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$$x[n]=2\sin\left(0.2\pi n-\frac{\pi}{2}\right)+4\cos(0.5\pi n-\pi)+\cos(0.8\pi n)$$ This signal is the result of sampling $x(t)$ with a frequency of 40Hz.

I have to see if this frequency can reconstruct the signal without aliasing. I understand that to verify this, we need $\omega_s > 2 \omega_{\max}$.

From $x[t]$ I get this data:

$\Omega_0=0.1\pi$ rad/s

$N=20$ s

$\omega_0=4\pi$ rad/s

In the resolution that I'm following I have that $\omega_s>2\cdot8\cdot4\pi$ so $\omega_{\max}=8\times4\pi$.

This is what I don't understand:

  1. Why is $\omega_s>2\times8\times4\pi$?

  2. Where does that $8$ come from?

Can someone clarify me this situation?

Mhenni Benghorbal
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1 Answers1

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There are several parts of your question that do not make sense, so I am going to make some educated guesses as to what you are trying to do.

  1. Figure out the maximum frequency in $x(t)$ assuming that the $x[n]$ is not aliased.
  2. Use this frequency to get the minimum required sampling frequency, $\omega_s$.

By inspection, the max frequency of $x[n]$ is $0.8 \pi$ (radians/sample). To put this in terms of radians per second, note that if you have $\phi$ radians for every sample, and you have $1$ sample for every $1/f_s$ seconds, then you have $\phi / (1/f_s) = \phi f_s$ radians per second. With the numbers you give, this means that

$$ 0.8\pi \cdot f_s = \pi \dfrac{4}{5} \cdot 40 = \pi 4 \cdot 8 = 32\pi $$

Thus, $w_{s} > 2 \cdot 32\pi = 2 \cdot 8 \cdot 4\pi$.

AnonSubmitter85
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