Network Theory MCQs – Elements of Realizability and Synthesis of One-Port Networks MCQs ( Network Theory ) MCQs
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Network Theory MCQs – Elements of Realizability and Synthesis of One-Port Networks MCQs ( Network Theory ) MCQs
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Hurwitz Polynomials
1. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless?
a) all odd terms are missing
b) all even terms are missing
c) all even or odd terms are missing
d) all even and odd terms are missing
Answer: c
Explanation: All the quotients in the polynomial P(s) are positive. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless all even or odd terms are missing.For example P(s) = s3+3s is Hurwitz because all quotient terms are positive and all even terms are missing.
2. The roots of the odd and even parts of a Hurwitz polynomial P (s) lie on ____________
a) right half of s plane
b) left half of s-plane
c) on jω axis
d) on σ axis
Answer: c
Explanation: The roots of the odd and even parts of a Hurwitz polynomial P (s) lie on jω axis not on right half of s plane or on left half of s-plane.
3. If the polynomial P (s) is either even or odd, then the roots of P (s) lie on __________
a) on σ axis
b) on jω axis
c) left half of s-plane
d) right half of s plane
Answer: b
Explanation: If the polynomial P (s) is either even or odd, then the roots of P (s) lie on jω axis not on right half of s plane or on left half of s-plane.
4. If the ratio of the polynomial P (s) and its derivative gives a continued fraction expansion with ________ coefficients, then the polynomial P (s) is Hurwitz.
a) all negative
b) all positive
c) positive or negative
d) positive and negative
Answer: b
Explanation: If the ratio of the polynomial P (s) and its derivative P‘(s) gives a continued fraction expansion with all positive coefficients, then the polynomial P (s) is Hurwitz. If all the quotients in the continued fraction expansion are positive, the polynomial P(s) is positive.
5. Consider the polynomial P(s)=s4+3s2+2. The given polynomial P (s) is Hurwitz.
a) True
b) False
Answer: a
Explanation: P(s)=s4+3s2+2 => P‘ (s)=4s3+6s
After doing the continued fraction expansion, we get all the quotients as positive. So, the polynomial P (s) is Hurwitz.
6. When s is real, the driving point impedance function is _________ function and the driving point admittance function is _________ function.
a) real, complex
b) real, real
c) complex, real
d) complex, complex
Answer: b
Explanation: When s is real, the driving point impedance function is real function and the driving point admittance function is real function because the quotients of the polynomials P(s) and Q(s) are real. When Z(s) is determined from the impedances of the individual branches, the quotients are obtained by adding together, multiplying or dividing the branch parameters which are real.
7. The poles and zeros of driving point impedance function and driving point admittance function lie on?
a) left half of s-plane only
b) right half of s-plane only
c) left half of s-plane or on imaginary axis
d) right half of s-plane or on imaginary axis
Answer: c
Explanation: The poles and zeros of driving point impedance function and driving point admittance function lie on left half of s-plane or on imaginary axis of the s-plane.
8. For real roots of sk, all the quotients of s in s2+ωk2 of the polynomial P (s) are __________
a) negative
b) non-negative
c) positive
d) non-positive
Answer: b
Explanation: For real roots of sk, all the quotients of s in s2+ωk2 of the polynomial P (s) are non-negative. So by multiplying all factors in P(s) we find that all quotients are positive.
9. The real parts of the driving point function Z (s) and Y (s) are?
a) positive and zero
b) positive
c) zero
d) positive or zero
Answer: d
Explanation: The real parts of the driving point impedance function Z (s) and driving point admittance function Y (s) are positive or zero.
10. For the complex zeros to appear in conjugate pairs the poles of the network function are ____ and zeros of the network function are ____________
a) complex, complex
b) complex, real
c) real, real
d) real, complex
Answer: c
Explanation: P(s) and Q(s) are real when s is real. So the poles of the network function are real and zeros of the network function are real, the complex zeros to appear in conjugate pairs.
