MIT 6.300 — Lecture 1

Fourier Series

§ Lecture: Pitch vs. Timbre

Consider the following graphs of air pressure over time detected by a microphone listening to different musical instruments.

Notice how each graph has the same period of $\frac{1}{262}$ seconds. Consequently, each instrument's sound has the same pitch: 262 Hz to be precise (or approximately C4). However, each instrument's sound has a different timbre. This is because the “shape” of each graph across a single period differs greatly.

Signal processing, spiritually, is the study of looking beyond just the pitch of a signal. How can we quantitatively describe the “shape” of a signal across one of its periods?

§ Lecture: Fourier Series

Here's the driving mathematical fact behind Fourier series.

Theorem. (Fourier Analysis) Any (sufficiently well-behaved) signal $f(t)$ with period $T$ can be expressed as a sum of sinusoids.

f(t) = \sum_{k = 0}^{\infty} (c_k \cos (k \omega_0 t) + d_k \sin (k \omega_0 t))

In the above, $\omega_0 := \frac{2\pi}{T}$ is the fundamental frequency, and the coefficients $\{c_k\}_{k = 0}^{\infty}$ and $\{d_k\}_{k = 0}^{\infty}$ are the following.

c_0 = \dfrac{1}{T} \int_0^T f(t) \, \mathrm{d}t ~~~ \parallel ~~~ c_{\ell} = \dfrac{2}{T} \int_0^T f(t) \cos(\ell \omega_0 t) \, \mathrm{d}t ~~~ \parallel ~~~ d_{\ell} = \dfrac{2}{T} \int_0^T f(t) \sin( \ell \omega_0 t) \, \mathrm{d} t.

Proof: Handwaving: assume that $f(t)$ can be expressed as a sum of sinusoids, then derive the formulas for $\{c_k\}_{k = 0}^{\infty}$ and $\{d_k\}_{k = 0}^{\infty}$ .

The $c_0$ coefficient is obvious. For the others, recall for any $A, B \in \mathbb{Z}^+$ that:

\int_0^T \cos(A \omega_0 t) \cos(B \omega_0 t) \, \mathrm{d}t = \int_0^T \dfrac{1}{2} \cos((A - B)\omega_0 t) + \dfrac{1}{2} \cos((A + B)\omega_0 t) \, \mathrm{d}t = \begin{cases} T/2 & \text{ if } A = B. \\ 0 & \text{ otherwise. } \end{cases}

\int_0^T \sin(A \omega_0 t) \cos(B \omega_0 t) \, \mathrm{d}t = \int_0^T \dfrac{1}{2} \sin((A - B) \omega_0 t) + \dfrac{1}{2} \sin((A + B) \omega_0 t) \, \mathrm{d}t = 0.

\int_0^T \sin(A \omega_0 t) \sin(B \omega_0 t) \, \mathrm{d}t = \int_0^T \dfrac{1}{2} \cos((A - B) \omega_0 t) - \dfrac{1}{2} \cos((A + B) \omega_0 t) \, \mathrm{d}t = \begin{cases} T/2 & \text{ if } A = B. \\ 0 & \text{ otherwise. } \end{cases}

These imply the formulas. In some sense, there's a sort of orthogonal basis here or whatever. $\blacksquare$

Remark. The relationship $\omega_0 = \frac{2\pi}{T}$ must not be broken! For example, if a signal has fundamental frequency $\omega_0 = 10\pi$ , then we must have $T = 0.2$ , so its Fourier coefficients must be computed by integrating over an interval of length $0.2$ .

The fundamental frequency of a signal captures its pitch, but the coefficients $\{c_k\}_{k = 0}^{\infty}$ and $\{d_k\}_{k = 0}^{\infty}$ capture its timbre. So we might visualize the harmonic structure of a signal by plotting $m_k := \sqrt{c_k^2 + d_k^2}$ against the harmonic number $k$ .

This tells us that the second harmonic of this signal is about twice as strong as its first harmonic (or fundamental).

Remark. The choice of plotting $m_k := \sqrt{c_k^2 + d_k^2}$ comes from the following rewrite.

\begin{align*}f(t) = \ & \sum_{k = 0}^{\infty} (c_k \cos (k \omega_0 t) + d_k \sin (k \omega_0 t)) \\ = \ & \sum_{k = 0}^{\infty} m_k \cos(k \omega_0 t - \phi_k) \text{ where } m_k := \sqrt{c_k^2 + d_k^2} \text{ and } \phi_k := \tan^{-1}\left(\dfrac{d_k}{c_k}\right).\end{align*}

So the measurement $m_k$ captures the strength of the part of the signal with frequency $k \omega_0$ .

Thus, there are two ways we can look at a signal:

As a function of time: $f(t)$ .
As a function of frequency: $k \mapsto (\omega_0, m_k, \phi_k)$ .

§ Recitation: Computing Fourier Series

Recall that $\cos(t)$ is even and $\sin(t)$ is odd. Therefore,

If $f(t)$ is even, then $f(t) \sin(k\omega_0 t)$ is odd for all $k$ , so $d_k = 0$ for all $k$ .
If $f(t)$ is odd, then $f(t) \cos(k\omega_0 t)$ is odd for all $k$ , so $c_k = 0$ for all $k$ .

Thus, if a signal is known to be even or odd, its Fourier series can be computed in half the usual number of steps.

Problem. Determine the Fourier coefficients of $f(t) = |\sin(t)|$ .

Solution: This is even with period $T = \pi$ and fundamental frequency $\omega_0 = 2$ . Thus, $d_k = 0$ and:

c_k = \dfrac{2}{\pi} \int_{0}^{\pi} \sin(t) \cos(2kt) \, \mathrm{d}t = \dfrac{1}{\pi} \int_0^{\pi} \sin((1 + 2k)t) + \sin((1 - 2k)t) \, \mathrm{d}t = \dfrac{1}{\pi} \left[ \dfrac{2}{1 + 2k} + \dfrac{2}{1 - 2k} \right] = \dfrac{-4}{\pi(4k^2 - 1)}.

Separately, we compute $c_0 = \frac{2}{\pi}$ as well.