# Network Theory Questions and Answers – Problems on Initial and Final Value Theorem

This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Problems on Initial and Final Value Theorem”.

1. What is the steady state value of F (t), if it is known that F(s) = $$\frac{2}{s(S+1)(s+2)(s+3)}$$?
a) $$\frac{1}{2}$$
b) $$\frac{1}{3}$$
c) $$\frac{1}{4}$$
d) Cannot be determined

Explanation: From the equation of F(s), we can infer that, a simple pole is at origin and all other poles are having negative real part.
∴ F(∞) = lims→0 s F(s)
= lims→0 $$\frac{2s}{s(S+1)(s+2)(s+3)}$$
= $$\frac{2}{(s+1)(s+2)(s+3)}$$
= $$\frac{2}{6} = \frac{1}{3}$$.

2. What is the steady state value of F (t), if it is known that F(s) = $$\frac{1}{(s-1)(s+2)}$$?
a) 1
b) –$$\frac{1}{2}$$
c) $$\frac{1}{2}$$
d) Cannot be determined

Explanation: The steady state value of this Laplace transform is cannot be determined since; F(s) has a pole s = 1. Hence the answer is that it cannot be determined.

3. What is the steady state value of F (t), if it is known that F(s) = $$\frac{1}{(s+2)^2 (s+4)}$$?
a) $$\frac{1}{16}$$
b) Cannot be determined
c) 0
d) $$\frac{1}{8}$$

Explanation: The steady state value of F(s) exists since all poles of the given Laplace transform have negative real part.
∴F(∞) = lims→0 s F(s)
= lims→0 $$\frac{s}{(s+2)^2 (s+4)}$$
= 0.

4. What is the steady state value of F (t), if it is known that F(s) = $$\frac{10}{(s+1)(s^2+1)}$$?
a) -5
b) 5
c) 10
d) Cannot be determined

Explanation: The steady state value of this Laplace transform is cannot be determined since; F(s) is having two poles on the imaginary axis (j and –j). Hence the answer is that it cannot be determined.

5. What is the steady state value of F (t), if it is known that F(s) = b/(s(s+1)(s+a)), where a>0?
a) $$\frac{b}{a}$$
b) $$\frac{a}{b}$$
c) 1
d) Cannot be determined

Explanation: F (∞) = lims→0 s F(s)
= lims→0 $$\frac{sb}{s(s+1)(s+a)}$$
= lims→0 $$\frac{b}{(s+1)(s+a)}$$
= $$\frac{b}{a}$$.
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6. The inverse Laplace transform of F(s) = $$\frac{2}{s^2+3s+2}$$ is ______________
a) -2e-2t + 2e-t
b) 2e-2t + 2e-t
c) -2e-2t – 2e-t
d) 2e-t + e-2t

Explanation: s2 + 3s + 2 = (s+2) (s+1)
Now, F(s) = $$\frac{A}{(s+2)} + \frac{B}{(s+1)}$$
Hence, A = (s+2) F(s) |s=-2
= $$\frac{2}{s+1}$$|s=-2 = -2
And, B = (s+1) F(s) |s=-1
= $$\frac{2}{s+2}$$|s=-1 = 2
∴ F(s) = $$\frac{-2}{(s+2)} + \frac{2}{(s+1)}$$
∴ F (t) = L-1{F(s)}
= -2e-2t + 2e-t for t≥0.

7. The inverse Laplace transform of F(s) = $$\frac{2}{s+c}e^{-bs}$$ is _____________
a) 2e-k (t+b) u (t+b)
b) 2e-k (t-b) u (t-b)
c) 2ek (t-b) u (t-b)
d) 2ek (t-b) u (t+b)

Explanation: Let G(s) = $$\frac{2}{s+c}$$
Or, G (t) = L-1{G(s)} = 2e-ct
∴ F (t) = L-1{G(s) e-bs}
= 2e-k (t-b) u (t-b).

8. The inverse Laplace transform of F(s) = $$\frac{3s+5}{s^2+7}$$ is _____________
a) 3 sin ($$\sqrt{7}$$t) – $$\frac{5}{\sqrt{7}}$$ cos ($$\sqrt{7}$$ t)
b) 3 sin ($$\sqrt{7}$$t) + $$\frac{5}{\sqrt{7}}$$ cos ($$\sqrt{7}$$ t)
c) 3 cos ($$\sqrt{7}$$t) + $$\frac{5}{\sqrt{7}}$$ sin ($$\sqrt{7}$$ t)
d) 3 cos ($$\sqrt{7}$$t) – $$\frac{5}{\sqrt{7}}$$ sin ($$\sqrt{7}$$ t)

Explanation: F (t) = L-1{F(s)}
= L-1$$\Big\{\frac{3s}{s^2+7} + \frac{5}{s^2+7}\Big\}$$
= L-1$$\Big\{\frac{3s}{s^2 + \sqrt{7}^2} + \frac{5}{\sqrt{7}} \frac{\sqrt{7}}{s^2 +\sqrt{7}^2}\Big\}$$
The inverse Laplace transform is = 3 cos ($$\sqrt{7}$$t) + $$\frac{5}{\sqrt{7}}$$ sin ($$\sqrt{7}$$ t).

