Given h[n] = {1, 2, 3, 2, 1}. It is a linear phase FIR filter because
A. h[n] = h[N - 1 - n]
B. h[n] has finite number of samples
C. h[n] = -h[N - 1 - n]
D. h[n] = h[N - n]
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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