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351.	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 10 rad/sec?(a) \(\frac{100}{s^2+10s+100}\)(b) \(\frac{s^2}{s^2+s+1}\)(c) \(\frac{s^2}{s^2+10s+100}\)(d) None of the mentionedThis question was addressed to me during an internship interview.My question is from Design of Low Pass Butterworth Filters in portion Digital Filters Design of Digital Signal Processing
Answer» The CORRECT OPTION is (C) \(\FRAC{s^2}{s^2+10s+100}\) Best explanation: The low pass-to-high pass transformation is s→Ωu/s Hence the required low pass filter is Ha(s)=H(s)\|s→10/s =\(\frac{s^2}{s^2+10s+100}\)

352.	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?(a) \(\frac{100}{s^2+10s+100}\)(b) \(\frac{s^2}{s^2+s+1}\)(c) \(\frac{s^2}{s^2+10s+100}\)(d) None of the mentionedI have been asked this question in an interview for internship.The doubt is from Design of Low Pass Butterworth Filters in section Digital Filters Design of Digital Signal Processing
Answer» Right choice is (B) \(\FRAC{s^2}{s^2+s+1}\) Easiest explanation: The low PASS-to-high pass TRANSFORMATION is s→Ωu/s Hence the REQUIRED high pass filter is Ha(s)= H(s)\|s→1/s =\(\frac{s^2}{s^2+s+1}\)

353.	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?(a) \(\frac{100}{s^2+10s+100}\)(b) \(\frac{s^2}{s^2+s+1}\)(c) \(\frac{s^2}{s^2+10s+100}\)(d) None of the mentionedThe question was posed to me by my college director while I was bunking the class.My doubt stems from Design of Low Pass Butterworth Filters topic in section Digital Filters Design of Digital Signal Processing
Answer» The correct option is (a) \(\frac{100}{s^2+10s+100}\) The BEST EXPLANATION: The LOW pass-to-low pass transformation is s→s/Ωu Hence the required low pass filter is Ha(s)=H(s)\|s→s/10 =\(\frac{100}{s^2+10s+100}\).

354.	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?(a) \(\frac{1}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)(b) \(\frac{1}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+1}\)(c) \(\frac{2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)(d) None of the mentionedI got this question during an online exam.The doubt is from Design of Low Pass Butterworth Filters in chapter Digital Filters Design of Digital Signal Processing
Answer» Right answer is (c) \(\frac{2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\) EASY explanation: From the given question, KP=-1 dB, ΩP=4 rad/sec, KS=-20 dB and ΩS=8 rad/sec We find out order as N=5 and ΩC=4.5787 rad/sec We know that for a 5th order normalized low pass Butterworth FILTER, system equation is given as H5(s)=\(\frac{1}{(s+1)(s^2+0.618s+1)(s^2+1.618s+1)}\) The specified low pass filter is OBTAINED by applying low pass-to-low pass TRANSFORMATION on the normalized low pass filter. That is, Ha(s)=H5(s)\|s→s/Ωc =H5(s)\|s→s/4.5787 upon calculating, we get Ha(s)=\({2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)

355.	What is the lowest order 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?(a) 4(b) 5(c) 6(d) 3I had been asked this question in my homework.I'm obligated to ask this question of Design of Low Pass Butterworth Filters in portion Digital Filters Design of Digital Signal Processing
Answer» Right ANSWER is (b) 5 The explanation: We know that the EQUATION for the order of the Butterworth filter is given as N=\(\frac{log⁡[(10^{-K_P/10}-1)/(10^{-K_s/10}-1)]}{2 log⁡(\frac{Ω_P}{Ω_S})}\) From the given question, KP=-1 dB, ΩP= 4 rad/sec, KS=-20 dB and ΩS= 8 rad/sec Upon substituting the VALUES in the above equation, we get N=4.289 Rounding off to the NEXT largest integer, we get N=5.

