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The OP states that for a (deterministic) power signal $x(t)$, the autocorrelation function is defined as $$R_x(\tau) = \lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt\tag{1}$$ and then wonders whether in the case when $x(t)$ happens to be a periodic signal with period $T_0$, then it is true that $$R_x(\tau) = \frac{1}{T_0}\int_0^{T_0} ...


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Assuming that the signal $x(t)$ is periodic then it can be described by its complex Fourier series as $$x(t) = \sum_{n=-\infty}^{\infty }c_ne^{j2{\pi}nf_0t}$$ Where $c_n$ are the complex Fourier coefficients. There is a Fourier transform pair that states $$\mathcal{F}(e^{j2{\pi}f_0t}) = \delta(f - f_0)$$ Applying this to the Fourier series of $x(t)$ element ...


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