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properties regarding time averages. For any symbol, the empirical frequency of that symbol approaches 1/|A|: for each each b E A \lim _{N \rightarrow \infty} \#\left\{n \in Z: a_{n}=b_{i}-N \leq n \leq N\right\} } \\{2 N+1}=\frac{1}{|\mathcal{A}|^{3}} where #S denotes the number of elements in the set S. For any pair of symbols and any spacing k + 0,the empirical frequency of appearances of the pair at the spacing approaches 1/|A|²: for b,c € A and k + 0 \lim _{N \rightarrow \infty} \frac{\#\left\{n \in Z: a_{n} k=b, a_{n}=c_{1}-N \leq n \leq N\right\}}{2 N+1}=\frac{1}{|\mathcal{A}|^{2}} . Let f : A → C map this alphabet to a signal constellation in C and let xn = f(an). Show that the sequence x has the following time average properties: \lim _{N \rightarrow \infty} \frac{1}{2 N+1} \sum_{n=-N}^{N}\left|x_{n}\right|^{2}=\frac{1}{|\mathcal{A}|} \sum_{b \in \mathcal{A}}|f(b)|^{2} \lim _{N \rightarrow \infty} \frac{1}{2 N+1} \sum_{n=N}^{N} x_{n-k x_{n}^{*}}=\left|\frac{1}{|\mathcal{A}|} \sum_{b \in \mathcal{A}} f(b)\right| \lim _{N \rightarrow \infty} \frac{1}{2 N+1} \sum_{n=-N}^{N} x_{n-k} x_{n}^{*}=\left|\frac{1}{|\mathcal{A}|} \sum_{b \in \mathcal{A}} f(b)\right|^{2}

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