Fourier series and Fourier transforms notes — Unit 4

A Fourier Series is an expansion of a periodic function $f(x)$ into an infinite sum of sines and cosines.

Unlike Taylor Series which expands functions into polynomials (and only works near a specific point), Fourier Series works for periodic functions over an entire interval, even if the function has discontinuities (breaks).

To find the Fourier series, we must calculate three constants: $a_0, a_n$, and $b_n$. These are called Euler's formulas:

• $$a_0 = \frac{1}{\pi} \int_c^{c+2\pi} f(x) \\, dx$$
• $$a_n = \frac{1}{\pi} \int_c^{c+2\pi} f(x) \cos(nx) \\, dx$$
• $$b_n = \frac{1}{\pi} \int_c^{c+2\pi} f(x) \sin(nx) \\, dx$$

If the interval is $(0, 2\pi)$, the integration limits are $0$ to $2\pi$.
If the interval is $(-\pi, \pi)$, the integration limits are $-\pi$ to $\pi$.

Not every function can be expanded into a Fourier series. A function $f(x)$ must satisfy Dirichlet's conditions in the interval $(c, c+2\pi)$:

• Single-Valued: $f(x)$ must have only one value for any given $x$.
• Bounded: $f(x)$ must be finite. It cannot go to infinity.
• Piecewise Continuous: $f(x)$ can have discontinuities, but only a finite number of them.
• Finite Max/Min: $f(x)$ can only have a finite number of maxima and minima.

If these conditions are met, the series converges to $f(x)$ at all points of continuity.

What happens if we evaluate the Fourier series exactly at a point where the function breaks (a discontinuity at $x=a$)?

Dirichlet proved that at a point of discontinuity, the Fourier series converges to the average of the left-hand limit and right-hand limit.

$$f(a) = \frac{f(a^-) + f(a^+)}{2}$$

Exam Tip: If a problem asks you to deduce a series sum (like $1 - \frac{1}{3} + \frac{1}{5} = \frac{\pi}{4}$) after finding the Fourier series, you just plug in a specific value of $x$. If that $x$ is at a break, use the average formula.

Using symmetry can cut your integration work in half. This ONLY applies when the interval is symmetric around zero: $(-\pi, \pi)$ or $(-L, L)$.

⇒Even Function: $f(-x) = f(x)$

(Example: $\cos x, x^2, |x|$). The graph is symmetric about the y-axis.

⇒Odd Function: $f(-x) = -f(x)$

(Example: $\sin x, x^3, x$). The graph is symmetric about the origin.

If the interval is $(-\pi, \pi)$:

⇒If $f(x)$ is EVEN:

• $b_n = 0$ (No sine terms!)
• $a_0 = \frac{2}{\pi} \int_0^{\pi} f(x) \\, dx$
• $a_n = \frac{2}{\pi} \int_0^{\pi} f(x) \cos(nx) \\, dx$

⇒If $f(x)$ is ODD:

• $a_0 = 0$ and $a_n = 0$ (No cosine terms!)
• $b_n = \frac{2}{\pi} \int_0^{\pi} f(x) \sin(nx) \\, dx$

You will repeatedly evaluate sine and cosine terms at integer multiples of $\pi$. Memorize these:

• $\sin(n\pi) = 0$
• $\cos(n\pi) = (-1)^n$ (Alternates: -1 for n=1, 1 for n=2)
• $\sin(2n\pi) = 0$
• $\cos(2n\pi) = 1$
• $\sin(0) = 0, \cos(0) = 1$

Sometimes you are given a function only in the half interval $(0, \pi)$ or $(0, L)$ and asked to expand it.

You can intentionally extend the function to $(-\pi, \pi)$ in two ways:

• Half Range Cosine Series: Extend it as an EVEN function. As a result, $b_n=0$. You only calculate $a_0$ and $a_n$.
• Half Range Sine Series: Extend it as an ODD function. As a result, $a_0=0$ and $a_n=0$. You only calculate $b_n$.

Real-world signals don't have periods of exactly $2\pi$. They have arbitrary periods like $2L$ (e.g., period of 10 sec, where $2L=10 \Rightarrow L=5$).

