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When I want to calculate the discrete Fourier transform of $\frac{1}{3}^{-k}$ with $k\le-1$ I get that: $$ \sum_{k=-{\infty}}^{-1}\left(\frac{1}{3}\right)^{-k}e^{-j\omega k} = \sum_{k'=1}^{\infty} \left(\frac{1}{3}e^{j\omega}\right)^{k'} = \frac{\frac{1}{3}e^{j\omega}}{1-\frac{1}{3}e^{j\omega}} $$ But if I change the variable before finding the Fourier transform I have the function $\left(\frac{1}{3}\right)^k$ with $k\ge1$ (both functions are the same). The discrete Fourier transform of the new function is as follows: $$ \sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}e^{-j\omega k} = \sum_{k=1}^{\infty}\left(\frac{1}{3}e^{-j\omega}\right)^{k} = \frac{\frac{1}{3}e^{-j\omega}}{1-\frac{1}{3}e^{-j\omega}} $$ Are these two the same?

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  • $\begingroup$ The sequence $\{(1/3)^{-k}\}$ for $k\leq -1$ is not the same sequence as $\{(1/3)^{m}\}$ for $m \geq 1$. $\endgroup$
    – Atul Ingle
    Nov 30, 2016 at 23:02
  • $\begingroup$ @AtulIngle you mean the order of the sequence? but the results of them in a definite point are equal, why is the order important? $\endgroup$
    – user137927
    Nov 30, 2016 at 23:07
  • $\begingroup$ See plots here. imgur.com/a/gVXpI The two sequences are mirror images in time i.e. flipped. This explains why their DFTs has a sign change on the frequency variable. $\endgroup$
    – Atul Ingle
    Nov 30, 2016 at 23:11
  • $\begingroup$ @AtulIngle Oh you're right, I got it, I really appreciate your time. $\endgroup$
    – user137927
    Nov 30, 2016 at 23:14

1 Answer 1

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As Atul has pointed out, your both sequences are not the same, but they are time-reversed versions of each other:

k = np.arange(-15, 15)

s1 = (1/3.)**(-k)*(k <= -1);
s2 = (1/3.)**(+k)*(k >= +1)


plt.stem(k, s1, 'r')
plt.stem(k, s2, 'g')

program output

Accordingly, what you see in your calculations is the time-reversal property of the Discrete-Time Fourier Transform:

$$x(t)=y(-t) \Leftrightarrow X(f)=Y(-f)$$

I.e. the frequency changes its sign in your calculations, as is expected from the time-reversal property.

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