Fourier Series

Contents

**Recall Periodic and Orthogonal Functions** 1

**Describe Dirichlet’s Conditions for existence of Fourier Series** 3

**Determine Fourier Coefficients using Euler’s Formula** 5

**Determine the Fourier Series of a Function** 7

**Determine Fourier Series of a Defined Function** 9

**Find Fourier Series of a Discontinuous Function** 11

**Determine Fourier Series of an Even-Function** 13

**Determine Fourier Series of an Odd-Function** 14

**Determine Fourier Series of a Function having Arbitrary Periods** 15

**Recall Half-range Fourier Series** 17

**Construct Half-range Sine Fourier Series and Cosine Fourier Series** 18

**Describe Parseval’s Formula** 20

**Apply Parseval’s Formula in Fourier Series** 23

**Determine Complex Form of Fourier Series** 25

**Describe Harmonic Analysis of Fourier Series** 27

**Recall Periodic and Orthogonal Functions**

Periodic functions are functions that repeat themselves over a fixed interval, known as the period. Mathematically, a function f(x) is periodic with period T if f(x + T) = f(x) for all x. A common example of a periodic function is the sine and cosine functions, which have a period of 2π.

Orthogonal functions are functions that are perpendicular to each other with respect to a certain inner product. Mathematically, two functions f(x) and g(x) are orthogonal on the interval [a,b] if their inner product is zero, i.e., . A common example of orthogonal functions are the Legendre polynomials, which are used to solve differential equations and other problems in physics and engineering.

Periodic and orthogonal functions are important in various areas of mathematics and its applications. In particular, they play a crucial role in Fourier analysis, which is a mathematical tool used to decompose a function into a series of simpler functions. The Fourier series representation of a function can be expressed as a linear combination of periodic and orthogonal functions, such as sine and cosine functions, or other sets of orthogonal functions, depending on the problem at hand.

The use of periodic and orthogonal functions in Fourier analysis allows for the efficient representation of signals and other types of data in terms of a finite number of terms, which makes it easier to analyze and process them. For example, the Fourier transform is used in signal processing to convert a signal from the time domain to the frequency domain, where it can be analyzed and filtered more easily.

In summary, periodic and orthogonal functions are important concepts in mathematics and its applications, particularly in Fourier analysis. They provide a powerful tool for representing complex functions and signals in a simpler form, which facilitates their analysis and processing.

**Recall Fourier Series**

Fourier series is a way of representing a periodic function as a sum of sines and cosines with various amplitudes and frequencies. It is named after Joseph Fourier, who first introduced the concept in the early 19th century.

A periodic function is a function that repeats itself over a fixed interval called the period. A function f(x) is said to be periodic with period P if for all x, f(x+P) = f(x).

Orthogonal functions are functions that satisfy certain mathematical properties that make them useful in applications such as Fourier series. Two functions f(x) and g(x) are said to be orthogonal if the integral of their product over the interval [a,b] is equal to zero, i.e., f(x)g(x) dx = 0.

Fourier series is a way of representing a periodic function f(x) as a sum of an infinite series of sines and cosines:

f(x) = a_{0} + [a_{n }*cos(nωx) + *b_{n} sin(nωx)]

where ω = 2π/P, P is the period of the function, and a_{0}, a_{n}, and b_{n} are coefficients that can be found using integration.

The coefficients of the Fourier series can be found using the formulas:

a_{0} = (1/P) f(x) dx

a_{n} = (2/P) f(x) cos(nωx) dx

b_{n} = (2/P) f(x) sin(nωx) dx

where f(x) is the periodic function, and the integrals are taken over one period of the function.

For example, the square wave function with period 2π can be represented by a Fourier series as:

f(x) = (4/π) [1/(2n-1) sin((2n-1)x)]

This series contains only odd harmonics, and each term is orthogonal to all the other terms. The series converges to the square wave function as the number of terms in the sum approaches infinity.

In summary, the Fourier series is a powerful tool for representing periodic functions as an infinite sum of sines and cosines with varying amplitudes and frequencies. It is widely used in many branches of mathematics, science, and engineering, including signal processing, electrical engineering, and physics.

**Describe Dirichlet’s Conditions for existence of Fourier Series**

Dirichlet’s conditions are a set of mathematical requirements that must be met for a periodic function to have a Fourier series. In other words, these conditions ensure that a given function can be expressed as a sum of sinusoidal functions.

Describe Dirichlet’s Conditions for existence of Fourier Series

- Periodicity: The function must be periodic, which means it must repeat itself over a fixed interval. Mathematically, if f(x) is periodic, then it satisfies the condition:

f(x + T) = f(x)

where T is the period of the function.

- Finite Number of Discontinuities: The function must have a finite number of discontinuities over one period. A discontinuity occurs when a function has a sudden change in value or a jump discontinuity. A finite number of discontinuities means that the function has only a finite number of such abrupt changes in one period.

