4.2 Fourier Series, Fourier Transforms & Frequency-Domain Analysis

4.2 Fourier Series, Fourier Transforms & Frequency-Domain Analysis

Frequency-domain spectral analysis is indispensable in electrical power systems, telecommunications, and signal processing. While Laplace transforms are optimized for transient IVP analysis with initial conditions, Fourier analysis decomposes arbitrary periodic and non-periodic signals into discrete spectral harmonics or continuous frequency distributions. On the PRC Electrical Engineering Board Exam, Fourier series expansion, RMS evaluation of non-sinusoidal currents, and Total Harmonic Distortion (THD) calculations are primary recurring topics.

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1. Trigonometric Fourier Series

Any periodic function $f(t)$ with fundamental period $T$ and fundamental angular frequency $\omega_0 = \frac{2\pi}{T}$ that satisfies the Dirichlet conditions (single-valued, bounded, finite number of discontinuities and extrema within period $T$) can be expressed as a Trigonometric Fourier Series: $f(t) = a_0 + \sum_{n=1}^\infty \left[ a_n \cos(n \omega_0 t) + b_n \sin(n \omega_0 t) \right]$

Euler-Fourier Coefficient Formulas

• DC Component $a_0$ (Average Value): $a_0 = \frac{1}{T} \int_0^T f(t) \, dt$
• Cosine Coefficients $a_n$: $a_n = \frac{2}{T} \int_0^T f(t) \cos(n \omega_0 t) \, dt \quad \text{for } n = 1, 2, 3, \dots$
• Sine Coefficients $b_n$: $b_n = \frac{2}{T} \int_0^T f(t) \sin(n \omega_0 t) \, dt \quad \text{for } n = 1, 2, 3, \dots$

Amplitude-Phase Form

Alternatively, combining sine and cosine terms at each harmonic frequency yields: $f(t) = A_0 + \sum_{n=1}^\infty A_n \cos(n \omega_0 t - \theta_n)$ where peak harmonic amplitude $A_n = \sqrt{a_n^2 + b_n^2}$ and phase angle $\theta_n = \arctan\left(\frac{b_n}{a_n}\right)$.

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2. Waveform Symmetry Rules

Recognizing geometric waveform symmetry eliminates unnecessary integration during board exam computations:

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3. Exponential (Complex) Fourier Series & Parseval's Power Theorem

Using Euler's identity $e^{j\theta} = \cos\theta + j\sin\theta$, the trigonometric series transforms into a compact complex exponential representation: $f(t) = \sum_{n=-\infty}^\infty c_n e^{j n \omega_0 t}$ where complex coefficients $c_n$ are given by: $c_n = \frac{1}{T} \int_0^T f(t) e^{-j n \omega_0 t} \, dt$ Relationships to trigonometric coefficients: $c_0 = a_0, \quad c_n = \frac{a_n - j b_n}{2}, \quad c_{-n} = c_n^* = \frac{a_n + j b_n}{2}$

Parseval's Power Theorem

Parseval's identity states that the total average power of a periodic signal $f(t)$ equals the sum of the average powers of its individual harmonic components: $P_{\text{avg}} = \frac{1}{T} \int_0^T |f(t)|^2 \, dt = \sum_{n=-\infty}^\infty |c_n|^2 = a_0^2 + \sum_{n=1}^\infty \frac{a_n^2 + b_n^2}{2} = A_0^2 + \sum_{n=1}^\infty \frac{A_n^2}{2}$

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4. RMS Values of Non-Sinusoidal Signals & Total Harmonic Distortion (THD)

