This set of Digital Signal Processing quiz focuses on “Chebyshev Filters – 2”.
1. What is the value of |TN(±1)|?
a) 0
b) -1
c) 1
d) None of the mentioned
View Answer
Explanation: We know that a chebyshev polynomial of degree N is defined as
TN(x) = cos(Ncos-1x), |x|≤1
cosh(Ncosh-1x), |x|>1
Thus |TN(±1)|=1.
2. The chebyshev polynomial is oscillatory in the range |x|<∞.
a) True
b) False
View Answer
Explanation: The chebyshev polynomial is oscillatory in the range |x|≤1 and monotonic outside it.
3. If NB and NC are the orders of the Butterworth and Chebyshev filters respectively to meet the same frequency specifications, then which of the following relation is true?
a) NC=NB
b) NC<NB
c) NC>NB
d) Cannot be determined
View Answer
Explanation: The equi-ripple property of the chebyshev filter yields a narrower transition band compared with that obtained when the magnitude response is monotone. As a consequence of this, the order of a chebyshev filter needed to achieve the given frequency domain specifications is usually lower than that of a Butterworth filter.
4. The chebyshev-I filter is equi-ripple in pass band and monotonic in the stop band.
a) True
b) False
View Answer
Explanation: There are two types of chebyshev filters. The Chebyshev-I filter is equi-ripple in the pass band and monotonic in the stop band and the chebyshev-II filter is quite opposite.
5. What is the equation for magnitude frequency response |H(jΩ)| of a low pass chebyshev-I filter?
a) \(\frac{1}{\sqrt{1-ϵ T_N^2 (\frac{Ω}{Ω_P})}}\)
b) \(\frac{1}{\sqrt{1+ϵ T_N^2 (\frac{Ω}{Ω_P})}}\)
c) \(\frac{1}{\sqrt{1-ϵ^2 T_N^2 (\frac{Ω}{Ω_P})}}\)
d) \(\frac{1}{\sqrt{1+ϵ^2 T_N^2 (\frac{Ω}{Ω_P})}}\)
View Answer
Explanation: The magnitude frequency response of a low pass chebyshev-I filter is given by
|H(jΩ)|=\(\frac{1}{\sqrt{1+ϵ^2 T_N^2(\frac{Ω}{Ω_P})}}\)
where ϵ is a parameter of the filter related to the ripple in the pass band and TN(x) is the Nth order chebyshev polynomial.
6. What is the number of minima’s present in the pass band of magnitude frequency response of a low pass chebyshev-I filter of order 4?
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: In the magnitude frequency response of a low pass chebyshev-I filter, the pass band has 2 maxima and 2 minima(order 4=2 maxima+2 minima).
7. What is the number of maxima present in the pass band of magnitude frequency response of a low pass chebyshev-I filter of order 5?
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: In the magnitude frequency response of a low pass chebyshev-I filter, the pass band has 3 maxima and 2 minima(order 5=3 maxima+2 minima).
8. The sum of number of maxima and minima in the pass band equals the order of the filter.
a) True
b) False
View Answer
Explanation: In the pass band of the frequency response of the low pass chebyshev-I filter, the sum of number of maxima and minima is equal to the order of the filter.
9. Which of the following is the characteristic equation of a Chebyshev filter?
a) 1+ϵ2TN2(s/j)=0
b) 1-ϵ2TN2(s/j)=0
c) 1+ϵ TN2(s/j)=0
d) None of the mentioned
View Answer
Explanation: We know that for a chebyshev filter, we have
|H(jΩ)|=\(\frac{1}{\sqrt{1+ϵ^2 T_N^2(\frac{Ω}{Ω_P})}}\)
=>|H(jΩ)|2=\(\frac{1}{\sqrt{1+ϵ^2 T_N^2(\frac{Ω}{Ω_P})}}\)
By replacing jΩ by s and consequently Ω by s/j in the above equation, we get
=>|HN(s)|2=\(\frac{1}{1+ϵ^2 T_N^2 (s/j)}\)
The poles of the above equation is given by the equation 1+ϵ2TN2(s/j)=0 which is called as the characteristic equation.
10. The poles of HN(s).HN(-s) are found to lie on ___________
a) Circle
b) Parabola
c) Hyperbola
d) Ellipse
View Answer
Explanation: The poles of HN(s).HN(-s) is given by the characteristic equation 1+ϵ2TN2(s/j)=0.
The roots of the above characteristic equation lies on ellipse, thus the poles of HN(s).HN(-s) are found to lie on ellipse.
11. If the discrimination factor ‘d’ and the selectivity factor ‘k’ of a chebyshev I filter are 0.077 and 0.769 respectively, then what is the order of the filter?
a) 2
b) 5
c) 4
d) 3
View Answer
Explanation: We know that the order of a chebyshev-I filter is given by the equation,
N=cosh-1(1/d)/cosh-1(1/k)=4.3
Rounding off to the next large integer, we get N=5.
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