Consider the following system
h(n) is a filter with frequency response, then
A. \[H\left( {{e^{j\omega }}} \right) = \left\{ \begin{array}{l} 1,\,\,\,\,\left| \omega \right| \le \pi /D\\ 0,\,\,\,\,{\rm{otherwise}} \end{array} \right.\]
B. \[H\left( {{e^{j\omega }}} \right) = \left\{ \begin{array}{l} {T_s},\,\,\,\,\left| \omega \right| \le \pi /D\\ 0,\,\,\,\,{\rm{otherwise}} \end{array} \right.\]
C. \[H\left( {{e^{j\omega }}} \right) = \left\{ \begin{array}{l} \frac{1}{{{T_s}}},\,\,\,\,\left| \omega \right| \le \pi /D\\ 0,\,\,\,\,{\rm{otherwise}} \end{array} \right.\]
D. \[H\left( {{e^{j\omega }}} \right) = \left\{ \begin{array}{l} {e^{ - j3\omega }},\,\,\,\,\left| \omega \right| \le \pi /D\\ 0,\,\,\,\,{\rm{otherwise}} \end{array} \right.\]
Answer: Option A
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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