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Ok, I found out from a book that linear prediction always results in minimum phase filters, which implies stability. This can be shown using the mean-squared error (MSE) criterion which is to be minimized in the autocorrelation method. The proof is done by contradiction; if an all-pole filter had a single pole outside of the unit circle, the derivative of ...


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There are two mistakes in your code/method. The first is the term $\sqrt{\Delta t}$ in your second formula; 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 is complex-valued, and ...


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You have the definition of the random process as : $$ X(t) = a \sin(\omega_0 t + \Theta) \tag{1} $$ where $a$ and $\omega_0$ are deterministic constants and $\Theta$ is a continuous R.V. uniformly distributed in $[0,2\pi]$. According to the indexed-set of RV interpretation of a R.P., for each index $t$ you have a new R.V denoted as $X_t$ which has its own ...


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You have a single random variable $\Theta$ in this example, so taking the expectation with respect to that random variable results in a single integral. The formula for $R_X(\tau)$ in your question is rather messy and full of mistakes; it should be as follows: $$\begin{align}R_X(\tau)&=E[X(t+\tau)X(t)]\\&=E[a\sin(\omega_0t+\omega_0\tau+\Theta)\cdot ...


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to add to hot's answer, when using autocorrelation or an equivalent method like AMDF or ASDF, although the autocorrelation $R_x[k]$ is evaluated only at integer values of lag $k$, which means the peak value and initial period length will be an integer number of samples, you can use quadratic interpolation using the point at the peak and the two neighboring ...


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For higher resolution periodicity estimation, given properly band-limited data, the data can be upsampled (interpolated) before autocorrelation, and/or, the autocorrelation results can be interpolated (Sinc kernel or some order of polynomial). This, of course, assumes the signal has a high enough S/N, and that the periodicity itself is unmodulated in ...


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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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Bcause the FFT is computed usingn $N\log N$ arithmetic operations, whereas the DFT using $N^2$ operations, where $N$ is the number of points in your discrete-time signal. So the number of operations is much lower if $N$ is particularly high, which is often the case ... Watch this video if you want to know more about this https://www.youtube.com/watch?v=...


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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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