What is the final result obtained by substituting Fc=kB-B/2, T= 1/2B and say n = 2m i.e., for even and n=2m-1 for odd in equation x(nT)= \(u_c (nT)cos⁡2πF_c nT-u_s (nT)sin⁡ 2πF_c nT\)?(a) \((-1)^m u_c (mT_1)-u_s\)(b) \(u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}\)(c) None(d) \((-1)^m u_c (mT_1)- u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}\)The question was posed to me by my school principal while I was bunking the class.Question is from Sampling of Band Pass Signals in division Sampling and Reconstruction of Signals of Digital Signal Processing

What is the final result obtained by substituting Fc=kB-B/2, T= 1/2B a

1.	What is the final result obtained by substituting Fc=kB-B/2, T= 1/2B and say n = 2m i.e., for even and n=2m-1 for odd in equation x(nT)= \(u_c (nT)cos⁡2πF_c nT-u_s (nT)sin⁡ 2πF_c nT\)?(a) \((-1)^m u_c (mT_1)-u_s\)(b) \(u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}\)(c) None(d) \((-1)^m u_c (mT_1)- u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}\)The question was posed to me by my school principal while I was bunking the class.Question is from Sampling of Band Pass Signals in division Sampling and Reconstruction of Signals of Digital Signal Processing
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