If $$L\left[ {f\left( t \right)} \right] = {{2\left( {s + 1} \right)} \over {{s^2} + 2s + 5}},$$ then f(0+) and f(∞) are given by
A. 0, 2 respectively
B. 2, 0 respectively
C. 0, 1 respectively
D. $${2 \over 5},$$ 0 respectively
Answer: Option B
A. 0, 2 respectively
B. 2, 0 respectively
C. 0, 1 respectively
D. $${2 \over 5},$$ 0 respectively
Answer: Option B
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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