356.	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?(a) 3.5787 rad/sec(b) 1.069 rad/sec(c) 6 rad/sec(d) 4.5787 rad/secI got this question during an interview.The origin of the question is Design of Low Pass Butterworth Filters in portion Digital Filters Design of Digital Signal Processing
Answer» Right option is (d) 4.5787 rad/sec The explanation: We KNOW that the equation for the CUTOFF frequency of a Butterworth FILTER is GIVEN as ΩC = \(\FRAC{\Omega_P}{(10^{-K_P/10}-1)^{1/2N}}\) We know that KP=-1 dB, ΩP=4 rad/sec and N=5 Upon substituting the values in the above equation, we get ΩC=4.5787 rad/sec.

357.	The cutoff frequency of the low pass Butterworth filter is the arithmetic mean of the two cutoff frequencies as found above.(a) True(b) FalseThis question was addressed to me in an interview.The doubt is from Design of Low Pass Butterworth Filters topic in section Digital Filters Design of Digital Signal Processing
Answer» CORRECT choice is (a) True To elaborate: The arithmetic MEAN of the two cutoff frequencies as found above is the final cutoff frequency of the LOW PASS Butterworth filter.

358.	What is the expression for cutoff frequency in terms of stop band gain?(a) \(\frac{\Omega_S}{(10^{-K_S/10}-1)^{1/2N}}\)(b) \(\frac{\Omega_S}{(10^{-K_S/10}+1)^{1/2N}}\)(c) \(\frac{\Omega_S}{(10^{K_S/10}-1)^{1/2N}}\)(d) None of the mentionedI have been asked this question by my school principal while I was bunking the class.My query is from Design of Low Pass Butterworth Filters in division Digital Filters Design of Digital Signal Processing
Answer» The CORRECT answer is (C) \(\frac{\Omega_S}{(10^{K_S/10}-1)^{1/2N}}\) The explanation is: We know that, \([\frac{Ω_S}{Ω_C}]^{2N} = 10^{-K_S/10}-1\) => \(Ω_C = \frac{Ω_S}{(10^{-K_S/10}-1)^{1/2N}}\).

359.	What is the expression for cutoff frequency in terms of pass band gain?(a) \(\frac{\Omega_P}{(10^{-K_P/10}-1)^{1/2N}}\)(b) \(\frac{\Omega_P}{(10^{-K_P/10}+1)^{1/2N}}\)(c) \(\frac{\Omega_P}{(10^{K_P/10}-1)^{1/2N}}\)(d) None of the mentionedI have been asked this question in examination.This key question is from Design of Low Pass Butterworth Filters topic in portion Digital Filters Design of Digital Signal Processing
Answer» RIGHT answer is (a) \(\frac{\Omega_P}{(10^{-K_P/10}-1)^{1/2N}}\) Explanation: We know that, \([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_P/10}-1\) => \(Ω_C = \frac{Ω_P}{(10^{-K_P/10}-1)^{1/2N}}\).

360.	Which of the following equation is True?(a) \([\frac{\Omega_S}{\Omega_C} ]^{2N} = 10^{-K_S/10}+1\)(b) \([\frac{\Omega_S}{\Omega_C} ]^{2N} = 10^{K_S/10}+1\)(c) \([\frac{\Omega_S}{\Omega_C} ]^{2N} = 10^{-K_S/10}-1\)(d) None of the mentionedThe question was asked in an international level competition.The question is from Design of Low Pass Butterworth Filters topic in section Digital Filters Design of Digital Signal Processing
Answer» The correct answer is (b) \([\FRAC{\Omega_S}{\Omega_C} ]^{2N} = 10^{K_S/10}+1\) To EXPLAIN I WOULD say: We KNOW that, KP=-10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\) =>\([\frac{Ω_S}{Ω_C}]^{2N} = 10^{\frac{-K_S}{10}}-1\)

