lec 1 full note

lec 5 u academy online playlist, p.1

ion. The reas

Fourier Series

1.1. Introduction

In engineering, systems are frequently subjected to time-varying periodic forces.

• Static vs. Dynamic Systems: A static system remains under constant forces (e.g., a bridge at rest). A dynamic system varies over time or vibrates (e.g., an RLC circuit oscillating, or a bridge vibrating as a vehicle crosses).
• Core Goal of Fourier Series: To decompose a periodic, non-sinusoidal signal into a linear combination of sines and cosines. This allows engineers to analyze complex time-varying waveforms as a superposition of simple harmonic components.

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1.2. Mathematical Definition

For a periodic function $f(x)$ with period $2l$, the Fourier Series is defined as:

$$f(x)=\frac{a_0}{2}+\sum _{n=1}^{\infty }{a}_n\cos \left(\frac{n\pi }{l}x\right)+\sum _{n=1}^{\infty }{b}_n\sin \left(\frac{n\pi }{l}x\right)\text{\hspace{1em}(1.1)}$$

1.2.1. Periodicity

A function $f(x)$ is periodic if there exists a positive constant $T=2l$ (period or wavelength) such that: $f(x+2l)=f(x)\text{for all }x$

left = -6; right = 6;
top = 2; bottom = -2;
grid = true
---
f(x) = \sin(\frac{\pi x}{2}) + 0.3\cos(\pi x)
f(x) | BLUE | SOLID
(0, 0.3) | open | RED | label:f(x)
(4, 0.3) | open | RED | label:f(x + 2l)
(2, -0.3) | open | BLACK | label:Period T = 2l = 4

Figure 1: A periodic wave demonstrating $f(x+2l)=f(x)$ with period $2l=4$ ($l=2$).

1.2.2. Orthogonality & Orthonormality

We treat signals as infinite-dimensional vectors. The inner product (“dot product”) of two continuous functions $p(x)$ and $q(x)$ over the interval $[\alpha ,\beta ]$ is:

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

• Orthogonality: Two functions are orthogonal if their inner product is zero: $${\int }_{\alpha }^{\beta }p(x)q(x)dx=0$$
• Orthonormality: A function is normalized (orthonormal) if: $${\int }_{\alpha }^{\beta }p(x{)}^{2}dx=1$$

The set of functions $\{1,\cos (\frac{n\pi }{l}x),\sin (\frac{n\pi }{l}x)\}$ forms an orthogonal basis over $[-l,l]$. This is established by the following integral identities (for all integers $m,n$):

$${\int }_{-l}^l\sin \left(\frac{m\pi }{l}x\right)\cos \left(\frac{n\pi }{l}x\right)dx=0\text{\hspace{1em}(1.2)}$$ $${\int }_{-l}^l\sin \left(\frac{m\pi }{l}x\right)\sin \left(\frac{n\pi }{l}x\right)dx=\{\begin{array}{ll}0& m\ne n\\ l& m=n\end{array}\text{\hspace{1em}(1.3)}$$ $${\int }_{-l}^l\cos \left(\frac{m\pi }{l}x\right)\cos \left(\frac{n\pi }{l}x\right)dx=\{\begin{array}{ll}0& m\ne n\\ l& m=n\end{array}\text{\hspace{1em}(1.4)}$$

NOTE

Proof of Orthogonality (Equation 1.4 for $m\ne n$ and $m=n$) Using the product-to-sum identity $\cos (A)\cos (B)=\frac{1}{2}[\cos (A+B)+\cos (A-B)]$:

$${\int }_{-l}^l\cos \left(\frac{m\pi }{l}x\right)\cos \left(\frac{n\pi }{l}x\right)dx=\frac{1}{2}{\int }_{-l}^l\left[\cos \left(\frac{(m+n)\pi }{l}x\right)+\cos \left(\frac{(m-n)\pi }{l}x\right)\right]dx$$

• Case $m\ne n$: Both terms are cosine waves integrated over full integer periods, evaluating to $0$.
• Case $m=n$: The second term becomes $\cos (0)=1$, yielding: $$\frac{1}{2}{\int }_{-l}^l\left[\cos \left(\frac{2n\pi }{l}x\right)+1\right]dx=\frac{1}{2}{\left[0+x\right]}_{-l}^l=\frac{1}{2}(2l)=l$$

left = -\pi; right = \pi;
top = 1.2; bottom = -1.2;
grid = true
---
f(x) = \sin(x)\cos(2x)
f(x) | PURPLE | SOLID

Figure 2: Visualizing Orthogonality. The product $g(x)=\sin (x)\cos (2x)$ shows perfect odd symmetry over $[-\pi ,\pi ]$, meaning its integral cancels out to exactly 0.

