Digital Signal Processing

Frequency Transformations

Question 1
Marks : +2 | -2
Pass Ratio : 50%
If A=\\(\\frac{Ω_1 (Ω_u-Ω_l)}{-Ω_1^2+Ω_u Ω_l}\\) and B=\\(\\frac{Ω_2 (Ω_u-Ω_l)}{Ω_2^2-Ω_u Ω_l}\\), then which of the following is the backward design equation for a low pass-to-band stop transformation?
ΩS=Max{|A|,|B|}
ΩS=Min{|A|,|B|}
ΩS=|B|
ΩS=|A|
Explanation:
If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band stop filter and Ω1 and Ω2 are the lower and upper cutoff stop band frequencies of the desired band stop filter, then the backward design equation is
Question 2
Marks : +2 | -2
Pass Ratio : 25%
Which of the following is a low pass-to-high pass transformation?
s → s / Ωu
s → Ωu / s
s → Ωu.s
none of the mentioned
Explanation:
The low pass-to-high pass transformation is simply achieved by replacing s by 1/s. If the desired high pass filter has the pass band edge frequency Ωu, then the transformation is
Question 3
Marks : +2 | -2
Pass Ratio : 25%
Which of the following is the backward design equation for a low pass-to-high pass transformation?
\\(\\Omega_S=\\frac{\\Omega_S}{\\Omega_u}\\)
\\(\\Omega_S=\\frac{\\Omega_u}{\\Omega’_S}\\)
\\(\\Omega’_S=\\frac{\\Omega_S}{\\Omega_u}\\)
\\(\\Omega_S=\\frac{\\Omega’_S}{\\Omega_u}\\)
Explanation:
If Ωu is the desired pass band edge frequency of new high pass filter, then the transfer function of this new high pass filter is obtained by using the transformation s → Ωu /s. If ΩS and Ω’S are the stop band frequencies of prototype and transformed filters respectively, then the backward design equation is given by
Question 4
Marks : +2 | -2
Pass Ratio : 50%
Which of the following is a low pass-to-high pass transformation?
s → s / Ωu
s → Ωu/s
s → Ωu.s
none of the mentioned
Explanation:
The low pass-to-high pass transformation is simply achieved by replacing s by 1/s. If the desired high pass filter has the pass band edge frequency Ωu, then the transformation is
Question 5
Marks : +2 | -2
Pass Ratio : 25%
If H(s) is the transfer function of a analog low pass normalized filter and Ωu is the desired pass band edge frequency of new low pass filter, then which of the following transformation has to be performed?
s → s/Ωu
s → s.Ωu
s → Ωu/s
none of the mentioned
Explanation:
If Ωu is the desired pass band edge frequency of new low pass filter, then the transfer function of this new low pass filter is obtained by using the transformation s → s/Ωu.
Question 6
Marks : +2 | -2
Pass Ratio : 25%
Which of the following is a low pass-to-band stop transformation?
s→\\(\\frac{s(Ω_u-Ω_l)}{s^2+Ω_u Ω_l}\\)
s→\\(\\frac{s(Ω_u+Ω_l)}{s^2+Ω_u Ω_l}\\)
s→\\(\\frac{s(Ω_u-Ω_l)}{s^2-Ω_u Ω_l}\\)
none of the mentioned
Explanation:
If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band stop filter, then the transformation to be performed on the normalized low pass filter is
Question 7
Marks : +2 | -2
Pass Ratio : 25%
Which of the following is a low pass-to-band pass transformation?
s→\\(\\frac{s^2+Ω_u Ω_l}{s(Ω_u+Ω_l)}\\)
s→\\(\\frac{s^2-Ω_u Ω_l}{s(Ω_u-Ω_l)}\\)
s→\\(\\frac{s^2+Ω_u Ω_l}{s(Ω_u-Ω_l)}\\)
s→\\(\\frac{s^2-Ω_u Ω_l}{s(Ω_u+Ω_l)}\\)
Explanation:
If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band pass filter, then the transformation to be performed on the normalized low pass filter is
Question 8
Marks : +2 | -2
Pass Ratio : 25%
What is the pass band edge frequency of an analog low pass normalized filter?
0 rad/sec
0.5 rad/sec
1 rad/sec
1.5 rad/sec
Explanation:
Let H(s) denote the transfer function of a low pass analog filter with a pass band edge frequency ΩP equal to 1 rad/sec. This filter is known as analog low pass normalized prototype.
Question 9
Marks : +2 | -2
Pass Ratio : 25%
Which of the following is the backward design equation for a low pass-to-low pass transformation?
\\(\\Omega_S=\\frac{\\Omega_S}{\\Omega_u}\\)
\\(\\Omega_S=\\frac{\\Omega_u}{\\Omega’_S}\\)
\\(\\Omega’_S=\\frac{\\Omega_S}{\\Omega_u}\\)
\\(\\Omega_S=\\frac{\\Omega’_S}{\\Omega_u}\\)
Explanation:
If Ωu is the desired pass band edge frequency of new low pass filter, then the transfer function of this new low pass filter is obtained by using the transformation s → s/Ωu. If ΩS and Ω’S are the stop band frequencies of prototype and transformed filters respectively, then the backward design equation is given by
Question 10
Marks : +2 | -2
Pass Ratio : 25%
If A=\\(\\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}\\) and B=\\(\\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}\\), then which of the following is the backward design equation for a low pass-to-band pass transformation?
ΩS=|B|
ΩS=|A|
ΩS=Max{|A|,|B|}
ΩS=Min{|A|,|B|}
Explanation:
If Ωu and Ωl are the upper and lower cutoff pass band frequencies of the desired band pass filter and Ω1 and Ω2 are the lower and upper cutoff stop band frequencies of the desired band pass filter, then the backward design equation is