This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Laplace Transform of a Periodic Function”.
1. For a network having 1 Ω resistor and 1 F capacitor in series, the impedance Z(s) is ____________
a) \(\frac{s+1}{s}\)
b) \(\frac{s+2}{s}\)
c) \(\frac{s}{s+1}\)
d) \(\frac{s}{s+2}\)
View Answer
Explanation: We know that the impedance, Z(s) is given by,
Z(s) = resistance + 1/capacitor
= 1 + \(\frac{1}{s}\)
= \(\frac{s+1}{s}\).
2. A system function H(s) = \(\frac{25}{s^2+4s+100}\), the resonant frequency in rad/sec and bandwidth in rad/sec is ____________
a) 5, 10
b) 5, 4
c) 10, 4
d) 10, 10
View Answer
Explanation: Given that, H(s) = \(\frac{25}{s^2+4s+100}\)
So, we can infer that, ωn = (100)0.5
= 10
And hence, ωr ≃ 10
So, resonant frequency = 10 rad/sec
And Bandwidth = 10 rad/sec.
3. Given a series RLC circuit. The impedance Z(s) of the circuit will be?
a) \(\frac{5(s^2+5)}{s^2+s+1}\)
b) \(\frac{5(s^2-5)}{s^2-s-1}\)
c) \(\frac{s^2-10s-100}{s}\)
d) \(\frac{s^2+10s+100}{s}\)
View Answer
Explanation: The impedance Z(s) = R + sL + \(\frac{1}{sC}\)
= \(\frac{s^2 LC+RCs+1}{sC}\)
This is similar to \(\frac{s^2+10s+100}{s}\).
4. A system function H(s) = \(\frac{V(s)}{I(s)} = \frac{s}{(s+4)}\). For i(t) = u(t) and value of system is 0 for t<0. Then v(t) is ___________
a) 1 – e-4t
b) e-4t
c) e4t
d) 1 + e-4t
View Answer
Explanation: Given that, V(s) = \(\frac{s}{s+4}.\frac{1}{s}\)
= \(\frac{1}{s+4}\)
So, v (t) = e-4t.
5. A system is at rest for t < 0. It is given by \(\frac{dy}{dt}\) + 2y = u(t)sin(2t+A). If steady state is reached at t = 0, then the value of angle A is ___________
a) 0°
b) 45°
c) -45°
d) ∞
View Answer
Explanation: Given that, \(\frac{dy}{dt}\) + 2y = u(t)sin(2t+A)
Or, y(s).s + 2y(s) = \(\frac{ω}{s^2+4}\)
Or, s + 2 = 0
Or, j ω + 2 = 0
Or, j2 + 2 = 0
Or, 1 + j = 0
Or, tan-1(1) = 45°.
6. The value of 2 [u (t – 1) – u (1 – 2t)] (u ( t + 1) + u (t)) at t = 3 sec is ____________
a) 0
b) 4
c) ∞
d) 1
View Answer
Explanation: Putting t=3 in the given equation, we get,
2[u (2) – u (1 – 6)] [u (4) + u (3)]
= 2 [1 – 0] [1 + 1]
= 4.
7. Barlett’s Bisection Theorem is applicable to ___________
a) Unsymmetrical networks
b) Symmetrical networks
c) Both unsymmetrical and symmetrical networks
d) Neither to unsymmetrical nor to symmetrical networks
View Answer
Explanation: A symmetrical network can be split into two halves.
So the z parameters of the network are symmetrical as well as reciprocal of each other. Hence Barlett’s Bisection Theorem is applicable to Symmetrical networks.
8. The values of z11 and z21 for a T circuit having resistances 20 Ω each is _____________
a) 40, 20
b) 40, 60
c) 60, 40
d) 40, 40
View Answer
Explanation: To determine the values of z11 and z21, we apply a voltage source V1 to the input port and leave the output port open.
Thus, z11 = \(\frac{V_1}{I_1} = \frac{(20+20) I_1}{I_1}\)
= 40 Ω
Now, z21 is the input impedance at port 1.
So, z21 = \(\frac{V_2}{I_1}\) = 20 Ω.
