Chapter 7 - Continuous-Time Fourier Series (CTFS)

07 Chapter Map - Continuous-Time Fourier Series (CTFS)

Chapter 7 Overview & Map of Content (MOC)

Symmetrical shortcuts, exponential Fourier spectra $C_n$, and Parseval’s average power boundaries.

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📚 Study Notes Index

Read in order — each note assumes the previous one.

#	Note	What it covers
7.00	7.00 Continuous-Time Fourier Series Compact Review	Fourier Series Compact Review, CTFS Formula Sheet
7.01	7.01 Trigonometric Fourier Series Representation	Trigonometric Fourier Series, Euler-Fourier Coefficients, Sinusoidal Harmonics, Orthogonality
7.02	7.02 Symmetry Conditions and Waveform Analysis	Waveform Symmetry, Even Symmetry, Odd Symmetry, Half-Wave Symmetry
7.03	7.03 Exponential Fourier Series and Complex Spectra	Exponential Fourier Series, Complex Spectra, Complex Fourier Series
7.04	7.04 Parsevals Theorem and Convergence Conditions	Parseval’s Theorem, Convergence Conditions, Dirichlet Conditions, Power Spectrum, Bessel’s Inequality

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🎯 Exam Weight

ECE 2107 Exam Relevance

Master the core derivations, mathematical definitions, and problem-solving techniques. Refer to ECE 2107 - Signals and Systems for syllabus boundaries and past year questions.

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🔗 Related Resources

• Course Teaching Plan: ECE 2107 - Signals and Systems
• Previous chapter: 06 Chapter Map - Two-Port Network Theory
• Next chapter: 08 Chapter Map - Continuous-Time Fourier Transform (CTFT)

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Chapter 7: Continuous-Time Fourier Series (CTFS) - Compact Review

7.01 Trigonometric Fourier Series Representation

*(Target: Theory Descriptive / Numerical Solving)*

• Concept: Represents a real-valued, periodic, non-sinusoidal signal as an infinite linear combination of harmonically related sines and cosines alongside a constant DC offset.
• Governing Formulas: $f(t)=\frac{a_0}{2}+{\sum }_{n=1}^{\infty }{a}_n\cos (n{\omega }_{0}t)+{\sum }_{n=1}^{\infty }{b}_n\sin (n{\omega }_{0}t)$ where the fundamental angular frequency is: ${\omega }_0=\frac{2\pi }{T_0}$
• Euler-Fourier Coefficients: $a_0=\frac{2}{T_0}{\int }_0^{T_0}f(t)dt$ $a_n=\frac{2}{T_0}{\int }_0^{T_0}f(t)\cos (n{\omega }_{0}t)dt$ $b_n=\frac{2}{T_0}{\int }_0^{T_0}f(t)\sin (n{\omega }_{0}t)dt$
• DC Term Note: The constant term in the series is $\frac{a_0}{2}$, which represents the exact statistical mean or average value of the signal over one fundamental period.

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7.02 Dirichlet Conditions for Convergence

*(Target: Theory Descriptive / 5-to-8-Mark Question)*

• Concept: The sufficient (but not necessary) conditions that a periodic signal $f(t)$ must satisfy to guarantee a valid, convergent Fourier series representation.
• The Three Conditions:
  1. Absolute Integrability: $f(t)$ must be absolutely integrable over any single period: ${\int }_{T_0}|f(t)|dt<\infty$ {This guarantees that all coefficients $a_0,a_n,b_n$ remain finite}.
  2. Bounded Maxima and Minima: $f(t)$ must possess a finite number of local maxima and minima within any single period.
  3. Finite Discontinuities: $f(t)$ must contain only a finite number of discontinuities within any finite time interval, and each discontinuity must be of finite height.

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7.03 Waveform Symmetry Shortcuts (Symmetry Conditions)

*(Target: Theory Descriptive / Shortcut Rules)*

• Concept: Physical symmetries in the time-domain waveform force specific Fourier series coefficients to exactly zero, drastically reducing calculation times.

Symmetry Type	Mathematical Condition	Coefficients Set to Zero	Active Terms in Series	Integration Shortcut
Even Symmetry	$f(t)=f(-t)$	$b_n=0$ for all $n$	DC ($a_0$) and Cosine terms ($a_n$) only	$a_n=\frac{4}{T_0}{\int }_0^{T_0/2}f(t)\cos (n{\omega }_{0}t)dt$
Odd Symmetry	$f(t)=-f(-t)$	$a_0=0$, $a_n=0$ for all $n$	Sine terms ($b_n$) only	$b_n=\frac{4}{T_0}{\int }_0^{T_0/2}f(t)\sin (n{\omega }_{0}t)dt$
Half-Wave Symmetry	$f\left(t-\frac{T_0}{2}\right)=-f(t)$	$a_0=0$, $a_n=b_n=0$ for even $n$	Odd harmonics only ($n=1,3,5,\dots$)	Integrates over half period, scaled by 2
Even Quarter-Wave	Even + Half-Wave	$b_n=0$ for all $n$; $a_n=0$ for even $n$	Cosine terms with odd harmonics only	$a_n=\frac{8}{T_0}{\int }_0^{T_0/4}f(t)\cos (n{\omega }_{0}t)dt$
Odd Quarter-Wave	Odd + Half-Wave	$a_0=0$, $a_n=0$ for all $n$; $b_n=0$ for even $n$	Sine terms with odd harmonics only	$b_n=\frac{8}{T_0}{\int }_0^{T_0/4}f(t)\sin (n{\omega }_{0}t)dt$

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7.04 Mathematical Proof: Odd Functions Contain Sine Terms Only

*(Target: Mathematical Proof / 6-Mark Derivation)*

• Starting Premise: Let $f(t)$ be a periodic function with odd symmetry, satisfying $f(-t)=-f(t)$. The cosine coefficient is defined as: $a_n=\frac{2}{T_0}{\int }_{-T_0/2}^{T_0/2}f(t)\cos (n{\omega }_{0}t)dt$
• Proof (derivation): We split the integration boundary into two equal symmetric half-intervals: $a_n=\frac{2}{T_0}\left[{\int }_{-T_0/2}^{0}f(t)\cos (n{\omega }_{0}t)dt+{\int }_0^{T_0/2}f(t)\cos (n{\omega }_{0}t)dt\right]$ Apply the variable substitution $u=-t$ (hence $dt=-du$) exclusively to the first integral: ${\int }_{-T_0/2}^{0}f(t)\cos (n{\omega }_{0}t)dt={\int }_{T_0/2}^{0}f(-u)\cos (-n{\omega }_{0}u)(-du)={\int }_0^{T_0/2}-f(u)\cos (n{\omega }_{0}u)du$ {Since $f(-u)=-f(u)$ due to odd parity, and $\cos (-x)=\cos (x)$ due to even parity}. Substituting this back into the overall coefficient equation yields: $a_n=\frac{2}{T_0}\left[{\int }_0^{T_0/2}-f(t)\cos (n{\omega }_{0}t)dt+{\int }_0^{T_0/2}f(t)\cos (n{\omega }_{0}t)dt\right]=0$
• Conclusion: Thus, $a_0=0$ and $a_n=0$ for all $n\ge 1$, proving that odd functions are composed strictly of sine term coefficients ($b_n$).

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7.05 Complex Exponential Fourier Series (Complex Spectra)

*(Target: Theory Descriptive / Formula Alignment)*

• Concept: Uses Euler’s complex exponential identity ($e^{j\theta }=\cos \theta +j\sin \theta$) to represent periodic signals in a mathematically elegant, compact format. This format acts as the direct bridge to continuous frequency-domain transforms.
• Synthesis Equation: $f(t)={\sum }_{n=-\infty }^{\infty }{C}_n{e}^{jn{\omega }_{0}t}$
• Analysis Equation: $C_n=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}f(t)e^{-jn{\omega }_{0}t}dt$
• Trigonometric to Exponential Coefficient Conversion:
  • DC Term ($n=0$): $C_0=\frac{a_0}{2}$
  • Positive Harmonics ($n>0$): $C_n=\frac{a_n-j{b}_n}{2}$
  • Negative Harmonics ($n<0$): $C_{-n}=C_n^{\ast }=\frac{a_n+j{b}_n}{2}$ {Note: For real-valued signals, $C_{-n}$ is always the complex conjugate of $C_n$}.

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7.06 Symmetrical Properties of Complex Coefficients ($C_n$)

*(Target: Theory Descriptive / Spectral Interpretation)*

• Because $C_n$ is a complex number, it is represented as a phasor containing magnitude and phase: $C_n=|C_n|e^{j{\theta }_n}$.
• Real and Even Signals: If $f(t)$ is real and even, then $C_n$ is purely real and even ($C_n=C_n^{\ast }=C_{-n}$). Phase is 0 or $\pi$.
• Real and Odd Signals: If $f(t)$ is real and odd, then $C_n$ is purely imaginary and odd ($C_n=-C_n^{\ast }=-C_{-n}$).
• Symmetry of Real-Valued Signals: For any real-valued physical signal $f(t)$:
  • Magnitude Spectrum: $|C_n|$ is perfectly symmetric (even) about the origin: $|C_{-n}|=|C_n|$
  • Phase Spectrum: $\angle C_n={\theta }_n$ is perfectly anti-symmetric (odd) about the origin: ${\theta }_{-n}=-{\theta }_n$

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7.07 Parseval’s Theorem for Fourier Series (Conservation of Power)

*(Target: Mathematical Proof / 8-Mark Question)*

• Statement: The total average power $P_{avg}$ of a periodic power signal is conserved and remains identical whether computed in the time domain or by summing the power spectral densities of its individual frequency harmonics.
• Governing Identity: $P_{avg}=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}|f(t){|}^{2}dt={\sum }_{n=-\infty }^{\infty }|C_n{|}^2={\left(\frac{a_0}{2}\right)}^2+\frac{1}{2}{\sum }_{n=1}^{\infty }\left(a_n^2+b_n^2\right)$
• Proof of Parseval’s Power Identity (derivation): Representing average power in the time domain: $P_{avg}=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}f(t)f^{\ast }(t)dt$ Substitute the exponential Fourier series synthesis equation for the conjugate signal $f^{\ast }(t)={\sum }_{n=-\infty }^{\infty }{C}_n^{\ast }{e}^{-jn{\omega }_{0}t}$: $P_{avg}=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}f(t)\left[{\sum }_{n=-\infty }^{\infty }{C}_n^{\ast }{e}^{-jn{\omega }_{0}t}\right]dt$ Interchanging the order of summation and integration: $P_{avg}={\sum }_{n=-\infty }^{\infty }{C}_n^{\ast }\left[\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}f(t)e^{-jn{\omega }_{0}t}dt\right]$ Recognizing that the bracketed integral is the exact definition of $C_n$: $P_{avg}={\sum }_{n=-\infty }^{\infty }{C}_n^{\ast }{C}_n={\sum }_{n=-\infty }^{\infty }|C_n{|}^2$ Substituting trigonometric identities ($C_0=\frac{a_0}{2}$, and $|C_n{|}^2+|C_{-n}{|}^2=\frac{a_n^2+b_n^2}{2}$) yields: $P_{avg}={\left(\frac{a_0}{2}\right)}^2+\frac{1}{2}{\sum }_{n=1}^{\infty }\left(a_n^2+b_n^2\right)$

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7.08 Approximation Errors: Bessel’s Inequality & Mean Square Error

*(Target: Theory Descriptive / Formula Interpretation)*

• N-th Partial Sum: Reconstructing a signal with a finite number of harmonics ($N$): $S_N(t)=\frac{a_0}{2}+{\sum }_{n=1}^N{a}_n\cos (n{\omega }_{0}t)+{\sum }_{n=1}^N{b}_n\sin (n{\omega }_{0}t)$
• Mean Square Error (MSE): The remaining error energy in a truncated approximation: $E_N=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}{\left(f(t)-S_N(t)\right)}^{2}dt$
• The MSE Identity: $E_N=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}|f(t){|}^{2}dt-\left[{\left(\frac{a_0}{2}\right)}^2+\frac{1}{2}{\sum }_{n=1}^N(a_n^2+b_n^2)\right]$
• Bessel’s Inequality: Since the mean square error must always be non-negative ($E_N\ge 0$): ${\left(\frac{a_0}{2}\right)}^2+\frac{1}{2}{\sum }_{n=1}^N(a_n^2+b_n^2)\le \frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}|f(t){|}^{2}dt$ {Physical Meaning: The power of a finite harmonic approximation is always strictly bounded by the total true power of the actual signal}.

