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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 ...

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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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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 ...

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