What is the reconstruction formula for the bandpass signal x(t) with samples taken at the rate of 2B samples per second?

Question

TuteeHUB · Accepted Answer

$\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t-mT)$

TuteeHUB · Answer

$\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t-mT)$

TuteeHUB · Answer

$\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t-mT)$

TuteeHUB · Answer

$\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t+mT)$

TuteeHUB · Answer

$\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t+mT)$

1.	What is the reconstruction formula for the bandpass signal x(t) with samples taken at the rate of 2B samples per second?
A.	\(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t-mT)\)
B.	\(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t-mT)\)
C.	\(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t+mT)\)
D.	\(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t+mT)\)
Answer» B. \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t-mT)\)

Discussion