Frequency Response of Reactive One-Ports
1. Based on the location of zeros and poles, a reactive one-port can have ____________ types of frequency response.
a) 1
b) 2
c) 3
d) 4
Answer: d
Explanation: A reactive one-port can have four types of frequency response based on the location of zeros and poles.
(i) frequency response with two external poles
(ii) frequency response with two external zeros
(iii) frequency response with an external zero ω = 0 and an external poles at ω = ∞
(iv) frequency response with an external zero ω = ∞ and an external poles at ω = 0.
2. A driving point impedance with poles at ω = 0, ω = ∞ must have ___________ term in the denominator polynomial.
a) s
b) s+1
c) s+2
d) s+3
Answer: a
Explanation: As there is a pole at ω = 0, (s-jω)=s. Poles are written in the denominator of Z(s). So there will be s term in the denominator polynomial in a driving point impedance function Z(s).
3. A driving point impedance with poles at ω = 0, ω = ∞ must have excess ___________ term in the numerator polynomial.
a) s1+ωn1
b) s1+ωn2
c) s2+ωn2
d) s2+ωn1
Answer: c
Explanation: The driving point impedance of the one-port is infinite, and it will not pass either direct current or alternating current of an infinitely high frequency.
4. A driving point impedance with zeros at ω = 0, ω = ∞ must have ___________ term in the numerator polynomial.
a) s+3
b) s+2
c) s+1
d) s
Answer: d
Explanation: As there is a zero at ω = 0, (s-jω)=s. Zeros are written in the numerator of Z(s). So there will be s term in the numerator polynomial in a driving point impedance function Z(s).
5. A driving point impedance with zeros at ω = 0, ω = ∞ must have an excess ___________ term in the denominator polynomial.
a) s2+ωn1
b) s2+ωn2
c) s1+ωn2
d) s1+ωn1
Answer: b
Explanation: The driving point impedance of the one-port is zero, and it will pass both direct current and an alternating current of an infinitely high frequency.
6. A driving point impedance with zero at ω = 0 and pole at ω = ∞ must have ___________ term in the numerator polynomial.
a) s+1
b) s
c) s+3
d) s+2
Answer: b
Explanation: As ω = 0, (s-jω)=s. The numerator of Z(s) contains poles and denominator contains zeros. So there will be s term in the numerator polynomial.
7. A driving point impedance with zero at ω = 0 and pole at ω = ∞ must have ___________ term in the numerator polynomial.
a) s1+ωn1
b) s2+ωn1
c) s1+ωn2
d) s2+ωn2
Answer: d
Explanation: If a pole is at ω = ∞, there will be an equal number of s2+ωn2 type terms in the numerator polynomial and the denominator polynomial.
8. A driving point impedance with zero at ω = 0 and pole at ω = ∞ must have ___________ term in the denominator polynomial.
a) s2+ωn2
b) s1+ωn1
c) s2+ωn1
d) s1+ωn2
Answer: a
Explanation: If there is a zero at ω = 0 and pole at ω = ∞, the one-port will pass direct current and block the alternating current of an infinitely high frequency.
9. A driving point impedance with pole at ω = 0 and zero at ω = ∞ must have ___________ term in the denominator polynomial.
a) s
b) s+3
c) s+1
d) s+2
Answer: a
Explanation: s-jω = (s-j(0)) = s. As pole is at ω = 0, there will be s term in the denominator polynomial.
10. A driving point impedance with pole at ω = 0 and zero at ω = ∞ must have ____________ term in the numerator and denominator.
a) s1+ωn2
b) s2+ωn2
c) s1+ωn1
d) s2+ωn1
Answer: b
Explanation: If a pole at ω = 0 and zero at ω = ∞, the one-port will block the direct current and pass the alternating current of an infinitely high frequency.
