Autocorrelation

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Problem 9.21a

Show that the autocorrelation $ \phi _{gg}(i-j) $ is given by


$ {\begin{aligned}\phi _{gg}(i-j)=\mathop {\sum } \limits _{t}^{}g_{t-j}g_{t-i}.\end{aligned}} $ (9.21a)

Solution

The autocorrelation of $ g_{t} $ is given by equation (9.8e):

$ {\begin{aligned}\phi _{gg}(\tau )=\mathop {\sum } \limits _{k}^{}g_{k}\ g_{k+\tau }.\end{aligned}} $

If we equate $ g_{t-j} $ in equation (9.21a) to $ g_{k} $ in equation (9.8e), we have $ k=t-j $, so equation (9.21a) becomes

$ {\begin{aligned}\mathop {\sum } \limits _{k}^{}g_{k}\ g_{k+i-j}=\phi _{gg}(i-j).\end{aligned}} $

Figure 9.21a.  Autocorrelation diagram.

Problem 9.21b

Calculate $ \phi _{ff} $ for the right triangle shown in Figure 9.21a.

$ {\begin{aligned}f(t)=0,\qquad t\leq 0,\\=(1-t)\;,\quad 0\leq t\leq 1,\\=0,\qquad t\geq 1.\end{aligned}} $

Solution

Let $ t=OA $, $ \tau =DO=BC $, $ AB=AF=1-\tau -t $, $ AG=1-t $:

$ {\begin{aligned}\phi _{ff}(\tau )={\int }_{-\infty }^{+\infty }f(t)f(t+\tau ){\rm {d}}t={\int }_{0}^{B}AG\times AF\ {\rm {d}}t\\={\int }_{}^{1-\tau }(1-t)(1-\tau -t){\rm {d}}t={\int }_{0}^{1-\tau }[(1-t)^{2}-\tau (1-t)]{\rm {d}}t\\=\left[(-1/3+t-t^{2}+t^{3}/3){\frac {\tau }{2}}-\tau t+\tau t^{2}/2\right]|_{0}^{1-\tau }\\=1/3-\tau /2+\tau ^{3}/6.\\\end{aligned}} $

Figure 9.21a shows that a displacement of $ -\tau $ gives the same result.

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