Explore topic-wise MCQs in Digital Signal Processing.

This section includes 13 Mcqs, each offering curated multiple-choice questions to sharpen your Digital Signal Processing knowledge and support exam preparation. Choose a topic below to get started.

1.

What 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)?

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) = \(\frac{1}{2} [X_l (F-F_c)+X_l^* (-F-F_
C. +X_l^* (F-F_c)]\) 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» E.
Discussion
2.

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) = \(\frac{1}{2} [X_l (F-F_c)+X_l^* (-F-F_
C. +X_l^* (F-F_c)]\) 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» E.
Discussion
3.

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» B. \(\sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin⁡(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}\)
Discussion
4.

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» C. All of the mentioned
Discussion
5.

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» C. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1-\frac{T_1}{2})}{(\frac{π}{T_1})(t-mT_1-\frac{T_1}{2})}\)
Discussion
6.

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» B. \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin⁡(\frac{π}{T_1}) (t-mT_1+T_1/2)}{(\frac{π}{T_1})(t-mT_1+\frac{T_1}{2})}\)
Discussion
7.

What is the new centre 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» C. Fc‘= Fc-B/2-B’/2
Discussion
8.

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» B. \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t-mT)\)
Discussion
9.

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» C. uc & us – lowpass signal component
Discussion
10.

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. \((-1)^m u_c (mT_1)- u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}\)
Answer» E.
Discussion
11.

The frequency shift can be achieved by multiplying the band pass signal as given in 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» B. False
Discussion
12.

What is the new centre frequency for the increased bandwidth signal ?

A. F<sub>c</sub>‘= F<sub>c</sub>+B/2+B’/2
B. F<sub>c</sub>‘= F<sub>c</sub>+B/2-B’/2
C. F<sub>c</sub>‘= F<sub>c</sub>-B/2-B’/2
D. None of the mentioned
Answer» C. F<sub>c</sub>‚Äö√Ñ√∂‚àö√ë‚àö‚â§= F<sub>c</sub>-B/2-B‚Äö√Ñ√∂‚àö√ë‚àö¬•/2
Discussion
13.

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

A. u<sub>c</sub>– lowpass signal component
B. u<sub>s</sub>– lowpass signal component
C. u<sub>c</sub> & u<sub>s</sub> – lowpass signal component
D. None of the mentioned
Answer» B. u<sub>s</sub>‚Äö√Ñ√∂‚àö√ë‚àö¬® lowpass signal component
Discussion