Synthesis of Reactive One-Ports by Foster’s Method
1. The driving point impedance of a one-port reactive network is given by Z(s)=5(s2+4)(s2+25)/s(s2+16). After taking the partial fractions, find the coefficient of 1/s.
a) 25/4
b) 50/4
c) 100/4
d) 125/4
Answer: d
Explanation: Since there is an extra term in the numerator compared to the denominator and also an s term in the denominator, the two poles exists at 0 and infinity. Therefore the network consists of first element and last element.
By taking the partial fraction expansion of Z(s), we obtain A=5(s2+4)(s2+25)/((s2+16))|s=0
=(5×4×25)/16=125/4.
2. The driving point impedance of a one-port reactive network is given by Z(s)=5(s2+4)(s2+25)/s(s2+16). After taking the partial fractions, find the coefficient of (s + j4).
a) 135/4
b) 145/4
c) 155/4
d) 165/4
Answer: a
Explanation: On taking the partial fraction expansion,
B=5 (s2+4)(s2+25)/s(s-j4)|s=-j4
= 135/8.
3. The driving point impedance of a one-port reactive network is given by Z(s)=5(s2+4)(s2+25)/s(s2+16). After taking the partial fractions, what is the value of H from Z (s)?
a) 3
b) 4
c) 5
d) 6
Answer: c
Explanation: By inspection, the value of H is 5.
So H = 5.
4. The driving point impedance of a one-port reactive network is given by Z(s)=5(s2+4)(s2+25)/s(s2+16). After taking the partial fractions, what is the value of C0?
a) 1/125
b) 4/125
c) 2/125
d) 3/125
Answer: b
Explanation: The coefficient of 1/s is 125/4.
And C0=1/P0
= 1/(125/4)=4/125 Farad.
5. The driving point impedance of a one-port reactive network is given by Z(s)=5(s2+4)(s2+25)/s(s2+16). After taking the partial fractions, what is the value of L∞?
a) 2
b) 3
c) 4
d) 5
Answer: d
Explanation: We know L∞ = H
= 5 H.
6. The driving point impedance of a one-port reactive network is given by Z(s)=5(s2+4)(s2+25)/s(s2+16). After taking the partial fractions, what is the value of C2?
a) 4/270
b) 8/270
c) 12/270
d) 16/270
Answer: b
Explanation: We know C2 = 1/2P2
= 8/(2×135)=8/270 F.
7. The driving point impedance of a one-port reactive network is given by Z(s)=5(s2+4)(s2+25)/s(s2+16). After taking the partial fractions, what is the value of L2?
a) 135/60
b) 135/62
c) 135/64
d) 135/66
Answer: c
Explanation: We know L2 = 2P2/ωn2
= (2×135)/(16×8)=135/64 H.
8. For performing second Foster form, after splitting the Z (S) in The driving point impedance of a one-port reactive network is given by Z(s)=5(s2+4)(s2+25)/s(s2+16), what is the coefficient of s/((s2+4)) is?
a) 1/35
b) 2/35
c) 3/35
d) 4/35
Answer: b
Explanation: The value of A is (1/5)(s(s2+16)))/(s-j2)(s2+25) at s=-j2
On solving we get the value of A as 2/35.
So the coefficient of s/((s2+4)) is 2/35.
9. For performing second Foster form, after splitting the Z (S) in The driving point impedance of a one-port reactive network is given by Z(s)=5(s2+4)(s2+25)/s(s2+16), what is the coefficient of s/((s2+4)) is?
a) 4/35
b) 3/35
c) 2/35
d) 1/35
Answer: c
Explanation: The coefficient of s/((s2+4)) is B=(1/5)((s(s2+16)))/(s-j5)(s2+4) at s=-j5. On solving we get B = 2/35.
10. Determine the value of L1 by performing second foster form for Z(s)=5(s2+4)(s2+25)/s(s2+16).
a) 35/4
b) 35/3
c) 35/2
d) 35
Answer: a
Explanation: P1 = 2/35.
We know L1 = 1/2P1
= 35/4 H.
Synthesis of Reactive One-Ports by Cauer Method
1. The driving point impedance of an LC network is given by Z(s)=(2s5+12s3+16s)/(s4+4s2+3). By taking the continued fraction expansion using first Cauer form, find the value of L1.
a) s
b) 2s
c) 3s
d) 4s
Answer: b
Explanation: The first Cauer form of the network is obtained by taking the continued fraction expansion of given Z(s). And we get he first quotient as 2s.
So, L1 = 2s.
2. Find the first reminder obtained by taking the continued fraction expansion in the driving point impedance of an LC network is given by Z(s)=(2s5+12s3+16s)/(s4+4s2+3). By taking the continued fraction expansion using first Cauer form.
a) 4s3+10s
b) 12s3+10s
c) 4s3+16s
d) 12s3+16s
Answer: a
Explanation: On taking the continued fraction expansion, the first reminder obtained is 4s3+10s.
3. The driving point impedance of an LC network is given by Z(s)=(2s5+12s3+16s)/(s4+4s2+3). By taking the continued fraction expansion using first Cauer form, find the value of C2.
a) 1
b) 1/2
c) 1/3
d) 1/4
Answer: d
Explanation: The second quotient obtained on taking the continued fraction expansion is s/4 and this is the value of sC2. So the value of C2 = 1/4.
4. The driving point impedance of an LC network is given by Z(s)=(2s5+12s3+16s)/(s4+4s2+3). By taking the continued fraction expansion using first Cauer form, find the value of L3.
a) 8
b) 8/3
c) 8/5
d) 8/7
Answer: b
Explanation: By taking the continued fraction expansion, the third quotient is 8s/3.
sL3 = 8s/3.
So L3 = 8/3H.
5. The driving point impedance of an LC network is given by Z(s)=(2s5+12s3+16s)/(s4+4s2+3). By taking the continued fraction expansion using first Cauer form, find the value of C4.
a) 1/2
b) 1/4
c) 3/4
d) 1
Answer: c
Explanation: We get the fourth quotient as 3s/4.
So sC4 = 3s/4.
C4 = 3/4F.
6. The driving point impedance of an LC network is given by Z(s)=(2s5+12s3+16s)/(s4+4s2+3). By taking the continued fraction expansion using first Cauer form, find the value of L5.
a) 2
b) 2/5
c) 2/7
d) 2/3
Answer: d
Explanation: On taking the continued fraction expansion fifth quotient is 2s/3.
sL5 = 2s/3
So L5 = 2/3H.
7. The driving point impedance of an LC network is given by Z(s)=(s4+4s2+3)/(s3+2s). By taking the continued fraction expansion using second Cauer form, find the value of C1.
a) 2/3
b) 2/2
c) 1/2
d) 4/2
Answer: a
Explanation: To obtain the second Cauer form, we have to arrange the numerator and the denominator of given Z(s) in ascending powers of s before starting the continued fraction expansion.
By taking the continued fraction expansion we get the first quotient as 3/2s.
So 1/sC1 = 3/2s
C1 = 2/3F.
8. The driving point impedance of an LC network is given by Z(s)=(s4+4s2+3)/(s3+2s). By taking the continued fraction expansion using second Cauer form, find the value of L2.
a) 1/5
b) 2/5
c) 3/5
d) 5/4
Answer: d
Explanation: On taking the continued fraction expansion the second quotient is 4/5s.
1/sL2 = 4/5s
So L2 = 5/4H.
9. The driving point impedance of an LC network is given by Z(s)=(s4+4s2+3)/(s3+2s). By taking the continued fraction expansion using second Cauer form, find the value of C3.
a) 25/s
b) 2/25s
c) 25/3s
d) 25/4s
Answer: b
Explanation: The third quotient is 25/2s.
1/sC3 = 25/2s.
C3 = 2/25F.
10. The driving point impedance of an LC network is given by Z(s)=(s4+4s2+3)/(s3+2s). By taking the continued fraction expansion using second Cauer form, find the value of L4.
a) 5
b) 2/5
c) 3/5
d) 4/5
Answer: a
Explanation: We obtain the fourth quotient as 1/5s.
1/sL4 = 1/5s
L4 = 5H.
Synthesis of R-L Network by the Foster Method
1. Consider a function Z(s)=5(s+1)(s+4)/(s+3)(s+5). Find the value of R1 after performing the first form of Foster method.
a) 1/3
b) 2/3
c) 3/3
d) 4/3
Answer: d
Explanation: After splitting the given function into partial according to the properties of first form of Foster method, we get
Z(s)=4/3+(5/3)(s/(s+3))+2s/(s+5).
So, R1 = 4/3Ω.
2. Consider a function Z(s)=5(s+1)(s+4)/(s+3)(s+5). Find the value of R1.
a) 4/3
b) 5/3
c) 3/5
d) 3/4
Answer: b
Explanation: On taking the partial fractions, we get P1 as 5/3 and we know R1=P1. So the value of R1 is 5/3Ω.
R1 = 5/3Ω.
3. Consider a function Z(s)=5(s+1)(s+4)/(s+3)(s+5). Find the value of L1 after performing the first form of Foster method.
a) 5/9
b) 9/5
c) 4/9
d) 9/4
Answer: a
Explanation: We know R1=(3)(L1) and as R1 is 5/3Ω. So the value of L1 is 5/9 H.
L1 = 5/9 H.
4. Consider a function Z(s)=5(s+1)(s+4)/(s+3)(s+5). Find the value of R2 after performing the first form of Foster method.
a) 1
b) 2
c) 3
d) 4
Answer: b
Explanation: We obtain P2 as 2.
R2=P2.
So the value of R2 is 2Ω.
R2 = 2Ω.
5. Consider a function Z(s)=5(s+1)(s+4)/(s+3)(s+5). Find the value of L2 after performing the first form of Foster method.
a) 4/5
b) 3/5
c) 2/5
d) 1/5
Answer: c
Explanation: As R2=(5)(L2)
So the value of L2 is 2/5 H.
L2 = 2/5 H.
6. Consider the admittance function, Y(s)=((2s2+16s+30))/(s2+6s+8). Determine the value of L1 after performing the second form of Foster method.
a) 1/3
b) 2/3
c) 3/3
d) 4/3
Answer: a
Explanation: After splitting the given function into partial according to the properties of first form of Foster method, we get Y(s)=2+3/(s+2)+1/(s+4).
We know L1=1/P1 and as P1 = 3, the value of L1 is 1/3H.
L1 = 1/3H.
7. Consider the admittance function, Y(s)=((2s2+16s+30))/(s2+6s+8). Determine the value of R1 after performing the second form of Foster method.
a) 4/3
b) 3/3
c) 2/3
d) 1/3
Answer: c
Explanation: R1 = 2/P1 and as P1 is 3, the value of R1 is 2/3Ω.
R1 = 2/3Ω.
8. Consider the admittance function, Y(s)=((2s2+16s+30))/(s2+6s+8). Determine the value of R2 after performing the second form of Foster method.
a) 1
b) 2
c) 3
d) 4
Answer: d
Explanation: On taking the partial fractions we get P2 as 1. And we know R2 = 4/P2. So the value of R2 is 4Ω.
R2 = 4Ω.
9. Consider the admittance function, Y(s)=((2s2+16s+30))/(s2+6s+8). Determine the value of L2 after performing the second form of Foster method.
a) 4
b) 1
c) 2
d) 3
Answer: b
Explanation: We got P2 = 1
And L2=1/P2
So the value of L2 is 1H.
L2 = 1H.
10. Consider the admittance function, Y(s)=((2s2+16s+30))/(s2+6s+8). Determine the value of R∞ after performing the second form of Foster method.
a) 3
b) 1
c) 2
d) 4
Answer: c
Explanation: On performing partial fractions we get the value of R∞ is 2Ω.