For example, the function defined as:

f(x) = 1 for -π ≤ x ≤ 0

-1 for 0 < x ≤ π

has a jump discontinuity at x = 0, which is the only point of discontinuity. Thus, this function satisfies the second condition.

- Finite Total Variation: The function must have finite total variation over one period. Total variation is a measure of how much a function deviates from its average value. Mathematically, if f(x) is a function with period T, then its total variation over one period is given by:

TV(f) = |f'(x)| dx

where f'(x) is the derivative of the function f(x). The condition for finite total variation requires that this integral be finite.

For example, the function defined as:

f(x) = x for -π ≤ x ≤ 0

-x for 0 < x ≤ π

has a derivative f'(x) that is discontinuous at x = 0. However, the absolute value of the derivative is a constant function, which means that the total variation over one period is finite. Thus, this function satisfies the third condition.

In summary, for a function to have a Fourier series, it must be periodic, have a finite number of discontinuities, and have finite total variation over one period. These conditions are essential in ensuring that the Fourier series converges to the function and does not produce any undesired effects such as the Gibbs phenomenon.

**Determine Fourier Coefficients using Euler’s Formula**

Fourier series is a powerful mathematical tool for representing periodic functions as a sum of sinusoidal functions. The Fourier coefficients are the coefficients that multiply the sines and cosines in the Fourier series. These coefficients can be determined using Euler’s formula, which relates the exponential function to trigonometric functions.

- Fourier Series: The Fourier series of a function f(x) with period T is given by:

f(x) = a_{0} + [a_{n} cos (nωx) + b_{n} sin(nωx)]

where ω = 2π/T is the angular frequency, a_{0}, a_{n}, and b_{n} are the Fourier coefficients, and n is the order of the harmonics.

- Euler’s Formula: Euler’s formula relates the exponential function to the trigonometric functions as follows:

e^{(iθ)} = cos(θ) + i sin(θ)

where i is the imaginary unit and θ is the angle in radians.

- Determining Fourier Coefficients: To determine the Fourier coefficients using Euler’s formula, we can use the following formulas:

a_{n} = (2/T) f(x) cos(nωx) dx

b_{n} = (2/T) f(x) sin(nωx) dx

where the integrals are taken over one period of the function.

By expanding the cos and sin functions in terms of the exponential function using Euler’s formula, we get:

cos(nωx) = (1/2) ^{(}e^{+ }e⁻^{)}

sin(nωx) = (1/2i) (e^{+ }e⁻)

Substituting these expressions into the Fourier coefficients formulas, we obtain:

a_{n} = (1/T) f(x) [(e^{+ }e⁻)]dx

b_{n} = (1/T) [(e^{+ }e⁻)] dx

Using the linearity of the integral, we can separate the integrals into two parts:

an = (1/T) [ f(x) e dx + f(x) e⁻ dx]

bn = (1/T)[ f(x) e dx – f(x) e⁻ dx]

The first integral in each formula is the complex conjugate of the second integral, which means that we can simplify the expressions as follows:

an = (2/T) Re[f(x) e dx]

bn = (2/T) Im[f(x) e dx]

where Re[.] and Im[.] denote the real and imaginary parts of a complex number, respectively.

In summary, Euler’s formula is a powerful tool for determining Fourier coefficients. By expressing the sines and cosines in terms of the exponential function, we can simplify the calculations and obtain a more elegant and compact representation of the Fourier series.

**Determine the Fourier Series of a Function**

Fourier series is a powerful mathematical tool used to represent periodic functions as a sum of sinusoidal functions. The Fourier series allows us to approximate a periodic function by a sum of sines and cosines, which is useful in many applications, including signal processing, control systems, and electrical engineering.

Determine the Fourier Series of a Function

- Definition of Fourier Series: A Fourier series is a representation of a periodic function f(x) as an infinite sum of sines and cosines of varying frequencies and amplitudes, given by the following formula:

f(x) = a_{n} + [a_{n }cos(nωx) + b_{n} sin(nωx)]

where ω = 2π/T is the angular frequency, a_{0}, a_{n}, and b_{n} are the Fourier coefficients, and n is the order of the harmonics.

- Determining the Fourier Coefficients: To determine the Fourier coefficients, we can use the formulas derived from Dirichlet’s conditions and Euler’s formula. The coefficients an and bn can be calculated as follows:

a_{n} = (2/T) f(x) cos (nωx) dx

b_{n} = (2/T) f(x) sin (nωx) dx

where the integrals are taken over one period of the function.

- Example: Let’s consider the square wave function f(x) shown below, which is periodic with a period of 2π.

