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I am a bit confused between the following two formulas:

$$ f\le \frac{f_s}{2} \tag{1}$$

and:

$$ f = \frac{f_s}{N} \tag{2}$$

Now I am given an ACF Plot of a sine/cosine wave and I am asked to find the frequency of the signal if the signal is sampled at 1,000 samples/second. I am asked to use the information given in the ACF plot and the sampling rate of 1,000 samples/second.

In the book "Probability and Stochastic Processes" I find the formula (1) i.e., Nyquist Theorem and at another site, here, I find the formula (2) in use.

I have the above two formulas but I am confused in which formula to use. I tried asking this question before but could not explain it clearly. I hope I have explained it properly this time.

Reproduced figure.

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1 Answer 1

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As you say,

$$ f\le \frac{f_s}{2} \tag{1}$$

is the Nyquist criterion or frequency. It means that, to avoid aliasing (frequency domain folding), a signal must have frequencies below $\frac{f_s}{2}$. This probably won't help you determine the frequency of the ACF.

This equation $$ f = \frac{f_s}{N} \tag{2}$$ is saying something different. The value $\frac{1}{f_s}$ is the time between samples, or the sampling period. When a signal has $N$ samples in one period, that means the period of the signal is $\frac{N}{f_s}$.

That means that the frequency of the signal is given by formula (2).


With reference to your (now added) ACF: this is not just a sinusoid. It's a damped sinusoid (it is a sinusoid with, probably, a linear decay).

That means it doesn't make sense to look at the peaks because the peak position will change because of the decay.

The peak that won't change is at Lag $0$. Start there.

Then look at the zeros. The zero positions won't change because of the decay envelope. I see three zeros: 10, 30, and 50.

All of these give a period of $$ 10/0.25 = 30/0.75 = 50/1.25 = 40\ \text{samples}.$$

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  • $\begingroup$ So I count the number of samples from the graph and along with the value $1,000$ put the values in second formula in order to find the frequency of the signal. right? $\endgroup$ May 26 at 13:15
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    $\begingroup$ @MuhammadAhmad Yes, pretty much! Often the period won't be an integer number of samples long. Then you need to do something much more complicated to get the frequency. $\endgroup$
    – Peter K.
    May 26 at 13:19
  • $\begingroup$ check my edit and kindly clarify it $\endgroup$ May 26 at 13:58
  • $\begingroup$ @MuhammadAhmad See update to answer. $\endgroup$
    – Peter K.
    May 26 at 14:05
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    $\begingroup$ @MuhammadAhmad And it says the same thing: 10/0.25 = 40 samples. $\endgroup$
    – Peter K.
    May 26 at 14:24

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