If \[x\left( n \right) = \left\{ \begin{array}{l} {\left( {\frac{1}{2}} \right)^n};\,0 \le n \le \left( {N - 1} \right)\\ 0;\,{\rm{otherwise}} \end{array} \right.\] then x(z) is equal to
A. $$\frac{{1 - {2^{ - N}}{z^{ - N}}}}{{1 - \left( {\frac{1}{2}} \right){z^{ - 1}}}}$$
B. $$\frac{{1 - {{\left( {\frac{1}{2}} \right)}^{ - N}}{z^{ - N}}}}{{1 - 2{z^{ - 1}}}}$$
C. $$\frac{{1 - {2^N}{z^{ - N}}}}{{1 - \left( {\frac{1}{2}} \right){z^{ - 1}}}}$$
D. $$\frac{{1 - {{\left( {\frac{1}{2}} \right)}^N}{z^{ - N}}}}{{1 - \left( {\frac{1}{2}} \right){z^{ - 1}}}}$$
Answer: Option D
This Question Belongs to Electronics And Communications Engineering >> Signal Processing
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