To determine the Fourier series of this function, we need to first calculate the Fourier coefficients. The DC component a0 can be calculated as the average value of the function over one period, which is:

a_{0} = (1/π) f(x) dx

= (1/π) [ f(x)1 dx – 1 dx]

= 0

The Fourier coefficients an can be calculated as:

an = (2/π) f(x) dx cos(n.x) dx

= (2/π) [ cos(nx) dx – cos(nx) dx]

= (4/πn) sin(nπ/2), if n is odd and 0 if n is even

The Fourier coefficients bn can be calculated as:

bn = (2/π) f(x) sin(nx) dx

= (2/π) [sin(nx) dx -sin(nx) dx]

= (-2/πn) cos(nπ/2), if n is odd and 0 if n is even

Using these coefficients, the Fourier series of the square wave function can be written as:

f(x) = [(4/πn) sin(nπx/2)], if x is in [0,π)

f(x) = [(4/πn) sin(nπ/2) cos(nπx/2)], if x is in [π,2π)

This means that the square wave function can be approximated as a sum of sines with varying amplitudes and frequencies.

**Determine Fourier Series of a Defined Function**

This Learning Outcome asks students to determine the Fourier series of a defined function. The Fourier series is a mathematical technique that allows us to represent a periodic function as an infinite sum of sines and cosines. The series can be used to approximate a function over a given interval by using the sum of the first n terms of the series.

To determine the Fourier series of a defined function, we must follow these steps:

- Determine the period of the function: The period of a function is the smallest positive value of T for which the function f(x) repeats itself for all x.
- Express the function as a periodic function: The function f(x) can be made periodic by defining it over an interval of length T, such that f(x) = f(x+T). This allows us to extend the function over an infinite range by repeating it periodically.

Determine the coefficients of the Fourier series: The coefficients of the Fourier series are given by the formulas:

a_{0} = (1/T) * ∫f(x)dx

a_{n} = (2/T) * ∫f(x)cos(nωx)dx

- b
_{n}= (2/T) * ∫f(x)sin(nωx)dx

where ω = 2π/T and n is a positive integer.

The coefficient a_{0}represents the average value of the function over one period, while coefficients an and bn represent the amplitudes of the cosine and sine functions, respectively. - Write the Fourier series as an infinite sum: The Fourier series of the function f(x) is given by:

f(x) = a_{0}/2 + Σ[a_{n}*cos(nωx) + b*sin(nωx)]_{n}

where the sum is taken over all positive integers n.

Example: Determine the Fourier series of the function f(x) = x, defined over the interval -π to π.

- The period of the function is T = 2π.
- We can extend the function over an infinite range by defining it as a periodic function with period T as:

f(x) = f(x + 2π) = x + 2πk, where k is any integer.

To determine the coefficients of the Fourier series, we use the formulas:

a_{0} = (1/T) * ∫f(x)dx = (1/2π) * x dx = 0

a_{n} = (2/T) * ∫f(x) cos(nωx)dx = (2/π) * x cos(nx) dx = 0

- b
_{n}= (2/T) * ∫f(x) sin(nωx) dx = (2/π) * x sin(nx) dx = (2/π) * [(-1)^{n}– 1]/n

Therefore, the Fourier series of f(x) is:

f(x) = Σ[(2/π) * [(-1)^{n}– 1]/n * sin(nx)]

where the sum is taken over all positive integers n.

We can approximate f(x) by using the sum of the first n terms of the series, which becomes more accurate as n increases.

**Find Fourier Series of a Discontinuous Function**

This Learning Outcome requires students to find the Fourier series of a discontinuous function. A discontinuous function is one that is not continuous at one or more points in its domain. Finding the Fourier series of a discontinuous function involves determining the coefficients of the Fourier series for each continuous piece of the function and then finding the Fourier series for the entire function by combining the series for each piece.

To find the Fourier series of a discontinuous function, we must follow these steps:

- Determine the period of the function: The period of a discontinuous function is the same as that of its continuous pieces.
- Express the function as a periodic function: The discontinuous function can be expressed as a periodic function by defining it over an interval of length T, such that the function repeats itself for all x in the interval.
- Determine the Fourier series for each continuous piece of the function: We must find the Fourier series for each continuous piece of the function separately. The coefficients of the Fourier series for each continuous piece can be found using the formulas:

a_{0}= (1/T) * ∫f(x)dx

a_{n} = (2/T) * ∫f(x)cos(nωx)dx

b_{n} = (2/T) * ∫f(x)sin(nωx)dx

where ω = 2π/T and n is a positive integer.

- Combine the Fourier series of each continuous piece of the function: We can combine the Fourier series of each continuous piece of the function by using the principle of superposition, which states that the sum of two or more functions is the sum of their individual Fourier series.
- Determine the Fourier series for the entire function: We can determine the Fourier series for the entire function by combining the Fourier series for each continuous piece of the function. The resulting series will be piecewise-defined, with a different series for each continuous piece of the function.

Example: Find the Fourier series of the function f(x) = {1 for -π < x < 0, -1 for 0 < x < π}.

