Digital Signal Processing

Design of Low Pass Butterworth Filters

Question 1
Marks : +2 | -2
Pass Ratio : 100%
If H(s)=\\(\\frac{1}{s^2+s+1}\\) represent the transfer function of a low pass filter (not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a stop band filter with a stop band of 2 rad/sec and a center frequency of 10 rad/sec?
\\(\\frac{(s^2+100)^2}{s^4+2s^3+204s^2+200s+10^4}\\)
\\(\\frac{(s^2+10)^2}{s^4+2s^3+204s^2+200s+10^4}\\)
\\(\\frac{(s^2+10)^2}{s^4+2s^3+400s^2+200s+10^4}\\)
None of the mentioned
Explanation:
The low pass-to- band stop transformation is
Question 2
Marks : +2 | -2
Pass Ratio : 100%
Which of the following equation is True?
\\([\\frac{\\Omega_P}{\\Omega_C}]^{2N} = 10^{-K_P/10}+1\\)
\\([\\frac{\\Omega_P}{\\Omega_C}]^{2N} = 10^{K_P/10}+1\\)
\\([\\frac{\\Omega_P}{\\Omega_C}]^{2N} = 10^{-K_P/10}-1\\)
None of the mentioned
Explanation:
We know that,
Question 3
Marks : +2 | -2
Pass Ratio : 100%
What is the order of the normalized low pass Butterworth filter used to design a analog band pass filter with -3.0103dB upper and lower cutoff frequency of 50Hz and 20KHz and a stop band attenuation 20dB at 20Hz and 45KHz?
2
3
4
5
Explanation:
Given information is
Question 4
Marks : +2 | -2
Pass Ratio : 100%
Which of the following is a frequency domain specification?
0 ≥ 20 log|H(jΩ)|
20 log|H(jΩ)| ≥ KP
20 log|H(jΩ)| ≤ KS
All of the mentioned
Explanation:
We are required to design a low pass Butterworth filter to meet the following frequency domain specifications.
Question 5
Marks : +2 | -2
Pass Ratio : 100%
The cutoff frequency of the low pass Butterworth filter is the arithmetic mean of the two cutoff frequencies as found above.
True
False
Explanation:
The arithmetic mean of the two cutoff frequencies as found above is the final cutoff frequency of the low pass Butterworth filter.
Question 6
Marks : +2 | -2
Pass Ratio : 100%
Which of the following equation is True?
\\([\\frac{\\Omega_S}{\\Omega_C} ]^{2N} = 10^{-K_S/10}+1\\)
\\([\\frac{\\Omega_S}{\\Omega_C} ]^{2N} = 10^{K_S/10}+1\\)
\\([\\frac{\\Omega_S}{\\Omega_C} ]^{2N} = 10^{-K_S/10}-1\\)
None of the mentioned
Explanation:
We know that,
Question 7
Marks : +2 | -2
Pass Ratio : 100%
If H(s)=\\(\\frac{1}{s^2+s+1}\\) represent the transfer function of a low pass filter (not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a low pass filter with a pass band 10 rad/sec?
\\(\\frac{100}{s^2+10s+100}\\)
\\(\\frac{s^2}{s^2+s+1}\\)
\\(\\frac{s^2}{s^2+10s+100}\\)
None of the mentioned
Explanation:
The low pass-to-low pass transformation is
Question 8
Marks : +2 | -2
Pass Ratio : 100%
Which of the following condition is true?
N ≤ \\(\\frac{log⁡(\\frac{1}{k})}{log⁡(\\frac{1}{d})}\\)
N ≤ \\(\\frac{log⁡(k)}{log⁡(d)}\\)
N ≤ \\(\\frac{log⁡(d)}{log⁡(k)}\\)
N ≤ \\(\\frac{log⁡(\\frac{1}{d})}{log⁡(\\frac{1}{k})}\\)
Explanation:
If ‘d’ is the discrimination factor and ‘K’ is the selectivity factor, then
Question 9
Marks : +2 | -2
Pass Ratio : 100%
What is the expression for cutoff frequency in terms of stop band gain?
\\(\\frac{\\Omega_S}{(10^{-K_S/10}-1)^{1/2N}}\\)
\\(\\frac{\\Omega_S}{(10^{-K_S/10}+1)^{1/2N}}\\)
\\(\\frac{\\Omega_S}{(10^{K_S/10}-1)^{1/2N}}\\)
None of the mentioned
Explanation:
We know that,
Question 10
Marks : +2 | -2
Pass Ratio : 100%
What is the stop band frequency of the normalized low pass Butterworth filter used to design a analog band pass filter with -3.0103dB upper and lower cutoff frequency of 50Hz and 20KHz and a stop band attenuation 20dB at 20Hz and 45KHz?
2 rad/sec
2.25 Hz
2.25 rad/sec
2 Hz
Explanation:
Given information is