Fourier series vs transform; DTFT of aⁿ

Fourier series vs Fourier transform

DTFT of $x[n] = a^n$, $|a| < 1$

Taken literally for all $n$, $a^n$ grows without bound as $n \to -\infty$ (for $|a|<1$), so it is not absolutely summable and its DTFT does not exist. The intended (textbook) signal is the causal one, $x[n] = a^n u[n]$.

$$\begin{aligned} X(e^{j\omega}) &= \sum_{n=0}^{\infty} a^n e^{-j\omega n} = \sum_{n=0}^{\infty}\left(ae^{-j\omega}\right)^n \\ &= \frac{1}{1 - ae^{-j\omega}}, \qquad |a| < 1 \end{aligned}$$

Magnitude and phase:

$$|X(e^{j\omega})| = \frac{1}{\sqrt{1 + a^2 - 2a\cos\omega}}, \qquad \angle X(e^{j\omega}) = -\tan^{-1}\frac{a\sin\omega}{1-a\cos\omega}$$

For $0 < a < 1$ the magnitude is largest at $\omega = 0$ ($\frac{1}{1-a}$) and smallest at $\omega = \pi$ ($\frac{1}{1+a}$), so the sequence is low-pass. For $-1<a<0$ it is the other way round (high-pass).

If the two-sided signal $x[n] = a^{|n|}$ is meant:

$$\begin{aligned} X(e^{j\omega}) &= \sum_{n=0}^{\infty} a^n e^{-j\omega n} + \sum_{n=1}^{\infty} a^n e^{j\omega n} \\ &= \frac{1}{1-ae^{-j\omega}} + \frac{ae^{j\omega}}{1-ae^{j\omega}} = \frac{1 - a^2}{1 - 2a\cos\omega + a^2} \end{aligned}$$

which is real and even, as expected for a real even sequence.