- The period of the function is T = 2π.
- We can express the function as a periodic function by defining it over an interval of length T as:

f(x) = {1 for -π < x < 0, -1 for 0 < x < π, 1 for π < x < 2π, -1 for -2π < x < -π, …}

To find the Fourier series for each continuous piece of the function, we use the formulas:

a_{0} = (1/T) * ∫f(x)dx = 0

a_{n} = (2/T) * ∫f(x)cos(nωx)dx = (2/π) * cos(nx) dx = 0

b_{n} = (2/T) * ∫f(x)sin(nωx)dx = (2/π) * sin(nx) dx = (4/πn) * [1 – (-1)^{n}]

Therefore, the Fourier series for each continuous piece of the function is:

f(x) = {Σ[(4/πn) * [1 – (-1)^{n}] * sin(nx)] for -π < x < 0,

- Σ[(4/πn) * [1 – (-1)
^{n}] * sin(nx)] for 0 < x < π}

**Determine Fourier Series of an Even-Function**

This Learning Outcome requires students to determine the Fourier series of an even function. An even function is a function that is symmetric about the y-axis, which means that for every point (x,y) on the graph of the function, there is a corresponding point (-x,y).

To find the Fourier series of an even function, we can follow these steps:

- Determine the period of the function: The period of an even function is 2π, which is the distance between the consecutive peaks of the function.
- Express the even function as a Fourier cosine series: An even function can be expressed as a Fourier cosine series, which contains only cosine terms and no sine terms. This is because a cosine function is even, which means that cos(-x) = cos(x) for all x.
- Determine the coefficients of the Fourier cosine series: The coefficients of the Fourier cosine series can be found using the formula:

a_{n}= (2/π) * ∫f(x)cos(nx)dx

where n is a positive integer.

Since an even function has no sine terms, b_{n}= 0. - Write the Fourier cosine series: The Fourier cosine series for the even function can be written as:

f(x) = a_{0}/2 + Σ[a_{n}cos(nx)]

where n is a positive integer.

Example: Find the Fourier series of the even function f(x) = cos(x).

- The period of the function is 2π.
- The even function can be expressed as a Fourier cosine series:

f(x) = a_{0}/2 + Σ[a_{n}cos(nx)] - To find the coefficients of the Fourier cosine series, we use the formula:

a_{n}= (2/π) * ∫f(x)cos(nx)dx = (2/π) * ∫cos(x)cos(nx)dx

Using the identity cos(a)cos(b) = (cos(a+b) + cos(a-b))/2, we can simplify the integral:

a_{n}= (1/π) * [∫cos((n+1)x)dx + ∫cos((n-1)x)dx] = (1/π) * [(1/(n+1))sin((n+1)x) + (1/(n-1))sin((n-1)x)]

Since a_{0} = 2/π * ∫f(x)dx = 2/π * ∫cos(x)dx = 2/π, the Fourier series for the even function is:

f(x) = π/2 + Σ[(1/π) * [(1/(n+1))sin((n+1)x) + (1/(n-1))sin((n-1)x)]] cos(nx)

where n is a positive integer.

**Determine Fourier Series of an Odd-Function**

This Learning Outcome requires students to determine the Fourier series of an odd function. An odd function is a function that is symmetric about the origin, which means that for every point (x,y) on the graph of the function, there is a corresponding point (-x,-y).

To find the Fourier series of an odd function, we can follow these steps:

- Determine the period of the function: The period of an odd function is 2π, which is the distance between the consecutive peaks of the function.
- Express the odd function as a Fourier sine series: An odd function can be expressed as a Fourier sine series, which contains only sine terms and no cosine terms. This is because a sine function is odd, which means that sin(-x) = -sin(x) for all x.
- Determine the coefficients of the Fourier sine series: The coefficients of the Fourier sine series can be found using the formula:

b_{n}= (2/π) * ∫f(x)sin(nx)dx

where n is a positive integer.

Since an odd function has no cosine terms, a_{n}= 0. - Write the Fourier sine series: The Fourier sine series for the odd function can be written as:

f(x) = Σ[b_{n}sin(nx)]

where n is a positive integer.

Example: Find the Fourier series of the odd function f(x) = sin(x).

- The period of the function is 2π.
- The odd function can be expressed as a Fourier sine series:

f(x) = Σ[b_{n}sin(nx)] - To find the coefficients of the Fourier sine series, we use the formula:

b_{n}= (2/π) * ∫f(x)sin(nx)dx = (2/π) * ∫sin (x) sin (nx) dx

Using the identity sin(a)sin(b) = (cos (a-b) – cos (a+b))/2, we can simplify the integral:

b_{n}= (1/π) * [∫cos ((n-1)x)dx – ∫cos ((n+1)x)dx] = (1/π) * [(1/(n-1)) cos ((n-1)x) – (1/(n+1)) cos ((n+1)x)]

The Fourier series for the odd function is:

f(x) = Σ[(1/π) * [(1/(n-1)) cos ((n-1)x) – (1/(n+1)) cos ((n+1)x)]] sin(nx)

where n is a positive integer.

