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I am new in this community. So excuse me if it seems easy.

I want to calculate the amplitude of a specific frequency in a signal. The frequency precision is (relevant question) $$ df = \frac{fs}{NP} $$ $$ fs=1/dt$$ NP is the number of points in the signal time seri. The index of the frequency in the signal is $$ ind = \frac{f}{df} $$ the length of power spectrum array of signal is length(pxx) = NP/2-1

Example 1: for small frequencies

frequency = 1.0/128.0    
dt=0.4
df = 0.001
NP = fs/df=2500
index of the signal = 8

lenght of pxx is 1249 so no problem.

Example 2: for large frequencies 

freq = 2.0    
dt=0.4
NP = fs/df=2500
index of the signal = 2000

The index of the frequency is larger than the number of points in pxx. If I double the number of points in the signal, the frequency precision halves and the index of frequency also doubles. Is there a frequency limit or I am doing something wrong?

Thanks for any guide.

Abolfazl
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1 Answers1

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Any Discrete-Time Fourier Transform (with constant sampling frequency) will only give information about the signal up to a frequency of

$$f = \frac{f_s}{2} = f_N$$

\$f_N\$ is called the Nyquist frequency. Frequencies in the signal higher than \$f_N\$ are subject to aliasing (related to the Nyquist-Shannon sampling theorem).

So if you want to find information about a certain frequency \$f_x\$, you will need at least a sampling frequency of \$f_s > 2\cdot f_x\$ and extra to accommodate for an anti-alias filter.

Increasing the number of points will only increase the resolution, without changing the Nyquist frequency.

Sven B
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