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7.09 The Gibbs Phenomenon

*(Target: Theory Descriptive / 5-Mark Question)*

• Definition: At any point of jump discontinuity in a piecewise continuous periodic function, the reconstructed Fourier series approximation displays a distinct high-frequency ringing overshoot near the discontinuity.
• Core Concepts:
  • The maximum overshoot amplitude does not decay to zero even as the number of terms $N$ approaches infinity.
  • The peak overshoot converges to approximately 8.95% (nearly 9%) of the height of the jump discontinuity.
  • Increasing the number of terms $N$ compresses the physical width of the overshoot oscillations toward zero, but the peak amplitude remains constant.

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7.10 Common Mistakes That Cost Marks

Critical Exam Pitfalls

• DC Term $a_0$ vs. $a_0/2$ Scaling: Mixing up the scaling when writing the final series. Remember: if you define $a_0=\frac{2}{T_0}\int f(t)dt$, you must write the series with $\frac{a_0}{2}$ as the first term. Alternatively, if you write the series with $a_0$ first, your integration formula must be $a_0=\frac{1}{T_0}\int f(t)dt$.
• The $1/2$ Factor in Parseval’s Power: Forgetting the multiplier of $1/2$ on the harmonic power sum when converting from trigonometric coefficients: $P_{avg}=C_0^2+\frac{1}{2}\sum (a_n^2+b_n^2)$.
• Complex Exponent Conjugation Signs: Forgetting that the forward complex coefficient $C_n$ utilizes a negative exponent ($e^{-jn{\omega }_{0}t}$), whereas the reconstruction synthesis sum utilizes a positive exponent ($e^{jn{\omega }_{0}t}$).
• Harmonic Bounds Violation: Applying trigonometric symmetries without verifying if the signal is purely even or odd. If a signal is asymmetrical, both $a_n$ and $b_n$ terms must be computed.

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7.11 PYQ Bank — Verbatim Questions & Answer Plans

Q1: Define Fourier Series and state Dirichlet conditions [KUET 2022, 2018, 2017]

• Answer Plan:
  1. State the verbal definition of Fourier Series.
  2. Provide the standard trigonometric representation equations along with Euler-Fourier coefficient integrals.
  3. Formally list the three Dirichlet conditions (Absolute integrability, finite extrema, finite discontinuities).

Q2: Justify that odd functions have only sine term coefficients [KUET 2018, 2017]

• Answer Plan:
  1. Define odd symmetry mathematically: $f(-t)=-f(t)$.
  2. Set up the integral for $a_n$ and split boundaries into $[-T_0/2,0]$ and $[0,T_0/2]$.
  3. Perform variable substitution $u=-t$ and show that the two integrals cancel each other out to zero, proving $a_n=0$.

Q3: State and prove Parseval’s Identity for Fourier Series [KUET 2020, 2019]

• Answer Plan:
  1. Formulate the physical meaning of Parseval’s identity (power conservation between time and frequency domains).
  2. Write out the exponential and trigonometric power equations.
  3. Exhaustively write out the step-by-step proof using complex conjugates and interchanging integration and summation.

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7.12 Self-Check Before Moving On

• Can you write down the Trigonometric Fourier Series equations and Euler-Fourier coefficient integrals from memory?
• Do you know how to convert trigonometric coefficients ($a_n,b_n$) directly into complex exponential coefficients ($C_n$)?
• Can you outline the exact steps to prove that an odd function contains no cosine terms?
• Do you understand the physical significance of Parseval’s theorem in terms of real-world physical power?
• Can you describe what happens to the overshoot of a Gibbs phenomenon as the harmonic order $N\to \infty$?

Source: Signals and Systems (K. Deergha Rao) Chapter 3, Continuous-Time Signals and Systems (Senior Notes Ch 7).

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7.00 Chapter Map - Continuous-Time Fourier Series | 7.02 Symmetry Conditions & Waveform Analysis

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7.01 Trigonometric Fourier Series Representation

Core Idea

Continuous-Time Fourier Series (CTFS) decomposes any continuous-time periodic signal $f(t)$ into a linear combination of infinite, harmonically related sinusoidal components {sines and cosines}. Because sinusoidal waves represent the natural modes of linear time-invariant (LTI) systems, this trigonometric decomposition simplifies complex system analysis by shifting our analytical focus from a complicated time-domain waveform to a discrete, manageable set of frequency-domain coefficients {Euler-Fourier amplitudes}.

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1. Mathematical Definitions & Foundations

1.1 Periodicity and Fundamental Parameters

A continuous-time signal $f(t)$ is classified as periodic if there exists a positive, non-zero constant $T_0$ such that:

$f(t+T_0)=f(t)\text{for all }t\in \mathbb{R}$

The minimum positive value of $T_0$ that satisfies this condition is the fundamental period {the duration of one complete repetitive wave cycle}. From this, we extract two fundamental frequency parameters:

• Fundamental Cyclic Frequency ($f_0$): $f_0=\frac{1}{T_0}\text{[Hz]}$
• Fundamental Angular Frequency (${\omega }_0$): ${\omega }_0=2\pi f_0=\frac{2\pi }{T_0}\text{[rad/s]}$

Harmonics are defined as integer multiples of this fundamental frequency ($n{\omega }_0$ for $n=1,2,3,\dots$).

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1.2 Signal Vector Spaces & Orthogonality

To mathematically justify the summation of sinusoidal components, we treat continuous signals as vectors in an infinite-dimensional function space. The inner product {which measures the overlap or similarity} between two continuous functions $p(t)$ and $q(t)$ over an arbitrary interval $[\alpha ,\beta ]$ is defined as:

$⟨p,q⟩={\int }_{\alpha }^{\beta }p(t)q(t)dt$

Mathematical Property	Mathematical Condition	Physical Signal Meaning
Orthogonality	${\int }_{\alpha }^{\beta }p(t)q(t)dt=0(p\ne q)$	The signals share zero common frequency information; they are completely independent.
Orthonormality	${\int }_{\alpha }^{\beta }p(t{)}^{2}dt=1\text{and}⟨p,q⟩=0$	Symmetrical independence where each basis signal is scaled to contain exactly unit energy.

The Trigonometric Orthogonal Basis

The set of functions: $\left\{1,\cos ({\omega }_{0}t),\cos (2{\omega }_{0}t),\dots ,\sin ({\omega }_{0}t),\sin (2{\omega }_{0}t),\dots \right\}$ forms an orthogonal basis over any interval of length $T_0=\frac{2\pi }{{\omega }_0}$. This is proven by three fundamental integral identities for all non-zero integer harmonics $m$ and $n$:

1. \int_{t_0}^{t_0 + T_0} \cos(m\omega_0 t) \sin(n\omega_0 t) \, dt = 0 \quad \text{for all } m, n \tag{Identity 1}
2. \int_{t_0}^{t_0 + T_0} \cos(m\omega_0 t) \cos(n\omega_0 t) \, dt = \begin{cases} 0, & m \neq n \\ \frac{T_0}{2}, & m = n \end{cases} \tag{Identity 2}
3. \int_{t_0}^{t_0 + T_0} \sin(m\omega_0 t) \sin(n\omega_0 t) \, dt = \begin{cases} 0, & m \neq n \\ \frac{T_0}{2}, & m = n \end{cases} \tag{Identity 3}

[GRAPH: Visualizing Orthogonality. Plot of g(t) = sin(w_0 t)cos(2w_0 t) highlighting perfect odd symmetry over [-T_0/2, T_0/2], illustrating why the net integrated area under the curve cancels out to exactly zero] [193]

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2. Trigonometric Fourier Series Representation

Any periodic signal $f(t)$ satisfying the convergence criteria can be expanded into its Trigonometric Fourier Series:

f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 t) + \sum_{n=1}^{\infty} b_n \sin(n\omega_0 t) \tag{1}

where the constant and harmonic amplitudes are evaluated using the Euler-Fourier Analysis Equations over one period $T_0$ [157, 195, 248-250]:

• DC / Average Value ($a_0$): a_0 = \frac{2}{T_0} \int_{t_0}^{t_0+T_0} f(t) \, dt \tag{2}
• Cosine Coefficients ($a_n$): a_n = \frac{2}{T_0} \int_{t_0}^{t_0+T_0} f(t) \cos(n\omega_0 t) \, dt \tag{3}
• Sine Coefficients ($b_n$): b_n = \frac{2}{T_0} \int_{t_0}^{t_0+T_0} f(t) \sin(n\omega_0 t) \, dt \tag{4}

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2.1 Rigorous Mathematical Proof of Euler-Fourier Coefficients

We derive the coefficients by applying our trigonometric orthogonality identities to the general synthesis equation (Eq. 1) [196, 212-213].

Part A: Derivation of the DC Term ($a_0$)

To isolate $a_0$, integrate both sides of Eq. 1 over a full fundamental period $[0,T_0]$:

${\int }_0^{T_0}f(t)dt={\int }_0^{T_0}\left[\frac{a_0}{2}+{\sum }_{n=1}^{\infty }{a}_n\cos (n{\omega }_{0}t)+{\sum }_{n=1}^{\infty }{b}_n\sin (n{\omega }_{0}t)\right]dt$

Distributing the integration across individual terms yields:

${\int }_0^{T_0}f(t)dt=\frac{a_0}{2}{\int }_0^{T_0}dt+{\sum }_{n=1}^{\infty }{a}_n\underset{=0}{\underbrace{{\int }_0^{T_0}\cos (n{\omega }_{0}t)dt}}+{\sum }_{n=1}^{\infty }{b}_n\underset{=0}{\underbrace{{\int }_0^{T_0}\sin (n{\omega }_{0}t)dt}}$

Since sinusoidal waves integrated over an exact integer multiple of their period evaluate to zero, the harmonic summation terms vanish completely:

${\int }_0^{T_0}f(t)dt=\frac{a_0}{2}[t{]}_0^{T_0}=\frac{a_0{T}_0}{2}$

Isolating $a_0$ yields the official average component formula:

$a_0=\frac{2}{T_0}{\int }_0^{T_0}f(t)dt■$

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Part B: Derivation of the Cosine Amplitudes ($a_n$)

To isolate $a_n$ for a specific harmonic $n=m$, multiply Eq. 1 by $\cos (m{\omega }_{0}t)$ and integrate over $[0,T_0]$:

${\int }_0^{T_0}f(t)\cos (m{\omega }_{0}t)dt={\int }_0^{T_0}\left[\frac{a_0}{2}+{\sum }_{n=1}^{\infty }{a}_n\cos (n{\omega }_{0}t)+{\sum }_{n=1}^{\infty }{b}_n\sin (n{\omega }_{0}t)\right]\cos (m{\omega }_{0}t)dt$

${\int }_0^{T_0}f(t)\cos (m{\omega }_{0}t)dt=\frac{a_0}{2}\underset{=0}{\underbrace{{\int }_0^{T_0}\cos (m{\omega }_{0}t)dt}}+{\sum }_{n=1}^{\infty }{a}_n{\int }_0^{T_0}\cos (n{\omega }_{0}t)\cos (m{\omega }_{0}t)dt+{\sum }_{n=1}^{\infty }{b}_n\underset{=0\text{ (Identity 1)}}{\underbrace{{\int }_0^{T_0}\sin (n{\omega }_{0}t)\cos (m{\omega }_{0}t)dt}}$

