Mathematical Derivations

For a square wave $v(t)$ with amplitude $V$ and fundamental period $T$, we define the angular frequency as $\omega_0 = \frac{2\pi}{T}$. The waveform over one period $T$ is defined as:

1. Conventional (Trigonometric) Form

The trigonometric Fourier series is given by:

• DC Component ($a_0$): Because the positive and negative halves are symmetrical, the average value is zero. $$ a_0 = 0 $$
• Cosine Coefficients ($a_n$): The waveform is an odd function ($v(-t) = -v(t)$), which mathematically yields zero when integrated over the defined bounds for cosine. $$ a_n = 0 $$
• Sine Coefficients ($b_n$):
  $$ b_n = \frac{2}{T} \int_{0}^{T} v(t) \sin(n\omega_0 t) dt $$ $$ b_n = \frac{2}{T} \left[ \int_{0}^{T/2} V \sin(n\omega_0 t) dt + \int_{T/2}^{T} (-V) \sin(n\omega_0 t) dt \right] $$ Evaluating the integrals gives: $$ b_n = \frac{2V}{n\omega_0 T} \left[ -\cos(n\omega_0 t) \right]_0^{T/2} - \frac{2V}{n\omega_0 T} \left[ -\cos(n\omega_0 t) \right]_{T/2}^T $$ Since $\omega_0 T = 2\pi$, this simplifies to: $$ b_n = \frac{V}{n\pi} \left[ 1 - \cos(n\pi) - \cos(n\pi) + \cos(2n\pi) \right] $$ $$ b_n = \frac{2V}{n\pi} \left( 1 - \cos(n\pi) \right) $$
  For even $n$, $\cos(n\pi) = 1$, so $b_n = 0$.
  For odd $n$, $\cos(n\pi) = -1$, so $b_n = \frac{4V}{n\pi}$.

Final Trigonometric Series:

Note: The normalized harmonic magnitude plotted in the spectrum graph above is calculated as the ratio of the $n$-th harmonic magnitude to the fundamental magnitude, which evaluates exactly to $\frac{1}{n}$.

2. Complex (Exponential) Form

The complex Fourier series is given by:

Where the complex coefficients $c_n$ are calculated as:

For even $n$, $\cos(n\pi) = 1$, so $c_n = 0$.
For odd $n$, $\cos(n\pi) = -1$, so $c_n = \frac{2V}{j n \pi}$.

Final Complex Series: