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If $x(t)$ is a real (aperiodic) power signal, i.e. \begin{equation} 0<\lim_{T\rightarrow\infty} \frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2 dt<\infty \end{equation} $x_T (t)$ is a truncated version of $x(t)$, $R_{xx}^T(\tau)$ is the autocorrelation function of $x_T (t)$, then \begin{equation} \int_{-\infty}^{\infty} R_{xx}^T (\tau) e^{-i2\pi f \tau} d\tau=\frac{1}{T} \left|X_T (f)\right|^2 \end{equation} and \begin{equation} \int_{-\infty}^{\infty} R_{xx} (\tau) e^{-i2\pi f \tau} d\tau=\lim_{T\rightarrow\infty}\frac{1}{T} \left|X_T (f)\right|^2 \ \ \ \ \ \ \ \ \ \ \ (1) \end{equation} where \begin{equation} R_{xx}=\lim_{T\rightarrow\infty}R_{xx} ^T \end{equation} As the equation (1) becomes if $x$ is a sequence (discrete signal)? Thank you very much.

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What is equation (1)? And what is your precise question? – Giuseppe Negro Jan 29 '13 at 18:45
I edited the question. Now the symbol (1) is more readable. I would like to arrive at an equation similar to (1) in the case x is a discrete signal. – Mark Jan 29 '13 at 18:50
up vote 0 down vote accepted

I would replace the integral with a sum. You get $$ \sum R(n) e^{-i 2\pi f n} = \lim_{N\to \infty} \frac{|X_N(f)|^2}{N},$$ where $X_N(f)$ is the discrete Fourier transform of a truncated version of the signal.

Note that all the $X_N(f)$ are periodic functions with the period $1$ so their limit, if it exists, will be as well.

You can also compute $\frac{|X_N(f)|^2}{N}$ using the auto-correlation of the truncated sequence as in the continuous case.

Two notes: if $x$ is generated by a stationary stochastic process it is most likely that the limit will not exist. And you might get better answer to questions like this by asking at

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