By Identity 2, the cosine product integral evaluates to zero for all terms where $n\ne m$. The only term that survives the infinite summation is $n=m$:

${\int }_0^{T_0}f(t)\cos (n{\omega }_{0}t)dt=a_n\left(\frac{T_0}{2}\right)$

Isolating $a_n$ yields the cosine amplitude formula:

$a_n=\frac{2}{T_0}{\int }_0^{T_0}f(t)\cos (n{\omega }_{0}t)dt■$

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Part C: Derivation of the Sine Amplitudes ($b_n$)

Similarly, to isolate $b_n$ for a specific harmonic $n=m$, multiply Eq. 1 by $\sin (m{\omega }_{0}t)$ and integrate over $[0,T_0]$:

${\int }_0^{T_0}f(t)\sin (m{\omega }_{0}t)dt={\int }_0^{T_0}\left[\frac{a_0}{2}+{\sum }_{n=1}^{\infty }{a}_n\cos (n{\omega }_{0}t)+{\sum }_{n=1}^{\infty }{b}_n\sin (n{\omega }_{0}t)\right]\sin (m{\omega }_{0}t)dt$

Applying Identity 1 and Identity 3, all terms evaluate to zero except the sine product term where $n=m$:

${\int }_0^{T_0}f(t)\sin (n{\omega }_{0}t)dt=b_n\left(\frac{T_0}{2}\right)$

Isolating $b_n$ yields the sine amplitude formula:

$b_n=\frac{2}{T_0}{\int }_0^{T_0}f(t)\sin (n{\omega }_{0}t)dt■$

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3. Dirichlet Conditions for Series Convergence

A periodic signal $f(t)$ can be represented as a convergent trigonometric Fourier series if and only if it satisfies the following Dirichlet Conditions {the mathematical gatekeepers of Fourier existence}:

Dirichlet's Three Criteria

1. Absolute Integrability: $f(t)$ must be absolutely integrable over any single period: \int_{t_0}^{t_0 + T_0} |f(t)| \, dt < \infty \tag{Dirichlet 1} This guarantees that every computed coefficient $a_0,a_n,b_n$ remains finite.
2. Finite Extrema: $f(t)$ must have a finite number of local maxima and minima within any single period. This rules out extremely oscillatory functions like $\sin (1/t)$ near the origin.
3. Finite Discontinuities: $f(t)$ must have a finite number of discontinuities within any single period, and each discontinuity must be of finite height.

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4. High-Yield Solved Numericals

4.1 Example 1: The Symmetrical Square Wave [PYQ 2022 / 2017 - 12 Marks]

Question: Obtain the trigonometric Fourier series components of the periodic square wave signal of amplitude $A$ and period $T_0$ which is symmetrical with respect to the vertical axis at time $t=0$.

               x(t)
                |
          +-----+-----+                 +-----+
          |  A  |     |                 |     |
    ------|     |     +-----------------+     |-------> t
   -T0/2  |     |     |   -A            |     |  T0/2
    +-----+-----+-----+-----------------+-----+
         -T0/4  0    T0/4              3T0/4

Step-by-Step Mathematical Solution:

1. Formulate the Piecewise Waveform Equation: Over a single symmetrical period $\left[-\frac{T_0}{2},\frac{T_0}{2}\right]$, the signal is defined as: $f(t)=\{\begin{array}{ll}A,& -\frac{T_0}{4}<t<\frac{T_0}{4}\\ -A,& \frac{T_0}{4}<|t|<\frac{T_0}{2}\end{array}$

2. Evaluate the DC Coefficient ($a_0$): Calculate the average value over the period: $a_0=\frac{2}{T_0}{\int }_{-T_0/2}^{T_0/2}f(t)dt=\frac{2}{T_0}\left[{\int }_{-T_0/2}^{-T_0/4}-Adt+{\int }_{-T_0/4}^{T_0/4}Adt+{\int }_{T_0/4}^{T_0/2}-Adt\right]$ $a_0=\frac{2}{T_0}\left[-A\left(\frac{T_0}{4}\right)+A\left(\frac{T_0}{2}\right)-A\left(\frac{T_0}{4}\right)\right]=0$ Note: Because the positive area balances the negative area perfectly, the DC term is zero.

3. Analyze Symmetry: Notice that $f(-t)=f(t)$. The waveform is an even function. This immediately dictates that the sine coefficients must vanish: $b_n=0\text{for all }n$

4. Evaluate the Cosine Coefficients ($a_n$): Using the even symmetry shortcut to integrate over the half-period $[0,T_0/2]$: $a_n=\frac{4}{T_0}{\int }_0^{T_0/2}f(t)\cos (n{\omega }_{0}t)dt=\frac{4}{T_0}\left[{\int }_0^{T_0/4}A\cos (n{\omega }_{0}t)dt+{\int }_{T_0/4}^{T_0/2}-A\cos (n{\omega }_{0}t)dt\right]$ $a_n=\frac{4A}{T_0}\left[{\left[\frac{\sin (n{\omega }_{0}t)}{n{\omega }_0}\right]}_0^{T_0/4}-{\left[\frac{\sin (n{\omega }_{0}t)}{n{\omega }_0}\right]}_{T_0/4}^{T_0/2}\right]$ Substitute ${\omega }_0=\frac{2\pi }{T_0}$: $a_n=\frac{4A}{2\pi n}\left[\sin \left(\frac{n\pi }{2}\right)-\sin (0)-\sin (n\pi )+\sin \left(\frac{n\pi }{2}\right)\right]$ Since $\sin (n\pi )=0$ for all integers $n$: a_n = \frac{4A}{n\pi} \sin\left(\frac{n\pi}{2}\right) \tag{5}

5. Evaluate Harmonic Cases: We evaluate Eq. 5 for different integer values of $n$:

   • For Even $n$ ($n=2,4,6,\dots$): $\sin (n\pi /2)=\sin (\pi )=0⟹a_n=0$.
   • For $n=1,5,9,\dots$: $\sin (n\pi /2)=1⟹a_n=\frac{4A}{n\pi }$.
   • For $n=3,7,11,\dots$: $\sin (n\pi /2)=-1⟹a_n=-\frac{4A}{n\pi }$.

6. Write the Final Trigonometric Series: $f(t)=\frac{4A}{\pi }\left[\cos ({\omega }_{0}t)-\frac{1}{3}\cos (3{\omega }_{0}t)+\frac{1}{5}\cos (5{\omega }_{0}t)-\dots \right]■$

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4.2 Example 2: The Continuous Symmetrical Triangular Wave [PYQ 2021 - 3 Marks]

Question: Obtain the trigonometric Fourier series expansion for the periodic continuous triangular waveform of period $T_0=2$ and peak amplitude $A=2$ shown below.

               f(t)
                |
          _/\_  2  _/\_
         / \  \ | /  / \
    ----/---\--\|/--/---\----> t
       -1   -0.5| 0.5   1

Step-by-Step Mathematical Solution:

1. Formulate the Piecewise Waveform Equation: With period $T_0=2$, we find ${\omega }_0=\frac{2\pi }{2}=\pi$. Over one symmetrical period $[-1,1]$, the triangle wave with peak 2 is formulated as: $f(t)=\{\begin{array}{ll}-4t,& -0.5\le t<0\\ 4t,& 0\le t<0.5\end{array}$ Note: This waveform is an odd function when centered about its slope changes, but if we shift it to have odd symmetry $f(-t)=-f(t)$, the series contains only sine terms. Let us evaluate the standard odd triangular wave where $a_0=0$ and $a_n=0$. The sine coefficients are: $b_n=\frac{4}{T_0}{\int }_0^{T_0/2}f(t)\sin (n{\omega }_{0}t)dt=\frac{4}{2}{\int }_0^{0.5}4t\sin (n\pi t)dt$ $b_n=8{\int }_0^{0.5}t\sin (n\pi t)dt$

2. Apply Integration by Parts ($\int udv=uv-\int vdu$): Let $u=t⟹du=dt$, and let $dv=\sin (n\pi t)dt⟹v=-\frac{\cos (n\pi t)}{n\pi }$: $b_n=8\left[{\left[-\frac{t\cos (n\pi t)}{n\pi }\right]}_0^{0.5}-{\int }_0^{0.5}-\frac{\cos (n\pi t)}{n\pi }dt\right]$ $b_n=8\left[-\frac{0.5\cos (n\pi /2)}{n\pi }+0+{\left[\frac{\sin (n\pi t)}{n^2{\pi }^2}\right]}_0^{0.5}\right]$ $b_n=-\frac{4\cos (n\pi /2)}{n\pi }+\frac{8\sin (n\pi /2)}{n^2{\pi }^2}$ For odd harmonics where $\cos (n\pi /2)=0$: b_n = \frac{16}{n^2\pi^2} \sin\left(\frac{n\pi}{2}\right) \tag{6}

3. Evaluate Harmonic Amplitudes:

   • $b_1=\frac{16}{{\pi }^2}$
   • $b_3=-\frac{16}{9{\pi }^2}$
   • $b_5=\frac{16}{25{\pi }^2}$

4. Write the Final Trigonometric Series: f(t) = \frac{16}{\pi^2} \left[ \sin(\pi t) - \frac{1}{9}\sin(3\pi t) + \frac{1}{25}\sin(5\pi t) - \dots \right] \quad \blacksquare \tag{7}

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5. ECE 2108 Laboratory Realization

In the ECE 2108 Signals & Systems-1 lab, we verify CTFS convergence computationally. As the number of summated harmonics $N$ approaches infinity ($N\to \infty$), the Fourier approximation matches the ideal signal.

[GRAPH: 3D frequency visualization showing periodic time-domain square wave decomposing into separate sinusoidal harmonics, showcasing how adding progressive sinusoidal components builds up a perfect square edge] [157]

5.1 MATLAB Implementation for a Symmetrical Pulse Train

This MATLAB script constructs a periodic rectangular pulse train using a specified number of harmonics:

clc; 
clear all; 
N = 1000;                  % Number of points in simulation
t = linspace(0, 1, N);    % Time vector
A = 5;                     % Peak Amplitude
res = A/2;                 % Initialize with DC component (a_0/2) [122]
M = input('Enter number of harmonics for reconstruction = ');
 
for i = 1:2:M
    e = (2*A/pi)*(1/i)*sin(i*2*pi*1*t); % Calculate single harmonic term
    res = res + e;                      % Accumulate into running sum
    
    subplot(3, 1, 1);
    plot(t, e); hold on; title('Individual Sinusoidal Harmonics');
    
    subplot(3, 1, 2);
    plot(t, res); hold on; title('Accumulated Fourier Series Summation');
end
hold off;
 
subplot(3, 1, 3);
plot(t, res, 'k', 'linewidth', 2);
title(['Resultant Pulse Train with ', num2str(M), ' Harmonics']);
xlabel('Time (s)'); ylabel('Amplitude (V)');

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6. Common Mistakes That Cost Marks

The "a0 / 2" DC Division Trap

In the standard Fourier equation, the constant term is written as $\frac{a_0}{2}$. However, the integration formula calculates $a_0$ directly. A very common exam error is forgetting to divide the calculated $a_0$ value by 2 when writing down the final series. This doubles the DC offset, resulting in a 4 to 5 mark deduction.

• If $a_0=\frac{2}{T_0}\int f(t)dt$, you must write $\frac{a_0}{2}$ in the final expansion.
• Alternatively, some textbooks define $a_0=\frac{1}{T_0}\int f(t)dt$ and write $a_0$ in the series. Always stick to your instructor’s textbook convention to prevent penalty!

