1. Fourier Series

1.1 Choices: Welcome Aboard

• The question of convergence of Fourier series, believe it or not, led G. Cantor near the turn of the 20th century to investigate and invent the theory of infinite sets and to distinguish different cardinalities of infinite sets.

1.2

1.2.1. Time and space.

The frequency and wavelength are related through the equation $v = λν$, which is the same as speed = distance/time.

More on spatial periodicity.

• a crystal has a regular, repeating pattern of atoms in space; the arrangement of atoms is called a lattice. The function that describes the electron density distribution of the crystal is then a periodic function of the spatial variables, in three dimensions, that describes the crystal.

1.2.2. Definitions, examples, and things to come.

$f(t)$ is defined on $\mathbb{R}$. The smallest $T$ such that

$$f(t+T) = f(t)$$

is called the fundamental period of the function.

• Is the sum of two periodic functions periodic?

No, consider $\cos t$ and $\cos \sqrt[]{2}t$.

1.2.3. The building blocks: A few more examples.

The state of the harmonic oscillator system is described by a single sinusoid, say of the form

$$A \sin (2 \pi \nu t + \phi )$$

The period is $\nu$.

When Fourier consider the problem when a ring is heated up, he came to the idea that the distribution of temperature can be modeled as a sum of sinusoids

$$\sum_{n=1}^{N} A_n \sin (n \theta + \phi_n)$$

1.3 It All Adds Up

It's a little bit awkward to use the above equation

$$\sum_{n=1}^{N} A_n \sin (2 \pi n t + \phi_n)$$

It’s more common to write a general trigonometric sum as

$$\tag{1-3-1} \frac{a_0}{2} + \sum_{n = 1}^{N} a_n \cos (2πnt) + b_n \sin (2πnt)$$

The above 2 equations are the same, because we have $\sin(α+β) = \sinα\cosβ +\cosα\sinβ$.

$$\sin (2 \pi n t + \phi_n) = \\ \sin (2 \pi n t) \cos (\phi_n) + \cos (2 \pi n t) \sin (\phi_n)$$

And let $a_n = \sin (\phi_n), b_n = \cos (\phi_n)$, then we can see they are the same.

Using complex exponentials.

Now from Appendix B, we know

$$\cos (2 \pi n t) = \frac{ e^{2πint} + e^{-2πint} }{2}, \quad \sin (2 \pi n t) = \frac{ e^{2πint} - e^{-2πint} }{2i}$$

Then we can rearrange and simplify the equation (1-3-1).

$$\sum_{n = -N}^{N} c_n e^{2πint} \\ c_0 = \frac{a_0}{2} \\ c_n = a_n - i b_n, \text{ for } n > 0,\\ c_n = a_n + i b_n, \text{ for } n < 0.\\ $$

Let

$$f(t) = \sum_{n = -N}^{N} c_n e^{2πint}$$

Note, $c_n$ can be complex numbers. If $f(t)$ is real, then

$$\overline{f(t)} = \sum_{n = -N}^{N} \overline{c_n e^{2πint}} = \sum_{n = -N}^{N} \overline{c_n} e^{-2πint} = f(t)$$

So we have

$$\overline{c_n} = c_{-n}$$

On the other hand, if $c_n$ satisfy $\overline{c_n} = c_{-n}$,

$$f(t) = \sum_{n = -N}^{N} c_n e^{2πint} \\ = c_0 + \sum_{n = 1}^{N} c_n e^{2πint} + c_{-n} e^{-2πint} \\ = c_0 + \sum_{n = 1}^{N} 2\Re(c_n e^{2πint}) \\ = c_0 + \Re\left\{ \sum_{n = 1}^{N} c_n e^{2πint} \right\}$$

$\square$

1.3.1. Lost at c.

Let

$$f(t) = \sum_{n = -N}^{N} c_n e^{2πint}$$

and we want to solve $c_k$. Rearrange it a little bit

$$c_k = e^{-2πikt} f(t) - e^{-2πikt} \sum_{n \neq k} c_n e^{2πint} \\ = e^{-2πikt} f(t) - \sum_{n \neq k} c_n e^{2πi(n-k)t}$$

This is as far as algebra will take us, and when algebra is exhausted, the desperate mathematician will turn to calculus, i.e., to differentiation or integration. Here’s a hint: differentiation won’t get you anywhere.

Another idea is needed, and that idea is integrating both sides from 0 to 1.

