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

41. What is the Fourier transform of x(t)?

A. X (F) = \(\frac{1}{2} [X_l (F-F_C.+X_l^* (F-F_C.]\)

B. X (F) = \(\frac{1}{2} [X_l (F-F_C.+X_l^* (F+F_C.]\)

C. X (F) = \(\frac{1}{2} [X_l (F+F_C.+X_l^* (F-F_C.]\)

D. X (F) = \(\frac{1}{2} [X_l (F-F_C.+X_l^* (-F-F_C.]\)

Answer: D

X (F) = \(\int_{-\infty}^∞ x(t)e^{-j2πFt} dt\)

=\(\int_{-\infty}^∞ \{Re[x_l (t) e^{j2πF_c t}]\}e^{-j2πFt} dt\)

Using the identity, Re(ε)=1/2(ε+ε^*)

X (F) = \(\int_{-\infty}^∞ [x_l (t) e^{j2πF_c t}+x_l^* (t)e^{-j2πF_c t}] e^{-j2πFt} dt\)

=\(\frac{1}{2}[X_l (F-F_C.+X_l^* (-F-F_C.]\).

42. What is the basic relationship between the spectrum of the real bandpass signal x(t) and the spectrum of the equivalent low pass signal xl(t)?

A. X (F) = \(\frac{1}{2} [X_l (F-F_C.+X_l^* (F-F_C.]\)

B. X (F) = \(\frac{1}{2} [X_l (F-F_C.+X_l^* (F+F_C.]\)

C. X (F) = \(\frac{1}{2} [X_l (F+F_C.+X_l^* (F-F_C.]\)

D. X (F) = \(\frac{1}{2} [X_l (F-F_C.+X_l^* (-F-F_C.]\)

Answer: D

X(F) = \(\frac{1}{2} [X_l (F-F_C.+X_l^* (-F-F_C.]\), where Xl(F) is the Fourier transform of xl(t). This is the basic relationship between the spectrum o f the real band pass signal x(t) and the spectrum of the equivalent low pass signal xl(t).

43. Which of the following is the right way of representing of the equation that contains only the positive frequencies in a given x(t) signal?

A. X+(F)=4V(F)X(F)
B. X+(F)=V(F)X(F)
C. X+(F)=2V(F)X(F)
D. X+(F)=8V(F)X(F)

Answer: C

In a real-valued signal x(t), has a frequency content concentrated in a narrow band of frequencies in the vicinity of a frequency Fc. Such a signal which has only positive frequencies can be expressed as X+(F)=2V(F)X(F)

Where X+(F) is a Fourier transform of x(t) and V(F) is the unit step function.

44. What is the equivalent time-domain expression of X+(F)=2V(F)X(F)?

A. F(+1)[2V(F)]*F(+1)[X(F)]
B. F(-1)[4V(F)]*F(-1)[X(F)]
C. F(-1)[V(F)]*F(-1)[X(F)]
D. F(-1)[2V(F)]*F(-1)[X(F)]

Answer: D

Given Expression, X+(F)=2V(F)X(F).It can be calculated as follows

\(x_+ (t)=\int_{-∞}^∞ X_+ (F)e^{j2πFt} dF\)

=\(F^{-1} [2V(F)]*F^{-1} [X(F)]\)

45. In time-domain expression, \(x_+ (t)=F^{-1} [2V(F)]*F^{-1} [X(F)]\). The signal x+(t) is known as

A. Systematic signal
B. Analytic signal
C. Pre-envelope of x(t)
D. Both Analytic signal & Pre-envelope of x(t)

Answer: D

In time-domain expression, \(x_+ (t)=F^{-1} [2V(F)]*F^{-1} [X(F)]\). The signal x+(t) is known as

\(x_+ (t)=F^{-1} [2V(F)] * F^{-1}[X(F)]\).

46. In equation \(x_+ (t)=F^{-1} [2V(F)]*F^{-1} [X(F)]\), if \(F^{-1} [2V(F)]=δ(t)+j/πt\) and \(F^{-1} [X(F)]\) = x(t). Then the value of ẋ(t) is?

A. \(\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t+τ} dτ\)

B. \(\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t-τ} dτ\)

C. \(\frac{1}{π} \int_{-\infty}^\infty \frac{2x(t)}{t-τ} dτ\)

D. \(\frac{1}{π} \int_{-\infty}^\infty \frac{4x(t)}{t-τ} dτ\)

Answer: B

\(x_+ (t)=[δ(t)+j/πt]*x(t)\)

\(x_+ (t)=x(t)+[j/πt]*x(t)\)

\(ẋ(t)=[j/πt]*x(t)\)

=\(\frac{1}{π} \int_{-\infty}^\infty \frac{x(t)}{t-τ} dτ\) Hence proved.

47. If the signal ẋ(t) can be viewed as the output of the filter with impulse response h(t) = 1/πt, -∞ < t < ∞ when excited by the input signal x(t) then such a filter is called as __________

A. Analytic transformer
B. Hilbert transformer
C. Both Analytic & Hilbert transformer
D. None of the mentioned

Answer: B

The signal ẋ(t) can be viewed as the output of the filter with impulse response h(t) = 1/πt, -∞ < t < ∞ when excited by the input signal x(t) then such a filter is called a Hilbert transformer.

48. What is the frequency response of a Hilbert transform H(F)=?

A. \(\begin{cases}&-j (F>0) \\ & 0 (F=0)\\ & j (F<0)\end{cases}\)

B. \(\left\{\begin{matrix}-j & (F<0)\\0 & (F=0) \\j & (F>0)\end{matrix}\right. \)

C. \(\left\{\begin{matrix}-j & (F>0)\\0 &(F=0) \\j & (F<0)\end{matrix}\right. \)

D. \(\left\{\begin{matrix}j&(F>0)\\0 & (F=0)\\j & (F<0)\end{matrix}\right. \)

Answer: A

H(F) =\(\int_{-∞}^∞ h(t)e^{-j2πFt} dt\)

=\(\frac{1}{π} \int_{-∞}^∞ 1/t e^{-2πFt} dt\)

=\(\left\{\begin{matrix}-j& (F>0)\\0&(F=0) \\ j& (F<0)\end{matrix}\right.\)

We Observe that │H (F)│=1 and the phase response ⊙(F) = -1/2π for F > 0 and ⊙(F) = 1/2π for F < 0.

49. What is the equivalent lowpass representation obtained by performing a frequency translation of X+(F) to Xl(F)=?

A. X+(F+FC.
B. X+(F-FC.
C. X+(F*FC.
D. X+(Fc-F)

Answer: A

The analytic signal x+(t) is a bandpass signal. We obtain an equivalent lowpass representation by performing a frequency translation of X+(F).

50. What is the equivalent time domain relation of xl(t) i.e., lowpass signal?

A. \(x_l (t)=[x(t)+j ẋ(t)]e^{-j2πF_c t}\)

B. x(t)+j ẋ(t) = \(x_l (t) e^{j2πF_c t}\)

C. \(x_l (t)=[x(t)+j ẋ(t)]e^{-j2πF_c t}\) & x(t)+j ẋ(t) = \(x_l (t) e^{j2πF_c t}\)

D. None of the mentioned

Answer: C

The equivalent time domain relation of xl(t) i.e., lowpass signal is

\(x_l (t)=x_+ (t) e^{-j2πF_c t}\)

=\([x(t)+j ẋ(t)] e^{-j2πF_c t}\)

Or equivalently, x(t)+j ẋ(t) =\(x_l (t) e^{j2πF_c t}\).

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