**Determine Fourier Series of a Function having Arbitrary Periods**

This Learning Outcome requires students to determine the Fourier series of a function having arbitrary periods. The Fourier series of a function with an arbitrary period can be found by transforming the function into a periodic function with a period of 2π, which is the standard period for Fourier series.

To find the Fourier series of a function with an arbitrary period, we can follow these steps:

- Express the function as a periodic function with a period of 2π: We can do this by repeating the function over its period, and then extending it to be periodic with a period of 2π.
- Find the Fourier series of the periodic function with a period of 2π: We can use the standard Fourier series formula to find the Fourier coefficients of the periodic function.
- Express the Fourier series in terms of the original period: We can express the Fourier series in terms of the original period by using the formula:

an = (2/period) * ∫f(x)cos(nπx/period)dx

bn = (2/period) * ∫f(x)sin(nπx/period)dx

where period is the original period of the function. - Write the Fourier series in terms of the original function: We can write the Fourier series in terms of the original function by replacing x with (x – k*period) for each term in the series, where k is an integer.

Example: Find the Fourier series of the function f(x) = x, where the period is 2.

- Express the function as a periodic function with a period of 2π:

g(x) = f(x/π) = x/π, where the period of g(x) is 2π. - Find the Fourier series of the periodic function with a period of 2π:

Using the standard Fourier series formula, we can find the Fourier coefficients:

a_{0}= (1/π) * ∫g(x) dx = (1/π) * ∫x/π dx = 0

a_{n }= (2/π) * ∫g(x) cos (nx)dx = (2/π) * ∫x/π cos(nx) dx = 0

b_{n}= (2/π) * ∫g(x) sin (nx)dx = (2/π) * ∫x/π sin(nx) dx = (4/πn) * (-1)^{n} - Express the Fourier series in terms of the original period:

a_{n}= (2/2) * ∫f(x) cos (nπx/2) dx = (1/π) * ∫x cos(nπx/2) dx

b_{n}= (2/2) * ∫f(x)sin(nπx/2)dx = (2/π) * ∫x sin(nπx/2)dx - Write the Fourier series in terms of the original function:

The Fourier series for the function f(x) = x with period 2 is:

f(x) = Σ[(4/πn) * (-1)^{n}sin(nπx/2)]

where n is a positive odd integer.

**Recall Half-range Fourier Series**

This Learning Outcome requires students to recall the concept of Half-range Fourier Series. Half-range Fourier Series is a technique used to find the Fourier series of a function that is defined only on a half interval. The technique is particularly useful in finding the Fourier series of even or odd functions defined on the interval [0, L], where L is a positive number.

To find the Half-range Fourier Series of a function f(x) defined on the interval [0, L], we can follow these steps:

- Determine whether the function is even or odd: If the function is even, we can use the even Half Range Fourier Series formula. If the function is odd, we can use the odd Half Range Fourier Series formula. If the function is neither even nor odd, we cannot use Half Range Fourier Series.
- Express the function as a Fourier series on the interval [0, L]: We can use the standard Fourier series formula to express the function as a Fourier series on the interval [0, L].
- Use the even or odd Half-range Fourier Series formula: Depending on whether the function is even or odd, we can use the appropriate Half Range Fourier Series formula to find the Fourier series for the function on the interval [0, L].

Even Half-range Fourier Series formula:

f(x) = a_{0}/2 + Σ[a_{n} cos(nπx/L)], where n is a positive integer.

Odd Half-range Fourier Series formula:

f(x) = Σ[b_{n} sin(nπx/L)], where n is a positive integer.

- Simplify the Half-range Fourier Series: The Half-range Fourier Series can be simplified by using the Fourier coefficients from the standard Fourier series formula.

Example: Find the Half-range Fourier Series for the function f(x) = x^{2} on the interval [0, L].

- Determine whether the function is even or odd: The function is even.
- Express the function as a Fourier series on the interval [0, L]:

Using the standard Fourier series formula, we can find the Fourier coefficients:

a_{0}= (2/L) * ∫f(x)dx = (2/L) * dx = L/3

a_{n}= (2/L) * ∫f(x)cos(nπx/L)dx = (2/L) * ∫x^{2}cos(nπx/L)dx = (2/L) * [L^{2}*(-1)^{n }+ 4*(-1)^{n}n^{2}π^{2}] / (n^{3π³})

b_{n}= 0 (since the function is even) - Use the even Half Range Fourier Series formula:

f(x) = a_{0}/2 + Σ[a_{n }*cos(nπx/L)] = (L/6) + Σ[(2/L) * [L*(-1^{2}^{n}+ 4(-1)^{n}n^{2π}] / (n^{3π³}) * cos(nπx/L)] - Simplify the Half Range Fourier Series:

