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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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As hinted to by @StanleyPawlukiewicz in comments, the correct definition to use is, using your style of notation: $$\rm dBV_{pk} = 10\log_{10}\left({V_{pk}}^2\right) = 20\log_{10}\left(\sqrt{{V_{pk}}^2}\right) = 20\log_{10}(|V_{pk}|).$$ Clearly, $\rm dBV_{pk}$ will not contain information about the sign of $\rm V_{pk}$. You cannot safely fiddle with the ...

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Logically, think of what an autocorrelation can be used for: it takes a signal, and looks for repetitive patterns by comparing the original signal to a shifter version of the signal. Since that’s the case, the zero time lag of just about any signal will have a non-zero value, since it essentially multiples all signal components with themselves, and then ...

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You can try solving the problem as a hypothesis testing problem. Consider the following: \begin{align} \mbox{Under }\mathcal{H}_0&: y[n] = s[n] + w[n] \\ \mbox{Under }\mathcal{H}_1&: y[n] = w[n]. \end{align} Then, you can find a test statistic to determine if your window $\Omega_N$ which contains $N$ samples of the measured signal contains signal or ...

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I would like to give another take on @DanielSank's answer. We first suppose that $v_{n} \sim \mathcal{CN}(0, \sigma^{2})$ and is i.i.d. Its Discrete Fourier Transform is then: $$V_{k} = \frac{1}{N} \sum_{n=0}^{N-1} v_{n} e^{-j 2 \pi \frac{n}{N} k}$$. We want to calculate the distribution of $V_{k}$ To start, we note that since $v_{n}$ is white Gaussian ...

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It depends on what you mean by SNR. It's a common joke in the DSP community to spell it out as "something to noise ratio", referring to the fact that there is no unique definition of SNR, so the term by itself means nothing. Define it yourself and use it appropriately. What's common is to define it as ${\rm SNR} = \frac{P_{\rm s}}{P_{\rm n}}$ where $P_{\rm ... 2 ...And now for a differing opinion.... The OP's representation of bandpass white noise as $$n(t) = n_I \cos(2\pi f_ct) - n_Q \sin(2\pi f_ct)\tag{1}$$ is inadequate; because each sample path of this noise process is a pure sinusoid of fixed frequency$f_c$Hz which is not noise-like at all. Why so? Well, a sample path is what one gets when all the random ... 0 $$-\frac{\pi}{2}\le \mathrm{arctan}(x)\le \frac{\pi}{2} \quad x\in (-\infty , \infty)$$ so as asked, no. Perhaps $$Y(f)=|H(f)|^2 X(f)$$ where$H(f)$is your narrow band filter, would imply that the random phase of the input$x(t)\$ is the same nature at the output.

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