7.02 Symmetry Conditions and Waveform Analysis

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.