Power Inverter Harmonic & SHE Analyzer

Interactive Selective Harmonic Elimination (SHE) for Multilevel CHB Topologies

Total Harmonic Distortion (THD): 12.50%
Harmonic Spectrum (Normalized Magnitude)

Mathematical Derivations

1. Standard Square Wave

The trigonometric Fourier series for an ideal square wave of amplitude $V$ is given by:

$$ v_{sq}(t) = \sum_{n=1,3,5,\dots}^{\infty} \frac{4V}{n\pi} \sin(n\omega_0 t) $$

2. Cascaded H-Bridge (CHB) Step Waveform General Derivation

A multi-level CHB inverter utilizes $k$ isolated DC sources per phase leg to form an $m$-level output where $m = 2k + 1$. Because the resulting staircase waveform possesses quarter-wave odd symmetry, the DC component ($a_0$) and all even harmonics ($a_n$, $b_{even}$) are zero.

The sine coefficients ($b_n$) are evaluated over the first quarter cycle ($0$ to $\frac{\pi}{2}$). Each H-bridge cell adds a voltage step $V_{dc}$ at switching angle $\theta_k$. Thus, the integration is segmented:

$$ b_n = \frac{4}{\pi} \left[ \int_{\theta_1}^{\pi/2} V_{dc} \sin(n\theta) d\theta + \int_{\theta_2}^{\pi/2} V_{dc} \sin(n\theta) d\theta + \dots + \int_{\theta_k}^{\pi/2} V_{dc} \sin(n\theta) d\theta \right] $$

Evaluating the integral $\int_{\alpha}^{\pi/2} \sin(n\theta) d\theta = \frac{\cos(n\alpha)}{n}$ (for odd $n$), we obtain the general coefficient formula for a $k$-cell inverter:

$$ b_n = \frac{4V_{dc}}{n\pi} \sum_{i=1}^{k} \cos(n\theta_i) \quad \text{for } n = 1, 3, 5, \dots $$

Derivation for 5-Level CHB ($k=2$):

With 2 cells, we have two degrees of freedom ($\theta_1, \theta_2$). The coefficient simplifies to:

$$ b_n = \frac{4V_{dc}}{n\pi} \left( \cos(n\theta_1) + \cos(n\theta_2) \right) $$

Derivation for 7-Level CHB ($k=3$):

With 3 cells, we have three degrees of freedom ($\theta_1, \theta_2, \theta_3$). The coefficient expands to:

$$ b_n = \frac{4V_{dc}}{n\pi} \left( \cos(n\theta_1) + \cos(n\theta_2) + \cos(n\theta_3) \right) $$

3. Selective Harmonic Elimination (SHE) Optimization

In SHE, specific lower-order harmonics are eliminated by forcing their respective Fourier coefficients ($b_n$) to exactly zero.

Eliminating the 3rd Harmonic (7-Level CHB):

To eliminate the 3rd harmonic, we set $b_3 = 0$:

$$ \cos(3\theta_1) + \cos(3\theta_2) + \cos(3\theta_3) = 0 $$

A valid analytical solution is $\theta_1 = 12^\circ$, $\theta_2 = 30^\circ$, and $\theta_3 = 48^\circ$.
Proof: $\cos(36^\circ) + \cos(90^\circ) + \cos(144^\circ) = 0.809 + 0 - 0.809 = 0$.

Simultaneous Elimination of the 5th and 7th Harmonics (7-Level CHB):

In three-phase systems, triplen harmonics (3rd, 9th, 15th) naturally cancel out between phases. Therefore, the primary targets for elimination are the 5th and 7th harmonics. With three variables ($\theta_1, \theta_2, \theta_3$), we can form a system of non-linear transcendental equations to set the fundamental voltage to a specific Modulation Index ($M$) while forcing $b_5 = 0$ and $b_7 = 0$:

$$ \begin{cases} \cos(\theta_1) + \cos(\theta_2) + \cos(\theta_3) = M \\ \cos(5\theta_1) + \cos(5\theta_2) + \cos(5\theta_3) = 0 \\ \cos(7\theta_1) + \cos(7\theta_2) + \cos(7\theta_3) = 0 \end{cases} $$

This system cannot be solved algebraically and requires numerical methods (such as Newton-Raphson). For an operational modulation index of $M = 2.37$, the numerical solution yields:

$$ \theta_1 \approx 11.6^\circ, \quad \theta_2 \approx 30.6^\circ, \quad \theta_3 \approx 58.2^\circ $$

Substituting these into the spectrum equations proves the elimination:
$5^{\text{th}}$ Harmonic: $\cos(58^\circ) + \cos(153^\circ) + \cos(291^\circ) \approx 0.529 - 0.891 + 0.358 \approx 0$
$7^{\text{th}}$ Harmonic: $\cos(81.2^\circ) + \cos(214.2^\circ) + \cos(407.4^\circ) \approx 0.153 - 0.827 + 0.676 \approx 0$