Properties of Fourier Coefficients

Fourier series representation allows us to express a periodic function as a sum of sine and cosine waves. The coefficients of these waves, known as Fourier coefficients, carry crucial information about the function's characteristics. Understanding their properties helps us analyze the function more effectively and can simplify calculations.

Linearity Property

If we have two periodic functions, $f(x)$ and $g(x)$, with the same period $2L$, and their Fourier series are given by: $f(x) = \frac{a_{f0}}{2} + \sum_{n=1}^{\infty} (a_{fn} \cos(\frac{n\pi x}{L}) + b_{fn} \sin(\frac{n\pi x}{L}))$ $g(x) = \frac{a_{g0}}{2} + \sum_{n=1}^{\infty} (a_{gn} \cos(\frac{n\pi x}{L}) + b_{gn} \sin(\frac{n\pi x}{L}))$ Then, for any constants $c_1$ and $c_2$, the Fourier series of the linear combination $c_1 f(x) + c_2 g(x)$ is: $c_1 f(x) + c_2 g(x) = \frac{c_1 a_{f0} + c_2 a_{g0}}{2} + \sum_{n=1}^{\infty} ((c_1 a_{fn} + c_2 a_{gn}) \cos(\frac{n\pi x}{L}) + (c_1 b_{fn} + c_2 b_{gn}) \sin(\frac{n\pi x}{L}))$ This means the Fourier coefficients of a linear combination of functions are the linear combinations of their respective Fourier coefficients.

Symmetry Properties (Even and Odd Functions)

The symmetry of a function significantly simplifies its Fourier series representation.

Even Functions

If $f(x)$ is an even function, meaning $f(-x) = f(x)$, then its Fourier series over an interval $[-L, L]$ contains only cosine terms. The sine coefficients $b_n$ will be zero. If $f(x)$ is even, then: $a_{f0} = \frac{2}{L} \int_{0}^{L} f(x) dx$ $a_{fn} = \frac{2}{L} \int_{0}^{L} f(x) \cos(\frac{n\pi x}{L}) dx$ $b_{fn} = 0$ for all $n \ge 1$. The Fourier series becomes a Fourier cosine series: $f(x) = \frac{a_{f0}}{2} + \sum_{n=1}^{\infty} a_{fn} \cos(\frac{n\pi x}{L})$ Example: $f(x) = x^2$ over $[-\pi, \pi]$. This is an even function.

Odd Functions

If $f(x)$ is an odd function, meaning $f(-x) = -f(x)$, then its Fourier series over an interval $[-L, L]$ contains only sine terms. The constant term $a_0$ and the cosine coefficients $a_n$ will be zero. If $f(x)$ is odd, then: $a_{f0} = 0$ $a_{fn} = 0$ for all $n \ge 1$. $b_{fn} = \frac{2}{L} \int_{0}^{L} f(x) \sin(\frac{n\pi x}{L}) dx$ The Fourier series becomes a Fourier sine series: $f(x) = \sum_{n=1}^{\infty} b_{fn} \sin(\frac{n\pi x}{L})$ Example: $f(x) = x^3$ over $[-\pi, \pi]$. This is an odd function.

Periodicity Property

If a function $f(x)$ has a period of $2L$, its Fourier series also has the same period $2L$. This is inherent in the definition of the Fourier series, which uses $\cos(\frac{n\pi x}{L})$ and $\sin(\frac{n\pi x}{L})$ terms, each having a period of $\frac{2L}{n}$, and thus the fundamental period of the series is $2L$.

Shifting Property

A time shift in the function $f(x)$ does not change its Fourier coefficients directly. Instead, it introduces a phase shift in the sinusoidal components. If $f(x)$ has Fourier series $S(x)$, then $f(x-c)$ will have a Fourier series where the coefficients are modified by a phase factor. Specifically, if $f(x) = \sum c_n e^{in\omega_0 x}$, then $f(x-c) = \sum c_n e^{in\omega_0 (x-c)} = \sum (c_n e^{-in\omega_0 c}) e^{in\omega_0 x}$. The coefficients $c_n$ are replaced by $c_n e^{-in\omega_0 c}$.

Convolution Property

The convolution of two functions in the time domain corresponds to the multiplication of their Fourier transforms in the frequency domain. While this is more directly related to Fourier Transforms, a similar concept applies to Fourier Series. The coefficients of the convolution of two series are related to the products of the coefficients of the individual series, but the relationship is more complex than simple multiplication due to the discrete nature of the series coefficients.