RMS Value of Complex Non-Sinusoidal Waveforms

When a voltage or current contains multiple orthogonal harmonic frequencies: $v(t) = V_0 + V_{1,m} \cos(\omega_0 t + \theta_1) + V_{2,m} \cos(2\omega_0 t + \theta_2) + \dots$ The overall Effective (RMS) Voltage $V_{\text{rms}}$ is: $V_{\text{rms}} = \sqrt{V_0^2 + \frac{V_{1,m}^2}{2} + \frac{V_{2,m}^2}{2} + \dots + \frac{V_{n,m}^2}{2}} = \sqrt{V_0^2 + V_{1,\text{rms}}^2 + V_{2,\text{rms}}^2 + \dots + V_{n,\text{rms}}^2}$

Total Harmonic Distortion (THD)

Total Harmonic Distortion measures waveform degradation caused by non-linear loads (such as variable frequency drives, rectifiers, and switching power supplies): $\text{THD}_V = \frac{\sqrt{\sum_{n=2}^\infty V_{n,\text{rms}}^2}}{V_{1,\text{rms}}} = \frac{\sqrt{V_{2,\text{rms}}^2 + V_{3,\text{rms}}^2 + V_{4,\text{rms}}^2 + \dots}}{V_{1,\text{rms}}}$ Expressed as a percentage, $\text{THD}% = \text{THD}_V \times 100%$.

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5. Continuous Fourier Transform

For non-periodic signals ($T \to \infty$), the discrete frequency spectrum becomes continuous. The continuous Fourier transform pairs continuous time signal $f(t)$ with continuous frequency spectrum $F(\omega)$:

Direct Continuous Fourier Transform

$\mathcal{F}\{f(t)\} = F(\omega) = \int_{-\infty}^\infty f(t) e^{-j \omega t} \, dt$

Inverse Continuous Fourier Transform

$\mathcal{F}^{-1}\{F(\omega)\} = f(t) = \frac{1}{2\pi} \int_{-\infty}^\infty F(\omega) e^{j \omega t} \, d\omega$

Operational Properties of Fourier Transform

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Step-by-Step Worked Numerical Examples

Example 4.2.1: Fourier Series Coefficients of a Symmetric Square Wave

Problem: A symmetric square wave voltage $v(t)$ alternates between $+100\text{ V}$ and $-100\text{ V}$ with period $T = 2\pi\text{ s}$ (fundamental frequency $\omega_0 = 1\text{ rad/s}$). Specifically, $v(t) = 100\text{ V}$ for $0 < t < \pi$ and $v(t) = -100\text{ V}$ for $\pi < t < 2\pi$. Calculate the Fourier series coefficients $a_0$, $a_n$, and $b_n$, and write the trigonometric series expansion.

Solution:

1. Check DC component $a_0$: $a_0 = \frac{1}{2\pi} \left[ \int_0^\pi 100 \, dt + \int_\pi^{2\pi} (-100) \, dt \right] = \frac{1}{2\pi} [100\pi - 100\pi] = 0$
2. Check symmetry: The wave exhibits odd symmetry $v(-t) = -v(t)$ and half-wave symmetry $v(t + \pi) = -v(t)$. Therefore, $a_n = 0$ for all $n$, and $b_n = 0$ for even $n$.
3. Compute $b_n$ for odd $n$ using quarter-wave formula: $b_n = \frac{4}{T} \int_0^{T/2} v(t) \sin(n t) \, dt = \frac{4}{2\pi} \int_0^\pi 100 \sin(n t) \, dt = \frac{200}{\pi} \left[ -\frac{\cos(n t)}{n} \right]_0^\pi$ $b_n = \frac{200}{n \pi} [1 - \cos(n \pi)]$
   • For even $n$, $\cos(n\pi) = 1 \implies b_n = 0$.
   • For odd $n$, $\cos(n\pi) = -1 \implies b_n = \frac{200}{n \pi} [1 - (-1)] = \frac{400}{n \pi}$.
4. Evaluate numerical values for first three odd harmonics:
   • $b_1 = \frac{400}{\pi} \approx 127.32\text{ V}$
   • $b_3 = \frac{400}{3\pi} \approx 42.44\text{ V}$
   • $b_5 = \frac{400}{5\pi} \approx 25.46\text{ V}$
5. Write Fourier series expansion: $v(t) = \frac{400}{\pi} \left( \sin(t) + \frac{1}{3}\sin(3t) + \frac{1}{5}\sin(5t) + \dots \right) \text{ V}$