R∞ = 2Ω.
Synthesis of R-L Network by Cauer Method
1. Consider the impedance function, Z(s)=((s+4)(s+8))/((s+2)(s+6)). Find the value of R1 after converting into first Cauer form.
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: To find out the first Cauer form, we have to take the continued fraction expansion of Z (s).
On solving, we get the first quotient as 1. So the value of R1 as 1Ω.
2. Consider the impedance function, Z(s)=((s+4)(s+8))/((s+2)(s+6)). Find the value of L2 after converting into first Cauer form.
a) 1
b) 1/2
c) 1/4
d) 1/8
Answer: c
Explanation: On taking the continued fraction expansion of Z (s), the second quotient is s/4. So the value of L2 is 1/4 H.
L2 = 1/4 H.
3. Consider the impedance function, Z(s)=((s+4)(s+8))/((s+2)(s+6)). Find the value of R2 after converting into first Cauer form.
a) 1/4
b) 2/4
c) 3/4
d) 4/4
Answer: c
Explanation: We get the third quotient on performing continued fraction expansion of Z (s) as 4/3 and is the value of 1/R2. So the value of R2 is 3/4Ω .
R2 = 3/4Ω .
4. Consider the impedance function, Z(s)=((s+4)(s+8))/((s+2)(s+6)). Find the value of L3 after converting into first Cauer form.
a) 4/3
b) 3/4
c) 4/5
d) 5/4
Answer: b
Explanation: The fourth quotient obtained is 3s/4. And this is the value of sL3. So the value of L3 is 3/4 H.
L3 = 3/4 H.
5. Consider the impedance function, Z(s)=((s+4)(s+8))/((s+2)(s+6)). Find the value of R3 after converting into first Cauer form.
a) 4
b) 3
c) 2
d) 1
Answer: b
Explanation: The value of 1/R3 (fourth quotient) obtained by continued fraction expansion 1/3. So the value of R3 is 3Ω.
6. Consider the impedance function, Z(s)=(2s2+8s+6)/(s2+8s+12). Find the value of R1 after converting into second Cauer form.
a) 1
b) 3/4
c) 1/2
d) 1/4
Answer: c
Explanation: To find out the second Cauer form, we have to write the impedance function in ascending powers and by taking the continued fraction expansion of Z (s).
On solving, the first quotient obtained is 1/2. So the value of R1 as 1/2 Ω.
7. Consider the impedance function, Z(s)=(2s2+8s+6)/(s2+8s+12). Find the value of L1 after converting into second Cauer form.
a) 1/3
b) 2/3
c) 3/3
d) 4/3
Answer: a
Explanation: The second quotient obtained is 3/s and this is the value of 1/sL1. So the value of L1 is 1/3 H.
L1 = 1/3 H.
8. Consider the impedance function, Z(s)=(2s2+8s+6)/(s2+8s+12). Find the value of R2 after converting into second Cauer form.
a) 6/7
b) 7/6
c) 7/8
d) 8/7
Answer: d
Explanation: On taking the continued fraction expansion, we get third quotient as 8/7. So the value of R2 is 8/7Ω.
R2 = 8/7Ω.
9. Consider the impedance function, Z(s)=(2s2+8s+6)/(s2+8s+12). Find the value of L2 after converting into second Cauer form.
a) 5/50
b) 10
c) 5/49
d) 49/5
Answer: c
Explanation: We obtain the fourth quotient i.e., 1/sL2 as 49/5s. So the value of L2 is 5/49 H.
L2 = 5/49H.
10. Consider the impedance function, Z(s)=(2s2+8s+6)/(s2+8s+12). Find the value of R3 after converting into second Cauer form.
a) 1/5
b) 14/5
c) 5/14
d) 5
Answer: b
Explanation: The value of R3 is 14/5Ω as the fifth quotient obtained is 5/14.