Parseval's Theorem (for Fourier Series)

Parseval's theorem relates the energy of a signal to the sum of the squares of its Fourier coefficients. For a function $f(x)$ with period $2L$, and its Fourier series: $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(\frac{n\pi x}{L}) + b_n \sin(\frac{n\pi x}{L}))$ Parseval's theorem states: $\frac{1}{L} \int_{-L}^{L} [f(x)]^2 dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2)$ This theorem is extremely useful for calculating the total power or energy of a periodic signal represented by its Fourier series. It shows that the total energy is distributed among the various frequency components (represented by the coefficients).

Fourier Transform on L²(-D, D)

The Fourier Transform is an extension of the Fourier Series concept to non-periodic functions. While Fourier Series decomposes a periodic function into a discrete sum of sinusoids, the Fourier Transform decomposes any function into a continuous spectrum of frequencies. The space $L^2(-D, D)$ refers to the space of square-integrable functions on the interval $[-D, D]$. This space is important in the context of Fourier analysis because it ensures that the function has finite energy.

Definition of Fourier Transform

For a function $f(x)$ that is absolutely integrable (i.e., $\int_{-\infty}^{\infty} |f(x)| dx < \infty$), its Fourier Transform, denoted by $\hat{f}(\xi)$ or $F(\xi)$, is defined as: $F(\xi) = \mathcal{F}\{f(x)\} = \int_{-\infty}^{\infty} f(x) e^{-i 2\pi \xi x} dx$ Here, $\xi$ represents frequency. The term $e^{-i 2\pi \xi x}$ can be written using Euler's formula as $\cos(2\pi \xi x) - i \sin(2\pi \xi x)$. This shows that the Fourier Transform decomposes the function into its complex sinusoidal components.

Inverse Fourier Transform

The Inverse Fourier Transform allows us to reconstruct the original function $f(x)$ from its Fourier Transform $F(\xi)$: $f(x) = \mathcal{F}^{-1}\{F(\xi)\} = \int_{-\infty}^{\infty} F(\xi) e^{i 2\pi \xi x} d\xi$ This pair of transform and inverse transform forms the basis of Fourier analysis.

Fourier Transform on L²(-D, D)

The concept of Fourier Transform is typically defined for functions over $(-\infty, \infty)$. However, the question specifies "Fourier transform on $L^2(-D, D)$". This phrasing might imply a few things: 1. Considering functions that are zero outside $[-D, D]$. 2. Considering the Fourier Series of a function extended periodically from $[-D, D]$ to $(-\infty, \infty)$, and then examining its behavior. 3. Or, more technically, considering the restriction of the Fourier Transform of a function defined on $(-\infty, \infty)$ to the domain of functions in $L^2(-D, D)$. Let's consider the case where we are interested in the Fourier Transform of a function $f(x)$ which is non-zero only on $[-D, D]$, i.e., $f(x) = 0$ for $|x| > D$. Such a function is automatically square-integrable over $(-\infty, \infty)$ if it's integrable over $[-D, D]$. If $f(x)$ is defined on $[-D, D]$ and is zero elsewhere, its Fourier Transform is: $F(\xi) = \int_{-D}^{D} f(x) e^{-i 2\pi \xi x} dx$ The inverse transform remains the same: $f(x) = \int_{-\infty}^{\infty} F(\xi) e^{i 2\pi \xi x} d\xi$ However, for a function that is zero outside $[-D, D]$, its Fourier Transform $F(\xi)$ will be continuous. The relationship between the Fourier Series coefficients and the Fourier Transform is that the Fourier Series coefficients can be seen as samples of the Fourier Transform of the corresponding periodic function at discrete frequencies. If we consider the Fourier Series of $f(x)$ on $[-L, L]$: $f(x) = \sum_{n=-\infty}^{\infty} c_n e^{i n \frac{\pi}{L} x}$ where $c_n = \frac{1}{2L} \int_{-L}^{L} f(x) e^{-i n \frac{\pi}{L} x} dx$. If we let $L \to \infty$, this process resembles the transition from Fourier Series to Fourier Transform. In the context of $L^2(-D, D)$, functions in this space are square-integrable. The Fourier Transform maps $L^2(\mathbb{R})$ to $L^2(\mathbb{R})$ (Plancherel's Theorem). If we consider a function $f \in L^2(-D, D)$, it is implicitly assumed to be zero outside $[-D, D]$ for the purpose of standard Fourier Transform definitions on $\mathbb{R}$. The key properties of Fourier Transforms that are relevant include: * **Linearity:** $\mathcal{F}\{af(x) + bg(x)\} = aF(\xi) + bG(\xi)$ * **Time Shifting:** $\mathcal{F}\{f(x-x_0)\} = e^{-i 2\pi \xi x_0} F(\xi)$ * **Frequency Shifting:** $\mathcal{F}\{e^{i 2\pi \xi_0 x} f(x)\} = F(\xi - \xi_0)$ * **Scaling:** $\mathcal{F}\{f(ax)\} = \frac{1}{|a|} F(\frac{\xi}{a})$ * **Differentiation:** $\mathcal{F}\{\frac{d^n f(x)}{dx^n}\} = (i 2\pi \xi)^n F(\xi)$ * **Convolution Theorem:** $\mathcal{F}\{(f * g)(x)\} = F(\xi) G(\xi)$, where $(f * g)(x) = \int_{-\infty}^{\infty} f(\tau) g(x-\tau) d\tau$. For functions restricted to $L^2(-D, D)$, if we interpret this as functions that are zero outside $(-D, D)$, then the integral for the Fourier Transform is effectively limited to the interval $[-D, D]$: $F(\xi) = \int_{-D}^{D} f(x) e^{-i 2\pi \xi x} dx$ This is a finite integral. The inverse transform still integrates over all frequencies $(-\infty, \infty)$.

