Relation between DTFT and DFT

Question

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?

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. — Matt L., Jan 30 '24 at 12:54
Is this actually homework? It's certainly an interesting question with an interesting answer. — Hilmar, Jan 30 '24 at 13:16
Now you have me curious @Hilmar if you see something besides the obvious (or what I think is the obvious) knowing the DFT is samples of the DTFT. — Dan Boschen, Jan 30 '24 at 18:19
This is just a response to the title of the question. Here is a mathematical comparison of the DTFT and DFT - how they are related. — robert bristow-johnson, Jan 30 '24 at 21:31
@MattL thanks I see the interesting part now. Nice answer RBJ (+1) — Dan Boschen, Jan 31 '24 at 00:01
The answers to this question show the solution to this problem. — Matt L., Jan 31 '24 at 10:37
@DanBoschen: I don't know exactly what you deem obvious. Spoiler alert: I worked the answer out to be $a[k] = \frac{3^{6-k}}{3^6-1} $ which wasn't obvious to me . — Hilmar, Feb 01 '24 at 00:36
@Hilmar Matt cleared up my confusion, I was thinking if it was a finite length then A[k] would simply be the DFT and the answer would be the first 6 samples of x[n] in that case—- or am I missing something else? — Dan Boschen, Feb 01 '24 at 10:41
@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. — Hilmar, Feb 02 '24 at 01:23
@Hilmar yes agreed- a very good example; it’s really interesting when you have x[n] infinite and I wasn’t thinking about that at all- and in that case not so obvious what the answer would be. — Dan Boschen, Feb 02 '24 at 06:27
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 — Hilmar, Feb 03 '24 at 14:59

score 2 · Accepted Answer · answered Jan 30 '24 at 13:13

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.

Relation between DTFT and DFT

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