Lesson 19

Transform of Periodic Sequences

Here we study the DTFT of periodic sequences. We'll start by looking at the Fourier Series expansion, analogous to what we did in continuous time. Then we will derive the same result using a different approach that will lead us into the Discrete Fourier Transform for finite length sequences.

Recall that for continuous time periodic signals, we found the Fourier transform by first doing a Fourier series expansion

$\displaystyle x(t) = \sum_{k=-\infty}^\infty a_k e^{jk\omega_0 t}$ $\textstyle \qquad$ $\displaystyle \mbox{synthesis equation}$ (1)
$\displaystyle a_k = \frac{1}{T} \int_T x(t)e^{-jk\omega_0 t}dt$ $\textstyle \qquad$ $\displaystyle \mbox{analysis equation}$ (2)

then using the fact that a complex exponential in time transforms to an impulse in the frequency domain

\begin{displaymath}e^{j\omega_0 t} \longleftrightarrow 2\pi\delta(\omega - \omega_0)\end{displaymath}

and linearity of the Fourier transform, we get that the CTFT of a periodic signal is made up of harmonically-related impulses with area $2\pi a_k$

\begin{displaymath}X(\omega) = 2\pi \sum_{k=-\infty}^\infty a_k \delta(\omega - k\omega_0)\end{displaymath}

Discrete-time periodic signals can also be described by a Fourier Series expansion:

$\displaystyle x[n] = \sum_{k\in <N>} a_k e^{jk\Omega_0 n}$ $\textstyle \qquad$ $\displaystyle \mbox{synthesis equation}$ (3)
$\displaystyle a_k = \frac{1}{N} \sum_{n \in <N>} x[n]e^{-jk\Omega_0 n}$ $\textstyle \qquad$ $\displaystyle \mbox{analysis equation}$ (4)

As one would expect, the integral in time goes to a sum. However, there is one more key difference: the sum in the synthesis equation is finite! (over an interval the length of one period).

First, recall that $\Omega_0 = \frac{2\pi}{N}$. Then since $ e^{jk\Omega_0 n} = e^{\frac{j k 2 \pi n}{N}} =
e^{\frac{j(k+N)2 \pi n}{N}}$ (since $e^{j2 \pi n} = 1, \forall n$), the $a_k$'s are periodic with period $N$ and only $N$ terms are needed in the sum.