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				| 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 u2061( /2T) (t-mT)}{( /2T)(t-mT)} cos u20612 F_c (t-mT) ) | 
| B. | ( sum_{m=- infty}^{ infty}x(mT) frac{sin u2061( /2T) (t+mT)}{( /2T)(t+mT)} cos u20612 F_c (t-mT) ) | 
| C. | ( sum_{m=- infty}^{ infty}x(mT) frac{sin u2061( /2T) (t-mT)}{( /2T)(t-mT)} cos u20612 F_c (t+mT) ) | 
| D. | ( sum_{m=- infty}^{ infty}x(mT) frac{sin u2061( /2T) (t+mT)}{( /2T)(t+mT)} cos u20612 F_c (t+mT) ) | 
| Answer» B. ( sum_{m=- infty}^{ infty}x(mT) frac{sin u2061( /2T) (t+mT)}{( /2T)(t+mT)} cos u20612 F_c (t-mT) ) | |