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I am reading a DSP textbook and I am on the sampling portion. I understand that a discrete time frequency greater than 1/2 will alias to frequencies in the interval -0.5 to 0.5 and other frequencies outside of that interval but the book just states which frequency it aliases to without providing the method by which they found the frequency.

For instance, one of the frequencies in $3\cos(600\pi t) + 2\cos(1800\pi t)$ when it is sampled at $F = 1000$ samples/second is $f = 0.9$. The book states that this aliases to $f = 0.1$, but I can't figure out how they got that result.

lennon310
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Michael
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1 Answers1

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Let $x(t) = 3\cos(600\pi t) + 2\cos(1800\pi t)$

With $F_s = 1000$ samples/sec we have $T_s= 1/1000$. So $x[n] = x_c(nT_s)$ is as follows:

$$x[n] = 3\cos(600\pi nT_s) + 2\cos(1800 \pi nT_s)$$

$$ x[n] = 3\cos(0.6\pi n) + 2\cos(1.8 \pi n)$$

so $\Omega_1$ from first sinusoidal is $0.6\pi$ and $\Omega_2$ from second sinusoidal is $1.8\pi$.

Now let's try and reconstruct it using Shannon reconstruction:

Let $\omega_r = (\omega_c + \frac{2\pi L}{T_s})$.

Choose $L$ such that $\omega_r \in (\frac{-\pi}{T_s}, \frac{\pi}{T_s})$. If that $L\neq 0$ we have aliasing.

So now we have:

$$\omega_{r_1} = 600\pi + 2000\pi L$$ so we choose $L =0$ to make $\omega_{r_1} \in (-1000\pi, 1000\pi)$.

and

$$\omega_{r_2} = 1800\pi + 2000\pi L$$

so now we have to choose $L=-1$ to make $\omega_{r_2} \in (-1000\pi, 1000\pi).$

So we have confirmed aliasing in the second sinusoidal. And the folded frequency is:

$$1800\pi -2000\pi = -200\pi$$

which corresponds to $f= \pm 100 \: \texttt{Hz}$ ($\texttt{cosine}$ is even, folding is happening from negative frequency as well). If I scale it with respect to the sampling frequency we have $f=0.1$.

Ahsan Yousaf
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