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I'm having trouble on computing the autocorrelation of the sinc function
I want to compute $$R_{hh}(\tau)=\int_{-\infty}^{\infty}\operatorname{sinc}(t) \ \operatorname{sinc}(t-\tau) \ \mathrm{d}t$$
where $$\operatorname{sinc}(t) \triangleq \begin{cases} \dfrac{\sin(\pi t)}{\pi t} \qquad & t \ne 0\\ \\ \quad \ \ 1 & t = 0 \end{cases}$$

Is the result another sinc function?

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  • $\begingroup$ Hint: Since sinc is symmetrical, convolution and correlation will yield the same result. Second hint : What is the fourier transform of a sinc? Third hint : Convolution (or cross-correlation in this case) is equivalent to multiplication in the frequency domain. $\endgroup$ – Ben Apr 1 at 2:26
  • $\begingroup$ The fourier transform of a sinc is a rectangular pulse in frequency domain. So multiplying two rect functions i get another rectangular pulse, and then its antitransform is sinc? Is that right? $\endgroup$ – Yu Hao Apr 1 at 2:32
  • $\begingroup$ yes. and directly integrating the above leads only to headache and sorrow. when converting autocorrelation to power spectrum or energy spectrum, be careful to dot your t's and cross your i's. $\endgroup$ – robert bristow-johnson Apr 1 at 3:34
  • $\begingroup$ Yes, that's right. The Fourier transform of $x(t) = \operatorname{sinc}(t)$ is $X(f)=\operatorname{rect}(f)$ and thus, $|X(f)|^2 = \operatorname{rect}(f)$. $\endgroup$ – Dilip Sarwate Apr 1 at 3:37
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Because of the symmetry of the $\operatorname{sinc}(t)$ we have
$$R_{hh}(\tau)=\int_{-\infty}^{\infty}\operatorname{sinc}(t) \ \operatorname{sinc}(t-\tau) \ \mathrm{d}t=\int_{-\infty}^{\infty}\operatorname{sinc}(t) \ \operatorname{sinc}(\tau-t) \ \mathrm{d}t = \operatorname{sinc}(\tau) \ast \operatorname{sinc}(\tau) $$

Analyzing in the frequency domain we have $$\mathcal{F}(R_{hh}(\tau)) = \mathcal{F}(\operatorname{sinc}(\tau))^2=\operatorname{rect}(f)^2=\operatorname{rect}(f)$$

$$\Rightarrow R_{hh}(\tau)=\mathcal{F}^{-1}(\mathcal{F}(R_{hh}(\tau))) = \mathcal{F}^{-1}(\operatorname{rect}(f))=\operatorname{sinc}(\tau)$$

where $\operatorname{rect}(f)$ denotes the rectangular function in the frequency domain

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  • $\begingroup$ Good job! Now try finding the autocorrelation function of $\displaystyle \operatorname{sinc}\left(\frac tT\right)$ $\endgroup$ – Dilip Sarwate Apr 2 at 15:32

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