Confusing Period (T) with Half-Period (l) Limits

If your function uses spatial coordinates $[-l,l]$ instead of time $[0,T_0]$, the fundamental frequency is ${\omega }_0=\frac{\pi }{l}$ instead of $\frac{2\pi }{T_0}$. Mixing these up in integration limits and harmonic coefficients results in incorrect frequency scaling and causes the entire derivation to fail.

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7. PYQ Bank — Verbatim Questions & Answer Plans

7.1 PYQ 2022 / 2018 [8 Marks]

Question: Define Fourier series. State necessary and sufficient conditions for the existence of the Fourier series representation for a signal.

• Answer Plan:
  1. Define Fourier Series: Write out the formal synthesis equation (Eq. 1) and state that it decomposes any periodic, non-sinusoidal signal into a linear combination of harmonically related sines, cosines, and a constant DC offset.
  2. State Existence Conditions: List the three Dirichlet Conditions in full mathematical detail (Absolute Integrability, Finite Extrema, Finite Discontinuities) as structured in Section 3.

7.2 PYQ 2016 [4 Marks]

Question: “A periodic signal can be represented as a summation of a number of sinusoidal wave with different frequencies” — Justify the statement.

• Answer Plan:
  1. Write the Trigonometric Fourier Series equation (Eq. 1).
  2. Explain that the basis set of sines and cosines is orthogonal over a period $T_0$, meaning no individual harmonic component overlaps or contains energy from any other harmonic.
  3. Outline the Euler-Fourier coefficient formulas (Eqs. 2–4), demonstrating that we can uniquely isolate and compute the exact amplitude contribution ($a_n,b_n$) at each specific discrete frequency $n{\omega }_0$.

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8. Self-Check Before Moving On

• Can you write down the general trigonometric Fourier series synthesis equation from memory?
• Do you know how to prove that the set of sines and cosines is mathematically orthogonal over a fundamental period?
• Can you rigorously derive the integration formula for the cosine harmonic coefficient $a_n$ using inner product integration?
• Are you able to state all three Dirichlet conditions for CTFS convergence?
• Do you know how to check the final series to ensure the DC coefficient is properly scaled by $1/2$?

---

Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 01 Fourier Series.pdf, ECE 2108 Signal & Systems-1 (1).pdf (Experiment 3), Rabiul sir class note.pdf.

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7.01 Trigonometric Fourier Series Representation | 7.03 Exponential Fourier Series & Complex Spectra

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7.02 Symmetry Conditions & Waveform Analysis

Core Idea

Calculating trigonometric Fourier series coefficients ($a_0,a_n,b_n$) by direct integration is computationally expensive and error-prone. By analyzing the geometric symmetry of a periodic waveform, we can predict beforehand which coefficients will be identically zero. This note covers the mathematical proofs, conditions, and waveforms for Even, Odd, Half-Wave (Even/Odd), and Quarter-Wave symmetries, establishing powerful shortcut methods that save crucial time in exams.

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1. Symmetry Classification Matrix

By examining how a signal behaves under time-reversal ($t\to -t$) or half-period shifts ($t\to t\pm T_0/2$), we classify its symmetry:

Symmetry Type	Mathematical Condition	Coefficients Set to Zero	Surviving Terms
Even Symmetry	$f(-t)=f(t)$	$b_n=0$ for all $n$	DC ($a_0$) and Cosines ($a_n$)
Odd Symmetry	$f(-t)=-f(t)$	$a_0=0,a_n=0$ for all $n$	Sines ($b_n$) only
Half-Wave Symmetry	$f\left(t\pm \frac{T_0}{2}\right)=-f(t)$	$a_0=0$, and $a_n=b_n=0$ for even $n$	Odd harmonics only ($n=1,3,5,\dots$)
Even Harmonics	$f\left(t\pm \frac{T_0}{2}\right)=f(t)$	$a_n=b_n=0$ for odd $n$	Even harmonics only ($n=2,4,6,\dots$)

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2. Rigorous Mathematical Proofs

2.1 Prove that Odd Functions Contain Only Sine Coefficients [PYQ 2018, 2017 - 6 Marks]

Theorem: If a periodic signal $f(t)$ is odd, its Fourier series contains no DC offset ($a_0=0$) and no cosine terms ($a_n=0$), leaving only sine terms ($b_n$).

Proof: By definition, a signal is odd if $f(-t)=-f(t)$.

Step 1: Proof for $a_0=0$

The DC coefficient is given by: $a_0=\frac{2}{T_0}{\int }_{-T_0/2}^{T_0/2}f(t)dt$ Split the integration interval into the negative half and positive half: $a_0=\frac{2}{T_0}\left[{\int }_{-T_0/2}^{0}f(t)dt+{\int }_0^{T_0/2}f(t)dt\right]$ In the first integral, perform the change of variable $\tau =-t⟹dt=-d\tau$. The limits of integration change from $[-T_0/2,0]$ to $[T_0/2,0]$: ${\int }_{-T_0/2}^{0}f(t)dt={\int }_{T_0/2}^{0}f(-\tau )(-d\tau )={\int }_0^{T_0/2}f(-\tau )d\tau$ Substitute the odd condition $f(-\tau )=-f(\tau )$: ${\int }_0^{T_0/2}f(-\tau )d\tau ={\int }_0^{T_0/2}-f(\tau )d\tau =-{\int }_0^{T_0/2}f(t)dt$ Substitute this back into the expression for $a_0$: $a_0=\frac{2}{T_0}\left[-{\int }_0^{T_0/2}f(t)dt+{\int }_0^{T_0/2}f(t)dt\right]=0■$

Step 2: Proof for $a_n=0$

The cosine coefficient is given by: $a_n=\frac{2}{T_0}{\int }_{-T_0/2}^{T_0/2}f(t)\cos (n{\omega }_{0}t)dt$ Because $f(t)$ is odd and $\cos (n{\omega }_{0}t)$ is even, let us analyze the parity of their product: $g(t)=f(t)\cos (n{\omega }_{0}t)$ $g(-t)=f(-t)\cos (-n{\omega }_{0}t)=-f(t)\cos (n{\omega }_{0}t)=-g(t)$ The integrand $g(t)$ is purely odd. Since the definite integral of any odd function over symmetrical limits is zero: $a_n=0■$

Step 3: Derivation of the Simplified $b_n$ Formula

The sine coefficient is given by: $b_n=\frac{2}{T_0}{\int }_{-T_0/2}^{T_0/2}f(t)\sin (n{\omega }_{0}t)dt$ The product of two odd functions is even: $h(t)=f(t)\sin (n{\omega }_{0}t)⟹h(-t)=f(-t)\sin (-n{\omega }_{0}t)=[-f(t)][-\sin (n{\omega }_{0}t)]=f(t)\sin (n{\omega }_{0}t)=h(t)$ Using the symmetric integration rule for even functions: $b_n=\frac{4}{T_0}{\int }_0^{T_0/2}f(t)\sin (n{\omega }_{0}t)dt■$

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2.2 Prove that Half-Wave Symmetric Signals Have No Even Harmonics [PYQ 2020, 2019 - 4 Marks]

Theorem: If a periodic signal satisfies $f\left(t-\frac{T_0}{2}\right)=-f(t)$, then its Fourier coefficients $a_n=b_n=0$ for all even values of $n$ ($n=2,4,6,\dots$).

Proof: The complex exponential Fourier coefficient is: $C_n=\frac{1}{T_0}{\int }_0^{T_0}f(t)e^{-jn{\omega }_{0}t}dt$ Split the integration over the two halves of the period: $C_n=\frac{1}{T_0}\left[{\int }_0^{T_0/2}f(t)e^{-jn{\omega }_{0}t}dt+{\int }_{T_0/2}^{T_0}f(t)e^{-jn{\omega }_{0}t}dt\right]$ For the second integral, apply the shift variable $\tau =t-\frac{T_0}{2}⟹t=\tau +\frac{T_0}{2}$ and $dt=d\tau$. The limits $[T_0/2,T_0]$ map to $[0,T_0/2]$: ${\int }_{T_0/2}^{T_0}f(t)e^{-jn{\omega }_{0}t}dt={\int }_0^{T_0/2}f\left(\tau +\frac{T_0}{2}\right)e^{-jn{\omega }_0(\tau +T_0/2)}d\tau$ Apply the half-wave symmetry condition $f\left(\tau +\frac{T_0}{2}\right)=-f(\tau )$: ${\int }_0^{T_0/2}-f(\tau )e^{-jn{\omega }_0\tau }{e}^{-jn{\omega }_0{T}_0/2}d\tau$ Since ${\omega }_0{T}_0=2\pi$, the exponential factor scales to: $e^{-jn{\omega }_0{T}_0/2}=e^{-jn(2\pi )/2}=e^{-jn\pi }=(-1{)}^n$ Substitute this back: ${\int }_{T_0/2}^{T_0}f(t)e^{-jn{\omega }_{0}t}dt=-(-1{)}^n{\int }_0^{T_0/2}f(\tau )e^{-jn{\omega }_0\tau }d\tau$ Reassemble $C_n$: $C_n=\frac{1}{T_0}\left[{\int }_0^{T_0/2}f(t)e^{-jn{\omega }_{0}t}dt-(-1{)}^n{\int }_0^{T_0/2}f(t)e^{-jn{\omega }_{0}t}dt\right]$ $C_n=\frac{1-(-1{)}^n}{T_0}{\int }_0^{T_0/2}f(t)e^{-jn{\omega }_{0}t}dt$ Analyze the cases for $n$:

• Case 1: $n$ is Even $⟹(-1{)}^n=1$: $C_n=\frac{1-1}{T_0}{\int }_0^{T_0/2}f(t)e^{-jn{\omega }_{0}t}dt=0■$
• Case 2: $n$ is Odd $⟹(-1{)}^n=-1$: $C_n=\frac{2}{T_0}{\int }_0^{T_0/2}f(t)e^{-jn{\omega }_{0}t}dt$ Thus, even harmonics are identically zero. Since $C_n=\frac{a_n-j{b}_n}{2}$, this guarantees $a_n=b_n=0$ for all even $n$.

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3. High-Yield Worked Examples (The Exam Killers)

3.1 Symmetrical Rectangular Square Wave [PYQ 2022, 2017 - 12 Marks]

Question: Obtain the Fourier components of the periodic square wave signal which is symmetrical with respect to the vertical axis at $t=0$, as shown below:

             +A |
       +--------+--------+              T_0 = Period
       |        |        |              Duty Cycle = 50%
  -T/2 |        |        | T/2
  o----+--------+--------+----o t
  -T/4 |                 | T/4
       |                 |
       +-----------------+
             -A |

Step 1: Analyze Waveform Symmetry

• Even Symmetry: The waveform is symmetric across the vertical vertical axis: $f(-t)=f(t)$. Therefore, $b_n=0$.
• Zero average area: Since the positive area matches the negative area over one cycle, the DC term $a_0=0$.
• Half-Wave Symmetry: Shifting the waveform by $T_0/2$ and flipping it vertically yields the exact same wave: $f(t-T_0/2)=-f(t)$. Therefore, $a_n=0$ for all even $n$.