9. The inverse Laplace transform of F(s) = $$\frac{5}{s^2-9}$$ is ______________
a) $$\frac{1}{6}e^{3t} + \frac{5}{6e^{-3t}}$$
b) $$\frac{1}{6}e^{3t} – \frac{5}{6e^{-3t}}$$
c) $$\frac{5}{6}e^{3t} + \frac{5}{6e^{-3t}}$$
d) $$\frac{5}{6}e^{3t} – \frac{5}{6e^{-3t}}$$

Explanation: F (t) = L-1{F(s)}
= L-1{$$\frac{5}{s^2-9}$$}
= L-1$${\frac{A}{s-3} + \frac{B}{s+3}}$$
Hence, A = (s-3) $$\frac{5}{(s-3)(s+3)}|_{s=3} = \frac{5}{6}$$
And, B = (s+3) $$\frac{5}{(s-3)(s+3)}|_{s=-3} = -\frac{5}{6}$$
The inverse Laplace transform is $$\frac{5}{6}e^{3t} – \frac{5}{6e^{-3t}}$$.

10. The inverse Laplace transform of F(s) = $$\frac{e^{-3s}}{s(s^2+3s+2)}$$ is ______________
a) {0.5 + 0.5e-2(t+3) – e-(t+3)} u (t+3)
b) {0.5 + 0.5e-2(t+3) – e-(t-3)} u (t-3)
c) {0.5 – 0.5e-2(t-3) – e-(t-3)} u (t-3)
d) 0.5 + 0.5e-2t – e-t)

Explanation: Let G(s) = $$\frac{1}{s(s^2+3s+2)}$$
Or, F(s) = G(s) e-3s
G (t) = L-1{G(s)}
= L-1$${\frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+1}}$$
Solving we get, A = 0.5, B = 0.5, C = -1
So, G (t) = 0.5 + 0.5e-2t-e-t
The inverse Laplace transform is F (t) = {0.5 + 0.5e-2(t+3) – e-(t-3)} u (t-3).

11. Given a sinusoidal voltage that has a peak to peak value of 100 V. The RMS value of the sinusoidal voltage is ___________
a) 50 V
b) 70.7 V
c) 35.35 V
d) 141.41 V

Explanation: Given that,
Peak value = 50 V
We know that, RMS value = $$\frac{Peak \,Value}{\sqrt{2}}$$
= $$\frac{50}{\sqrt{2}}$$ = 35.35 V.

12. The Laplace transform of F(t) = sin(2t)cos(2t) is ______________
a) $$\frac{4}{2(s^2+16)}$$
b) $$\frac{1}{s+4} – \frac{2}{s+2} + \frac{1}{s}$$
c) $$\frac{2}{s^3}$$
d) $$\frac{1}{2s} + \frac{s}{2(s^2+36)}$$

Explanation: Using the Trigonometric Identity,
We get, sin (2t) cos (2t) = $$\frac{1}{2}$$ (sin (4t))
∴L {sin (2t) cos (2t)} = $$\frac{4}{2(s^2+16)}$$.

13. Convolution of step signal 100 times that is 100 convolution operations. The Laplace transform is ______________
a) $$\frac{1}{s^{100}}$$
b) $$\frac{1}{s^{50}}$$
c) 1
d) s100

Explanation: n times = u (t) * u (t) * …… * u (t)
Laplace transform of the above function = $$\frac{1}{s^n}$$, where n is number of convolutions.
∴ Laplace transform for 100 convolutions = $$\frac{1}{s^{100}}$$.

14. For the circuit given below, the Time-constant is __________

a) 1.5
b) 1.25
c) 2.5
d) 2.25

Explanation: We know that,
Time constant is given by Req.C
The equivalent resistance is given by,
Req = R || R
= $$\frac{R*R}{R+R}$$
= $$\frac{R}{2}$$
= 5 Ω
So, Time-constant = $$\frac{5X0.5}{2}$$
= 1.25.

15. In a dual slop integrating type digital voltmeter, the first integrating is carried out for 50 periods of the supply frequency of 50 Hz. If the reference voltage used is 10 V, the total conversion time for an input of 40 V is?
a) 3 s
b) 2 s
c) 4 s
d) 1 s

Explanation: In a dual slope integrating digital voltmeter,
$$(\frac{t_1}{t_2})$$ Vin = Vref
Where, t1 = first integration time = 50 × $$\frac{1}{50}$$ = 1
But Vin = 40 V and Vref = 10 V
∴ t2 = $$\frac{V_{in} t_1}{V_{ref}}$$ = 4 s.

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