# Fourier transformations of discrete time signals [closed]

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.

• The chances that your formula for $X(\omega)$ would work were pretty low to start with because it doesn't even use the values $x[n]$. One question, why don't you just look up the correct definition of the discrete-time Fourier transform and just fill in your numbers? – Matt L. Nov 9 '18 at 17:07
• I'm voting to close this question as off-topic because it's a homework type question with zero effort shown. – Matt L. Nov 9 '18 at 17:08

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) &= \sum_n x[n] e^{-j\omega n} \\ & = 1e^{-j\omega 0} + 0e^{-j\omega 1} + 2e^{-j\omega 2} \\ & = 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) &= 2 + 2e^{-j\omega 4} \\ &= 2e^{-j\omega 0} + 2e^{-j\omega 4} \\ &= 2e^{-j\omega 2} (e^{j\omega2 } + e^{-j\omega 2}) \\ &= 2e^{-j\omega 2} (2\cos(\omega2)) \\ &= 4e^{-j\omega 2} \cos(\omega2) \end{align} Hope this helps.

• The standard definition of the DTFT has a negative sign in the exponent. – Matt L. Nov 9 '18 at 17:25
• I see, the signals should therefore be $$X[\omega] = 9*e^{\omega *(-i)}$$ and $$Y[\omega]=19*e^{-i*\omega*2}+9*e^{-i*\omega*4}+9$$ Does this seem correct? – TootsieRoll Nov 9 '18 at 22:27
• @TootsieRoll, Yes, those are correct. – Trey Nov 10 '18 at 1:57
• @Trey Thanks for the help, but i was wondering if the $$Y[\omega]=19*e^{-i*\omega*2}+9*e^{-i*\omega*4}+9$$ answer could be simplified via the exponents as well, using the $$cos[\omega] = (e^{i*\omega}+e^{-i*\omega})$$ as you pointed out previously. Can this be done a second time since there are 3 factors, or can it be done in one step? – TootsieRoll Nov 11 '18 at 16:58
• @TootsieRoll, you can only do it once. You could write expressions like this in a number of different ways, but it looks like just combining the terms with the 9 will be the simplest. Do you expect a different answer (was a different answer given)? Also, don't forget the 1/2 factor, $$cos(\omega) = \frac{1}{2} (e^{j\omega} + e^{j\omega})$$ – Trey Nov 12 '18 at 17:28