluating the Fourier integral over a quarter cycle yields the coefficient for odd harmonics:
$$ b_n = \frac{4V}{n\pi} \quad \text{for } n = 1, 3, 5, \dots $$
2. Generalized Cascaded H-Bridge (CHB) Equation
A multi-level CHB inverter utilizes $k$ isolated DC sources per phase leg to synthesize an $m$-level stepped output, where $m = 2k + 1$. Because the staircase waveform has quarter-wave odd symmetry, the DC offset ($a_0$) and all even harmonics are zero.
The generalized Fourier sine coefficient $b_n$ with $k$ switching angles is derived by integrating the voltage steps over the first quarter cycle ($0$ to $\frac{\pi}{2}$):
$$ b_n = \frac{4V_{dc}}{n\pi} \sum_{i=1}^{k} \cos(n\theta_i) \quad \text{for } n = 1, 3, 5, \dots $$
3. Explicit $b_n$ Derivations by Topology
5-Level CHB ($k=2$ cells)
$$ b_n = \frac{4V_{dc}}{n\pi} \Big( \cos(n\theta_1) + \cos(n\theta_2) \Big) $$
7-Level CHB ($k=3$ cells)
$$ b_n = \frac{4V_{dc}}{n\pi} \Big( \cos(n\theta_1) + \cos(n\theta_2) + \cos(n\theta_3) \Big) $$
9-Level CHB ($k=4$ cells)
$$ b_n = \frac{4V_{dc}}{n\pi} \Big( \cos(n\theta_1) + \cos(n\theta_2) + \cos(n\theta_3) + \cos(n\theta_4) \Big) $$
11-Level CHB ($k=5$ cells)
$$ b_n = \frac{4V_{dc}}{n\pi} \Big( \cos(n\theta_1) + \cos(n\theta_2) + \cos(n\theta_3) + \cos(n\theta_4) + \cos(n\theta_5) \Big) $$
4. Selective Harmonic Elimination (SHE) Numerical Proofs
By treating the fundamental amplitude (Modulation Index, $M$) as one constraint, we can use the remaining degrees of freedom to equate specific higher-order $b_n$ coefficients to zero. Below are the numerical examples proving the elimination of target harmonics for each topology based on the slider presets.
For 5-Level (Eliminate 5th Harmonic)
$$ \begin{cases}
\cos(\theta_1) + \cos(\theta_2) = M \\
\cos(5\theta_1) + \cos(5\theta_2) = 0
\end{cases} $$
Substitution: $\theta_1 = 12.0^\circ, \theta_2 = 24.0^\circ$
• 5th Harmonic ($n=5$):
$\cos(5 \times 12^\circ) + \cos(5 \times 24^\circ)$
$= \cos(60^\circ) + \cos(120^\circ) = 0.5 + (-0.5) = 0$
For 7-Level (Eliminate 5th & 7th Harmonics)
$$ \begin{cases}
\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_3) = M \\
\cos(5\theta_1) + \cos(5\theta_2) + \cos(5\theta_3) = 0 \\
\cos(7\theta_1) + \cos(7\theta_2) + \cos(7\theta_3) = 0
\end{cases} $$
Substitution: $\theta_1 = 11.6^\circ, \theta_2 = 30.6^\circ, \theta_3 = 58.2^\circ$
• 5th Harmonic ($n=5$):
$\cos(58.0^\circ) + \cos(153.0^\circ) + \cos(291.0^\circ)$
$\approx 0.5299 - 0.8910 + 0.3584 \approx 0$
• 7th Harmonic ($n=7$):
$\cos(81.2^\circ) + \cos(214.2^\circ) + \cos(407.4^\circ)$
$\approx 0.1530 - 0.8271 + 0.6743 \approx 0$
For 9-Level (Eliminate 5th, 7th, 11th Harmonics)
$$ \begin{cases}
\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_3) + \cos(\theta_4) = M \\
\cos(5\theta_1) + \cos(5\theta_2) + \cos(5\theta_3) + \cos(5\theta_4) = 0 \\
\cos(7\theta_1) + \cos(7\theta_2) + \cos(7\theta_3) + \cos(7\theta_4) = 0 \\
\cos(11\theta_1) + \cos(11\theta_2) + \cos(11\theta_3) + \cos(11\theta_4) = 0
\end{cases} $$
Substitution: $\theta_1 = 11.3^\circ, \theta_2 = 26.9^\circ, \theta_3 = 46.1^\circ, \theta_4 = 64.3^\circ$
• 5th Harmonic ($n=5$):
$\cos(56.5^\circ) + \cos(134.5^\circ) + \cos(230.5^\circ) + \cos(321.5^\circ)$
$\approx 0.5519 - 0.7009 - 0.6361 + 0.7826 \approx 0$
• 7th Harmonic ($n=7$):
$\cos(79.1^\circ) + \cos(188.3^\circ) + \cos(322.7^\circ) + \cos(450.1^\circ)$
$\approx 0.1891 - 0.9895 + 0.7955 + 0.0017 \approx 0$
• 11th Harmonic ($n=11$):
$\cos(124.3^\circ) + \cos(295.9^\circ) + \cos(507.1^\circ) + \cos(707.3^\circ)$
$\approx -0.5635 + 0.4368 - 0.8406 + 0.9755 \approx 0$
For 11-Level (Eliminate 5th, 7th, 11th, 13th Harmonics)
$$ \begin{cases}
\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_3) + \cos(\theta_4) + \cos(\theta_5) = M \\
\cos(5\theta_1) + \dots + \cos(5\theta_5) = 0 \\
\cos(7\theta_1) + \dots + \cos(7\theta_5) = 0 \\
\cos(11\theta_1) + \dots + \cos(11\theta_5) = 0 \\
\cos(13\theta_1) + \dots + \cos(13\theta_5) = 0
\end{cases} $$
Substitution: $\theta_1 = 10.8^\circ, \theta_2 = 20.7^\circ, \theta_3 = 33.9^\circ, \theta_4 = 52.9^\circ, \theta_5 = 63.3^\circ$
• 5th Harmonic ($n=5$):
$\cos(54.0^\circ) + \cos(103.5^\circ) + \cos(169.5^\circ) + \cos(264.5^\circ) + \cos(316.5^\circ)$
$\approx 0.5878 - 0.2334 - 0.9833 - 0.0958 + 0.7254 \approx 0$
• 7th Harmonic ($n=7$):
$\cos(75.6^\circ) + \cos(144.9^\circ) + \cos(237.3^\circ) + \cos(370.3^\circ) + \cos(443.1^\circ)$
$\approx 0.2487 - 0.8181 - 0.5402 + 0.9839 + 0.1201 \approx 0$
• 11th Harmonic ($n=11$):
$\cos(118.8^\circ) + \cos(227.7^\circ) + \cos(372.9^\circ) + \cos(581.9^\circ) + \cos(696.3^\circ)$
$\approx -0.4818 - 0.6730 + 0.9748 - 0.7443 + 0.9157 \approx 0$
• 13th Harmonic ($n=13$):
$\cos(140.4^\circ) + \cos(269.1^\circ) + \cos(440.7^\circ) + \cos(687.7^\circ) + \cos(822.9^\circ)$
$\approx -0.7705 - 0.0157 + 0.1616 + 0.8415 - 0.2096 \approx 0$