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The power spectrum of a stationary discrete-time random signal is

$$\Phi_{xx}(e^{j\omega})=\begin{cases} 1 & |\omega|<\pi/2 \\ 0 & \pi/2 <|\omega| \le\pi \end{cases} $$

(a) Determine the autocorrelation function $\phi_{xx}[m]$

I know that the power spectrum is the DTFT of the ACF at 0, but how can I find the ACF from the power spectrum, given that we need to find the ACF for all possible n values?

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  • $\begingroup$ The power spectrum is the DTFT of the ACF. The ACF at 0 is a single sample. You can't really take a DTFT of a single sample (well, not an interesting one). $\endgroup$ – Peter K. Jun 26 '16 at 1:14
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The power spectral density is the fourier transformed ACF. So what you have to to is to do the inverse fourier transform of the spectrum. And since the spectrum is a rectangle in frequency, it will be a $\frac{\sin(x)}{x}$ in the time domain.

Additionally:

I know that the power spectrum is the DTFT of the ACF at 0

The signals power is given by $\phi_{xx}(\tau = 0)$: $$ \phi_{example}(\tau) = \int_{-\infty}^{\infty} x_{1}(t)* x_{2}(t-\tau)\mathrm{d}x $$ If you autocorrelate (correlate a signal with itself), your subscripts vanish ($ x_1, x_2$ become $x$) and if you calculate the ACF at $\tau = 0$ you will get the squared power of the signal: $$ \phi_{example}(\tau = 0) = \int_{-\infty}^{\infty} x(t)* x(t)\mathrm{d}x = \int_{-\infty}^{\infty} x(t)^2\mathrm{d}x $$

If your signal does not have a finite amount of energy, you will have to use the following definition of the ACF: $$ \phi_{example}(\tau) = \lim_{T \to \infty} \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} x(t)* x(t-\tau)\mathrm{d}x $$

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