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We're covering spectral analysis for discrete time stationary processes in lectures, and I can't really get my head around something which seemingly should be easy.

"Because time is discrete, we only need to consider frequencies in $[\frac{-1}{2},\frac{1}{2}]$"

I have absolutely no understanding of why this is so. From my understanding of frequency, it's the number of observations (i.e. measurements making your time series) per pre-defined time interval. So, for example, if the interval is a day and we take a measurement per hour, frequency would be 24. This seems to agree.

I don't know where my misunderstanding lies.

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  • $\begingroup$ Can't formulate an answer right now, but read on Nyquist-Shannon sampling theorem. $\endgroup$ – Firebug Oct 27 '17 at 15:08
  • $\begingroup$ In spectral analysis, frequency is not the number of observations per fundamental time interval. Best to re-read your spectral analysis notes. $\endgroup$ – Stephan Kolassa Oct 27 '17 at 15:26
  • $\begingroup$ @StephanKolassa Thank you for your response. I'm posting here because I have reread them (many times) and the answer to my question appears to be presumed knowledge. I've asked on here as a last resort. Could you give more insight into what the frequency 'means' in spectral analysis? $\endgroup$ – hhattiecc Oct 27 '17 at 15:55
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Let's say your time tick is 1 sec. You get one amplitude observation every second. What is the highest frequency you can detect? Your notes are telling you that it's 1/2 Hz.

Suppose you are sampling a perfect wave with a period 1 sec, its frequency is 1Hz: $\cos( 2\pi t)$. At time zero you measure $x_0=1$, then you get next few measurements: $$x_0=1,x_1=1,x_2=1,x_3=1,\dots$$ You got the point: this is undetectable.

On the other hand, let's look at 1/2 Hz perfect wave $\cos(\pi t)$ for $t=0,1,2,3,\dots$: $$x_t=1,-1,1,-1,\dots$$ - now we can catch this wave in the signal. That's the intuition, and Shannon has a proof.

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