The system function of the digital filter is

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A. $$H\left( Z \right) = \sum\limits_{K = 0}^N {\frac{{{C_K}}}{{1 - {e^P}{K^T}{Z^{ - 1}}}}} $$

B. $$H\left( Z \right) = \sum\limits_{K = 1}^N {\frac{{{C_K}}}{{1 - {e^P}{K^T}{Z^{ - 1}}}}} $$

C. $$H\left( Z \right) = \sum\limits_{K = - N}^N {\frac{{{C_K}}}{{1 - {e^P}{K^T}{Z^{ - 1}}}}} $$

D. $$H\left( Z \right) = \sum\limits_{K = 0}^\infty {\frac{{{C_K}}}{{1 - {e^P}{K^T}{Z^{ - 1}}}}} $$

Answer: Option B

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