R3 = 5/14Ω.
Synthesis of R-C Network by Foster Method
1. Consider the impedance function Z(s)=3(s+2)(s+4)/(s+1)(s+3). Find the value of R1 after realizing by first Foster method.
a) 9/2
b) 2/9
c) 9
d) 1/9
Answer: a
Explanation: The first Foster form can be realized by taking the partial fraction of Z (s).
On solving, we get
Z (s)=3+(9/2/(s+1))+(3/2/(s+3)).
So the value of R1 is 9/2Ω.
2. Consider the impedance function Z(s)=3(s+2)(s+4)/(s+1)(s+3). Find the value of C1 after realizing by first Foster method.
a) 1/9
b) 9
c) 2/9
d) 9/2
Answer: c
Explanation: After taking the partial fractions, as P1 = 1/C1, we get the 1/C1 value as 9/2. So the value of C1 is 2/9 F.
C1 = 2/9 F.
3. Consider the impedance function Z(s)=3(s+2)(s+4)/(s+1)(s+3). Find the value of C2 after realizing by first Foster method.
a) 1/3
b) 3
c) 3/2
d) 2/3
Answer: d
Explanation: We know P2 = 1/C2. The value of P2 is 3/2. So the value of C2 is 2/3 F.
C2 = 2/3 F.
4. Consider the impedance function Z(s)=3(s+2)(s+4)/(s+1)(s+3). Find the value of R2 after realizing by first Foster method.
a) 1
b) 1/2
c) 1/4
d) 1/8
Answer: b
Explanation: We got 1/R2C2 = 3. And the value of C2 is 2/3 F. So the value of R2 is 1/2 Ω.
R2 = 1/2 Ω.
5. Consider the impedance function Z(s)=3(s+2)(s+4)/(s+1)(s+3). Find the value of R∞ after realizing by first Foster method.
a) 1
b) 2
c) 3
d) 4
Answer: c
Explanation: On taking partial fraction of Z(s), we get the value of R∞ is 3Ω.
R∞ = 3Ω.
6. Consider the impedance function Y(s)=(s2+4s+3)/(3s2+18s+24). Find the value of R0 after realizing by second Foster method.
a) 4
b) 8
c) 12
d) 16
Answer: b
Explanation: The second Foster Form can be realized by taking the reciprocal of the impedance function and by taking partial fractions. So we get 1/R0 as 1/8. So the value of R0 is 8Ω.
R0 = 8Ω.
7. Consider the impedance function Y(s)=(s2+4s+3)/(3s2+18s+24). Find the value of R1 after realizing by second Foster method.
a) 16
b) 12
c) 8
d) 4
Answer: b
Explanation: On taking the partial fractions we get P1 as 1/12. And we know that R1=1/P1. So the value of R1 is 12Ω.
R1 = 12Ω.
8. Consider the impedance function Y(s)=(s2+4s+3)/(3s2+18s+24). Find the value of C1 after realizing by second Foster method.
a) 1/24
b) 1/12
c) 1/6
d) 1/3
Answer: a
Explanation: The value of C1 is 1/((2)(R1)). As R1 is 12Ω, the value of C1 is 1/24 F.
C1 = 1/24 F.
9. Consider the impedance function Y(s)=(s2+4s+3)/(3s2+18s+24). Find the value of R2 after realizing by second Foster method.
a) 5
b) 6
c) 7
d) 8
Answer: d
Explanation: We get the value of P2 on taking the partial fractions as 1/8. And we know R2 = 1/P2 So the value of R2 is 8Ω.
So, R2 = 8Ω.
10. Consider the impedance function Y(s)=(s2+4s+3)/(3s2+18s+24). Find the value of C2 after realizing by second Foster method.
a) 1/16
b) 1/8
c) 1/32
d) 1/64
Answer: c
Explanation: We obtained R2 as 8Ω and we know C2 = 1/((4)(R2)). So the value of C2 is 1/32 F.