The Half Range Fourier Series for the function f(x) = x^{2}on the interval [0, L] is:

f(x) = (L/6) + (4L^{2/π}³) * Σ[(-1)^{n}/(n^{3}) * cos(nπx/L)]

**Construct Half-range Sine Fourier Series and Cosine Fourier Series**

This Learning Outcome requires students to be able to construct Half-range Sine Fourier Series and Cosine Fourier Series. Half-range Fourier Series is a technique used to find the Fourier series of a function that is defined only on a half interval. The technique is particularly useful in finding the Fourier series of even or odd functions defined on the interval [0, L], where L is a positive number.

The Half-range Fourier Series can be represented as a sum of either only sine or only cosine functions, depending on whether the function is odd or even. The Half Range Sine Fourier Series and Cosine Fourier Series formulas are given below.

Half-range Sine Fourier Series formula:

f(x) = Σ[b_{n}*sin(nπx/L)], where n is a positive integer.

Half Range Cosine Fourier Series formula:

f(x) = a_{0}/2 + Σ[a_{n}*cos(nπx/L)], where n is a positive integer.

To construct the Half-range Sine Fourier Series or Cosine Fourier Series for a function defined on the interval [0, L], we can follow these steps:

- Determine whether the function is even or odd: If the function is even, we can use the even Half Range Cosine Fourier Series formula. If the function is odd, we can use the Half Range Sine Fourier Series formula. If the function is neither even nor odd, we cannot use the Half Range Fourier Series.

Find the Fourier coefficients: We can find the Fourier coefficients by using the standard Fourier series formula.

a_{0} = (2/L) * ∫f(x)dx

a_{n} = (2/L) * ∫f(x)cos(nπx/L)dx

b_{n} = (2/L) * ∫f(x)sin(nπx/L)dx

- Construct the Half-range Sine or Cosine Fourier Series:

Depending on whether the function is even or odd, we can use the appropriate Half Range Fourier Series formula to construct the Half-range Fourier Series for the function on the interval [0, L]. - Simplify the Half-range Fourier Series: The Half-range Fourier Series can be simplified by using the Fourier coefficients from the standard Fourier series formula.

Example: Construct the Half-range Sine Fourier Series for the function f(x) = x on the interval [0, π].

- Determine whether the function is even or odd: The function is odd.

Find the Fourier coefficients:

a_{0} = (2/π) * ∫f(x) dx = 0

a_{n} = (2/π) * ∫f(x) (nπx/π)dx = 0

b_{n} = (2/π) * ∫f(x)sin(nπx/π)dx

= (2/π) * ∫x*sin(nπx/π)dx*

*= (2/π) * [-x*cos(nπx/π)/(nπ) + ∫cos(nπx/π)/(nπ)dx] (Integration by parts)

= (2/π) * [-x*cos(nπx/π)/(nπ) + sin(nπx/π)/(n^{2π2})]

2. Construct the Half-range Sine Fourier Series:

Using the Half-range Sine Fourier Series formula, we get:

f(x) = Σ[b_{n }*sin(nπx/π)]*

*= Σ[(2/π) * [-x *cos(nπx/π)/(nπ) + sin(nπx/π)/(n^{2π})] * sin(nπx/π)]

**Describe Parseval’s Formula**

This Learning Outcome requires students to be able to describe Parseval’s Formula. Parseval’s formula relates the energy of a signal, as measured by the square of its magnitude, to the coefficients of its Fourier series.

Parseval’s Formula can be stated in the following way:

For a periodic function f(x) with period 2L, the total power P of the function is given by:

P = (1/2L) * ∫(f(x))^{2}dx

And, the sum of the squares of the Fourier coefficients is given by:

Σ[|c_{n}|^{2}] = (1/2L) * ∫(f(x))^{2}dx

Where c_{n} are the Fourier coefficients of f(x).

In other words, Parseval’s formula states that the total power of a function is equal to the sum of the squares of its Fourier coefficients. The formula is a useful tool for calculating the total power of a signal from its Fourier coefficients.

Parseval’s formula can also be used to show that the Fourier series of a function converges to the function in the mean-square sense. This means that the difference between the function and its Fourier series approaches zero as the number of terms in the series increases.

Example: Use Parseval’s formula to find the total power of the function f(x) = x on the interval [-1, 1].