1.2.3. Dirichlet Conditions for Convergence

A function $f(x)$ defined on $[-l,l]$ can be represented as a Fourier series if:

1. It is absolutely integrable over one period: ${\int }_{-l}^l|f(x)|dx<\infty$.
2. It has a finite number of maxima and minima within any single period.
3. It has a finite number of discontinuities (piecewise continuous) in any single period.

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1.3. Derivation of Euler-Fourier Coefficients

1. Deriving $a_0$ (DC Offset / Average)

Integrate both sides of the Fourier Series definition (Equation 1.1) from $-l$ to $l$:

$${\int }_{-l}^{l}f(x)dx={\int }_{-l}^l\frac{a_0}{2}dx+\sum _{n=1}^{\infty }{a}_n{\int }_{-l}^l\cos \left(\frac{n\pi }{l}x\right)dx+\sum _{n=1}^{\infty }{b}_n{\int }_{-l}^l\sin \left(\frac{n\pi }{l}x\right)dx$$

By orthogonality, the integrals of the pure sines and cosines over a full period $[-l,l]$ are 0. The equation simplifies to:

$${\int }_{-l}^{l}f(x)dx=\frac{a_0}{2}\cdot (2l)=a_{0}l⟹a_0=\frac{1}{l}{\int }_{-l}^{l}f(x)dx\text{\hspace{1em}(1.5)}$$

2. Deriving $a_n$

Multiply Equation 1.1 by $\cos \left(\frac{m\pi }{l}x\right)$ and integrate from $-l$ to $l$:

$${\int }_{-l}^{l}f(x)\cos \left(\frac{m\pi }{l}x\right)dx=\frac{a_0}{2}{\int }_{-l}^l\cos \left(\frac{m\pi }{l}x\right)dx+\sum _{n=1}^{\infty }{a}_n{\int }_{-l}^l\cos \left(\frac{n\pi }{l}x\right)\cos \left(\frac{m\pi }{l}x\right)dx+\sum _{n=1}^{\infty }{b}_n{\int }_{-l}^l\sin \left(\frac{n\pi }{l}x\right)\cos \left(\frac{m\pi }{l}x\right)dx$$

Using Equations 1.2 and 1.4, only the cosine product term where $n=m$ survives (evaluating to $l$):

$${\int }_{-l}^{l}f(x)\cos \left(\frac{n\pi }{l}x\right)dx=a_n\cdot l⟹a_n=\frac{1}{l}{\int }_{-l}^{l}f(x)\cos \left(\frac{n\pi }{l}x\right)dx\text{\hspace{1em}(1.6)}$$

3. Deriving $b_n$

Similarly, multiplying Equation 1.1 by $\sin \left(\frac{m\pi }{l}x\right)$ and integrating from $-l$ to $l$ isolates $b_n$ (using Equation 1.3):

$$b_n=\frac{1}{l}{\int }_{-l}^{l}f(x)\sin \left(\frac{n\pi }{l}x\right)dx\text{\hspace{1em}(1.7)}$$

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1.4. Worked Examples

Example 1: Continuous Triangular Wave

Find the Fourier Series for the periodic triangular wave:

$$f(x)=\{\begin{array}{ll}-x& -l\le x<0\\ x& 0\le x<l\end{array}\text{with}f(x+2l)=f(x)$$

1. Calculate $a_0$:

$$a_0=\frac{1}{l}{\int }_{-l}^{l}f(x)dx=\frac{1}{l}\left[{\int }_{-l}^0-xdx+{\int }_0^{l}xdx\right]=\frac{1}{l}\left[\frac{l^2}{2}+\frac{l^2}{2}\right]=l$$

2. Calculate $a_n$:

$$a_n=\frac{1}{l}\left[{\int }_{-l}^0-x\cos \left(\frac{n\pi }{l}x\right)dx+{\int }_0^{l}x\cos \left(\frac{n\pi }{l}x\right)dx\right]$$

TIP

Integration by Parts Recall Using $\int udv=uv-\int vdu$. For $\int x\cos (kx)dx$, set $u=x⟹du=dx$, and $dv=\cos (kx)dx⟹v=\frac{1}{k}\sin (kx)$:

$$\int x\cos (kx)dx=\frac{x}{k}\sin (kx)-\int \frac{1}{k}\sin (kx)dx=\frac{x}{k}\sin (kx)+\frac{1}{k^2}\cos (kx)$$

