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Let $x[n] = \cos(\frac{\pi}{8}n)$ and $X(k) = \text{DFT}_{16} \{ x[n]\}$

In order to find $X(k)$ I write out the DFT sum:

$$X(k) = \sum_{n=0}^{N-1} x[n] e^{-j2\pi kn/N} = \sum_{n=0}^{15} \cos(\frac{\pi}{8}n) e^{-j2\pi kn/16} $$

I could expand the $\cos(\cdot)$ as two complex exponentials inside the sum but that would make the computation rather tedious.

However, if we expand it before jumping into the summation we see that:

$$\cos(\frac{\pi}{8}n) = 0.5e^{j\pi n/8}+0.5e^{-j\pi n/8}$$

Which is just the 16 Pt IDFT of $0.5 \cdot \delta[k-1] + 0.5\cdot \delta[k+1]$ (Properties table in $\textbf{FSP 2014 Vetterli et al Pg 256}$).

At this stage I am not sure how to proceed with handling this IDFT and getting the $\text{DFT}_{16}$ of the original $x[n]$ although I do see some similarities between this and the duality property of the CTFT which makes me think that perhaps if I multiply my IDFT with $N=16$ I should get my desired DFT but I am not sure.

Sidenote: What if the function was changed to $\cos(\frac{\pi}{16}n)$ or $\cos(\frac{\pi}{4}n)$. Would we still handle it the same way?

$\textbf{Edit:}$ Attempt at a solution for the sidenote, $x[n] = \cos\frac{\pi}{16}n$:

Let $x[n] = \cos\frac{\pi}{16}n = 0.5e^{j\pi n/16}+0.5e^{-j\pi n/16}$

Split the DFT sum into two parts:

$$ \frac 12 \sum_{n=0}^{N-1}e^{j\frac{\pi}{16}n}e^{-j\pi/8 kn} = \frac 12 \sum_{n=0}^{N-1}e^{j\frac{\pi}{8}n(\frac 12 - k)} = \frac 12 \frac{1 - (e^{j\frac{\pi}{8}(\frac 12 - k)})^{16}}{1-e^{j\frac{\pi}{8}(\frac 12 - k)}}$$

$$ \frac 12 \sum_{n=0}^{N-1}e^{-j\frac{\pi}{16}n}e^{-j\pi/8 kn} = \frac 12 \sum_{n=0}^{N-1}e^{-j\frac{\pi}{8}n(\frac 12 + k)} = \frac 12 \frac{1 - (e^{-j\frac{\pi}{8}(\frac 12 + k)})^{16}}{1-e^{-j\frac{\pi}{8}(\frac 12 + k)}}$$

Upper one equals $\frac{1}{1-e^{j\pi /16}\cdot e^{-j\pi/8 k}}$

And lower one equals $\frac{1}{1-e^{j\pi /16}\cdot e^{j\pi/8 k}}$

How can I combine the two to get the final result?

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    $\begingroup$ Yes, your approach is correct. Here is a visual representation of the DFT in question. Just keep in mind that the shift to the left represented by $\delta[k+1]$ is a circular shift. Whatever falls off, comes back up on the other side! $\endgroup$
    – Ahsan Yousaf
    Apr 11 at 13:00
  • $\begingroup$ Thank you! By the way, shouldn't the third plot have $n$ as its x-axis label? $\endgroup$
    – user67157
    Apr 11 at 14:10
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    $\begingroup$ Yes you are correct. It should be 'n' instead of 'k' $\endgroup$
    – Ahsan Yousaf
    Apr 11 at 14:11
  • $\begingroup$ I see. Also, what about handling functions like $\cos(\frac{\pi}{16}n)$ in this fashion? $\endgroup$
    – user67157
    Apr 11 at 14:13
  • $\begingroup$ @AcerbicNarcissist hm, you're asking this for a reason. What do you think would be different? $\endgroup$
    – Marcus Müller
    Apr 11 at 14:27

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