Fourier transformations of discrete time signals How does one Fourier transform the following signals? $$x[n]=[0,9,0]$$ and $$y[n]=[9,0,19,0,9]$$
I've tried to get it along the way of $$X[\omega]=e^{-i\omega n}$$
but this seems incorrect. I'm not sure how the second one is supposed to be handled as well.
When finding the DTFT of short discrete sequences like this, you can directly compute the sum. 

For example, say you had x[n] = [1 0 2] for n=0,1,2. Notice that I defined the values for n. If you had different n's, then the answer would be different. 

You can compute the sum as
\begin{align}
X(\omega) &amp;= \sum_n x[n] e^{-j\omega n} \\
&amp; = 1e^{-j\omega 0} + 0e^{-j\omega 1} + 2e^{-j\omega 2} \\
&amp; = 1 + 2e^{-j\omega 2}
\end{align}

Another note, you can sometimes simplify the answer by combining two exponentials. Be on the lookout for exponentials with the same coefficient. An example would be (using $e^{-j0}$ = 1)

\begin{align}
X(\omega) &amp;= 2 + 2e^{-j\omega 4} \\
&amp;= 2e^{-j\omega 0} + 2e^{-j\omega 4} \\
&amp;= 2e^{-j\omega 2} (e^{j\omega2 } + e^{-j\omega 2}) \\
&amp;= 2e^{-j\omega 2} (2\cos(\omega2)) \\
&amp;= 4e^{-j\omega 2} \cos(\omega2)
\end{align}
Hope this helps.