Sampling and Reconstruction of Signal MCQ [Free PDF] – Objective Question Answer for Sampling and Reconstruction of Signal Quiz

31. The frequency shift can be achieved by multiplying the bandpass signal as given in the equation x(t) = \(u_c (t) cos⁡2π F_c t-u_s (t) sin⁡2π F_c t\) by the quadrature carriers cos[2πFct] and sin[2πFct] and lowpass filtering the products to eliminate the signal components of 2Fc.

A. True
B. False

Answer: A

It is certainly advantageous to perform a frequency shift of the bandpass signal by sampling the equivalent low pass signal. Such a frequency shift can be achieved by multiplying the bandpass signal as given in the above equation by the quadrature carriers cos[2πFct] and sin[2πFct] and low pass filtering the products to eliminate the signal components at 2Fc.

Clearly, the multiplication and the subsequent filtering are first performed in the analog domain and then the outputs of the filters are sampled.

 

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

D. None

Answer: C

\(x(nT)=u_c (nT)cos⁡ 2πF_c nT-u_s (nT)sin⁡2πF_c nT\) → 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.

 

33. Which low pass signal component occurs at the rate of B samples per second with even-numbered samples of x(t)?

A. uc-lowpass signal component
B. us-lowpass signal component
C. uc & us-lowpass signal component
D. none of the mentioned

Answer: A

The even-numbered samples of x(t), which occur at the rate of B samples per second, produce samples of the low pass signal component uc.

 

34. Which low pass signal component occurs at the rate of B samples per second with odd-numbered samples of x(t)?

A. uc – lowpass signal component
B. us – lowpass signal component
C. uc & us – lowpass signal component
D. none of the mentioned

Answer: B

The odd-numbered samples of x(t), which occur at the rate of B samples per second, produce samples of the low pass signal component us.

 

35. 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. \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t+mT)\)

Answer: A

The reconstruction formula for the bandpass signal x(t) with samples taken at the rate of 2B samples per second is

\(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t-mT)\)

where T=1/2B

 

36. What is the new center frequency for the increased bandwidth signal?

A. Fc‘= Fc+B/2+B’/2
B. Fc‘= Fc+B/2-B’/2
C. Fc‘= Fc-B/2-B’/2
D. None of the mentioned

Answer: B

A new center frequency for the increased bandwidth signal is

Fc‘ = Fc+B/2-B’/2

 

37. According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for uc(t) = ?

A. \(\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\)

B. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+T_1/2)}{(\frac{π}{T_1})(t-mT_1+\frac{T_1}{2})}\)

C. \(\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t+mT_1)}{(\frac{π}{T_1})(t+mT_1)}\)

D. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t+mT_1+\frac{T_1}{2})}{(\frac{π}{T_1})(t+mT_1+\frac{T_1}{2})}\)

Answer: A

To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B.
\(u_c (t)=\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\).

 

38. According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for us(t) = ?

A. \(\sum_{m=-∞}^∞ u_c (mT_1) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1)}{(\frac{π}{T_1})(t-mT_1)}\)

B. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+\frac{T_1}{2})}{(π/T_1)(t-mT_1+\frac{T_1}{2})}\)

C. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1-\frac{T_1}{2})}{(\frac{π}{T_1})(t-mT_1-\frac{T_1}{2})}\)

D. \(\sum_{m=-∞}^∞ u_c (mT_1) \frac{sin⁡(\frac{π}{T_1}) (t+mT_1)}{(\frac{π}{T_1})(t+mT_1)}\)

Answer: B

To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B.

\(u_s (t)=\sum_{m=-∞}^∞ u_s (mT_1-T_1/2) \frac{sin⁡(π/T_1) (t-mT_1+T_1/2)}{(π/T_1)(t-mT_1+T_1/2)}\)

 

39. What is the expression for low pass signal component uc(t) that can be expressed in terms of samples of the bandpass signal?

A. \(\sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\)

B. \(\sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}\)

C. All of the mentioned

D. None of the mentioned

Answer: B

The low pass signal components uc(t) can be expressed in terms of samples of the band pass signal as follows:

\(u_c (t) = \sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}\).

 

40. What is the expression for low pass signal component us(t) that can be expressed in terms of samples of the bandpass signal?

A. \(\sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\)

B. \(\sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}\)

C. All of the mentioned

D. None of the mentioned

Answer: A

The low pass signal components us(t) can be expressed in terms of samples of the band pass signal as follows:

\(u_s (t) = \sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\)

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