Digital Signal Processing

Design of Low Pass Butterworth Filters

Question 1
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 2
Marks : +2 | -2
Pass Ratio : 100%
What is the system function of the Butterworth filter with specifications as pass band gain KP=-1 dB at ΩP=4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS=8 rad/sec?
\\(\\frac{1}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\\)
\\(\\frac{1}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+1}\\)
\\(\\frac{2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\\)
None of the mentioned
Explanation:
From the given question,
Question 3
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 4
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 5
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 high pass filter with a cutoff frequency of 1rad/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-high pass transformation is
Question 6
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 7
Marks : +2 | -2
Pass Ratio : 100%
What is the order N of the low pass Butterworth filter in terms of KP and KS?
\\(\\frac{log⁡[(10^\\frac{K_P}{10}-1)/(10^\\frac{K_s}{10}-1)]}{2 log⁡(\\frac{\\Omega_P}{\\Omega_S})}\\)
\\(\\frac{log⁡[(10^\\frac{K_P}{10}+1)/(10^\\frac{K_s}{10}+1)]}{2 log⁡(\\frac{\\Omega_P}{\\Omega_S})}\\)
\\(\\frac{log⁡[(10^\\frac{-K_P}{10}+1)/(10^\\frac{-K_s}{10}+1)]}{2 log⁡(\\frac{\\Omega_P}{\\Omega_S})}\\)
\\(\\frac{log⁡[(10^\\frac{-K_P}{10}-1)/(10^\\frac{-K_s}{10}-1)]}{2 log⁡(\\frac{\\Omega_P}{\\Omega_S})}\\)
Explanation:
Question 8
Marks : +2 | -2
Pass Ratio : 100%
What is the value of gain at the stop band frequency, i.e., what is the value of KS?
-10 \\(log⁡[1+(\\frac{\\Omega_S}{\\Omega_C})^{2N}]\\)
-10 \\(log⁡[1-(\\frac{\\Omega_S}{\\Omega_C})^{2N}]\\)
10 \\(log⁡[1-(\\frac{\\Omega_S}{\\Omega_C})^{2N}]\\)
10 \\(log⁡[1+(\\frac{\\Omega_S}{\\Omega_C})^{2N}]\\)
Explanation:
We know that the formula for gain is
Question 9
Marks : +2 | -2
Pass Ratio : 100%
What is the expression for cutoff frequency in terms of pass band gain?
\\(\\frac{\\Omega_P}{(10^{-K_P/10}-1)^{1/2N}}\\)
\\(\\frac{\\Omega_P}{(10^{-K_P/10}+1)^{1/2N}}\\)
\\(\\frac{\\Omega_P}{(10^{K_P/10}-1)^{1/2N}}\\)
None of the mentioned
Explanation:
We know that,
Question 10
Marks : +2 | -2
Pass Ratio : 100%
What is the cutoff frequency of the Butterworth filter with a pass band gain KP=-1 dB at ΩP=4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS=8 rad/sec?
3.5787 rad/sec
1.069 rad/sec
6 rad/sec
4.5787 rad/sec
Explanation:
We know that the equation for the cutoff frequency of a Butterworth filter is given as