MIT 6.300 — Lecture 3

Complex Fourier Series

§ Lecture: Complex Fourier Series

Annoyingly, we denote $j := \sqrt{-1}$ . Fourier series look a lot simpler now.

f(t) = \sum_{k = 0}^{\infty} \left( c_k \cos(k \omega_0 t) + d_k \sin(k \omega_0 t) \right) = \sum_{k = -\infty}^{\infty} a_k e^{j k \omega_0t}.

The relationship between the two looks like this:

c_k = a_k + a_{(-k)} ~ \text{ and } ~ d_k = j\left(a_k - a_{(-k)}\right) ~ \text{ for } k \geq 1. ~~~ \parallel ~~~ a_k = \begin{cases} c_0 & \text{ for } k = 0. \\ \frac{1}{2}(c_k - jd_k) & \text{ for } k > 0. \\ \frac{1}{2}(c_{(-k)} + jd_{(-k)}) & \text{ for } k < 0. \end{cases}

And more directly, the coefficients $\{a_k\}_{k = -\infty}^{\infty}$ are generated like so.

Theorem. (Complex Fourier Coefficients) Given a signal $f(t)$ with period $T$ and fundamental frequency $\omega_0 := 2\pi / T$ , we have:

a_{\ell} = \dfrac{1}{T} \int_T f(t) e^{-j \omega_0 \ell t} \, \mathrm{d}t.

Here, $\int_T$ denotes an integral over any interval of length $T$ . Note that this holds even for $\ell = 0$ .

Proof: It's even simpler than last time.

\int_T e^{j k \omega_0 t} \cdot e^{-j\omega_0 \ell t} \, \mathrm{d}t = \int_T e^{j \omega_0 (k - \ell) t} \, \mathrm{d}t = \begin{cases} T & \text{ if } k = \ell. \\ 0 & \text{ otherwise.} \end{cases}

We can do the same “Transforming Fourier Series” analysis as before using complex numbers, too.

Problem. Given the Fourier coefficients $\{a_k\}_{k = -\infty}^{\infty}$ for $f(t)$ , determine the Fourier coefficients for $f(2t)$ and $f(t - T/2)$ .

Solution: For $f(2t)$ , the coefficients remain the same. Meanwhile, for $f(t - T/2)$ , we notice (recalling $\omega_0 = 2\pi / T$ ) that:

e^{j k \omega_0 (t - T/2)} = e^{-j k \omega_0 T/2} \cdot e^{j k \omega_0 t} = (-1)^k \cdot e^{j k \omega_0 t}.

So for $f(t - T/2)$ , the odd-indexed coefficients are negated, and all others remain the same.

The usefulness of complex numbers reveals itself when considering more complicated transformations.

Problem. Determine how the Fourier coefficients for $f(t)$ relate to those of $f(t - T/8)$ .

Solution: Sparing you the computational details, you can verify:

c_k' = \begin{cases} c_k & k \equiv 0 \pmod{8}. \\ \frac{\sqrt{2}}{2}(c_k - d_k) & k \equiv 1 \pmod{8}. \\ -d_k & k \equiv 2 \pmod{8}. \\ \frac{\sqrt{2}}{2}(-c_k - d_k) & k \equiv 3 \pmod{8}. \\ -c_k & k \equiv 4 \pmod{8}. \\ \frac{\sqrt{2}}{2}(-c_k + d_k) & k \equiv 5 \pmod{8}. \\ d_k & k \equiv 6 \pmod{8}. \\ \frac{\sqrt{2}}{2}(c_k + d_k) & k \equiv 7 \pmod{8}. \end{cases}

The expression for $d_k'$ is equally messy. In comparison, $a_k' = e^{-j k(2\pi / 8)} a_k$ ; the $k^{\text{th}}$ coefficient rotates clockwise by $(45k)^{\circ}$ .

§ Recitation: Complex Fourier Series

In some cases, finding the complex Fourier series can be really easy.

Problem. Determine the Fourier series coefficients for $f(t) = \cos^{100}(t)$ , taking $T = 2\pi$ .

Solution: This is just $\left(\frac{1}{2} e^{jt} + \frac{1}{2}e^{-jt}\right)^{100}$ , so $a_k = 2^{-100} \binom{100}{50 - k / 2}$ for all $k \in \{-100, -98, \dots, 98, 100\}$ , and $a_k = 0$ otherwise.

Problem. Determine the Fourier series coefficients for $f(t) = \dfrac{-1}{\left(\sin\left(\frac{3\pi}{2}t\right) + j\cos\left(\frac{3\pi}{2}t\right)\right)^2}$ , taking $T = 4$ .

Solution: It's just $a_6 = 1$ and everything else is zero.

As we saw in lecture, shifting a signal just multiplies each of its complex Fourier coefficients by a unit-magnitude constant.

Theorem. (Delay Property) If $f(t)$ has Fourier coefficients $\{a_k\}_{k = -\infty}^{\infty}$ , then $f(t - \tau)$ has Fourier coefficients $a_k' = a_k \cdot e^{-jk\omega_0 \tau}$ .

We can also say things about symmetry and antisymmetry.

A signal $f(t)$ is real iff $\{a_k\}$ is Hermitian symmetric: $a_{-k} = \overline{a_k}$ .
- A real $f(t)$ is symmetric iff every $a_k$ is purely real.
- A real $f(t)$ is antisymmetric iff every $a_k$ is purely imaginary.

Might as well share a fun fact.

Theorem. (Parseval's Theorem) For a signal with Fourier coefficients $\{a_k\}_{k = -\infty}^{\infty}$ , we have $\frac{1}{T} \int_T |f(t)|^2 \, \mathrm{d}t = \sum_{k = -\infty}^{\infty} |a_k|^2$ .

Proof: Just use $z \overline{z} = |z|^2$ and the fact that $\{e^{jk \omega_0 t}\}_{k = -\infty}^{\infty}$ forms an orthonormal basis under $\langle g, h \rangle := \frac{1}{T} \int_T g(t) \overline{h(t)} \, \mathrm{d}t$ . $\blacksquare$