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I have met this question, but cannot prove it using DTFT definition. Given: $g$ is a discrete sequence filter and: $$ g \in l^2(Z)$$ $$\langle g_n, g_{n-2k} \rangle = \delta_{k}$$ Prove: $$|G(e^{j \omega})|^2 + |G(e^{j(\omega + \pi)})|^2 = 2$$

Could you guys give me some direction on this one? Thank a lot

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  • $\begingroup$ just keep in mind that $$ G(e^{j(\omega+\pi)}) = G(e^{j\omega} e^{j\pi}) = G(-e^{j\omega}) $$ $\endgroup$ – robert bristow-johnson Feb 23 at 23:32
  • $\begingroup$ Hi @robertbristow-johnson I got stuck with 2k. I don't know how to interpret it in term of DTFT? Is it just a shift by 2k? then I add e^(jw2k), but this does not make any progress $\endgroup$ – hminle Feb 24 at 0:40
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Here a few hints to get you started:

  1. Define a sequence $x[k]=\langle g_n, g_{n-k} \rangle$ and write it as a convolution. From that it should be easy to find its DTFT $X(e^{j\omega})$ in terms of $G(e^{j\omega})$.

  2. Next note that the given sequence is just a downsampled version of $x[k]$, so its spectrum is an aliased version of $X(e^{j\omega})$. That's how you end up with those two terms on the left-hand side of the frequency-domain representation.

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  • $\begingroup$ Hi, I managed to follow your guide, so here what I have now: $X(e^{jw}) = |G(e^{jw})|^2$. And we have $x_{2k}$ is a downsampled version of $x_k$. Then I have: $$ 1/2(X(e^{(jw)/2}) + X(e^{j(w-2\pi)/2})) = 1 $$ $$ => |G(e^{jw/2})|^2 + |G(e^{j(w-2\pi)/2})|^2 = 2 $$ I can only come here, so close to the result, but cannot move further. You have any other ideas transform it $\endgroup$ – hminle Feb 24 at 16:45
  • $\begingroup$ Oh, could I do this? Let $$w' = w/2$$ Then $$ => |G(e^{jw'})|^2 + |G(e^{j(w'-\pi)})|^2 = 2 $$ and $$ |G(e^{j(w'-\pi)})| = |G(e^{j(w'+\pi)})| $$ Then I have the final result. $$ |G(e^{jw'})|^2 + |G(e^{j(w'+\pi)})|^2 = 2 $$ I am not sure I could do like this? Because $w'$ is different than $w$ $\endgroup$ – hminle Feb 24 at 18:13
  • $\begingroup$ @hminle: Yes, you can do that. The result is true for any $\omega$. $\endgroup$ – Matt L. Feb 24 at 18:41
  • $\begingroup$ wow, great sir. Thank you a lot $\endgroup$ – hminle Feb 24 at 21:21

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