According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for us(t) = ?

Question

TuteeHUB · Accepted Answer

$\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1-\frac{T_1}{2})}{(\frac{π}{T_1})(t-mT_1-\frac{T_1}{2})}$

TuteeHUB · Answer

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

TuteeHUB · Answer

$\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+\frac{T_1}{2})}{(π/T_1)(t-mT_1+\frac{T_1}{2})}$

TuteeHUB · Answer

$\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1-\frac{T_1}{2})}{(\frac{π}{T_1})(t-mT_1-\frac{T_1}{2})}$

TuteeHUB · Answer

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

1.	According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for us(t) = ?
A.	\(\sum_{m=-∞}^∞ u_c (mT_1) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(\frac{π}{T_1})(t-mT_1)}\)
B.	\(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+\frac{T_1}{2})}{(π/T_1)(t-mT_1+\frac{T_1}{2})}\)
C.	\(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1-\frac{T_1}{2})}{(\frac{π}{T_1})(t-mT_1-\frac{T_1}{2})}\)
D.	\(\sum_{m=-∞}^∞ u_c (mT_1) \frac{sin⁡(\frac{π}{T_1}) (t+mT_1)}{(\frac{π}{T_1})(t+mT_1)}\)
Answer» C. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1-\frac{T_1}{2})}{(\frac{π}{T_1})(t-mT_1-\frac{T_1}{2})}\)

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