Plancherel's Theorem for Fourier Transform

This theorem is crucial for $L^2$ spaces. It states that the Fourier Transform preserves the inner product (and thus the norm) of functions in $L^2(\mathbb{R})$. For $f, g \in L^2(\mathbb{R})$: $\int_{-\infty}^{\infty} f(x) \overline{g(x)} dx = \int_{-\infty}^{\infty} F(\xi) \overline{G(\xi)} d\xi$ If $g(x) = f(x)$, this becomes Parseval's theorem for Fourier Transforms: $\int_{-\infty}^{\infty} |f(x)|^2 dx = \int_{-\infty}^{\infty} |F(\xi)|^2 d\xi$ This means the total energy of the function $f(x)$ is equal to the total energy of its Fourier Transform $F(\xi)$. The left side is the energy in the time domain, and the right side is the energy in the frequency domain.

Fourier Integral Theorem

The Fourier Integral Theorem is a fundamental result that bridges the gap between Fourier Series and Fourier Transforms. It states that any function $f(x)$ that satisfies certain conditions (Dirichlet conditions) can be represented as a Fourier Integral. This representation is essentially the continuous analogue of the Fourier Series.

Statement of the Theorem

Let $f(x)$ be a function defined on $(-\infty, \infty)$ such that: 1. $f(x)$ is absolutely integrable, i.e., $\int_{-\infty}^{\infty} |f(x)| dx < \infty$. 2. $f(x)$ has a finite number of discontinuities and a finite number of maxima and minima in any finite interval. (Dirichlet conditions) Then, the Fourier Integral representation of $f(x)$ is given by: $f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) e^{-i \xi t} dt d\xi$ The inner integral $\int_{-\infty}^{\infty} f(t) e^{-i \xi t} dt$ is the Fourier Transform of $f(x)$, let's call it $F(\xi)$. So, the theorem can be written as: $f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\xi) e^{i \xi x} d\xi$ This looks very similar to the inverse Fourier transform formula, differing by a constant factor $\frac{1}{2\pi}$ and the variable of integration (often $\omega$ or $k$ instead of $\xi$). Different conventions exist for the placement of the $2\pi$ factor. Some common forms are: * $F(\xi) = \int_{-\infty}^{\infty} f(x) e^{-i \xi x} dx$ and $f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\xi) e^{i \xi x} d\xi$ * $F(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-i \omega x} dx$ and $f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(\omega) e^{i \omega x} d\omega$ (Symmetric form) * $F(k) = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} f(x) \cos(kx) dx$ and $f(x) = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} F(k) \cos(kx) dk$ (Cosine transform for even functions) The form $f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\xi) e^{i \xi x} d\xi$ signifies that $f(x)$ is obtained by integrating the product of its Fourier Transform $F(\xi)$ and the complex exponential $e^{i \xi x}$ over all frequencies $\xi$. This integral sums up all the frequency components, weighted by their respective amplitudes and phases as described by $F(\xi)$, to reconstruct the original function.