Applying Integration by Parts ($k=\frac{n\pi }{l}$):

$$a_n=\frac{1}{l}\left[{\left[-\frac{lx}{n\pi }\sin \left(\frac{n\pi }{l}x\right)-\frac{l^2}{n^2{\pi }^2}\cos \left(\frac{n\pi }{l}x\right)\right]}_{-l}^0+{\left[\frac{lx}{n\pi }\sin \left(\frac{n\pi }{l}x\right)+\frac{l^2}{n^2{\pi }^2}\cos \left(\frac{n\pi }{l}x\right)\right]}_0^l\right]$$

Using the evaluation identities $\sin (\pm n\pi )=0$, $\sin (0)=0$, $\cos (0)=1$, and $\cos (-n\pi )=\cos (n\pi )=(-1{)}^n$:

$$a_n=\frac{2l}{n^2{\pi }^2}(\cos (n\pi )-1)=\{\begin{array}{ll}-\frac{4l}{n^2{\pi }^2}& n=1,3,5,\dots \\ 0& n=2,4,6,\dots \end{array}$$

3. Calculate $b_n$:

$$b_n=\frac{1}{l}\left[{\int }_{-l}^0-x\sin \left(\frac{n\pi }{l}x\right)dx+{\int }_0^{l}x\sin \left(\frac{n\pi }{l}x\right)dx\right]=0(\text{since }f(x)\text{ is even})$$

Fourier Series Representation:

$$f(x)=\frac{l}{2}-\frac{4l}{{\pi }^2}\sum _{n=1,3,5,\dots }^{\infty }\frac{1}{n^2}\cos \left(\frac{n\pi }{l}x\right)=\frac{l}{2}-\frac{4l}{{\pi }^2}\sum _{n=1}^{\infty }\frac{1}{(2n-1{)}^2}\cos \left(\frac{(2n-1)\pi }{l}x\right)$$

left = -6; right = 6;
top = 4; bottom = -1;
grid = true
---
f(x) = |x - 6 \cdot \operatorname{round}(x/6)|
S_1(x) = 1.5 - \frac{12}{\pi^2} \cos(\frac{\pi x}{3})
S_3(x) = S_1(x) - \frac{12}{9\pi^2} \cos(\pi x)
S_5(x) = S_3(x) - \frac{12}{25\pi^2} \cos(\frac{5\pi x}{3})
 
f(x) | BLACK | SOLID
S_1(x) | RED | DASHED
S_3(x) | GREEN | DASHED
S_5(x) | BLUE | SOLID

Figure 3: Periodic triangular wave ($l=3$) with Fourier approximations $S_1(x)$ (1 component), $S_3(x)$ (2 components), and $S_5(x)$ (3 components) demonstrating fast convergence.

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Example 2: Piecewise Discontinuous Function

Find the Fourier Series for:

$$f(x)=\{\begin{array}{ll}x+1& -1<x<0\\ 1& 0\le x<1\end{array}\text{with}f(x+2)=f(x)(l=1)$$

1. Calculate $a_0$:

$$a_0=\frac{1}{1}\left[{\int }_{-1}^0(x+1)dx+{\int }_0^{1}1dx\right]={\left[\frac{(x+1{)}^2}{2}\right]}_{-1}^0+[x{]}_0^1=\frac{1}{2}+1=\frac{3}{2}$$

2. Calculate $a_n$:

$$a_n={\int }_{-1}^0(x+1)\cos (n\pi x)dx+{\int }_0^1\cos (n\pi x)dx$$

The second integral is ${\left[\frac{\sin (n\pi x)}{n\pi }\right]}_0^1=0$. Applying integration by parts to the first integral:

$$a_n={\left[\frac{(x+1)\sin (n\pi x)}{n\pi }\right]}_{-1}^0-{\int }_{-1}^0\frac{\sin (n\pi x)}{n\pi }dx=0+{\left[\frac{\cos (n\pi x)}{n^2{\pi }^2}\right]}_{-1}^0=\frac{1-\cos (n\pi )}{n^2{\pi }^2}=\{\begin{array}{ll}\frac{2}{n^2{\pi }^2}& n=1,3,5,\dots \\ 0& n=2,4,6,\dots \end{array}$$

