Digital Signal Processing

Sampling of Band Pass Signals

Question 1
Marks : +2 | -2
Pass Ratio : 100%
According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for us(t) = ?
\\(\\sum_{m=-∞}^∞ u_c (mT_1) \\frac{sin⁡(\\frac{π}{T_1}) (t-mT_1)}{(\\frac{π}{T_1})(t-mT_1)}\\)
\\(\\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})}\\)
\\(\\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})}\\)
\\(\\sum_{m=-∞}^∞ u_c (mT_1) \\frac{sin⁡(\\frac{π}{T_1}) (t+mT_1)}{(\\frac{π}{T_1})(t+mT_1)}\\)
Explanation:
To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B .
Question 2
Marks : +2 | -2
Pass Ratio : 100%
What is the expression for low pass signal component us(t) that can be expressed in terms of samples of the bandpass signal?
\\(\\sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \\frac{sin⁡(Ï€/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(Ï€/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\\)
\\(\\sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \\frac{sin⁡(Ï€/(2T^{‘})) (t-2nT^{‘})}{(Ï€/(2T^{‘}))(t-2nT^{‘})}\\)
All of the mentioned
None of the mentioned
Explanation:
The low pass signal components us(t) can be expressed in terms of samples of the
Question 3
Marks : +2 | -2
Pass Ratio : 100%
What is the expression for low pass signal component uc(t) that can be expressed in terms of samples of the bandpass signal?
\\(\\sum_{n=-∞}^∞ (-1)^{n+r+1} x(2nT^{‘}-T^{‘}) \\frac{sin⁡(Ï€/(2T^{‘})) (t-2nT^{‘}+T^{‘})}{(Ï€/(2T^{‘}))(t-2nT^{‘}+T^{‘})}\\)
\\(\\sum_{n=-∞}^∞ (-1)^n x(2nT^{‘}) \\frac{sin⁡(Ï€/(2T^{‘})) (t-2nT^{‘})}{(Ï€/(2T^{‘}))(t-2nT^{‘})}\\)
All of the mentioned
None of the mentioned
Explanation:
The low pass signal components uc(t) can be expressed in terms of samples of the
Question 4
Marks : +2 | -2
Pass Ratio : 100%
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\\)?
\\((-1)^m u_c (mT_1)-u_s\\)
\\(u_s (mT_1-\\frac{T_1}{2})(-1)^{m+k+1}\\)
None
\\((-1)^m u_c (mT_1)- u_s (mT_1-\\frac{T_1}{2})(-1)^{m+k+1}\\)
Explanation:
Question 5
Marks : +2 | -2
Pass Ratio : 100%
According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for uc(t) = ?
\\(\\sum_{m=-∞}^∞ u_c (mT_1)\\frac{sin⁡(\\frac{π}{T_1}) (t-mT_1)}{(π/T_1)(t-mT_1)}\\)
\\(\\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})}\\)
\\(\\sum_{m=-∞}^∞ u_c (mT_1)\\frac{sin⁡(\\frac{π}{T_1}) (t+mT_1)}{(\\frac{π}{T_1})(t+mT_1)}\\)
\\(\\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})}\\)
Explanation:
To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B.
Question 6
Marks : +2 | -2
Pass Ratio : 100%
Which low pass signal component occurs at the rate of B samples per second with odd numbered samples of x(t)?
uc – lowpass signal component
us – lowpass signal component
uc & us – lowpass signal component
none of the mentioned
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.
Question 7
Marks : +2 | -2
Pass Ratio : 100%
The frequency shift can be achieved by multiplying the band pass signal as given in equation
True
False
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[2Ï€Fct] and sin[2Ï€Fct] 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.
Question 8
Marks : +2 | -2
Pass Ratio : 100%
What is the reconstruction formula for the bandpass signal x(t) with samples taken at the rate of 2B samples per second?
\\(\\sum_{m=-\\infty}^{\\infty}x(mT)\\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t-mT)\\)
\\(\\sum_{m=-\\infty}^{\\infty}x(mT)\\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t-mT)\\)
\\(\\sum_{m=-\\infty}^{\\infty}x(mT)\\frac{sin⁡(π/2T) (t-mT)}{(π/2T)(t-mT)} cos⁡2πF_c (t+mT)\\)
\\(\\sum_{m=-\\infty}^{\\infty}x(mT)\\frac{sin⁡(π/2T) (t+mT)}{(π/2T)(t+mT)} cos⁡2πF_c (t+mT)\\)
Explanation:
\\(\\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
Question 9
Marks : +2 | -2
Pass Ratio : 100%
What is the new centre frequency for the increased bandwidth signal?
Fc‘= Fc+B/2+B’/2
Fc‘= Fc+B/2-B’/2
Fc‘= Fc-B/2-B’/2
None of the mentioned
Explanation:
A new centre frequency for the increased bandwidth signal is Fc‘ = Fc+B/2-B’/2
Question 10
Marks : +2 | -2
Pass Ratio : 100%
Which low pass signal component occurs at the rate of B samples per second with even numbered samples of x(t)?
uc-lowpass signal component
us-lowpass signal component
uc & us-lowpass signal component
none of the mentioned
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.