This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Sampling of Band Pass Signals”.
1. The frequency shift can be achieved by multiplying the band pass signal as given in equation
x(t) = \(u_c (t) cos2π F_c t-u_s (t) sin2π F_c t\) 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
View Answer
Explanation: It is certainly advantageous to perform a frequency shift of the band pass signal by and sampling the equivalent low pass signal. Such a frequency shift can be achieved by multiplying the band pass 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.
2. 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)= \(u_c (nT)cos2πF_c nT-u_s (nT)sin 2πF_c nT\)?
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) \((-1)^m u_c (mT_1)- u_s (mT_1-\frac{T_1}{2})(-1)^{m+k+1}\)
View Answer
Explanation:
\(x(nT)=u_c (nT)cos 2πF_c nT-u_s (nT)sin2πF_c nT\) → 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, \(x(2mT) ≡ xmT_{(1)} = u_c (mT_1)cosπm(2k-1)=(-1)^m u_c (mT_1)\)
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.
3. 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
View Answer
Explanation: With 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.
4. 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
View Answer
Explanation: With 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.
5. What is the reconstruction formula for the bandpass signal x(t) with samples taken at the rate of 2B samples per second?
a) \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin(π/2T) (t-mT)}{(π/2T)(t-mT)} cos2πF_c (t-mT)\)
b) \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin(π/2T) (t+mT)}{(π/2T)(t+mT)} cos2πF_c (t-mT)\)
c) \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin(π/2T) (t-mT)}{(π/2T)(t-mT)} cos2πF_c (t+mT)\)
d) \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin(π/2T) (t+mT)}{(π/2T)(t+mT)} cos2πF_c (t+mT)\)
View Answer
Explanation: \(\sum_{m=-\infty}^{\infty}x(mT)\frac{sin(π/2T) (t-mT)}{(π/2T)(t-mT)} cos2πF_c (t-mT)\), where T=1/2B
6. 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
View Answer
Explanation: A new centre frequency for the increased bandwidth signal is Fc‘ = Fc+B/2-B’/2
7. 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) \(\sum_{m=-∞}^∞ u_s (mT_1-\frac{T_1}{2}) \frac{sin(\frac{π}{T_1}) (t+mT_1+\frac{T_1}{2})}{(\frac{π}{T_1})(t+mT_1+\frac{T_1}{2})}\)
View Answer
Explanation: To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B.
\(u_c (t)=\sum_{m=-∞}^∞ u_c (mT_1)\frac{sin(\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\).
8. 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) \(\sum_{m=-∞}^∞ u_c (mT_1) \frac{sin(\frac{π}{T_1}) (t+mT_1)}{(\frac{π}{T_1})(t+mT_1)}\)
View Answer
Explanation: To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B .
\(u_s (t)=\sum_{m=-∞}^∞ u_s (mT_1-T_1/2) \frac{sin(π/T_1) (t-mT_1+T_1/2)}{(π/T_1)(t-mT_1+T_1/2)}\)
9. 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
View Answer
Explanation: The low pass signal components uc(t) can be expressed in terms of samples of the
band pass signal as follows:
\(u_c (t) = \sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \frac{sin(π/(2T^{‘})) (t-2nT^{‘})}{(π/(2T^{‘}))(t-2nT^{‘})}\).
10. 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
View Answer
Explanation: The low pass signal components us(t) can be expressed in terms of samples of the
band pass signal as follows:
\(u_s (t) = \sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \frac{sin(π/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(π/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\)
11. 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) = \(\frac{1}{2} [X_l (F-F_c)+X_l^* (-F-F_c)]\)
View Answer
Explanation:
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) = \(\int_{-\infty}^∞ [x_l (t) e^{j2πF_c t}+x_l^* (t)e^{-j2πF_c t}] e^{-j2πFt} dt\)
=\(\frac{1}{2}[X_l (F-F_c)+X_l^* (-F-F_c)]\).
12. 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) = \(\frac{1}{2} [X_l (F-F_c)+X_l^* (-F-F_c)]\)
View Answer
Explanation: 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).
Sanfoundry Global Education & Learning Series – Digital Signal Processing.
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