3. Calculate $b_n$:

$$b_n={\int }_{-1}^0(x+1)\sin (n\pi x)dx+{\int }_0^1\sin (n\pi x)dx$$

• Integral 2: ${\int }_0^1\sin (n\pi x)dx={\left[-\frac{\cos (n\pi x)}{n\pi }\right]}_0^1=\frac{1-\cos (n\pi )}{n\pi }$
• Integral 1: Set $u=x+1,dv=\sin (n\pi x)dx⟹du=dx,v=-\frac{\cos (n\pi x)}{n\pi }$: $${\int }_{-1}^0(x+1)\sin (n\pi x)dx={\left[-\frac{(x+1)\cos (n\pi x)}{n\pi }\right]}_{-1}^0+{\int }_{-1}^0\frac{\cos (n\pi x)}{n\pi }dx=-\frac{1}{n\pi }+0=-\frac{1}{n\pi }$$
• Total $b_n$: $$b_n=-\frac{1}{n\pi }+\frac{1-\cos (n\pi )}{n\pi }=-\frac{\cos (n\pi )}{n\pi }=\frac{(-1{)}^{n+1}}{n\pi }$$

Fourier Series Representation:

$$f(x)=\frac{3}{4}+\sum _{n=1}^{\infty }\left[\frac{2\cos ((2n-1)\pi x)}{(2n-1{)}^2{\pi }^2}+\frac{(-1{)}^{n+1}\sin (n\pi x)}{n\pi }\right]$$

left = -2; right = 2;
top = 2; bottom = -1;
grid = true
---
p(x) = x - 2 \cdot \operatorname{floor}(\frac{x+1}{2})
f(x) = \{p(x) < 0: p(x) + 1, 1\}
 
S_1(x) = 0.75 + \frac{2\cos(\pi x)}{\pi^2} + \frac{\sin(\pi x)}{\pi}
S_3(x) = S_1(x) + \frac{2\cos(3\pi x)}{9\pi^2} - \frac{\sin(2\pi x)}{2\pi} + \frac{\sin(3\pi x)}{3\pi}
S_7(x) = S_3(x) + \frac{2\cos(5\pi x)}{25\pi^2} - \frac{\sin(4\pi x)}{4\pi} + \frac{\sin(5\pi x)}{5\pi} - \frac{\sin(6\pi x)}{6\pi} + \frac{\sin(7\pi x)}{7\pi}
 
f(x) | BLACK | SOLID
S_1(x) | RED | DASHED
S_3(x) | GREEN | DASHED
S_7(x) | BLUE | SOLID

Figure 4: Reconstructing the piecewise signal. Ringing oscillations (Gibbs Phenomenon) occur around the jump discontinuities at $x=\pm 1,\pm 3,\dots$.

left = 0.7; right = 1.3;
top = 1.3; bottom = -0.3;
grid = true
---
p(x) = x - 2 \cdot \operatorname{floor}(\frac{x+1}{2})
f(x) = \{p(x) < 0: p(x) + 1, 1\}
 
S_{20}(x) = 0.75 + \sum_{n=1}^{20} (\frac{2\cos((2n-1)\pi x)}{(2n-1)^2 \pi^2} + \frac{(-1)^{n+1}\sin(n\pi x)}{n\pi})
 
f(x) | BLACK | SOLID
S_{20}(x) | BLUE | SOLID
(0.95, 1.09) | cross | RED | label:Overshoot (~9%)

Figure 5: Detailed view of the Gibbs Phenomenon at the discontinuity $x=1$. The overshoot height remains constant at $\approx 9\%$ of the jump size, compressing in width as harmonic order increases.

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1.5. Symmetry Analysis: Even & Odd Functions

Using symmetry properties saves significant calculation time by instantly setting certain coefficients to zero.

• Even Functions: $f(-x)=f(x)$ (y-axis reflection symmetry). E.g., $\cos (x),x^2$.
• Odd Functions: $f(-x)=-f(x)$ (origin rotation symmetry). E.g., $\sin (x),x^3$.

left = -3; right = 3;
top = 5; bottom = -5;
grid = true
---
E(x) = x^2
O(x) = 0.2x^3
 
E(x) | GREEN | SOLID | label:Even (y-axis symmetry)
O(x) | ORANGE | SOLID | label:Odd (origin symmetry)

Figure 6: Comparison of even parity ($y=x^2$) and odd parity ($y=0.2x^3$).

