Quasi-Square Wave Inverter Performance

Interactive Harmonic Analysis & Mathematical Derivation

30°
0° (Full Square Wave) 90° (Zero Output)

RMS Voltage (Vrms)

0.00 p.u.

Fundamental Peak (V1)

0.00 p.u.

Total Harmonic Distortion (THD)

0.0 %

Time Domain Waveform (1 Cycle)

Frequency Spectrum (Up to 29th Harmonic)

Dynamic Fourier Series Expression

The instantaneous output voltage evaluated for α = 30° (Assuming Vdc = 1 p.u.):

Mathematical Derivation

The generalized trigonometric Fourier series for a periodic function $v(\omega t)$ is given by:

$$ v(\omega t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(n\omega t) + b_n \sin(n\omega t) \right) $$

1. Symmetry Considerations

  • No DC Offset: The positive and negative half-cycles have equal areas, so the average value $a_0 = 0$.
  • Odd Symmetry: The waveform satisfies $v(-\omega t) = -v(\omega t)$. Therefore, all cosine coefficients evaluate to zero ($a_n = 0$).
  • Half-Wave Symmetry: The waveform satisfies $v(\omega t + \pi) = -v(\omega t)$. This implies all even harmonics are zero ($n = 2, 4, 6\dots$).

Thus, the series reduces to odd harmonic sine terms only ($n = 1, 3, 5\dots$).

2. Evaluating the Coefficients ($b_n$)

Using quarter-wave symmetry, we can integrate over the first quarter-cycle ($0$ to $\pi/2$) and multiply the result by $4$:

$$ b_n = \frac{4}{\pi} \int_{0}^{\pi/2} v(\theta) \sin(n\theta) d\theta $$

For the quasi-square wave with dead band $\alpha$, $v(\theta) = 0$ for $0 \le \theta \le \alpha$, and $v(\theta) = V_{dc}$ for $\alpha \le \theta \le \pi/2$. Substituting these limits:

$$ b_n = \frac{4}{\pi} \int_{\alpha}^{\pi/2} V_{dc} \sin(n\theta) d\theta $$

Evaluating the definite integral:

\begin{align} b_n &= \frac{4V_{dc}}{\pi} \left[ \frac{-\cos(n\theta)}{n} \right]_{\alpha}^{\pi/2} \\ &= \frac{4V_{dc}}{n\pi} \left( -\cos\left(\frac{n\pi}{2}\right) + \cos(n\alpha) \right) \end{align}

Since $n$ is an odd integer, $\cos(n\pi/2)$ is always exactly $0$. This leaves us with the final coefficient formula:

$$ b_n = \frac{4V_{dc}}{n\pi} \cos(n\alpha) \quad \text{for } n = 1, 3, 5\dots $$

Substituting $b_n$ back into the original sum yields the complete Fourier series representation for the instantaneous output voltage:

$$ v(\omega t) = \sum_{n=1,3,5\dots}^{\infty} \frac{4V_{dc}}{n\pi} \cos(n\alpha) \sin(n\omega t) $$