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I have used the energy-type signal autocorrelation function:

$$\mathcal{R}_{xx}(\tau)=\int_{-\infty}^{\infty}x(t)x^*(t+\tau)dt$$

I have rewritten the equation as: $$\begin{align} \int_{-\infty}^{\infty}\big[u(t)-u(t-1)\big]\big[u(t+\tau )-u(t+\tau-1)\big]dt \\ \end{align}$$

How do I simplify this equation?

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    $\begingroup$ This is the deterministic auto-correlation computation of a time-domain signal x(t). Can you interpret the first integral as a convolution of $x(t)$ with $h(t)$ given by $$ y(\tau) = \int_{-\infty}^{\infty} x(t)h(\tau-t)dt $$? What's $h(t)$ in this case? $\endgroup$ – Fat32 Feb 13 at 16:40
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    $\begingroup$ Whenever the difference of two $u(t)$ is involved it's a good idea to draw it. Than it becomes obvious what the actual integration boundaries need to be. $\endgroup$ – Hilmar Feb 13 at 18:37
  • $\begingroup$ @Hilmar I have tried drawing the graph, I am still unsure of the integration limits. am I right to say that the limits will change as tau goes from negative to positive? $\endgroup$ – Dugong98 Feb 14 at 6:56
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Don't make this more complicated than it really is. $x(t)$ is non-zero in the interval $t\in[0,1]$, and $x(t+\tau)$ is non-zero in the interval $t\in[-\tau,1-\tau]$. The integrand is non-zero only if the two functions overlap. There is no overlap for $1-\tau<0$ and $-\tau>1$, i.e., for $|\tau|>1$. So for $|\tau|>1$ the autocorrelation is zero.

For $-1<\tau<1$, we have, according to above considerations, the following integral:

$$\mathcal{R}_{xx}(\tau)=\int_{\max\{0,-\tau\}}^{\min\{1,1-\tau\}}dt=\begin{cases}\displaystyle\int_{0}^{1-\tau}dt,&0<\tau<1\\\displaystyle\int_{-\tau}^{1}dt,&-1<\tau<0\end{cases}$$

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