Parity Product Rules:

• $\text{Even}\times \text{Even}=\text{Even}$
• $\text{Odd}\times \text{Odd}=\text{Even}$ (e.g., $x\cdot \sin (x)$)
• $\text{Even}\times \text{Odd}=\text{Odd}$

Symmetric Limits of Integration:

Over a symmetric interval $[-l,l]$:

• For an odd function, the net area cancels: ${\int }_{-l}^l{f}_{\text{odd}}(x)dx=0$.
• For an even function, the area is doubled: ${\int }_{-l}^l{f}_{\text{even}}(x)dx=2{\int }_0^l{f}_{\text{even}}(x)dx$.

Simplified Coefficients:

• If $f(x)$ is Even: $$a_0=\frac{2}{l}{\int }_0^{l}f(x)dx,a_n=\frac{2}{l}{\int }_0^{l}f(x)\cos \left(\frac{n\pi }{l}x\right)dx,b_n=0\text{\hspace{1em}(1.8)}$$
• If $f(x)$ is Odd: $$a_0=0,a_n=0,b_n=\frac{2}{l}{\int }_0^{l}f(x)\sin \left(\frac{n\pi }{l}x\right)dx\text{\hspace{1em}(1.9)}$$

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1.6. Half-Range Expansions

When a function $f(x)$ is defined only on the domain $[0,l]$, it can be artificially extended to $[-l,l]$ to produce either a pure cosine or pure sine series:

1. Cosine Expansion (Even Extension): Set $f(x)=f(-x)$ for $x<0$. Yields a series with only cosine terms ($b_n=0$).
2. Sine Expansion (Odd Extension): Set $f(x)=-f(-x)$ for $x<0$. Yields a series with only sine terms ($a_0=a_n=0$).

Example 3: Cosine and Sine Half-Range Extensions

Given the function:

$$f(x)=\{\begin{array}{ll}1& 0<x<1\\ 0& 1<x<2\end{array}(l=2)$$

1. Cosine Expansion (Even Extension)

Extend $f(x)$ symmetrically about the y-axis to $[-2,2]$.

• Calculate $a_0$: $$a_0=\frac{2}{2}{\int }_0^{2}f(x)dx={\int }_0^{1}1dx=1$$
• Calculate $a_n$: $$a_n=\frac{2}{2}{\int }_0^{2}f(x)\cos \left(\frac{n\pi }{2}x\right)dx={\int }_0^1\cos \left(\frac{n\pi }{2}x\right)dx=\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)=\{\begin{array}{ll}0& n=2,4,\dots \\ \frac{2}{n\pi }& n=1,5,9,\dots \\ -\frac{2}{n\pi }& n=3,7,11,\dots \end{array}$$
• Cosine Series: $$f(x)=\frac{1}{2}+\sum _{n=1}^{\infty }\left[\frac{2}{(4n-3)\pi }\cos \left(\frac{(4n-3)\pi }{2}x\right)-\frac{2}{(4n-1)\pi }\cos \left(\frac{(4n-1)\pi }{2}x\right)\right]$$

left = -4; right = 4;
top = 1.5; bottom = -0.5;
grid = true
---
p(x) = |x - 4 \cdot \operatorname{round}(x/4)|
f(x) = \{p(x) < 1: 1, 0\}
 
S_5(x) = 0.5 + \sum_{n=1}^{5} (\frac{2\cos(0.5(4n-3)\pi x)}{(4n-3)\pi} - \frac{2\cos(0.5(4n-1)\pi x)}{(4n-1)\pi})
 
f(x) | BLACK | SOLID
S_5(x) | BLUE | SOLID

Figure 7: Cosine half-range expansion (even periodic extension of the pulse) with its 5-term Fourier approximation.

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2. Sine Expansion (Odd Extension)

Extend $f(x)$ anti-symmetrically about the origin to $[-2,2]$.

• $a_0=a_n=0$.
• Calculate $b_n$: $$b_n=\frac{2}{2}{\int }_0^{2}f(x)\sin \left(\frac{n\pi }{2}x\right)dx={\int }_0^1\sin \left(\frac{n\pi }{2}x\right)dx=\frac{2}{n\pi }\left[1-\cos \left(\frac{n\pi }{2}\right)\right]$$

NOTE

Evaluating $\cos (n\pi /2)$ Cases:

• If $n$ is odd ($n=1,3,5,\dots$): $\cos (n\pi /2)=0⟹b_n=\frac{2}{n\pi }$.
• If $n$ is even but not divisible by 4 ($n=2,6,10,\dots$): $\cos (n\pi /2)=-1⟹b_n=\frac{4}{n\pi }$.
• If $n$ is a multiple of 4 ($n=4,8,12,\dots$): $\cos (n\pi /2)=1⟹b_n=0$.

