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So, I am trying to solve this question.

Consider the signal, $$x=\cos\left(\left(\dfrac{2\pi}{3} \right)n\right)$$

The signal is downsampled by a factor of two. Indicate the frequency of the resulting output, normalized by $2\pi$

(E.g., if the frequency is $\dfrac{\pi}{2}$, write $\dfrac{1}{4}$)

So, what I understood is,

from the signal, $f = \dfrac{1}{N} \implies \dfrac{1}{3}$ or $\dfrac{2\pi}{3}$ if we consider angular frequency.

So after downsampling, $$f_d = \dfrac{f}{M} = \frac{1}{3}\cdot \frac{1}{2} \implies \frac{1}{6}$$.

or, if taking $\dfrac{2\pi}{3}$ and downsampling by factor $2$ then, $\dfrac{2\pi}{3}\cdot\dfrac{1}{2} = \dfrac{\pi}{3}$ or $\dfrac{1}{6}$

But the answer is wrong. How should I solve it?

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If the term downsampling is used in its correct meaning as anti-aliasing lowpass filtering followed by decimation, then your output should be zero, as the lowpass filter with a cutoff frequency of $\omega = \pi/2$ should remove the input $x[n] = \cos( \frac{2\pi}{3} n) $ ; i.e. , frequency of the cosine is out of the passband of the lowpass filter, and will be blocked by the filter.

However, if you simply mean decimation (or compression) by the term downsampling, then, the output will be: $y[n] = x[2n] = \cos(\frac{2\pi}{3} 2n) = \cos(\frac{4\pi}{3} n) = \cos(\frac{2\pi}{3} n) = x[n]$. This equivalence is due to the aliasing that occurs for omitting the anti-aliasing lowpass filter.

Hence, in this case the frequency of the output is the same as frequency of the input.

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