Step 2: Formulate the Piecewise Function over one period $[-T_0/2,T_0/2]$:

$f(t)=\{\begin{array}{ll}-A,& -\frac{T_0}{2}<t<-\frac{T_0}{4}\\ A,& -\frac{T_0}{4}\le t<\frac{T_0}{4}\\ -A,& \frac{T_0}{4}\le t<\frac{T_0}{2}\end{array}$

Step 3: Calculate the surviving odd Cosine Coefficients ($a_n$ for odd $n$)

Using the even symmetry shortcut: $a_n=\frac{4}{T_0}{\int }_0^{T_0/2}f(t)\cos (n{\omega }_{0}t)dt$ Split the integral from $0$ to $T_0/2$: $a_n=\frac{4}{T_0}\left[{\int }_0^{T_0/4}A\cos (n{\omega }_{0}t)dt+{\int }_{T_0/4}^{T_0/2}(-A)\cos (n{\omega }_{0}t)dt\right]$ $a_n=\frac{4A}{T_0}\left[{\frac{\sin (n{\omega }_{0}t)}{n{\omega }_0}|}_0^{T_0/4}-{\frac{\sin (n{\omega }_{0}t)}{n{\omega }_0}|}_{T_0/4}^{T_0/2}\right]$ Substitute ${\omega }_0=\frac{2\pi }{T_0}⟹{\omega }_0\frac{T_0}{4}=\frac{\pi }{2}$ and ${\omega }_0\frac{T_0}{2}=\pi$: $a_n=\frac{4A}{n{\omega }_0{T}_0}\left[\sin \left(\frac{n\pi }{2}\right)-0-\left(\sin (n\pi )-\sin \left(\frac{n\pi }{2}\right)\right)\right]$ Since ${\omega }_0{T}_0=2\pi$ and $\sin (n\pi )=0$ for all integers $n$: $a_n=\frac{4A}{2\pi n}\left[\sin \left(\frac{n\pi }{2}\right)+\sin \left(\frac{n\pi }{2}\right)\right]=\frac{4A}{n\pi }\sin \left(\frac{n\pi }{2}\right)$

Step 4: Evaluate the Harmonic cases

• For even $n$ ($n=2,4,6,\dots$) $⟹\sin (n\pi /2)=0$: $a_n=0\text{perfect agreement with Half-Wave Symmetry!}$
• For odd $n$ ($n=1,3,5,\dots$):
  • $n=1⟹a_1=\frac{4A}{\pi }\sin (\pi /2)=\frac{4A}{\pi }$
  • $n=3⟹a_3=\frac{4A}{3\pi }\sin (3\pi /2)=-\frac{4A}{3\pi }$
  • $n=5⟹a_5=\frac{4A}{5\pi }\sin (5\pi /2)=\frac{4A}{5\pi }$

Step 5: Write the final Trigonometric Fourier Series:

$f(t)=\frac{4A}{\pi }\left[\cos ({\omega }_{0}t)-\frac{1}{3}\cos (3{\omega }_{0}t)+\frac{1}{5}\cos (5{\omega }_{0}t)-\frac{1}{7}\cos (7{\omega }_{0}t)+\dots \right]$

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3.2 Half-Wave Rectified Sine Wave [Textbook Classic - Fig E2.3]

Question: Obtain the trigonometric Fourier series for the half-wave rectified sine wave shown below:

     Amplitude
        ^
      A |      _              _
        |    /   \          /   \
        |   /     \        /     \
      0 +--o-------o------+-------o----> t
        0        T/2      T     3T/2

Step 1: Analyze Symmetry Conditions

• Even/Odd Symmetries: The waveform is neither symmetric about the vertical vertical axis nor anti-symmetric about the origin. Thus, both $a_n$ and $b_n$ are expected to exist.
• Half-Wave Symmetry: Since $f(t-T/2)\ne -f(t)$ (the negative half-cycle is flat zero, not inverted), the waveform does not possess half-wave symmetry. Thus, even harmonics will exist.

Step 2: Formulate the piecewise definition over one period $[0,T_0]$:

$f(t)=\{\begin{array}{ll}A\sin ({\omega }_{0}t),& 0\le t<\frac{T_0}{2}\\ 0,& \frac{T_0}{2}\le t<T_0\end{array}$

Step 3: Evaluate the DC Term ($a_0$):

$a_0=\frac{2}{T_0}{\int }_0^{T_0/2}f(t)dt=\frac{2}{T_0}{\int }_0^{T_0/2}A\sin ({\omega }_{0}t)dt=\frac{2A}{T_0}{\left[\frac{-\cos ({\omega }_{0}t)}{{\omega }_0}\right]}_0^{T_0/2}$ Substitute ${\omega }_0{T}_0=2\pi ⟹{\omega }_0\frac{T_0}{2}=\pi$: $a_0=\frac{2A}{2\pi }[-\cos (\pi )+\cos (0)]=\frac{A}{\pi }[-(-1)+1]=\frac{2A}{\pi }$ The constant baseline DC offset is $\frac{a_0}{2}=\frac{A}{\pi }$.

Step 4: Calculate the Cosine Coefficients ($a_n$)

$a_n=\frac{2}{T_0}{\int }_0^{T_0/2}A\sin ({\omega }_{0}t)\cos (n{\omega }_{0}t)dt$ Using the trigonometric product-to-sum identity $\sin (x)\cos (y)=\frac{1}{2}[\sin (x+y)+\sin (x-y)]$: $a_n=\frac{A}{T_0}{\int }_0^{T_0/2}[\sin (1+n){\omega }_{0}t+\sin (1-n){\omega }_{0}t]dt$ For $n\ne 1$: $a_n=\frac{A}{T_0}{\left[\frac{-\cos (1+n){\omega }_{0}t}{(1+n){\omega }_0}+\frac{-\cos (1-n){\omega }_{0}t}{(1-n){\omega }_0}\right]}_0^{T_0/2}$ $a_n=\frac{A}{2\pi }\left[\frac{1-\cos (n+1)\pi }{n+1}-\frac{1-\cos (n-1)\pi }{n-1}\right]$

• If $n$ is odd: $\cos (n+1)\pi =1$ and $\cos (n-1)\pi =1⟹{\mathbf{a}}_{\mathbf{n}}=0$.
• If $n$ is even: $\cos (n+1)\pi =-1$ and $\cos (n-1)\pi =-1⟹$: $a_n=\frac{A}{2\pi }\left[\frac{2}{n+1}-\frac{2}{n-1}\right]=\frac{A}{\pi }\left[\frac{(n-1)-(n+1)}{n^2-1}\right]=-\frac{2A}{\pi ({\mathbf{n}}^2-1)}\text{for even }n$

Special Case: For $n=1$: $a_1=\frac{2}{T_0}{\int }_0^{T_0/2}A\sin ({\omega }_{0}t)\cos ({\omega }_{0}t)dt=\frac{A}{T_0}{\int }_0^{T_0/2}\sin (2{\omega }_{0}t)dt=\frac{A}{T_0}{\frac{-\cos (2{\omega }_{0}t)}{2{\omega }_0}|}_0^{T_0/2}=0$

Step 5: Calculate the Sine Coefficients ($b_n$)

$b_n=\frac{2}{T_0}{\int }_0^{T_0/2}A\sin ({\omega }_{0}t)\sin (n{\omega }_{0}t)dt$ Using product identity $\sin (x)\sin (y)=\frac{1}{2}[\cos (x-y)-\cos (x+y)]$: $b_n=\frac{A}{T_0}{\int }_0^{T_0/2}[\cos (1-n){\omega }_{0}t-\cos (1+n){\omega }_{0}t]dt$ For $n\ne 1$: b_n = \frac{A}{T_0} \left[ \frac{\sin(1-n)\omega_0 t}{(1-n)\omega_0} - \frac{\sin(1+n)\omega_0 t}{(1+n)\omega_0} \right]_{0}^{T_0/2} = 0 \quad \text{{since } \sin(k\pi) = 0}

Special Case: For $n=1$: $b_1=\frac{2}{T_0}{\int }_0^{T_0/2}A{\sin }^2({\omega }_{0}t)dt=\frac{A}{T_0}{\int }_0^{T_0/2}[1-\cos (2{\omega }_{0}t)]dt=\frac{A}{T_0}\left[\frac{T_0}{2}-0\right]=\frac{\mathbf{A}}{2}$

Step 6: Assemble the final half-wave rectified series:

$f(t)=\frac{A}{\pi }+\frac{A}{2}\sin ({\omega }_{0}t)-\frac{2A}{\pi }\left[\frac{1}{3}\cos (2{\omega }_{0}t)+\frac{1}{15}\cos (4{\omega }_{0}t)+\frac{1}{35}\cos (6{\omega }_{0}t)+\dots \right]$

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4. Common Mistakes That Cost Marks

The Half-Period Integration Boundary Oversight

For symmetrical integrations (like even/odd shortcut equations), the coefficient multipliers double (becoming $4/T_0$) while the limits of integration reduce by half (integrating from $0$ to $T_0/2$). A very common exam failure is writing the $4/T_0$ multiplier but integrating over the full period $T_0$ anyway, which doubles the correct value and yields a zero-grade for the question.

The Non-Symmetric Period Origin Shift Trap

Symmetries (even and odd) are defined relative to the vertical time origin $t=0$. If a periodic waveform is physically shifted in time such that its axis of symmetry is no longer at $t=0$, you cannot use the $a_n=0$ or $b_n=0$ shortcuts directly. You must either solve it with the full integrals or mathematically shift the time axis ($t^{\prime }=t-t_d$) to apply symmetry shortcuts.

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5. PYQ Bank — Verbatim Questions & Answer Plans

5.1 PYQ 2017 Question 4b [6 Marks]

Question: With regard to Fourier series representation, justify that odd function have only sine term coefficients.

• Answer Plan:
  1. Define the trigonometric Fourier series and state the odd function criterion $f(-t)=-f(t)$.
  2. Write down the full proof showing $a_0=0$ and $a_n=0$ as derived step-by-step in Section 2.1.
  3. Derivate the surviving simplified formula for $b_n=\frac{4}{T_0}{\int }_0^{T_0/2}f(t)\sin (n{\omega }_{0}t)dt$ to demonstrate completeness.

5.2 PYQ 2020 Question 4a [2 Marks]

Question: What is meant by Fourier series expansion? What are the “Symmetry Conditions” of Fourier series?

• Answer Plan:
  1. Define Fourier Series Expansion as the decomposition of any continuous-time periodic signal into a linear combination of mutually orthogonal sines, cosines, and a constant DC offset.
  2. Define Symmetry Conditions as the structural parity properties of periodic signals (Even, Odd, and Half-Wave) that force certain coefficients to zero.
  3. Reproduce the Symmetry Classification Matrix from Section 1 to secure full marks.

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6. Self-Check Before Moving On

• Can you prove mathematically why $a_n=0$ for any periodic function that is purely odd? [2.1]
• Do you know what happens to the Fourier coefficients if a signal has half-wave symmetry?
• Why does a half-wave rectified sine wave have both sine and cosine coefficients? [3.2]
• Can you evaluate the Fourier series of a symmetrical square wave on a whiteboard? [3.1]

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Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 01 Fourier Series.pdf, Rabiul sir class note.pdf.

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7.02 Symmetry Conditions & Waveform Analysis | 7.04 Parseval’s Theorem & Convergence Conditions

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7.03 Exponential Fourier Series & Complex Spectra

Core Idea

While the Trigonometric Fourier Series decomposes a periodic signal into real-valued sines and cosines, the Complex Exponential Fourier Series uses Euler’s relation to represent periodic waveforms using complex exponential functions. This formulation is much more compact and mathematically elegant because it consolidates both amplitude and phase information into a single set of complex coefficients $C_n$ {each representing a phasor at a specific harmonic frequency}. These coefficients extend across both positive and negative frequencies, yielding a symmetrical Double-Sided Complex Spectrum that serves as the perfect mathematical bridge to the continuous Fourier Transform.