• Sine Series: $$f(x)=\frac{2}{\pi }\sum _{n=1}^{\infty }\frac{1}{2n-1}\left[\sin \left(\frac{(2n-1)\pi }{2}x\right)+\sin ((2n-1)\pi x)\right]$$

left = -4; right = 4;
top = 1.5; bottom = -1.5;
grid = true
---
q(x) = x - 4 \cdot \operatorname{round}(x/4)
f(x) = \{0 < q(x) < 1: 1, -1 < q(x) < 0: -1, 0\}
 
S_5(x) = \frac{2}{\pi} \sum_{n=1}^{5} \frac{\sin(0.5(2n-1)\pi x) + \sin((2n-1)\pi x)}{2n-1}
 
f(x) | BLACK | SOLID
S_5(x) | RED | SOLID

Figure 8: Sine half-range expansion (odd periodic extension of the pulse) with its 5-term Fourier approximation.

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1.7. Engineering Interpretation & Signal Power

From an engineering perspective, Fourier Series decomposes a signal into:

1. DC component ($a_0/2$): The statistical mean value or average level of the signal.
2. AC component (Summation): Alternating signals at integer harmonics of the fundamental frequency: ${\omega }_0=\frac{\pi }{l}$

Power Spectrum

If $f(x)$ is a voltage signal, the power dissipated in a $1\Omega$ resistor is proportional to $f(x{)}^2$. The power of individual harmonics is given by:

• Amplitude of $n$-th harmonic: $C_n=\sqrt{a_n^2+b_n^2}$
• Power contribution of $n$-th harmonic: $P_n=a_n^2+b_n^2$

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1.8. Error, Convergence, & Parseval’s Theorem

To analyze reconstruction quality, we define:

• $N$-th Partial Sum: $$S_N(x)=\frac{a_0}{2}+\sum _{n=1}^N{a}_n\cos \left(\frac{n\pi }{l}x\right)+\sum _{n=1}^N{b}_n\sin \left(\frac{n\pi }{l}x\right)\text{\hspace{1em}(1.10)}$$
• Mean Square Error (MSE): $$E_N=\frac{1}{2l}{\int }_{-l}^l(f(x)-S_N(x){)}^{2}dx\text{\hspace{1em}(1.11)}$$

NOTE

Deriving the Mean Square Error Formula (Equation 1.12) Expanding the squared term in Equation 1.11:

$$E_N=\frac{1}{2l}{\int }_{-l}^l\left[f(x{)}^2-2f(x)S_N(x)+S_N(x{)}^2\right]dx$$

Substituting $S_N(x)$ and using the Euler-Fourier coefficient integrals:

$$\frac{1}{2l}{\int }_{-l}^l-2f(x)S_N(x)dx=-\left[\frac{a_0^2}{2}+\sum _{n=1}^N(a_n^2+b_n^2)\right]$$

Squaring $S_N(x)$ and integrating (cross-products vanish due to orthogonality):

$$\frac{1}{2l}{\int }_{-l}^l{S}_N(x{)}^{2}dx=\frac{1}{2}\left[\frac{a_0^2}{2}+\sum _{n=1}^N(a_n^2+b_n^2)\right]$$

Combining terms:

$$E_N=\frac{1}{2l}{\int }_{-l}^l(f(x){)}^{2}dx-\frac{a_0^2}{4}-\frac{1}{2}\sum _{n=1}^N(a_n^2+b_n^2)\text{\hspace{1em}(1.12)}$$

Bessel’s Inequality & Parseval’s Identity

Because $E_N\ge 0$, Equation 1.12 establishes Bessel’s Inequality:

$$\frac{a_0^2}{4}+\frac{1}{2}\sum _{n=1}^N(a_n^2+b_n^2)\le \frac{1}{2l}{\int }_{-l}^l(f(x){)}^{2}dx\text{\hspace{1em}(1.13)}$$

As $N\to \infty$, the mean square error converges to $0$ ($E_N\to 0$), turning the inequality into Parseval’s Identity:

$$\frac{a_0^2}{2}+\sum _{n=1}^{\infty }(a_n^2+b_n^2)=\frac{1}{l}{\int }_{-l}^l(f(x){)}^{2}dx\text{\hspace{1em}(1.14)}$$

Physical Meaning: The total average power of a periodic signal is equal to the sum of the power contributions of its DC component and all its AC harmonics. Power is conserved when moving from the time domain (right) to the frequency domain (left).