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Let $X \left( e^{j \omega} \right)$ be the DTFT of a discrete-time signal $x \left[ n \right]$ given as

$$ x \left[ n \right] = \left( \frac{1}{3} \right)^{n} u \left[ n \right] \tag{1} $$

Determine a $6$-point causal sequence $a \left[ n \right]$ whose $6$-point DFT $A \left[ k \right]$ is the $6$ samples of the DTFT $X \left( e^{j \omega} \right)$ such that,

$$ A \left[ k \right] = X \left( e^{j 2 \pi k/6} \right), ~~~~ k = 0, 1, \ldots, 5 \tag{2} $$

As DFT can be obtained by the uniform sampling of DTFT of the signal, therefore, $$ A \left[{k} \right]= X \left( e^{j 2 \pi k/6} \right)= 1/(1-(1/3)\left( e^{-j 2 \pi k/6} \right))) $$ and how to convert $A \left[{k} \right]$ back to time domain signal $a \left[{n} \right]$. I tried using IDFT but it's too long is there any alternative method?

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    $\begingroup$ To answer your question: yes, you can. But we're not here to do your homework. You can show what you've tried and ask a more specific question and you might get some help. $\endgroup$
    – Matt L.
    Jan 30 at 12:54
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    $\begingroup$ @DanBoschen: Only for finite length signals ... $\endgroup$
    – Matt L.
    Jan 30 at 21:24
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    $\begingroup$ The answers to this question show the solution to this problem. $\endgroup$
    – Matt L.
    Jan 31 at 10:37
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    $\begingroup$ @DanBoschen: That's basically it: I thought this is a very good example for time domain aliasing. It has a significant effect but it is also easy enough to work it out by hand. $\endgroup$
    – Hilmar
    Feb 2 at 1:23
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    $\begingroup$ Since the question was upvoted 4 times, had a some good comments, and any homework due date has probably expired by now, I will undelete my answer. Feel free to re-delete if that feels inappropriate $\endgroup$
    – Hilmar
    Feb 3 at 14:59

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Yes, you can.

The DTFT is continuous and the DTF is discrete in the frequency domain. The DFT in this problem is a sampled version of the DTFT. Sampling in one domain is equivalent to periodic repetition in the other domain.

To implement periodic repetition we can simply cut up the original impulse response into chunks of 6 and sum the up the chunks.

$$a[k] = \sum_{n=-\infty}^{\infty}x[k+6 \cdot n] \tag{1}$$

For the specific impulse response at hand we get

$$a[k] = \sum_{n=0}^{\infty} \left( \frac{1}{3} \right)^{k+6\cdot n} = 3^{-k} \sum_{n=0}^{\infty} 3^{-6 \cdot n} \tag{2}$$

We can easily see that each coefficient is simply a third of the previous one and we can simply work out all remaining coefficients from the first one as

$$a[k] = a[0] \cdot 3^{-k} \tag{3}$$

We can write the sum in the form of a geometric sum

$$ a[0] = \sum_{n=0}^{\infty} 3^{-6 \cdot n} = \sum_{n=0}^{\infty} \left( \frac{1}{3^6} \right) ^n = \frac{1}{1-\left( \frac{1}{3^6} \right)} = \frac{3^6}{3^6-1}\tag{4}$$

And with this we get the final impulse response

$$a[k] = 3^{-k} \frac{3^6}{3^6-1} = \frac{3^{6-k}}{3^6-1} \tag{5}$$

This is a prime example of "time domain aliasing". The sampling theorem applies equally to both domains. In this case you are under sampling in the frequency domain and hence the time domain aliases.

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