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.