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
The Fourier transform of a real valued time signal has
A. Odd symmetry
B. Even symmetry
C. Conjugate symmetry
D. No symmetry
A. $$V$$
B. $${{{T_1} - {T_2}} \over T}V$$
C. $${V \over {\sqrt 2 }}$$
D. $${{{T_1}} \over {{T_2}}}V$$
A. $$T = \sqrt 2 {T_s}$$
B. T = 1.2Ts
C. Always
D. Never
A. $${{\alpha - \beta } \over {\alpha + \beta }}$$
B. $${{\alpha \beta } \over {\alpha + \beta }}$$
C. α
D. β
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