Note

$$\int_{0}^{1} e^{2πi(n-k)t} dt = \\ \frac{1}{2πi(n-k)} e^{2πi(n-k)t} \bigg|_{0}^{1} = \\ \frac{1}{2πi(n-k)} (e^{2πi(n-k)} - 1) = 0$$

So

$$c_k = \int_{0}^{1} e^{-2πikt} f(t) dt$$

Let’s summarize and be careful to note what we’ve done here, and what we haven’t done. We’ve shown that if we can write a periodic function $f(t)$ of period $1$ as a sum

$$f(t) = \sum_{n = -N}^{N} c_n e^{2πint}$$

then the coefficients $c_n$ must be given by

$$c_n = \int_{0}^{1} e^{-2πint} f(t) dt.$$

We have not shown that every periodic function can be expressed this way.

If we assume $f(t)$ is real, then

$$\overline{c_n} = \overline{\int_{0}^{1} e^{-2πint} f(t) dt} = \int_{0}^{1} \overline{e^{-2πint} f(t) dt} = \\ \int_{0}^{1} \overline{e^{-2πint}} f(t) dt = \\ \int_{0}^{1} e^{2πint} f(t) dt = c_{-n}$$

$\square$

1.3.2. Fourier coefficients.

This sum is called Fourier Series

$$\sum_{n = -N}^{N} c_n e^{2πint}$$

Fourier coefficients, hip version

$$\hat{f}(n) = \int_{0}^{1} e^{-2πint} f(t) dt$$

The next step is to show given any $a$,

$$\hat{f}(n) = \int_{a}^{a+1} e^{-2πint} f(t) dt$$

In "Understanding Analysis", we learned that the 2nd part of The Fundamental Theorem of Calculus, if $g(x)$ is integrable, and $G(x) = \int_{a}^{x} g(x)$. If $g(x)$ is continuous at $x$, then $G'(x) = g(x)$.

Then let

$$G(a) = \int_{0}^{a} g(x) \\ H(a) = \int_{0}^{a+1} g(x)$$

Then

$$G'(a) = g(a), H'(a) = G'(a+1) = g(a+1)$$

So $\frac{d}{da} \int_{a}^{a+1} g(x) = g(a+1) - g(a)$.

We assume this is also true for the complex function. Then

$$\frac{d}{da} \hat{f}(n) = e^{-2πint} f(t) \bigg|_{a}^{a+1} \\ = e^{-2πin(a+1)} f(a+1) - e^{-2πina} f(a) \\ = e^{-2πin} e^{-2πina} f(a+1) - e^{-2πina} f(a) \\ = e^{-2πina} f(a+1) - e^{-2πina} f(a) \\ = 0 \\ $$

That means $\hat{f}(n)$ is independent of $a$.

There are times when the following is helpful — integrating over a symmetric interval, that is.

$$\int_{-1/2}^{1/2} e^{-2πint} f(t)dt$$

Symmetry relations.

If $f(t)$ is real, we have seen

$$\overline{c_n} = c_{-n} \Leftrightarrow \overline{\hat{f}(n)} = \hat{f}(-n) \\ \Rightarrow \\ |\overline{\hat{f}(n)}| = |\hat{f}(-n)|$$

So the magnitude of the Fourier coefficients is an even function of $n$.

Now assume $f(t)$ is even function We have

$$\begin{split} \hat{f}(-n) &= \int_{0}^{1} e^{-2πi(-n)t} f(t) dt \\ &= \int_{0}^{1} e^{-2πi(-n)t} f(-t) dt \end{split}$$

Now let $t = g(s) = -s$ and recall the change-of-variable formula

$$\int_{a}^{b} f(g(x)) g'(x) dx = \int_{g(a)}^{g(b)} f(t) dt$$

We have

$$\int_{0}^{1} e^{-2πi(-n)t} f(t) dt = \int_{0}^{-1} e^{-2πi(-n)(-s)} f(-s) g'(s) ds\\ = -\int_{0}^{-1} e^{-2πins} f(s) ds\\ = \int_{-1}^{0} e^{-2πins} f(s) ds\\ = \hat{f}(n)$$

So we have

$$\hat{f}(-n) = \hat{f}(n)$$

$\square$

• If $f(t)$ is even, then the Fourier coefficients $\hat{f}(n)$ are also even.
• If $f(t)$ is real and even, then the Fourier coefficients $\hat{f}(n)$ are also real and even.

