11. What is the cutoff frequency of the Butterworth filter with a passband gain KP=-1 dB at ΩP=4 rad/sec and stopband 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/sec
Answer: D
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.
12. What is the system function of the Butterworth filter with specifications as passband gain KP=-1 dB at ΩP=4 rad/sec and stopband 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 mentioned
Answer: C
From the given question,
KP=-1 dB, ΩP=4 rad/sec
KS=-20 dB and ΩS=8 rad/sec
We find out an order as N=5 and ΩC=4.5787 rad/sec
We know that for a 5th order normalized low pass Butterworth filter, the system equation is given as
13. If H(s)=\(\frac{1}{s^2+s+1}\) represents the transfer function of a low pass filter (not Butterworth) with a passband of 1 rad/sec, then what is the system function of a low pass filter with a passband 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 mentioned
Answer: A
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
=\(\frac{100}{s^2+10s+100}\).
14. If H(s)=\(\frac{1}{s^2+s+1}\) represents the transfer function of a low pass filter (not Butterworth) with a passband 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 mentioned
Answer: B
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
=\(\frac{s^2}{s^2+s+1}\)
15. If H(s)=\(\frac{1}{s^2+s+1}\) represents the transfer function of a low pass filter (not Butterworth) with a passband 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 mentioned
Answer: C
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
=\(\frac{s^2}{s^2+10s+100}\)
16. If H(s)=\(\frac{1}{s^2+s+1}\) represents the transfer function of a low pass filter (not Butterworth) with a passband of 1 rad/sec, then what is the system function of a bandpass filter with a passband of 10 rad/sec and a center frequency of 100 rad/sec?
A. \(\frac{s^2}{s^4+10s^3+20100s^2+10^5 s+1}\)
B. \(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+1}\)
C. \(\frac{s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\)
D. \(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\)
17. If H(s)=\(\frac{1}{s^2+s+1}\) represents the transfer function of a low pass filter (not Butterworth) with a passband 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?
A. \(\frac{(s^2+100)^2}{s^4+2s^3+204s^2+200s+10^4}\)
B. \(\frac{(s^2+10)^2}{s^4+2s^3+204s^2+200s+10^4}\)
C. \(\frac{(s^2+10)^2}{s^4+2s^3+400s^2+200s+10^4}\)
18. What is the stopband frequency of the normalized low pass Butterworth filter used to design an analog bandpass filter with -3.0103dB upper and lower cutoff frequency of 50Hz and 20KHz and a stopband attenuation 20dB at 20Hz and 45KHz?
A=\(\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}\) and B=\(\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}\)
=> A=2.51 and B=2.25
Hence ΩS=Min{|A|,|B|}=>ΩS=2.25 rad/sec.
19. What is the order of the normalized low pass Butterworth filter used to design an analog bandpass filter with -3.0103dB upper and lower cutoff frequency of 50Hz and 20KHz and a stopband attenuation of 20dB at 20Hz and 45KHz?
A. 2
B. 3
C. 4
D. 5
Answer: B
Given information is
Ω1=2π*20=125.663 rad/sec
Ω2=2π*45*103=2.827*105 rad/sec
Ωu=2π*20*103=1.257*105 rad/sec
Ωl=2π*50=314.159 rad/sec
We know that
A=\(\frac{-Ω_1^2+Ω_u Ω_l}{Ω_1 (Ω_u-Ω_l)}\) and B=\(\frac{Ω_2^2-Ω_u Ω_l}{Ω_2 (Ω_u-Ω_l)}\)
=> A=2.51 and B=2.25
Hence ΩS=Min{|A|,|B|}=> ΩS=2.25 rad/sec.
The order N of the normalized low pass Butterworth filter is computed as follows
Rounding off to the next large integer, we get, N=3.
20. Which of the following condition is true?
A. N ≤ \(\frac{log(\frac{1}{k})}{log(\frac{1}{d})}\)
B. N ≤ \(\frac{log(k)}{log(D.}\)
C. N ≤ \(\frac{log(D.}{log(k)}\)
D. N ≤ \(\frac{log(\frac{1}{d})}{log(\frac{1}{k})}\)
Answer: D
If ‘d’ is the discrimination factor and ‘K’ is the selectivity factor, then
N ≤ \(\frac{log(\frac{1}{d})}{log(\frac{1}{k})}\)
21. Which of the following defines a Chebyshev polynomial of order N, TN(x)?
A. cos(Ncos-1x) for all x
B. cosh(Ncosh-1x) for all x
C.cos(Ncos-1x), |x|-1x), |x|>1
D. None of the mentioned
Answer: C
In order to understand the frequency-domain behavior of Chebyshev filters, it is of utmost importance to define a Chebyshev polynomial and then its properties. A Chebyshev polynomial of degree N is defined as
TN(x) = cos(Ncos-1x), |x|≤1
cosh(Ncosh-1x), |x|>1.
22. What is the formula for chebyshev polynomial TN(x) in recursive form?
A. 2TN-1(x) – TN-2(x)
B. 2TN-1(x) + TN-2(x)
C. 2xTN-1(x) + TN-2(x)
D. 2xTN-1(x) – TN-2(x)
Answer: D
We know that a chebyshev polynomial of degree N is defined as
TN(x) = cos(Ncos-1x), |x|≤1
cosh(Ncosh-1x), |x|>1
From the above formula, it is possible to generate the Chebyshev polynomial using the following recursive formula
TN(x)= 2xTN-1(x)-TN-2(x), N ≥ 2.
23. What is the value of the Chebyshev polynomial of degree 0?
A. 1
B. 0
C. -1
D. 2
Answer: A
We know that a chebyshev polynomial of degree N is defined as
TN(x) = cos(Ncos-1x), |x|≤1
cosh(Ncosh-1x), |x|>1
For a degree 0 Chebyshev filter, the polynomial is obtained as
T0(x)=cos(0)=1.
24. What is the value of the Chebyshev polynomial of degree 1?
A. 1
B. x
C. -1
D. -x
Answer: B
We know that a chebyshev polynomial of degree N is defined as
TN(x) = cos(Ncos-1x), |x|≤1
cosh(Ncosh-1x), |x|>1
For a degree 1 Chebyshev filter, the polynomial is obtained as
T0(x)=cos(cos-1x)=x.
25. What is the value of the Chebyshev polynomial of degree 3?
A. 3x3+4x
B. 3x3-4x
C. 4x3+3x
D. 4x3-3x
Answer: D
We know that a Chebyshev polynomial of degree N is defined as