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1. The Physics of Complex Exponential Harmonics

A continuous-time periodic signal $x(t)$ with a fundamental period $T_0$ repeats itself for all time: $x(t)=x(t+T_0)$. Its fundamental angular frequency is defined as: \Omega_0 = rac{2\pi}{T_0} \quad ext{[rad/s]}

Under the complex exponential framework, we construct a set of harmonically related complex exponential signals of the form: ${\varphi }_n(t)=e^{jn{\Omega }_{0}t},n\in \mathbb{Z}\{0,\pm 1,\pm 2,\dots \}$

1.1 The Orthogonality of Complex Exponentials

Just as sines and cosines form an orthogonal basis, the complex exponential set is orthogonal over any full period $T_0$:

eq m \end{cases}$$ This means that multiplying any harmonic by the complex conjugate of another harmonic and integrating over a period yields zero if their frequencies differ, indicating that harmonically related complex exponentials share **zero mutual information**. --- ## 2. The Core Mathematical Formulation The **Complex Exponential Fourier Series** represents a periodic signal $x(t)$ as an infinite linear combination of these orthogonal complex exponential phasors: $$x(t) = \sum_{n=-\infty}^{\infty} C_n e^{j n \Omega_0 t} \quad ext{--- [The Synthesis Equation]}$$ To calculate the complex Fourier coefficients $C_n$, we use the corresponding analysis integral over any convenient single period interval of duration $T_0$: $$C_n = rac{1}{T_0} \int_{T_0} x(t) e^{-j n \Omega_0 t} \, dt \quad ext{--- [The Analysis Equation]}$$ ### 2.1 Tabular Breakdown of Formulation Variables | Variable | Mathematical Role | Physical Meaning | Dimension / Unit | | :--- | :--- | :--- | :--- | | **$x(t)$** | Left-hand side function | Continuous-time periodic wave | V, A, etc. | | **$C_n$** | Complex scaling multiplier | Harmonic phasor coefficient | Same as $x(t)$ | | **$n$** | Summation integer index | Harmonic index number (positive/negative) | Unitless | | **$\Omega_0$** | Fundamental angular frequency | Spacing between adjacent spectral lines | rad/s | | **$e^{j n \Omega_0 t}$** | Harmonically related basis function | Symmetrical rotating vector in complex plane | Unitless | --- ## 3. Rigorous Trigonometric-to-Exponential Derivation This is a **high-yield 6-mark theoretical derivation** that is frequently tested in Section B of ECE 2107 exams. ### The Proof: 1. **State the Trigonometric Fourier Series Baseline:** $$x(t) = rac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(n \Omega_0 t) + \sum_{n=1}^{\infty} b_n \sin(n \Omega_0 t) \quad ext{--- (Equation 1)}$$ 2. **Substitute Euler's Identities:** Replace the real-valued trigonometric terms with their complex exponential equivalents using Euler's formulas: $$\cos(n \Omega_0 t) = rac{e^{j n \Omega_0 t} + e^{-j n \Omega_0 t}}{2}$$ $$\sin(n \Omega_0 t) = rac{e^{j n \Omega_0 t} - e^{-j n \Omega_0 t}}{2j}$$ 3. **Incorporate into the Series expansion:** $$x(t) = rac{a_0}{2} + \sum_{n=1}^{\infty} a_n \left( rac{e^{j n \Omega_0 t} + e^{-j n \Omega_0 t}}{2} ight) + \sum_{n=1}^{\infty} b_n \left( rac{e^{j n \Omega_0 t} - e^{-j n \Omega_0 t}}{2j} ight)$$ 4. **Group the terms by exponent sign:** Recall that $rac{1}{j} = -j$. Re-algebraize the coefficients of the positive exponentials ($e^{j n \Omega_0 t}$) and the negative exponentials ($e^{-j n \Omega_0 t}$): $$x(t) = rac{a_0}{2} + \sum_{n=1}^{\infty} \left( rac{a_n - j b_n}{2} ight) e^{j n \Omega_0 t} + \sum_{n=1}^{\infty} \left( rac{a_n + j b_n}{2} ight) e^{-j n \Omega_0 t}$$ 5. **Define the Symmetrical Complex Coefficients $C_n$:** To simplify this expression into a single consolidated sum from $-\infty$ to $+\infty$, we define: * **For $n = 0$ (The DC offset):** $$C_0 = rac{a_0}{2}$$ * **For $n > 0$ (Positive harmonics):** $$C_n = rac{a_n - j b_n}{2}$$ * **For $n < 0$ (Negative harmonics):** Let $m = -n > 0$: $$C_n = C_{-m} = rac{a_m + j b_m}{2} = C_m^*$$ 6. **Assemble the Symmetrical Compact Sum:** By mapping these definitions, the negative exponent sum merges with the positive exponent sum to yield the single unified expression: $$x(t) = \sum_{n=-\infty}^{\infty} C_n e^{j n \Omega_0 t} \quad ext{(Q.E.D.)}$$ --- ## 4. Complex Spectral Plots (Magnitude & Phase) Because the complex coefficients $C_n$ are generally complex numbers, they must be represented in polar form to be plotted physically: $$C_n = |C_n| e^{j ngle C_n}$$ * **Magnitude Spectrum ($|C_n|$):** A plot of the magnitude of each harmonic phasor versus frequency $n \Omega_0$. For any real-valued signal, the magnitude spectrum is **strictly symmetric (even)**: $$|C_{-n}| = |C_n|$$ * **Phase Spectrum ($ngle C_n$):** A plot of the phase angle of each harmonic phasor versus frequency $n \Omega_0$. For any real-valued signal, the phase spectrum is **strictly anti-symmetric (odd)**: $$ngle C_{-n} = -ngle C_n$$ ``` Magnitude Spectrum |Cn| Phase Spectrum ngle Cn | | +pi/2 (90 deg) 0.75 (DC) | o | | | o | o 0.216 +----+----o---> n | | | -pi/2| | --+---+---+-----> n (Harmonics) (o) | -1 0 1 | ``` --- ## 5. High-Yield Worked Examples (The Exam Killers) ### 5.1 The 12-Mark Half-Wave Rectified Cosine [PYQ 2021] **Question:** Determine the complex exponential Fourier series for a half-wave rectified cosine signal with amplitude $A$ and fundamental period $T_0$. ``` Amplitude = A _ _ _ _ / \ / -----------+ +-----------+ +----------- -T0/4 T0/4 3T0/4 5T0/4 <-------- Period T0 ----------> ``` #### Step-by-Step Solution: 1. **Formulate the Piecewise Mathematical Model:** For a single period centered around the origin $t = 0$: $$x(t) = egin{cases} A \cos(\Omega_0 t), & -rac{T_0}{4} \le t \le rac{T_0}{4} \ 0, & -rac{T_0}{2} \le t < -rac{T_0}{4} \quad ext{and} \quad rac{T_0}{4} < t \le rac{T_0}{2} \end{cases}$$ 2. **Evaluate the Complex Fourier Integral:** Using the analysis equation over the interval $[-rac{T_0}{2}, rac{T_0}{2}]$: $$C_n = rac{1}{T_0} \int_{-T_0/4}^{T_0/4} A \cos(\Omega_0 t) e^{-j n \Omega_0 t} \, dt$$ Substitute $\cos(\Omega_0 t) = rac{e^{j \Omega_0 t} + e^{-j \Omega_0 t}}{2}$: $$C_n = rac{A}{2 T_0} \int_{-T_0/4}^{T_0/4} \left( e^{j \Omega_0 t} + e^{-j \Omega_0 t} ight) e^{-j n \Omega_0 t} \, dt$$ $$C_n = rac{A}{2 T_0} \int_{-T_0/4}^{T_0/4} \left[ e^{-j(n-1)\Omega_0 t} + e^{-j(n+1)\Omega_0 t} ight] \, dt$$ 3. **Evaluate for the Specific Case $n = \pm 1$:** Substitute $n = 1$: $$C_1 = rac{A}{2 T_0} \int_{-T_0/4}^{T_0/4} \left( 1 + e^{-j 2 \Omega_0 t} ight) \, dt$$ $$C_1 = rac{A}{2 T_0} \left[ t + rac{e^{-j 2 \Omega_0 t}}{-j 2 \Omega_0} ight]_{-T_0/4}^{T_0/4}$$ Recall that $\Omega_0 = rac{2\pi}{T_0} \implies 2 \Omega_0 \left(rac{T_0}{4} ight) = \pi$: $$C_1 = rac{A}{2 T_0} \left[ rac{T_0}{2} + rac{e^{-j \pi} - e^{j \pi}}{-j 2 \Omega_0} ight]$$ Since $e^{-j \pi} = e^{j \pi} = -1 \implies e^{-j\pi} - e^{j\pi} = 0$: $$C_1 = rac{A}{2 T_0} \left( rac{T_0}{2} + 0 ight) = rac{A}{4}$$ Since the signal is real and even, we are guaranteed that $C_{-1} = C_1^* = rac{A}{4}$. 4. **Evaluate for general $n eq \pm 1$:** $$C_n = rac{A}{2 T_0} \left[ rac{e^{-j(n-1)\Omega_0 t}}{-j(n-1)\Omega_0} + rac{e^{-j(n+1)\Omega_0 t}}{-j(n+1)\Omega_0} ight]_{-T_0/4}^{T_0/4}$$ Substitute the upper and lower limits $\Omega_0 \left(rac{T_0}{4} ight) = rac{\pi}{2}$: $$C_n = rac{A}{2 T_0 \Omega_0} \left[ rac{e^{-j(n-1)\pi/2} - e^{j(n-1)\pi/2}}{-j(n-1)} + rac{e^{-j(n+1)\pi/2} - e^{j(n+1)\pi/2}}{-j(n+1)} ight]$$ Substitute $T_0 \Omega_0 = 2\pi$ and simplify the complex fractions using $rac{e^{j heta} - e^{-j heta}}{2j} = \sin( heta)$: $$C_n = rac{A}{2\pi} \left[ rac{\sin((n-1)\pi/2)}{n-1} + rac{\sin((n+1)\pi/2)}{n+1} ight]$$ Applying trigonometric angle-addition identities: * $\sin\left(rac{n\pi}{2} - rac{\pi}{2} ight) = -\cos\left(rac{n\pi}{2} ight)$ * $\sin\left(rac{n\pi}{2} + rac{\pi}{2} ight) = \cos\left(rac{n\pi}{2} ight)$ $$C_n = rac{A \cos(n\pi/2)}{2\pi} \left[ -rac{1}{n-1} + rac{1}{n+1} ight]$$ $$C_n = rac{A \cos(n\pi/2)}{2\pi} \left[ rac{-(n+1) + (n-1)}{n^2 - 1} ight] = rac{A \cos(n\pi/2)}{2\pi} \left[ rac{-2}{n^2 - 1} ight]$$ $$\mathbf{C_n = rac{A \cos(n\pi/2)}{\pi(1 - n^2)}} \quad (n eq \pm 1)$$ 5. **Write out the Final Complex Exponential Fourier Series:** $$x(t) = rac{A}{4} e^{j \Omega_0 t} + rac{A}{4} e^{-j \Omega_0 t} + \sum_{n=-\infty, n eq \pm 1}^{\infty} \left[ rac{A \cos(n\pi/2)}{\pi(1 - n^2)} ight] e^{j n \Omega_0 t}$$ --- ### 5.2 The 3-Mark Spectral Identification Question [PYQ 2018] **Question:** The complex exponential Fourier representation of a signal $x(t)$ over the period $(0, T)$ is: $$x(t) = \sum_{n=-\infty}^{\infty} rac{3}{4+(n\pi)^2} e^{j 3 n \pi t}$$ Determine: (i) the fundamental period $T$, and (ii) plot the double-sided amplitude and phase spectrum. #### Step-by-Step Solution: 1. **Identify the Fundamental Frequency:** Compare the exponent term of the given equation directly to the standard Synthesis Equation: $$e^{j n \Omega_0 t} = e^{j 3 n \pi t} \implies n \Omega_0 t = 3 n \pi t \implies \Omega_0 = 3\pi ext{ rad/s}$$ 2. **Calculate the Period $T$:** $$\Omega_0 = rac{2\pi}{T} \implies 3\pi = rac{2\pi}{T} \implies \mathbf{T = rac{2}{3} ext{ s}}$$ 3. **Calculate Spectral Magnitude Values $|C_n|$:** The complex coefficient is purely real and positive for all $n$: $$C_n = rac{3}{4+(n\pi)^2} \implies |C_n| = rac{3}{4+(n\pi)^2}$$ * For $n=0$: $|C_0| = rac{3}{4} = \mathbf{0.75}$ * For $n = \pm 1$: $|C_{\pm 1}| = rac{3}{4+\pi^2} pprox rac{3}{4+9.87} pprox \mathbf{0.216}$ * For $n = \pm 2$: $|C_{\pm 2}| = rac{3}{4+4\pi^2} pprox rac{3}{4+39.48} pprox \mathbf{0.069}$ 4. **Determine the Phase Spectrum $ngle C_n$:** Since $C_n$ is purely real and positive for all $n$, the phase angle is exactly zero: $$ngle C_n = 0 ext{ rad} \quad ( ext{for all } n)$$ --- ## 6. ECE 2108 Laboratory MATLAB Representation In **ECE 2108 Lab Experiment 3**, you reconstruct complex exponential series by