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According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for uc(t) = ?(a) \(\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\)(b) \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+T_1/2)}{(\frac{π}{T_1})(t-mT_1+\frac{T_1}{2})}\)(c) \(\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t+mT_1)}{(\frac{π}{T_1})(t+mT_1)}\)(d) \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t+mT_1+\frac{T_1}{2})}{(\frac{π}{T_1})(t+mT_1+\frac{T_1}{2})}\)This question was posed to me in a job interview.This interesting question is from Sampling of Band Pass Signals in division Sampling and Reconstruction of Signals of Digital Signal Processing

Answer»

The correct option is (a) \(\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\)

Easiest EXPLANATION: To reconstruct the equivalent low PASS SIGNALS. Thus, according to the SAMPLING theorem for low pass signals with T1=1/B.

\(u_c (t)=\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\).


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