# Practice Question on Computing the Fourier Transform of a Discrete-time Signal

Compute the Fourier transform of the signal

$x[n] = 3^n u[-n].\$

${\mathcal X} (\omega) = \sum_{n=-\infty}^\infty x[n]e^{-j\omega n}=\sum_{n=-\infty}^\infty 3^n u[-n]e^{-j\omega n}=\sum_{n=-\infty}^0 3^n e^{-j\omega n}=\sum_{n=-\infty}^0 \Bigg(\frac{3}{e^{j\omega}}\Bigg)^n$ Let k=-n

$= \sum_{n=0}^\infty \Bigg(\frac{e^{j\omega}}{3}\Bigg)^k$

$\mathcal X (\omega) = \frac{1}{1-\frac{e^{j\omega}}{3}}$

--Cmcmican 19:42, 28 February 2011 (UTC)

TA's comments: The answer is correct. The geometric series converges because $\color{OliveGreen}{\left|\frac{e^{j\omega}}{3}\right|=\frac{1}{3}<1}$.