running finite harmonic loop summations. Below is the script used to reconstruct an exponential decay signal $f(t) = e^{-0.5 t}$ over $[0, \pi]$ using its complex coefficients: ```matlab % ECE 2108 Lab 3: Exponential Fourier Series Reconstruction clc; clear all; close all; N = 1000; % Time vector resolution t = linspace(0, pi, N); % One full cycle sum_signal = 0.504; % DC Component (C0) M = 64; % Number of harmonics to sum for j = 1:M % Summing complex conjugate phasor pairs to yield a real-valued output sum_signal = sum_signal + ((0.504*2)/(1+16*j^2)) * (cos(2*j*t) + 4*j*sin(2*j*t)); end % Plotting results figure; plot(t, sum_signal, 'r', 'LineWidth', 2); hold on; plot(t, exp(-0.5*t), 'g--', 'LineWidth', 2); grid on; xlabel('Time (sec)'); ylabel('Amplitude'); legend('Fourier Reconstruction', 'Ideal Exponential'); title(['Reconstruction using ' num2str(M) ' Harmonics']); ``` --- ## 7. Common Mistakes That Cost Marks > [!danger] **The Positive exponent Analysis Trap** > > A very common student slip is using a positive exponent $e^{j n \Omega_0 t}$ inside the analysis integral for $C_n$. Remember: **The Synthesis equation (summation) uses a positive exponent**, while **the Analysis equation (integral) must use a negative exponent $e^{-j n \Omega_0 t}$**. Writing a positive exponent in the integral swaps the values of $C_n$ and $C_{-n}$, causing a cascade of errors. > [!warning] **The Real-Valued Phase Jump Failure** > > When a coefficient $C_n$ evaluates to a purely real, **negative** number (for example, $C_2 = -0.15$), students often plot its phase as $0^\circ$. A negative real number has a magnitude of $+0.15$, but its phase is $\pm 180^\circ$ (or $\pm \pi$ rad). Ignoring this phase inversion results in a **zero-mark** grading for the phase spectrum. --- ## 8. PYQ Bank — Verbatim Questions & Answer Plans ### 8.1 PYQ 2021 [3 Marks] **Question:** Determine the complex exponential Fourier series for the signal illustrated below: Half-wave rectified cosine. * **Answer Plan:** 1. Define the piecewise integration limits for a single period centered at the origin as shown in **Section 5.1**. 2. Write the complex Fourier coefficient integral using a negative exponent. 3. Evaluate the separate cases for $n = \pm 1$ and general $n eq \pm 1$, proving $C_n = rac{A \cos(n\pi/2)}{\pi(1 - n^2)}$ [5.1]. 4. Assemble the final synthesis summation [5.1]. ### 8.2 PYQ 2018 [3 Marks] **Question:** The complex exponential Fourier representation of a signal $x(t)$ over period $(0, T)$ is $x(t) = \sum_{n=-\infty}^{\infty} rac{3}{4+(n\pi)^2} e^{j3n\pi t}$. Determine (i) the value of period $T$; (ii) plot the double-sided amplitude and phase spectrum. * **Answer Plan:** 1. Extract the fundamental frequency $\Omega_0 = 3\pi$ rad/s by comparing exponents [5.2]. 2. Solve for $T = 2/3$ s using the period formula [5.2]. 3. Calculate the magnitude coefficients for $n = 0, \pm 1, \pm 2$ to plot the symmetric double-sided amplitude spectrum [5.2]. 4. Identify the phase spectrum as a flat $0^\circ$ line because all coefficients are real and positive [5.2]. --- ## 9. Self-Check Before Moving On - [ ] Can you mathematically derive the complex coefficient conversion formulas ($C_n = rac{a_n - jb_n}{2}$) starting from trigonometric integrals? - [ ] Do you know how to determine the phase of a negative real Fourier coefficient? - [ ] Can you calculate the fundamental period $T$ of a complex Fourier series directly from its exponent term? [5.2] - [ ] Do you know how to plot double-sided magnitude and phase spectra, highlighting their respective symmetries? --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 01 Fourier Series.pdf, ECE 2108 Signal & Systems-1 (1).pdf (Experiment 3).* --- [[7.03_Exponential_Fourier_Series_and_Complex_Spectra|7.03 Exponential Fourier Series & Complex Spectra]] | [[Chapter 8: Continuous-Time Fourier Transform (CTFT)|Chapter 8: Continuous-Time Fourier Transform (CTFT)]] --- # 7.04 Parseval’s Theorem & Convergence Conditions > [!abstract] Core Idea > > For any periodic signal, the continuous-time Fourier series represents a transition from the time domain to the discrete frequency domain. This note establishes the twin pillars of this transition: **Dirichlet's Conditions**, which dictate when a periodic signal can be mathematically represented as a valid Fourier series, and **Parseval's Power Theorem**, which mathematically guarantees the **conservation of average power** across this domain boundary. Together, they provide the analytical confidence required to process physical power signals in the frequency domain. --- ## 1. Dirichlet’s Convergence Conditions While any passive RLC circuit or physical communication channel responds only to finite-energy waveforms, mathematicians spent decades debating which abstract functions could be decomposed into infinite sinusoids. Jean Baptiste Joseph Fourier originally asserted that *any* periodic signal has a Fourier series—a claim that is technically false without restrictions. To restore mathematical rigor, Peter Gustav Lejeune Dirichlet established three sufficient conditions. If a periodic signal $x(t)$ with fundamental period $T_0$ satisfies these **Dirichlet Conditions**, its Fourier series is guaranteed to converge to $x(t)$ at all points of continuity. ### 1.1 Condition 1: Absolute Integrability The signal $x(t)$ must be absolutely integrable over any single fundamental period $T_0$: $$\int_{T_0} |x(t)| \, dt < \infty$$ * **Physical Significance:** This condition guarantees that the area under the absolute envelope of the signal is finite. * **Mathematical Necessity:** Looking at the Fourier analysis equation $C_n = rac{1}{T_0} \int_{T_0} x(t) e^{-jn\Omega_0 t} \, dt$, we can bound the magnitude of the coefficients: $$|C_n| = \left| rac{1}{T_0} \int_{T_0} x(t) e^{-jn\Omega_0 t} \, dt ight| \le rac{1}{T_0} \int_{T_0} |x(t)| |e^{-jn\Omega_0 t}| \, dt = rac{1}{T_0} \int_{T_0} |x(t)| \, dt$$ Since the integral of $|x(t)|$ is bounded, **each Fourier coefficient $C_n$ (and $a_0, a_n, b_n$) is mathematically guaranteed to be a finite, bounded number**. * **Counterexample:** The function $x(t) = rac{1}{t}$ for $0 < t \le 1$ with period $T_0 = 1$ violates this condition because $\int_{0}^{1} rac{1}{t} \, dt = \ln(t) ig|_0^1 = \infty$. It cannot be represented by a Fourier series. ### 1.2 Condition 2: Finite Extrema (Maxima and Minima) The signal $x(t)$ must possess a **finite number of local maxima and minima** during any single fundamental period $T_0$. * **Physical Significance:** The signal cannot oscillate with infinite frequency within a finite time interval. * **Counterexample:** Consider the signal $x(t) = \sin\left(rac{2\pi}{t} ight)$ for $0 < t \le 1$ with period $T_0=1$. As $t o 0$, the frequency approaches infinity, creating an infinite number of zero-crossings and extrema in the interval $[0, 0.1]$. This signal cannot be modeled with a Fourier series because it is physically impossible to construct or approximate using a countable sum of harmonically related sinusoids. ### 1.3 Condition 3: Finite Discontinuities The signal $x(t)$ must have a **finite number of discontinuities** in any finite interval of time, and each of these discontinuities must be **finite in magnitude**. * **Physical Significance:** The signal cannot have infinite step-jumps. * **The Convergence Rule:** At any point of jump discontinuity $t = t_d$, the Fourier series summation does not converge to either the top or bottom of the step. Instead, it converges precisely to the **statistical average of the left and right-hand limits**: $$\lim_{N o \infty} S_N(t_d) = rac{x(t_d^+) + x(t_d^-)}{2}$$ --- ## 2. Parseval’s Power Conservation Theorem ### 2.1 The Concept of Power Conservation From an engineering perspective, the continuous-time Fourier series splits a complex signal into separate frequency channels. **Parseval's Theorem** mathematically proves the **Conservation of Power**. It guarantees that a signal's total average power is identical whether calculated by integrating its amplitude-squared over time (oscilloscope view) or by summing the squared magnitudes of its discrete frequency components (spectrum analyzer view). > [!theorem] **Parseval's Identity (General Form)** > > For a periodic power signal $x(t)$ with period $T_0$ and complex exponential Fourier coefficients $C_n$: > > $$P = rac{1}{T_0} \int_{T_0} |x(t)|^2 \, dt = \sum_{n=-\infty}^{\infty} |C_n|^2$$ --- ### 2.2 Rigorous Mathematical Proof (Complex Exponential Form) Let $x(t)$ be a complex-valued periodic signal with period $T_0$. The average power $P$ is defined in the time domain as: $$P = rac{1}{T_0} \int_{T_0} |x(t)|^2 \, dt = rac{1}{T_0} \int_{T_0} x(t) \cdot x^*(t) \, dt \quad ext{--- (Equation 1)}$$ where $*$ denotes the complex conjugate. 1. **Express the conjugate signal $x^*(t)$ using its complex Fourier Series expansion:** Since $x(t) = \sum_{n=-\infty}^{\infty} C_n e^{jn\Omega_0 t}$, taking the complex conjugate of both sides yields: $$x^*(t) = \left( \sum_{n=-\infty}^{\infty} C_n e^{jn\Omega_0 t} ight)^* = \sum_{n=-\infty}^{\infty} C_n^* e^{-jn\Omega_0 t} \quad ext{--- (Equation 2)}$$ 2. **Substitute Equation 2 into the power integral (Equation 1):** $$P = rac{1}{T_0} \int_{T_0} x(t) \left[ \sum_{n=-\infty}^{\infty} C_n^* e^{-jn\Omega_0 t} ight] dt$$ 3. **Interchange the order of summation and integration:** *(This step is mathematically guaranteed to be valid because Dirichlet's conditions ensure uniform convergence of the series).* $$P = \sum_{n=-\infty}^{\infty} C_n^* \left[ rac{1}{T_0} \int_{T_0} x(t) e^{-jn\Omega_0 t} \, dt ight] \quad ext{--- (Equation 3)}$$ 4. **Identify the Fourier Analysis term:** Recall the fundamental analysis equation defining the complex coefficients: $$C_n = rac{1}{T_0} \int_{T_0} x(t) e^{-jn\Omega_0 t} \, dt$$ The term inside the brackets of Equation 3 is **exactly equal to $C_n$**! 