9. The values of z12 and z22 for a T circuit having resistances 20 Ω each is _____________
a) 40, 20
b) 40, 60
c) 20, 40
d) 40, 40
View Answer
Explanation: To find z12 and z22, we apply a voltage source V2 to the output port and leave the input port open.
Thus, z12 = \(\frac{V_1}{I_2} = \frac{(20) I_2}{I_2}\)
= 20 Ω
Now, z22 is the input impedance at port 1.
So, z22 = \(\frac{V_2}{I_2} = \frac{(20+20) I_2}{I_2}\) = 40 Ω.
10. If y(t) = 120e10x(t), then the relation is _________
a) Dynamic
b) Static
c) Memory
d) Memoryless but not static
View Answer
Explanation: Given relation, y (t) = 12010x(t).
The system represented by the above relation is static because the present output of the system as well as memoryless as its present output does not depend on its past input. It is not a dynamic system since the value of the system increases exponentially.
11. The z parameters form a matrix of the form ___________
a) [z11 z12; z21 z22]
b) [z11 z12; z22 z21]
c) [z12 z11; z21 z22]
d) [z11 z22; z12 z21]
View Answer
Explanation: Z parameters are also called as the impedance parameters.
There are 4 main types of Z-parameters, z11, z12, z21, z22
They are arranged in the form of a matrix given by [z11 z12; z21 z22].
12. The Laplace transform of the function e-2tcos(3t) + 5e-2tsin(3t) is ____________
a) \(\frac{(s+2)-15}{(s+2)^2-9}\)
b) \(\frac{(s+2)+15}{(s+2)^2-9}\)
c) \(\frac{(s+2)+15}{(s+2)^2+9}\)
d) \(\frac{(s+2)-15}{(s+2)^2+9}\)
View Answer
Explanation: L {e-2tcos(3t) + 5e-2tsin(3t)} = \(\frac{(s+2)}{(s+2)^2+9} + \frac{5 3}{(s+2)^2+9}\)
= \(\frac{(s+2)+15}{(s+2)^2+9}\).
13. The Laplace transform of the function 6e5tcos(2t) – e7t is ______________
a) \(\frac{6(s-5)}{(s-5)^2+4} – \frac{1}{s-7}\)
b) \(\frac{6(s-5)}{(s-5)^2+4} + \frac{1}{s-7}\)
c) \(\frac{6(s+5)}{(s+5)^2+4} – \frac{1}{s-7}\)
d) \(\frac{6(s+5)}{(s+5)^2+4} + \frac{1}{s-7}\)
View Answer
Explanation: L {6e5tcos(2t) – e7t} = \(\frac{6(s-5)}{(s-5)^2+4} – \frac{1}{s-7}\).
14. The Laplace transform of the function cosh2(t) is ____________
a) \(\frac{s^2+2}{s(s^2+4)}\)
b) \(\frac{s^2-2}{s(s^2-4)}\)
c) \(\frac{s^2-2}{s(s^2+4)}\)
d) \(\frac{s^2+2}{s(s^2-4)}\)
View Answer
Explanation: L ((\(\frac{1}{2}\)(et – e-t))2)
= L \(\left(\frac{e^{2t}}{4} + \frac{1}{2} + \frac{e^{-2t}}{4}\right)\)
= \(\frac{1}{4} \frac{1}{s-2} + \frac{1}{2} \frac{1}{s} + \frac{1}{4} \frac{1}{s+2}\)
= \(\frac{s^2-2}{s(s^2-4)}\).
15. Given a system function H(s) = \(\frac{V(s)}{I(s)} = \frac{(s+4)}{(s+3)^2}\). And i(t) is a unit step, then V(t) in the steady state is ___________
a) \(\frac{4}{9}\)
b) \(\frac{4}{3}\)
c) 0
d) ∞
View Answer
Explanation: V(s) = \(\frac{(s+4)}{(s+3)^2}, I(s) = \frac{(s+4)}{s(s+3)^2}\)
At steady state, sV(s) = \(\frac{0+4}{(0+3)^2}\)
= \(\frac{4}{9}\).
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