Fourier transform of periodic sequences

A periodic sequence is not absolutely summable, so its DTFT does not exist in the ordinary sense. It is found by allowing impulses in the frequency domain: each Fourier series term becomes a train of impulses.

Step 1: DTFT of a single complex exponential

Consider the spectrum consisting of impulses of area $2\pi$ at $\omega_0$, repeated every $2\pi$:

$$X(e^{j\omega}) = \sum_{l=-\infty}^{\infty}2\pi\,\delta(\omega - \omega_0 - 2\pi l)$$

Its inverse DTFT, taken over one period of $2\pi$ that contains only the impulse at $\omega_0$, is

$$x[n] = \frac{1}{2\pi}\int_{2\pi}2\pi\,\delta(\omega - \omega_0)e^{j\omega n}d\omega = e^{j\omega_0 n}$$

Hence

$$e^{j\omega_0 n} \xleftrightarrow{DTFT} \sum_{l=-\infty}^{\infty}2\pi\,\delta(\omega - \omega_0 - 2\pi l)$$

Step 2: Periodic sequence as a sum of exponentials

A periodic $x[n]$ with period $N$ has the DTFS

$$x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}, \qquad a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]e^{-jk(2\pi/N)n}$$

Step 3: Apply linearity

Each term $a_k e^{jk(2\pi/N)n}$ gives impulses of area $2\pi a_k$ at $\omega = 2\pi k/N + 2\pi l$. Since $a_k$ is periodic in $k$ with period $N$, the impulses for all $k$ and $l$ can be combined into one sum over all integers $k$:

$$\boxed{X(e^{j\omega}) = \sum_{k=-\infty}^{\infty}2\pi\,a_k\,\delta\!\left(\omega - \frac{2\pi k}{N}\right)}$$

Procedure: find the DTFS coefficients $a_k$, then place an impulse of area $2\pi a_k$ at each frequency $2\pi k/N$.

 X(e^jw)
     2pi a0   2pi a1        2pi a0 (repeat)
       ^        ^      ^      ^
       |   ^    |      |      |
 ------+---+----+------+------+---- w
       0  2pi/N 4pi/N  ...   2pi

Example: periodic impulse train

$x[n] = \sum_k\delta[n - kN]$ has $a_k = 1/N$ for all $k$, so

$$X(e^{j\omega}) = \frac{2\pi}{N}\sum_{k=-\infty}^{\infty}\delta\!\left(\omega - \frac{2\pi k}{N}\right)$$

An impulse train in time gives an impulse train in frequency.

Example 2: $\cos\omega_0 n = \frac{1}{2}(e^{j\omega_0 n} + e^{-j\omega_0 n})$ gives $X(e^{j\omega}) = \sum_l \pi[\delta(\omega - \omega_0 - 2\pi l) + \delta(\omega + \omega_0 - 2\pi l)]$.