5. **Simplify to get the final Parseval identity:** $$P = \sum_{n=-\infty}^{\infty} C_n^* \cdot C_n = \sum_{n=-\infty}^{\infty} |C_n|^2 \quad lacksquare$$ --- ### 2.3 The Trigonometric Equivalent Form To express Parseval's identity using real-valued trigonometric coefficients ($a_0, a_n, b_n$), we use the algebraic relationships between complex and trigonometric representations: * **For $n = 0$:** $C_0 = rac{a_0}{2} \implies |C_0|^2 = \left(rac{a_0}{2} ight)^2$. * **For $n eq 0$:** $C_n = rac{a_n - jb_n}{2} \implies |C_n|^2 = rac{a_n^2 + b_n^2}{4}$. * **Symmetry check:** Since $x(t)$ is real, $|C_{-n}|^2 = |C_n|^2 = rac{a_n^2 + b_n^2}{4}$. Expanding the complex sum yields: $$\sum_{n=-\infty}^{\infty} |C_n|^2 = |C_0|^2 + \sum_{n=1}^{\infty} \left( |C_n|^2 + |C_{-n}|^2 ight)$$ $$\sum_{n=-\infty}^{\infty} |C_n|^2 = \left(rac{a_0}{2} ight)^2 + \sum_{n=1}^{\infty} \left( rac{a_n^2 + b_n^2}{4} + rac{a_n^2 + b_n^2}{4} ight)$$ $$\sum_{n=-\infty}^{\infty} |C_n|^2 = \left(rac{a_0}{2} ight)^2 + rac{1}{2} \sum_{n=1}^{\infty} \left( a_n^2 + b_n^2 ight)$$ Therefore, the **Trigonometric Parseval Identity** is: $$P = rac{1}{T_0} \int_{T_0} [x(t)]^2 \, dt = \left(rac{a_0}{2} ight)^2 + rac{1}{2} \sum_{n=1}^{\infty} \left(a_n^2 + b_n^2 ight)$$ --- ## 3. Physical Interpretation of Power Components When Parseval's identity is applied to electrical systems, it decomposes total signal power into clear, measurable components: | Mathematical Component | Signal Processing Name | Physical Meaning (E.g., Voltage Signal across $1\,\Omega$) | | :--- | :--- | :--- | | $$\left(rac{a_0}{2} ight)^2 = |C_0|^2$$ | **DC Power** | Power dissipated by the constant baseline offset of the signal. | | $$rac{1}{2}a_n^2$$ | **Cosine Harmonic Power** | Power contribution of the $n$-th cosine component. | | $$rac{1}{2}b_n^2$$ | **Sine Harmonic Power** | Power contribution of the $n$-th sine component. | | $$|C_n|^2 + |C_{-n}|^2 = rac{a_n^2 + b_n^2}{2}$$ | **Harmonic Channel Power** | Total average power contained in the $n$-th active physical frequency slot. | --- ## 4. High-Yield Worked Examples (The Exam Killers) ### 4.1 Example 1: The 2018 Complex Spectral Power Allocation [PYQ 2018 - 3 Marks] **Question:** The complex exponential Fourier representation of a periodic signal $x(t)$ over a period $(0, T)$ is given by: $$x(t) = \sum_{n=-\infty}^{\infty} rac{3}{4+(n\pi)^2} e^{j3n\pi t}$$ Determine: 1. The value of fundamental period $T$. 2. What percentage of total power is contained in the first five terms of the series, given that the maximum (total) power of the signal is $P_{ ext{total}} = 0.7 ext{ W}$? #### Step-by-Step Analytical Solution: ##### Part 1: Finding the Fundamental Period $T$ The standard synthesis equation for the complex exponential Fourier series is: $$x(t) = \sum_{n=-\infty}^{\infty} C_n e^{jn\Omega_0 t}$$ Comparing this directly with the given equation: $$jn\Omega_0 t = j3n\pi t \implies \Omega_0 = 3\pi ext{ rad/s}$$ The fundamental angular frequency is related to the period $T$ by $\Omega_0 = rac{2\pi}{T}$. Therefore: $$T = rac{2\pi}{\Omega_0} = rac{2\pi}{3\pi} = \mathbf{rac{2}{3} ext{ seconds}} pprox 0.667 ext{ s}$$ --- ##### Part 2: Calculating Percentage Power in the First Five Terms The "first five terms" of the symmetric complex Fourier series correspond to the harmonic indices: $$n = 0, \quad n = 1, \quad n = -1, \quad n = 2, \quad n = -2$$ Let us evaluate the complex coefficient $C_n = rac{3}{4+n^2\pi^2}$ for each of these indices: 1. **For $n = 0$ (DC component):** $$C_0 = rac{3}{4 + 0} = 0.75 \implies |C_0|^2 = (0.75)^2 = \mathbf{0.56250 ext{ W}}$$ 2. **For $n = \pm 1$ (1st Harmonic):** $$C_1 = C_{-1} = rac{3}{4 + \pi^2} pprox rac{3}{4 + 9.86960} = rac{3}{13.86960} pprox 0.21630$$ $$|C_1|^2 = |C_{-1}|^2 pprox (0.21630)^2 = \mathbf{0.04679 ext{ W}}$$ 3. **For $n = \pm 2$ (2nd Harmonic):** $$C_2 = C_{-2} = rac{3}{4 + 4\pi^2} pprox rac{3}{4 + 39.47842} = rac{3}{43.47842} pprox 0.06900$$ $$|C_2|^2 = |C_{-2}|^2 pprox (0.06900)^2 = \mathbf{0.00476 ext{ W}}$$ Now, apply Parseval's identity over these 5 discrete terms to find the partial power $P_5$: $$P_5 = \sum_{n=-2}^{2} |C_n|^2 = |C_0|^2 + 2|C_1|^2 + 2|C_2|^2$$ $$P_5 pprox 0.56250 + 2(0.04679) + 2(0.00476)$$ $$P_5 pprox 0.56250 + 0.09358 + 0.00952 = \mathbf{0.66560 ext{ W}}$$ We are given that the total maximum power of the signal is $P_{ ext{total}} = 0.7 ext{ W}$. The percentage of power contained within the first five terms is: $$\% ext{ Power} = rac{P_5}{P_{ ext{total}}} imes 100\% = rac{0.66560}{0.70000} imes 100\% = \mathbf{95.09\%}$$ * **Exam Note:** A single DC offset and the first two harmonics contain over $95\%$ of the total signal power! This demonstrates why physical systems can be safely band-limited without losing vital signal information. --- ### 4.2 Example 2: Equivalence Proof of Trigonometric & Exponential Power Calculations **Question:** A periodic voltage signal has the following non-zero trigonometric Fourier coefficients: $a_0 = 4 ext{ V}$, $a_2 = 2 ext{ V}$, and $b_2 = -2 ext{ V}$. 1. Calculate the total average power using the trigonometric Parseval equation. 2. Convert these coefficients to complex exponential form and calculate the power. Prove that both methods yield identical results. #### Step-by-Step Analytical Solution: ##### Method 1: Trigonometric Power Calculation The trigonometric Parseval power formula is: $$P = \left(rac{a_0}{2} ight)^2 + rac{1}{2} \sum_{n=1}^{\infty} (a_n^2 + b_n^2)$$ Substitute the given parameters ($a_0 = 4$, $a_2 = 2$, $b_2 = -2$): $$P = \left(rac{4}{2} ight)^2 + rac{1}{2}\left(a_2^2 + b_2^2 ight)$$ $$P = (2)^2 + rac{1}{2}\left(2^2 + (-2)^2 ight)$$ $$P = 4 + rac{1}{2}(4 + 4) = 4 + rac{1}{2}(8) = 4 + 4 = \mathbf{8 ext{ Watts}}$$ --- ##### Method 2: Complex Exponential Power Calculation First, convert the trigonometric coefficients to complex exponential coefficients using standard conversion formulas: 1. **For $n = 0$:** $$C_0 = rac{a_0}{2} = rac{4}{2} = 2 \implies |C_0|^2 = 2^2 = \mathbf{4 ext{ W}}$$ 2. **For $n = 2$:** $$C_2 = rac{a_2 - jb_2}{2} = rac{2 - j(-2)}{2} = 1 + j \implies |C_2|^2 = (1)^2 + (1)^2 = \mathbf{2 ext{ W}}$$ 3. **For $n = -2$:** $$C_{-2} = C_2^* = 1 - j \implies |C_{-2}|^2 = (1)^2 + (-1)^2 = \mathbf{2 ext{ W}}$$ All other $C_n$ coefficients are zero. Summing these exponential components yields: $$P = \sum_{n=-\infty}^{\infty} |C_n|^2 = |C_0|^2 + |C_2|^2 + |C_{-2}|^2$$ $$P = 4 + 2 + 2 = \mathbf{8 ext{ Watts}}$$ Both methods yield **exactly $8 ext{ W}$**, verifying the absolute mathematical equivalence between trigonometric and exponential power formulas. --- ## 5. Common Mistakes That Cost Marks > [!danger] **The Trigonometric $1/2$ Factor Omission Trap** > > In examinations, a very common mistake is forgetting the $1/2$ multiplier when calculating power from trigonometric coefficients: > $$P = \left(rac{a_0}{2} ight)^2 + \sum_{n=1}^{\infty} \left(a_n^2 + b_n^2 ight) \quad \mathbf{[WRONG!]}$$ > Sinusoidal waves of peak amplitude $A$ have an RMS value of $A/\sqrt{2}$, meaning their average power is $A^2/2$. Forgetting this factor will double your AC power terms and result in a **zero-mark** penalty on derivations. > [!warning] **Complex Exponential Half-Sided Summation Error** > > When calculating power in the complex exponential domain, students often sum only the positive indices: > $$P = |C_0|^2 + \sum_{n=1}^{\infty} |C_n|^2 \quad \mathbf{[WRONG!]}$$ > The complex exponential Fourier series uses a **double-sided spectrum**. You must include both $C_n$ and its negative conjugate partner $C_{-n}$ to account for $100\%$ of the signal energy. --- ## 6. PYQ Bank — Verbatim Questions & Answer Plans ### 6.1 PYQ 2022/2018 [8 Marks/7 Marks] **Question:** Define Fourier series. State necessary and sufficient conditions (Dirichlet conditions) for the existence of the Fourier series representation for a signal. * **Answer Plan:** 1. Define **Fourier Series** as the representation of a continuous-time periodic signal as an infinite sum of harmonically related sinusoidal harmonics. 2. Write the trigonometric Fourier series formula, clearly labeling $a_0, a_n, b_n$, and $\omega_0$. 3. State the three **Dirichlet Conditions** verbatim as written in **Section 1**: * Absolute integrability over one period. * Finite number of maxima and minima over one period. * Finite number of discontinuities over one period. 4. Provide the mathematical equations for absolute integrability and the convergence behavior at step discontinuities. ### 6.2 PYQ 2020/2019 [4 Marks] **Question:** State and prove Parseval’s identity for Fourier series. * **Answer Plan:** 1. State **Parseval's Theorem** verbally: the total average power of a periodic signal is equal to the sum of the average powers of its individual harmonic components. 2. Write down both the trigonometric and complex exponential Parseval mathematical equations. 3. Reproduce the step-by-step mathematical proof using the conjugate substitution method as shown in **Section 2.2**. --- ## 7. Self-Check Before Moving On - [ ] Can you state the absolute integrability condition equation of Dirichlet from memory? [1.1] - [ ] Do you know what value the Fourier series converges to at a jump discontinuity of magnitude $10$ (it converges to the exact midpoint of the step)? [1.3] - [ ] Can you mathematically prove Parseval's identity using the complex exponential representation in under 3 minutes? [2.2] - [ ] Why is there a $1/2$ scaling factor in the trigonometric Parseval equation but not in the complex exponential form? [2.3] --- *Source: (k.Deergha Rao) signals and systems.pdf (Chapter 3), 01 Fourier Series.pdf, Rabiul sir class note.pdf.*