What does the first and second zero crossing time of the autocorrelation function signify anything for a signal?

How do we get the period of the signal from its autocorrelation? Im trying it for a periodic signal with some low frequency component in it.

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    \$\begingroup\$ What autocorrelation function? \$\endgroup\$ – user17592 Jun 6 '13 at 10:45
  • \$\begingroup\$ I think it's a reasonable assumption the question is about autocorrelation in signal processing. \$\endgroup\$ – Phil Frost Jun 6 '13 at 11:38

For a signal \$x(t)\$ of finite duration (say nonzero only for \$t \in [0,T]\$), the (unnormalized) autocorrelation function is $$R_x(\tau) = \int_0^T x(t)x(t-\tau)\,\mathrm dt, ~\tau \geq 0$$ and of course \$R_x(\tau) = R_x(-\tau)\$ for \$\tau < 0\$. Since \$x(t-\tau)\$ is nonzero only when \$\tau \in [\tau, \tau+T]\$, the lower limit on the integral can be increased to \$\tau\$. Note that \$R_x(\tau) = 0\$ for \$|\tau| \geq T\$. If \$t_1 < T\$ is the smallest positive real number such that \$R_x(t_1) = 0\$, then this means that the signals \$x(t)\$ and \$x(t-t_1)\$ are orthogonal over the interval \$[\tau,T]\$, (or over \$[0,T]\$ if you like).

If \$x(t)\$ consists of \$n \geq 1\$ periods of a single-frequency sinusoid, that is, \$x(t) = \cos(2\pi nt/T + \theta)\$, then \$R_x(\tau) = \frac{1}{2}(T-|\tau|)\cos(2\pi n\tau/T)\$ for \$0 \leq \tau \leq T\$ and so the zero-crossings are at times \$t_i = \frac{i}{T}, 1 \leq i \leq n\$. If\$x(t)\$ also contains signals other than the single-frequency sinusoid mentioned, there can be other zero-crossings too.


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