I have a process $X$ with frequency spectra enter image description here

I sample this process with sampling frequency $f_s = 2$.

What will the frequency spectra $R_z(f)$ of the sampled process $Z$ look like? I realize that aliasing occurs since the original process contains frequencies above the Nyquist Frequency (in fact only contains frequencies above the Nyquist Frequency). The process has no frequencies within the interval where no aliasing occurs, i.e, $-1 \leq f \leq 1$. Therefore I reasoned that the frequency spectra of the sampled process would be $R_z(f) = 0, \quad 1\leq f\leq 1$ but the answer is $R_z(f) = 2, \quad 1\leq f\leq 1.$ Why?


Sampling in the time domain is equivalent to periodic repetition in the frequency domain with $1/T_s$ as the period. So formally you'd have something like $$R_{sampled}(f) = \sum_{n=-\infty}^{\infty}R_z(f-2\cdot n)$$

If you evaluate this, you actually get $R(f) = 2$. It's easier to see if you draw it: take the initial spectrum, shift it by -6,-4,-2,0,2,4,6... and sum it all graphically

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  • $\begingroup$ Alright, I see. But why are the frequencies within the nyquist range disturbed? Can't we accurately represent the frequencies within this range, even if some other frequencies are outside of it? $\endgroup$ – Heuristics Dec 27 '18 at 19:52

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