C2 = 1/32 F.
Synthesis of R-C Network by Cauer Method
1. Consider the impedance functionZ(s)=(s2+6s+8)/(s2+3s). Find the value of R1 after performing the first Cauer form.
a) 1
b) 2
c) 3
d) 4
Answer: a
Explanation: To find the first Cauer form, we take the continued fraction expansion by the divide, invert, divide procedure.
On performing this we get the first quotient is 1Ω.
So, R1 = 1Ω.
2. Consider the impedance functionZ(s)=(s2+6s+8)/(s2+3s). Find the first reminder obtained by taking the continued fraction expansion after performing the first Cauer form.
a) s + 8
b) 2s + 8
c) 3s + 8
d) 4s + 8
Answer: c
Explanation: The continued fraction expansion is done by the divide, invert, divide procedure. So the first reminder obtained is 3s + 8.
3. Consider the impedance functionZ(s)=(s2+6s+8)/(s2+3s). Find the value of R2 after performing the first Cauer form.
a) 4
b) 3
c) 6
d) 9
Answer: d
Explanation: On performing the continued fraction expansion we get the third quotient as 9.
So the value of R2 is 9Ω.
R2 = 9Ω.
4. Consider the impedance functionZ(s)=(s2+6s+8)/(s2+3s). Find the value of C1 after performing the first Cauer form.
a) 1/4
b) 1/3
c) 1/2
d) 1
Answer: b
Explanation: The second quotient of the continued fraction expansion is s/3.
So the value of C1 is 1/3 F.
C1 = 1/3 F.
5. Consider the impedance functionZ(s)=(s2+6s+8)/(s2+3s). Find the value of C2 after performing the first Cauer form.
a) 1/6
b) 1/12
c) 1/24
d) 1/48
Answer: c
Explanation: We obtain the fourth quotient on performing continued fraction expansion as s/24.
So the value of C2 is 1/24 F.
C2 = 1/24 F.
6. Consider the impedance function Z(s)=(s2+6s+8)/(s2+3s). Find the value of C1 after performing the second Cauer form.
a) 1/2
b) 3/8
c) 1/4
d) 1/8
Answer: b
Explanation: The second Cauer network can be obtained by arranging the numerator and denominator polynomials of Z(s) in ascending powers of s. After performing the continued fraction expansion, we get
C1 = 3/8 F.
7. Consider the impedance function Z(s)=(s2+6s+8)/(s2+3s). Find the first reminder obtained by taking the continued fraction expansion after performing the second Cauer form.
a) 10s/3+s2
b) s/3+s2
c) 10s/3+3s2
d) s/3+3s2
Answer: a
Explanation: On performing the continued fraction expansion after arranging the numerator and denominator polynomials of Z(s) in ascending powers of s, the first reminder is 10s/3+s2.
8. Consider the impedance function Z(s)=(s2+6s+8)/(s2+3s). Find the value of R1 after performing the second Cauer form.
a) 9/10
b) 10/9
c) 8/9
d) 9/8
Answer: b
Explanation: The second quotient on performing continued fraction expansion is 9/10. This is the value of 1/R1. So the value of R1 is 10/9Ω.
R1 = 10/9Ω.
9. Consider the impedance function Z(s)=(s2+6s+8)/(s2+3s). Find the value of C2 after performing the second Cauer form.
a) 3
b) 3/10
c) 3/100
d) 3/1000
Answer: c
Explanation: On performing continued fraction expansion, the third quotient is 100/3s. So the value of C2 is 3/100 F.
C2 = 3/100 F.
10. Consider the impedance function Z(s)=(s2+6s+8)/(s2+3s). Find the value of R1 after performing the second Cauer form.
a) 10
b) 1
c) 100
d) 1000
Answer: a
Explanation: The final quotient is 1/10. So the value of R1 is 1/1/10.So the value of R1 is 10Ω.
R1 = 10Ω.