Question

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

Answer 1

Right option is (d) $(-1)^m u_c (mT_1)- u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}$

Easy explanation: $x(nT)=u_c (nT)cos⁡ 2πF_c nT-u_s (nT)sin⁡2πF_c nT$ → equ1

=$u_c (nT)cos⁡\frac{πn(2k-1)}{2}-u_s(nT)sin⁡\frac{πn(2k-1)}{2}$ → equ2

On substituting the above values in equ1, we get say n=2m, $x(2mT) ≡ xmT_{(1)} = u_c (mT_1)cos⁡πm(2k-1)=(-1)^m u_c (mT_1)$

where $T_1=2T=\frac{1}{B}$. For n odd, say n=2m-1 in equ2 then we get the result as follows

$u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}$

Hence proved.