Find the Fourier coefficients of f(x):

The Fourier coefficients of f(x) are given by:

c_{0} = (1/2) * ∫(f(x))dx = 0

c_{n} = (1/πn) * ∫f(x)cos(nπx)dx

= (1/πn) * ∫x*cos(nπx)dx*

*= (1/πn ^{2}) * [x*sin(nπx) + (1/n)∫sin(nπx)dx] (Integration by parts)

= (1/πn^{2}) * [x*sin(nπx) – cos(nπx)/n^{2π}]

Calculate the total power P of f(x):

The total power of f(x) is given by:

P = (1/2) dx

= (1/2) dx

= 1/3

Calculate the sum of the squares of the Fourier coefficients:

The sum of the squares of the Fourier coefficients is given by:

Σ[|c_{n}|^{2}] = (1/2) * [(c_{0})^{2} + Σ[(c_{n})^{2}]]

= (1/2) * Σ[(c_{n})^{2}]

= (1/2) * Σ[(1/πn^{2})^{2} * [x sin(nπx) – cos(nπx)/n^{2π}]^{2}]

= (1/2π^{2}) * Σ[(x^{2}/n^{2}) + (1/n^{4})]

Verify Parseval’s formula:

We can verify Parseval’s formula by comparing the total power of f(x) to the sum of the squares of its Fourier coefficients:

P = (1/3)

Σ[|c_{n}^{2}] = (1/2π^{2}) * Σ[(x^{2}/n^{2}) + (1/n^{4})]

= (1/2π^{2}) * [(π^{2}/3) + (π^{4}/90)]

= 1/3

**Apply Parseval’s Formula in Fourier Series**

This Learning Outcome requires students to be able to apply Parseval’s Formula in Fourier Series. Parseval’s formula relates the energy of a signal, as measured by the square of its magnitude, to the coefficients of its Fourier series.

Parseval’s formula can be used to find the total power of a function and to verify that the Fourier series of a function converges to the function in the mean-square sense.

To apply Parseval’s formula in Fourier Series, the following steps can be followed:

- Find the Fourier coefficients of the function using the appropriate Fourier series (full range, half range, complex exponential, etc.).
- Calculate the total power of the function using the formula:

P = (1/T) * ∫(f(x))^{2}dx

Where T is the period of the function. - Calculate the sum of the squares of the Fourier coefficients using the formula:

Σ[|c_{n}^{2}] = (1/T) * ∫(f(x))^{2}dx - Verify Parseval’s formula by comparing the total power of the function to the sum of the squares of its Fourier coefficients. If the two values are equal, then Parseval’s formula is valid for the function.

Example: Use Parseval’s formula to find the total power of the function f(x) = x^{2} on the interval [-π, π] and verify that the Fourier series of f(x) converges to f(x) in the mean-square sense.

Find the Fourier coefficients of f(x):

The Fourier coefficients of f(x) are given by:

a_{0} = (1/π) ∫(f(x))dx = (1/3)π

a_{n} = (1/π) ∫f(x)cos(nx)dx

= (1/π) * ∫x^{2 }cos(nx)dx

= (2/(πn^{2})) * [(-1)^{n} – 1]

b_{n} = (1/π) * ∫f(x)sin(nx)dx

= 0

Calculate the total power P of f(x):

The total power of f(x) is given by:

P = (1/π) * dx

= (1/π) * dx

= (1/5)π^{2}

Calculate the sum of the squares of the Fourier coefficients:

The sum of the squares of the Fourier coefficients is given by:

Σ[|c_{n}|^{2}] = (1/π) * [(a_{0})^{2} + Σ[(a_{n})^{2} + (b_{n})^{2}]]

= (1/π) * [(1/3)π^{2} + Σ[(4/(πn^{4})) * [(-1)^{n }– 1]^{2}]]

= (1/π) * [(1/3)π^{2} + (32/π^{4}) * Σ[1/n^{4}]]

Verify Parseval’s formula:

We can verify Parseval’s formula by comparing the total power of f(x) to the sum of the squares of its Fourier coefficients:

P = (1/5)π^{2}

Σ[|c_{n}|^{2}] = (1/π) * [(1/3)π^{2} + (32/π^{4}) * Σ[1/n^{4}]]

= (1/5)π^{2 }

**Determine Complex Form of Fourier Series**

This Learning Outcome requires students to be able to determine the complex form of Fourier series. The complex form of the Fourier series is a more compact and convenient way to represent a periodic function in terms of its Fourier coefficients.

To determine the complex form of the Fourier series, the following steps can be followed:

- Find the Fourier coefficients of the function using the appropriate Fourier series (full range, half range, complex exponential, etc.).
- Write the complex exponential form of the Fourier series as:

f(x) = Σ[c_{n}* e^{(i n x)]}

Where c_{n}is the n^{th}Fourier coefficient and i is the imaginary unit.

Rewrite the series in terms of its real and imaginary parts using Euler’s formula:

f(x) = Σ[(a_{n – i} b_{n}) * e^{(i n x)}]

= Σ[a_{n} * cos(nx) – b_{n} * sin(nx)] + i Σ[b_{n }* cos(nx) + a_{n} * sin(nx)]

Where a_{n} and b_{n} are the real and imaginary parts of c_{n}, respectively.

