# Tag Info

## New answers tagged autocorrelation

0

Autocorrelation reveals periodicity patterns in the data. Taking an FFT will reveal the frequencies of the patterns. (don't understand why this is relevant enough ... will add more later)

1

And the power spectral density is the Fourier transform of the autocorrelation. I don't know about EEG data, but you can compute autocorrelation either in time domain or frequency domain. For instance if you need only a few samples of the autocorrelation it may be computed more efficiently using inner products than it would by $FFT^{-1}(|FFT(x)|^2)$. You ...

0

This reminds me of the days I was programming a correlation routine in C, working under MS DOS. All code and data had to fit in a memory space less than 640 kbytes. . . If you want a correlation output over a small time window, while you have very long records, you can split up these records and use the overlap-add method to do correlation in the fourier ...

0

Don't make this more complicated than it really is. $x(t)$ is non-zero in the interval $t\in[0,1]$, and $x(t+\tau)$ is non-zero in the interval $t\in[-\tau,1-\tau]$. The integrand is non-zero only if the two functions overlap. There is no overlap for $1-\tau<0$ and $-\tau>1$, i.e., for $|\tau|>1$. So for $|\tau|>1$ the autocorrelation is zero. ...

2

The OP's updated working is incorrect. Following up what Hilmar suggested gives \begin{align} Y(t) &= a\left(X(t)\right)^2\\ &= a\left(S(t) + N(t)\right)^2\\ &= a\left(S(t)\right)^2 + 2aS(t)N(t) + a\left(N(t)\right)^2\\ &{\large\Downarrow}\\ E[Y(t)]&= aE\left[\left(S(t)\right)^2 \right] + 2aE\left[S(t)N(t)\right] + aE\left[\left(S(t)\...

2

Put your second equations into your first equation, express $Y(t)$ as a function of $S(t)$ and $N(t)$ Apply the definition for mean and autocorrelation. Simplify and solve

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