# Find the autocorrelation function of signal $x(t) = u(t) - u(t-1)$

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?

• 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? – Fat32 Feb 13 at 16:40
• 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. – Hilmar Feb 13 at 18:37
• @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? – Dugong98 Feb 14 at 6:56

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