Connection to Fourier Series

Consider a periodic function $f(x)$ with period $T = 2L$. Its Fourier Series is: $f(x) = \sum_{n=-\infty}^{\infty} c_n e^{i n \omega_0 x}$, where $\omega_0 = \frac{2\pi}{T} = \frac{\pi}{L}$. The coefficients are $c_n = \frac{1}{T} \int_{-L}^{L} f(x) e^{-i n \omega_0 x} dx$. Let $F(\omega)$ be the Fourier Transform of $f(x)$ if it were treated as a non-periodic function (assuming it satisfies the conditions). The Fourier Series coefficients $c_n$ can be seen as samples of the Fourier Transform $F(\omega)$ at discrete frequencies $\omega = n \omega_0$. Specifically, $c_n = \frac{1}{T} \int_{-T/2}^{T/2} f(x) e^{-i n \omega_0 x} dx$. If we let $T \to \infty$ (so $L \to \infty$), then $\omega_0 \to 0$. The discrete frequencies $n \omega_0$ become a continuous variable, say $\omega$. The sum $\sum c_n$ becomes an integral. The term $c_n$ is related to $F(n\omega_0)$ by $c_n \approx \frac{1}{T} F(n\omega_0)$. The Fourier Series becomes: $f(x) = \sum_{n=-\infty}^{\infty} c_n e^{i n \omega_0 x} \approx \sum_{n=-\infty}^{\infty} \frac{F(n\omega_0)}{T} e^{i n \omega_0 x}$ As $T \to \infty$, let $\Delta\omega = \omega_0 = \frac{2\pi}{T}$. Then $T = \frac{2\pi}{\Delta\omega}$. $f(x) \approx \sum_{n=-\infty}^{\infty} \frac{F(n\Delta\omega)}{2\pi/\Delta\omega} e^{i n \Delta\omega x} = \sum_{n=-\infty}^{\infty} \frac{F(n\Delta\omega) e^{i n \Delta\omega x}}{2\pi} \Delta\omega$ As $\Delta\omega \to 0$, this sum becomes an integral: $f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i \omega x} d\omega$ This derivation shows how the Fourier Integral Theorem arises as a generalization of Fourier Series for non-periodic functions.

Dirichlet Conditions for Convergence

For the Fourier Integral representation to converge to $f(x)$, the function must satisfy the Dirichlet conditions (or similar regularity conditions). At points of discontinuity, the Fourier Integral converges to the average of the left-hand and right-hand limits: $\frac{1}{2\pi} \int_{-\infty}^{\infty} F(\xi) e^{i \xi x} d\xi = \begin{cases} f(x) & \text{if } f \text{ is continuous at } x \\ \frac{f(x^+) + f(x^-)}{2} & \text{if } f \text{ has a jump discontinuity at } x \end{cases}$

Example: Fourier Integral of a Rectangular Pulse

Consider the rectangular pulse function: $f(x) = \begin{cases} 1 & \text{if } |x| \le D \\ 0 & \text{if } |x| > D \end{cases}$ This function is absolutely integrable. Let's find its Fourier Transform $F(\xi)$: $F(\xi) = \int_{-\infty}^{\infty} f(x) e^{-i 2\pi \xi x} dx = \int_{-D}^{D} 1 \cdot e^{-i 2\pi \xi x} dx$ $F(\xi) = \left[ \frac{e^{-i 2\pi \xi x}}{-i 2\pi \xi} \right]_{-D}^{D} = \frac{e^{-i 2\pi \xi D} - e^{i 2\pi \xi D}}{-i 2\pi \xi}$ Using $e^{i\theta} - e^{-i\theta} = 2i \sin(\theta)$: $F(\xi) = \frac{-(e^{i 2\pi \xi D} - e^{-i 2\pi \xi D})}{-i 2\pi \xi} = \frac{-2i \sin(2\pi \xi D)}{-i 2\pi \xi} = \frac{\sin(2\pi \xi D)}{\pi \xi}$ If $\xi = 0$, $F(0) = \int_{-D}^{D} 1 dx = 2D$. The expression $\frac{\sin(2\pi \xi D)}{\pi \xi}$ can be written in terms of the sinc function, $\text{sinc}(u) = \frac{\sin(\pi u)}{\pi u}$. Let $u = 2\xi D$. Then $\xi = u/(2D)$. $F(\xi) = \frac{\sin(\pi u)}{\pi (u/2D)} = 2D \frac{\sin(\pi u)}{\pi u} = 2D \cdot \text{sinc}(2\xi D)$. So, $F(\xi) = \begin{cases} 2D & \text{if } \xi = 0 \\ \frac{\sin(2\pi \xi D)}{\pi \xi} & \text{if } \xi \neq 0 \end{cases}$ Now, using the Inverse Fourier Transform (or Fourier Integral Theorem): $f(x) = \int_{-\infty}^{\infty} F(\xi) e^{i 2\pi \xi x} d\xi$ (using the convention where $F(\xi)$ has $2\pi$ in the exponent and $f(x)$ has $1$ coefficient) $f(x) = \int_{-\infty}^{\infty} \frac{\sin(2\pi \xi D)}{\pi \xi} e^{i 2\pi \xi x} d\xi$ This integral, when evaluated, reconstructs the original rectangular pulse. This demonstrates the power of the Fourier Integral Theorem in representing non-periodic functions using a continuous spectrum of frequencies.