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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)= \(u_c (nT)cos2πF_c nT-u_s (nT)sin 2πF_c nT\)? |
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. | \((-1)^m u_c (mT_1)- u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}\) |
Answer» E. | |