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) \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t+mT)\)The question was asked in an interview for job.Origin of the question is Sampling of Band Pass Signals in division Sampling and Reconstruction of Signals of Digital Signal Processing

What is the reconstruction formula for the bandpass signal x(t) with s

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) \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t+mT)\)The question was asked in an interview for job.Origin of the question is Sampling of Band Pass Signals in division Sampling and Reconstruction of Signals of Digital Signal Processing
Answer» CORRECT answer is (a) \(\sum_{m=-\INFTY}^{\infty}x(MT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t-mT)\) EXPLANATION: \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t-mT)\), where T=1/2B

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