Reed Eaton

2022-06-30

Consider a signal that is a sum of sinusoids, e.g.
$x\left(t\right)=Asin\left(at\right)+Bcos\left(bt\right)$
Is there an easy and general way to get an analytical solution for the autocorrelation of x(t)?
Is the best way to simply plug x(t) into the autocorrelation formula?

Aaron Everett

Expert

Step 1
Consider a general sum of sinusoids
$x\left(t\right)=\sum _{i}{C}_{i}{f}_{i}\left({a}_{i}t\right)$
where ${f}_{i}\left({a}_{i}t\right)=sin\left({a}_{i}t\right)$ or ${f}_{i}\left({a}_{i}t\right)=cos\left({a}_{i}t\right)$ . The coefficients ${C}_{i}$ are simply constants. For example this sum could take the form
$x\left(t\right)=2cos\left(2t\right)-3cos\left(3t\right)+4cos\left(4t\right)-5sin\left(5t\right)$ .
Now, substitute this sum into the autocorrelation formula:
${R}_{xx}\left(\tau \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\left(t+\tau \right)x\left(t\right)dt$
And realize that each term in the integrand can be one of the following four cases:
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{C}_{i}{C}_{j}sin\left({a}_{i}\left(t+\tau \right)\right)sin\left({a}_{j}t\right)dt=\frac{{C}_{i}{C}_{j}}{2}cos\left({a}_{i}\tau \right)$
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{C}_{i}{C}_{j}sin\left({a}_{i}\left(t+\tau \right)\right)cos\left({a}_{j}t\right)dt=\frac{{C}_{i}{C}_{j}}{2}sin\left({a}_{i}\tau \right)$
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{C}_{i}{C}_{j}cos\left({a}_{i}\left(t+\tau \right)\right)sin\left({a}_{j}t\right)dt=-\frac{{C}_{i}{C}_{j}}{2}sin\left({a}_{i}\tau \right)$
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{C}_{i}{C}_{j}cos\left({a}_{i}\left(t+\tau \right)\right)cos\left({a}_{j}t\right)dt=\frac{{C}_{i}{C}_{j}}{2}cos\left({a}_{i}\tau \right)$
So the general autocorrelation can be written as a double sum:
${R}_{xx}\left(\tau \right)=\frac{1}{2}\sum _{ij}{C}_{i}{C}_{j}{r}_{ij}\left(\tau \right)$
where

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