Power Inverter Harmonic & SHE Analyzer
Interactive Selective Harmonic Elimination (SHE) for Multilevel CHB Topologies
Total Harmonic Distortion (THD): 17.50%
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:
$$ v_{sq}(t) = \sum_{n=1,3,5,\dots}^{\infty} \frac{4V}{n\pi} \sin(n\omega_0 t) $$
The normalized harmonic magnitude is directly proportional to $\frac{1}{n}$. Thus, higher-order harmonics decay relatively slowly, resulting in a high inherent THD (~48.3%).
2. Cascaded H-Bridge (5-Level) Step Waveform Derivation
A 5-level Cascaded H-Bridge (CHB) inverter utilizes two isolated DC sources per phase leg ($V_{dc1} = V_{dc2} = V_{dc}$). The total output voltage $v_{chb}(t)$ is the sum of the outputs from the two individual H-bridge cells: $v_{chb}(t) = v_1(t) + v_2(t)$.
Assuming quarter-wave symmetry, the DC component ($a_0$) and all even harmonics ($a_n$, $b_{even}$) are zero. The sine coefficients ($b_n$) are evaluated over a quarter cycle ($0$ to $\frac{\pi}{2}$):
$$ b_n = \frac{4}{\pi} \int_{0}^{\pi/2} v_{chb}(\omega t) \sin(n\omega t) d(\omega t) $$
The waveform steps to $V_{dc}$ at switching angle $\theta_1$, and steps again to $2V_{dc}$ at switching angle $\theta_2$. The integral can be split according to these steps:
$$ b_n = \frac{4V_{dc}}{\pi} \left[ \int_{\theta_1}^{\pi/2} \sin(n\theta) d\theta + \int_{\theta_2}^{\pi/2} \sin(n\theta) d\theta \right] $$
Evaluating the integration for odd $n$:
$$ \int_{\alpha}^{\pi/2} \sin(n\theta) d\theta = \left[ -\frac{\cos(n\theta)}{n} \right]_{\alpha}^{\pi/2} = 0 - \left( -\frac{\cos(n\alpha)}{n} \right) = \frac{\cos(n\alpha)}{n} $$
Substituting this result back into the $b_n$ equation yields the final Fourier coefficient:
$$ b_n = \frac{4V_{dc}}{n\pi} \left( \cos(n\theta_1) + \cos(n\theta_2) \right) \quad \text{for } n = 1, 3, 5, \dots $$
3. Selective Harmonic Elimination (SHE)
To eliminate a specific harmonic component, its corresponding Fourier coefficient $b_n$ must be forced to zero. For the $n$-th harmonic in a 5-level CHB, this requires:
$$ \cos(n\theta_1) + \cos(n\theta_2) = 0 $$
- Eliminate 3rd Harmonic ($n=3$): Setting $\cos(3\theta_1) + \cos(3\theta_2) = 0$ yields a solution where $3\theta_1 + 3\theta_2 = 180^\circ$, or $\theta_1 + \theta_2 = 60^\circ$. A valid optimized pair is $\theta_1 = 20^\circ$ and $\theta_2 = 40^\circ$.
- Eliminate 5th Harmonic ($n=5$): Setting $\cos(5\theta_1) + \cos(5\theta_2) = 0$ yields a solution where $5\theta_1 + 5\theta_2 = 180^\circ$, or $\theta_1 + \theta_2 = 36^\circ$. A valid optimized pair is $\theta_1 = 12^\circ$ and $\theta_2 = 24^\circ$.