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Example 4.2.2: Total RMS Current and THD of Industrial Load Current

Problem: An industrial load current contains a fundamental component and two higher harmonics: $i(t) = 10 \cos(\omega t) + 3 \cos(3\omega t + 30^\circ) + 1.5 \cos(5\omega t - 45^\circ) \text{ A}$ Calculate: (a) RMS value of fundamental current $I_{1,\text{rms}}$, (b) overall total RMS current $I_{\text{rms}}$, and (c) Current Total Harmonic Distortion ($\text{THD}_I$).

Solution:

1. Determine peak amplitudes: $I_{1,m} = 10\text{ A}, \quad I_{3,m} = 3\text{ A}, \quad I_{5,m} = 1.5\text{ A}$
2. Calculate individual component RMS values: $I_{1,\text{rms}} = \frac{10}{\sqrt{2}} \approx 7.0711\text{ A}$ $I_{3,\text{rms}} = \frac{3}{\sqrt{2}} \approx 2.1213\text{ A}$ $I_{5,\text{rms}} = \frac{1.5}{\sqrt{2}} \approx 1.0607\text{ A}$
3. Calculate total RMS current $I_{\text{rms}}$: $I_{\text{rms}} = \sqrt{I_{1,\text{rms}}^2 + I_{3,\text{rms}}^2 + I_{5,\text{rms}}^2} = \sqrt{(7.0711)^2 + (2.1213)^2 + (1.0607)^2}$ $I_{\text{rms}} = \sqrt{50 + 4.5 + 1.125} = \sqrt{55.625} \approx 7.4582\text{ A}$
4. Compute Total Harmonic Distortion ($\text{THD}_I$): $\text{THD}_I = \frac{\sqrt{I_{3,\text{rms}}^2 + I_{5,\text{rms}}^2}}{I_{1,\text{rms}}} = \frac{\sqrt{4.5 + 1.125}}{7.0711} = \frac{\sqrt{5.625}}{7.0711} = \frac{2.3717}{7.0711} \approx 0.3354$ Expressed as percentage: $\text{THD}_I = 33.54%$.

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Example 4.2.3: Continuous Fourier Transform of an Exponential Pulse

Problem: Find the continuous Fourier transform $F(\omega)$ of the causal single-sided exponential decay signal $f(t) = e^{-a t} u(t)$ with attenuation $a = 5\text{ s}^{-1}$. Calculate the magnitude spectrum $\vert F(\omega) \vert$ at angular frequency $\omega = 5\text{ rad/s}$.

Solution:

1. Apply continuous Fourier transform integral: $F(\omega) = \int_0^\infty e^{-5 t} e^{-j \omega t} \, dt = \int_0^\infty e^{-(5 + j \omega) t} \, dt$
2. Evaluate the definite integral: $F(\omega) = \left[ \frac{e^{-(5 + j \omega) t}}{-(5 + j \omega)} \right]_0^\infty = 0 - \left( \frac{1}{-(5 + j \omega)} \right) = \frac{1}{5 + j \omega}$
3. Determine magnitude formula $\vert F(\omega) \vert$: $\vert F(\omega) \vert = \frac{1}{\sqrt{5^2 + \omega^2}} = \frac{1}{\sqrt{25 + \omega^2}}$
4. Evaluate at $\omega = 5\text{ rad/s}$: $\vert F(5) \vert = \frac{1}{\sqrt{25 + 5^2}} = \frac{1}{\sqrt{50}} = \frac{1}{7.0711} \approx 0.1414$
5. Conclusion: $F(\omega) = \frac{1}{5 + j \omega}$ and $\vert F(5) \vert = 0.1414$.