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My suggestion would be to plot the frequency response by taking the FFT of a reasonable length to see the spectrum. Apart from the two spikes in FFT output (corresponding to sinewave) you could see the DFT of the periodic spur signal with approximate amplitude of $N$ where $N$ is the period of the combined signal. The DFT of spur is a constant amplitude ...

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take fft of data let's say data is x = [ 3 4 5 6 6 9 7 ] take fft y = fft(x); shift so -fs/2 to fs/2 y = fftshift(x); plot mag plot(abs(y)) see link for help fft Help

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no, this would not work, even if a technology advanced the state of the art of cameras to the wavelength resolution necessary to measure a shift. In order to measure a shift, you need to know what the unshifted frequency is. In astronomy, the spectral emission lines are known. You need both frequencies to measure shift. How does this scheme, know the ...

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Say the intensity signal at one of the color channels changes linearly from 1 to 0.5 relative to full scale when the wavelength changes by 5 nm near 580 nm, and you have a laser for illumination at that wavelength. For an object moving 100 km/hour or 28 m/s, the doppler shift is 28 m/s * 2 / speed of light = 0.0000002 times the wavelength, 580 nm * 0.0000002 ...

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Practically very difficult isn't it ? Speed of ordinary objects in an office environment will be so slow compared to light speed that the resulting shift in the color (wavelength) would be exremely small; much less than the unavoidable noise floor of the typical sensor...

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I think that there are two mistakes in your code/method. The first is the term $\sqrt{\Delta t}$ in your second formula; I think it should be replaced by $\Delta t$. The second is in the computation of the power spectrum from the estimated auto-correlation. What you do is square the result of the FFT Y to obtain mY, but that's not correct. First of all, Y ...

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If your question really is how to calculate the energy of a given signal $x(t)$ then the answer is $$E_x=\int_{-\infty}^{\infty}|x(t)|^2dt\tag{1}$$ but I wonder if that is actually what you're asking, because it would have been very easy to find that formula.

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The frequency response of a single FFT bin (when using the default rectangular window) is that of a periodic Sinc or Dirichlet function. If using a non-rectangular window (Hann or Hamming, etc.) the frequency response of each FFT bin is that of the transform of the window applied before the FFT. It's not the average of the signal energy within the bin width,...

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I'll start with a counter-example: Assume your continuous-time signal is a sinusoid at freqeuncy $f_0$. You then sample that signal, and use a rectangular window to select $N$ samples. Feed those $N$ samples to your DFT, and you will see non-zero values in the whole spectrum (unless in the special cases where $f_0$ is a multiple of $f_s/N$). So you start ...

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The "Effective" Number of Independent Observations in an Autocorrelated Time Series is a defined statistical term - https://www.jstor.org/stable/2983560?seq=1#page_scan_tab_contents The number of independent observations n' of n observations with a constant variance but having a lag 1 autocorrelation $\rho$ equals $n'=n\frac{1-\rho}{1+\rho}$ Also note ...

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You are on the right track. After upsampling the spectrum of X would look the same except the x axis would be divided by 2. So the limit of the spectrum would be at pi/4. Yes, that's correct. Since the filter has amplitude 2 would the amplitude of X be increased from 1 to 2 or will the amplitude of X remain unchanged? Recall that in the frequency ...

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You terminology is something I am not used to. $$\hat{x} = \mathcal{F}_{G} x$$ If by $\hat{x}$ you mean the DFT (complex except very special cases), $x$ is your input (real or complex), and $\mathcal{F}_{G}$ is the matrix composed of the sinusoidal basis vectors, then your assertion that $U$ is real is dubious. $\mathcal{F}_{G}$ is most certainly not. ...

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