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.


closed as off-topic by Matt L., Stanley Pawlukiewicz, MBaz, lennon310, Fat32 Nov 9 '18 at 21:26

  • This question does not appear to be about signal processing within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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? $\endgroup$ – Matt L. Nov 9 '18 at 17:07
  • 4
    $\begingroup$ I'm voting to close this question as off-topic because it's a homework type question with zero effort shown. $\endgroup$ – 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.

  • 1
    $\begingroup$ The standard definition of the DTFT has a negative sign in the exponent. $\endgroup$ – Matt L. Nov 9 '18 at 17:25
  • $\begingroup$ 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? $\endgroup$ – TootsieRoll Nov 9 '18 at 22:27
  • $\begingroup$ @TootsieRoll, Yes, those are correct. $\endgroup$ – Trey Nov 10 '18 at 1:57
  • $\begingroup$ @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? $\endgroup$ – TootsieRoll Nov 11 '18 at 16:58
  • $\begingroup$ @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, $$ cos(\omega) = \frac{1}{2} (e^{j\omega} + e^{j\omega}) $$ $\endgroup$ – Trey Nov 12 '18 at 17:28

Not the answer you're looking for? Browse other questions tagged or ask your own question.