361.	What is the value of gain at the stop band frequency, i.e., what is the value of KS?(a) -10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\)(b) -10 \(log⁡[1-(\frac{\Omega_S}{\Omega_C})^{2N}]\)(c) 10 \(log⁡[1-(\frac{\Omega_S}{\Omega_C})^{2N}]\)(d) 10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\)This question was posed to me in exam.My question is based upon Design of Low Pass Butterworth Filters in portion Digital Filters Design of Digital Signal Processing
Answer» The correct option is (a) -10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\) To explain: We know that the formula for GAIN is K = 20 log\|H(jΩ)\| We know that \(\|H(j \Omega)\|=\frac{1}{\SQRT{(1+(\frac{\Omega}{\Omega_C})^{2N}}}\) By applying 20log on both sides of above equation, we get K = 20 \(log\|H(j\Omega)\|=-20 [log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]]^{1/2}\) = -10 \(log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]\) We know that K= KS at Ω=ΩS => KS=-10 \(log⁡[1+(\frac{\Omega_S}{\Omega_C})^{2N}]\).

362.	What is the order N of the low pass Butterworth filter in terms of KP and KS?(a) \(\frac{log⁡[(10^\frac{K_P}{10}-1)/(10^\frac{K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)(b) \(\frac{log⁡[(10^\frac{K_P}{10}+1)/(10^\frac{K_s}{10}+1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)(c) \(\frac{log⁡[(10^\frac{-K_P}{10}+1)/(10^\frac{-K_s}{10}+1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)(d) \(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\)This question was posed to me by my college director while I was bunking the class.Question is taken from Design of Low Pass Butterworth Filters in portion Digital Filters Design of Digital Signal Processing
Answer» The correct choice is (d) \(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\) To EXPLAIN I would SAY: We know that, \([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_P/10}-1\) and \([\frac{Ω_P}{Ω_C}]^{2N} = 10^{-K_S/10}-1\). By dividing the above two EQUATIONS, we get => \([Ω_P/Ω_S]^{2N} = (10^{-K_S/10}-1)(10^{-K_P/10}-1)\) By taking log in both sides, we get => N=\(\frac{log⁡[(10^\frac{-K_P}{10}-1)/(10^\frac{-K_s}{10}-1)]}{2 log⁡(\frac{\Omega_P}{\Omega_S})}\).

363.	What is the value of gain at the pass band frequency, i.e., what is the value of KP?(a) -10 \(log⁡ [1-(\frac{\Omega_P}{\Omega_C})^{2N}]\)(b) -10 \(log⁡ [1+(\frac{\Omega_P}{\Omega_C})^{2N}]\)(c) 10 \(log⁡ [1-(\frac{\Omega_P}{\Omega_C})^{2N}]\)(d) 10 \(log⁡ [1+(\frac{\Omega_P}{\Omega_C})^{2N}]\)I had been asked this question in an interview.This interesting question is from Design of Low Pass Butterworth Filters in section Digital Filters Design of Digital Signal Processing
Answer» The correct answer is (B) -10 \(log⁡ [1+(\frac{\Omega_P}{\Omega_C})^{2N}]\) To explain: We know that the formula for gain is K = 20 log\|H(jΩ)\| We know that \(\|H(j\Omega)\|=\frac{1}{\sqrt{(1+(\frac{\Omega}{\Omega_C})^{2N}}}\) By APPLYING 20log on both SIDES of above equation, we get K = 20 \(log\|H(j \Omega)\|=-20 [log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]]^{1/2}\) = -10 \(log⁡[1+(\frac{\Omega}{\Omega_C})^{2N}]\) We know that K= KP at Ω=ΩP => KP=-10 \(log⁡[1+(\frac{\Omega_P}{\Omega_C})^{2N}]\).

364.	Which of the following is a frequency domain specification?(a) 0 ≥ 20 log\|H(jΩ)\|(b) 20 log\|H(jΩ)\| ≥ KP(c) 20 log\|H(jΩ)\| ≤ KS(d) All of the mentionedThe question was posed to me in a national level competition.I would like to ask this question from Design of Low Pass Butterworth Filters topic in chapter Digital Filters Design of Digital Signal Processing
Answer» Right CHOICE is (d) All of the mentioned Easy explanation: We are required to design a LOW pass Butterworth filter to meet the FOLLOWING frequency domain specifications. KP ≤ 20 log\|H(jΩ)\| ≤ 0 and 20 log\|H(jΩ)\| ≤ KS.

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