We modify the Euler formulas by replacing $x$ with $\frac{\pi x}{L}$.

$$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos\left(\frac{n\pi x}{L}\right) + \sum_{n=1}^\infty b_n \sin\left(\frac{n\pi x}{L}\right)$$

The new coefficients for interval $(c, c+2L)$:

• $$a_0 = \frac{1}{L} \int_c^{c+2L} f(x) \\, dx$$
• $$a_n = \frac{1}{L} \int_c^{c+2L} f(x) \cos\left(\frac{n\pi x}{L}\right) \\, dx$$
• $$b_n = \frac{1}{L} \int_c^{c+2L} f(x) \sin\left(\frac{n\pi x}{L}\right) \\, dx$$

All symmetry rules (Even/Odd) work exactly the same way for interval $(-L, L)$.

Parseval's identity connects the average power of the function to its Fourier coefficients. It states that the integral of the square of the function equals the sum of the squares of the coefficients.

Exam usage: They will ask you to find the Fourier series, and then say "Hence deduce that $\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} ...$" Parseval's theorem is the tool to solve these deductions.

Sometimes, instead of an equation $f(x)$, you are given raw data points (a table of $x$ and $y$ values from an experiment). Finding the Fourier series from numerical data is called Harmonic Analysis.

We use summation instead of integration:

• $a_0 = 2 \times \frac{\sum y}{N}$
• $a_n = 2 \times \frac{\sum y \cos(nx)}{N}$
• $b_n = 2 \times \frac{\sum y \sin(nx)}{N}$

Where $N$ is the number of data points (excluding the repeating last point).

Fourier Series is for periodic functions. But what if a function is non-periodic (like a single pulse)? We use the Fourier Transform, which extends the period to infinity ($L \to \infty$).

The Inverse Fourier Transform is:
$$f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(s) e^{isx} \\, ds$$

If the function is defined only on $(0, \infty)$, we use Sine and Cosine transforms.

⇒Fourier Cosine Transform (FCT):

$$F_c(s) = \int_0^\infty f(x) \cos(sx) \\, dx$$

⇒Fourier Sine Transform (FST):

$$F_s(s) = \int_0^\infty f(x) \sin(sx) \\, dx$$

(Note: Different textbooks might place a $\sqrt{\frac{2}{\pi}}$ coefficient in front. Check your university syllabus standard).

Like Laplace, Fourier transforms have shifting and scaling properties:

• Linearity: $F\\{af(x) + bg(x)\\} = aF(s) + bG(s)$
• Change of Scale: $F\\{f(ax)\\} = \frac{1}{a} F\left(\frac{s}{a}\right)$
• Shifting Theorem: $F\\{f(x-a)\\} = e^{-isa} F(s)$
• Modulation Theorem: $F\\{f(x) \cos(ax)\\} = \frac{1}{2} [F(s+a) + F(s-a)]$

The convolution of two functions $f(x)$ and $g(x)$ over the entire real line is defined as:

$$(f \ast g)(x) = \int_{-\infty}^\infty f(u) g(x-u) \\, du$$

Parseval's identity applies to transforms as well, stating that the total energy in the time domain equals the total energy in the frequency domain.

$$\int_{-\infty}^\infty |f(x)|^2 \\, dx = \frac{1}{2\pi} \int_{-\infty}^\infty |F(s)|^2 \\, ds$$

For Cosine and Sine transforms on $(0, \infty)$:

$$\int_0^\infty |f(x)|^2 \\, dx = \frac{2}{\pi} \int_0^\infty |F_c(s)|^2 \\, ds = \frac{2}{\pi} \int_0^\infty |F_s(s)|^2 \\, ds$$

To solve differential equations using Fourier transforms, we need the derivative property. Assuming $f(x) \to 0$ as $x \to \pm \infty$:

• $F\\{f'(x)\\} = (is) F(s)$
• $F\\{f''(x)\\} = (is)^2 F(s) = -s^2 F(s)$

This converts derivatives into simple multiplication by $-s^2$, turning boundary value PDEs (like heat conduction) into solvable ODEs.

Fourier Transforms are typically used to solve Partial Differential Equations (PDEs) defined on infinite or semi-infinite intervals.

1. 1.Take the Fourier transform of the PDE with respect to one variable (usually $x$).
2. 2.The PDE will reduce to an ODE in the other variable (usually $t$).
3. 3.Solve the resulting ODE.
4. 4.Take the Inverse Fourier Transform to find the final solution.

⇒Before the exam, ensure you can:

• State Dirichlet's conditions for Fourier expansion.
• Write the $a_0, a_n, b_n$ formulas for an arbitrary period $2L$.
• Explain what happens to a Fourier series at a point of discontinuity.
• Identify an Even vs Odd function to save time calculating $a_n$ or $b_n$.
• Use Parseval's identity to deduce a series sum.
• Write the definition of the Fourier Sine and Cosine transforms.