Let y[n] denote the convolution of h[n] and g[n], where h[n] = $${\left( {\frac{1}{2}} \right)^n}$$ u[n] and g[n] is a causal sequence. If y[0] = 1 and y[1] = $$\frac{1}{2},$$ then g[1] equals
A. 0
B. $${1 \over 2}$$
C. 1
D. $${3 \over 2}$$
Answer: Option A
Related Questions on 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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