Now assume $f(t)$ is odd function. We have

$$\begin{split} \hat{f}(-n) &= \int_{0}^{1} e^{-2πi(-n)t} f(t) dt \\ &= -\int_{0}^{-1} e^{-2πi(-n)(-s)} f(-s) dt \\ &= \int_{0}^{-1} e^{-2πins} f(s) dt \\ &= -\int_{-1}^{0} e^{-2πins} f(s) dt \\ &= - \hat{f}(n) \end{split}$$

• If $f(t)$ is odd, then the Fourier coefficients $\hat{f}(n)$ are also odd.
• If $f(t)$ is real and odd, then the Fourier coefficients $\hat{f}(n)$ are odd and purely imaginary.

If we think of $\hat{f}(n) = \int_{0}^{1} e^{-2πint}f(t)dt$ as a transform of $f$ to a new function $\hat{f}(n)$. While $f(t)$ is defined for a continuous variable $t$, the transformed function $\hat{f}(n)$ is defined on the integers. There are reasons for this that are much deeper than simply solving for the unknown coefficients in a Fourier series.

1.3.3. Period, frequencies, and spectrum.

We’re assuming that $f(t)$ is a real, periodic signal of period $1$ with a representation as the series

$$f(t) = \sum_{n = -\infty }^{\infty} \hat{f}(n) e^{2πint}$$

• Rather, being able to write f(t) as a Fourier series means that it is synthesized from many harmonics, many frequencies, positive and negative, perhaps an infinite number.
• The set of frequencies n that are present in a given periodic signal is the spectrum of the signal.

For a real signal, $\overline{\hat{f}(n)} = \hat{f}(-n)$. Then the coefficients $\hat{f}(n)$ and $\hat{f}(-n)$ are either both zero or both nonzero.

If the coefficients are all zero from some point on, say $\hat{f}(n) = 0 \text{ for } |n| > N$, it’s common to say that the signal has no spectrum from that point on. One also says in this case that the signal is bandlimited and that the bandwidth is $2N$.

$|\hat{f}(n)|^2$ is said to be the energy of the (positive and negative) harmonic $e^{\pm 2πint}$.

The sequence of squared magnitudes $|\hat{f}(n)|^2$ is called the energy spectrum or the power spectrum.

Rayleigh’s identity:

$$\int_{0}^{1} |f(t)|^2 dt = \sum_{n = -\infty }^{\infty} |\hat{f}(n)|^2$$

Viewing a signal: Why do musical instruments sound different?

• why do two instruments sound different even when they are playing the same note?

  • It’s because the note that an instrument produces is not a single sinusoid of a single frequency, not a pure A at 440 Hz, for example, but a sum of many sinusoids each of its own frequency and each contributing its own amount of energy.
  • two instruments sound different because of the different harmonics they produce and because of the different strengths of the harmonics.

• To oversimplify, your inner ear (the cochlea) finds the harmonics and their amounts (the key is resonance) and passes that data on to your brain to do the synthesis. Nature knew Fourier analysis all along!

1.3.4. Changing the period, and another reciprocal relationship.

Suppose $f(t)$ has period $T$. Then let $g(t) = f(Tt)$,

$$g(t+1) = f(T(t+1)) = f(Tt + T) = f(Tt) = g(t)$$

So $g$ has period of $1$.

$$g(t) = \sum_{n = -\infty}^{\infty} c_n e^{2πint}$$

Let $s = Tt$

$$f(s) = g(t) = \sum_{n = -\infty}^{\infty} c_n e^{2πint} = \sum_{n = -\infty}^{\infty} c_n e^{2πins/T}$$

Also

$$c_n = \int_{0}^{1} e^{-2πint} g(t) dt \\ = \frac{1}{T} \int_{0}^{T} e^{-2πin s/T} f(s) ds$$

Sometimes, it's useful to write

$$c_n = \frac{1}{T} \int_{-T/2}^{T/2} e^{-2πin s/T} f(s) ds$$

Or take the harmonics as

$$\frac{1}{\sqrt[]{T}} e^{2πint/T} \\ c_n = \frac{1}{\sqrt[]{T}} \int_{0}^{T} e^{-2πint/T} f(t) dt$$

Time domain – frequency domain reciprocity.

Given

$$f(t) = \sum_{n = -\infty}^{\infty} c_n e^{2πint/T}$$

• In the time domain the signal repeats after T seconds, while the points in the spectrum are $0, ±1/T, ±2/T, ...$, which are spaced $1/T$ apart.
• This is worth elevating to an aphorism:
  • The larger the period in time, the smaller the spacing in the spectrum. The smaller the period in time, the larger the spacing in the spectrum.