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?

  • $\begingroup$ 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 '16 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 '16 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 '16 at 23:11
  • $\begingroup$ @AtulIngle Oh you're right, I got it, I really appreciate your time. $\endgroup$ – user137927 Nov 30 '16 at 23:14

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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