3. Combine the real and imaginary parts into a single complex exponential using the identity:

e^{(iθ)} = cos(θ) + i sin(θ)

Thus, the complex form of the Fourier series is:

f(x) = Σ[c_{n }* e^{(i n x)}]

Where cn is a complex number given by:

c_{n} = a_{n – i} b_{n}

= (1/T) * ∫f(x) * e^{(-i n x}) dx

Example: Find the complex form of the Fourier series of the function f(x) = x on the interval [-π, π].

Find the Fourier coefficients of f(x):

The Fourier coefficients of f(x) are given by:

c_{n} = (1/π) * ∫f(x) * e^{(-i n x})dx

= (1/π) * ∫x * e^{(-i n x)} dx

= (1/π) * [(e^{(i n π)} – 1)/(in)^{2}]

Write the complex exponential form of the Fourier series:

f(x) = Σ[c_{n} * e^{(i n x)}]

= Σ[(e^{(i n π)} ^{– 1})/(in π) * e^{(i n x)}]

Rewrite the series in terms of its real and imaginary parts:

f(x) = Σ[c_{n} * e^{(i n x)}]

= Σ[(e^{(i n π) – 1)}/(in π) * (cos(nx) + i sin(nx))]

= Σ[(e^{(i n π) – 1})/(in π) * cos(nx)] + i Σ[(e^{(inπ) – 1})/(in π) * sin(nx)]

Combine the real and imaginary parts into a single complex exponential:

f(x) = Σ[c_{n} * e^{(i n x)}]

= Σ[(e^{(i n π) – 1})/(in π) * cos(nx)] + i Σ[(e^{(i n π) – 1})/(in π) * sin(nx)]

= Σ[(1 – e^{(i n π)})/(in π^{)} * cos(nx)] + i Σ[(1 – e^{(i n π)})/(in π) * sin(nx)]

**Describe Harmonic Analysis of Fourier Series**

This Learning Outcome requires students to be able to describe harmonic analysis of Fourier series. Harmonic analysis is the process of breaking down a complex periodic function into its constituent harmonics, which are simple periodic functions that have frequencies that are integer multiples of the fundamental frequency.

The harmonic analysis of a Fourier series involves analyzing the individual harmonics of the series in terms of their amplitudes and phases. The harmonics of a Fourier series are represented by the terms c_{n }* e^{(i n x),} where c_{n} is the nth Fourier coefficient and e^{(i n x)} is a complex exponential with frequency n times the fundamental frequency.

The amplitude of each harmonic is given by the magnitude of its Fourier coefficient:

|c_{n}| = √(a_{n}^{2}_{ }+ b_{n}^{2})

where a_{n }and b_{n} are the real and imaginary parts of c_{n} , respectively. The amplitude represents the strength of the harmonic in the original function.

The phase of each harmonic is given by the argument of its Fourier coefficient:

θ_{n }=_{ }tan(-1)(b_{n}/a_{n})

where θ_{n} is the phase angle of the n^{th} harmonic. The phase angle represents the shift of the harmonic in time with respect to the fundamental frequency.

The fundamental frequency of the Fourier series is the inverse of the period of the original function, and the harmonics are integer multiples of the fundamental frequency. The n^{th} harmonic has a frequency of n times the fundamental frequency and a period of 1/n times the period of the original function.

By analyzing the amplitudes and phases of the harmonics, we can gain insight into the properties of the original function. For example, if a particular harmonic has a large amplitude, it indicates that the original function has a strong periodic component at that frequency. If a particular harmonic has a phase angle of π/2 or -π/2, it indicates that the original function has a shift in time that is equivalent to a quarter of a period of the harmonic.

Example: Consider the function f(x) = sin(x) + 2 sin(2x) + 3 sin(3x) on the interval [-π, π]. Determine the amplitude and phase angle of each harmonic.

The Fourier series of f(x) is given by:

f(x) = Σ[c_{n} * e^{(i n x)}]

where

c_{n} = (1/π) * ∫f(x) * e^{(-i n x)}dx

The Fourier coefficients can be calculated as:

c_{0} = (1/π) * ∫f(x) dx = 0

c_{n} = (1/π) * ∫f(x) * e^{(-i n x)}dx

= (1/π) * ∫[sin(x) + 2 sin(2x) + 3 sin(3x)] * e^{(-i n x)}dx

= (1/π) * [(-1)^{n} – 2*(-1)^{n}/(2^{n}) + 3*(-1)^{n}/(3^{n})]

The amplitudes and phase angles of the harmonics can then be calculated as:

Amplitude of the first harmonic (n = 1):

|c_{1}| = √(a_{1}^{2} + b_{1}^{2}) = √(1^{2} + 0^{2}) = 1

Phase angle of the first harmonic (n = 1):

θ_{1 }= tan^{(-1)(b1/a1) }= 0