This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Definition of the Laplace Transform”.

1. The Laplace transform of a function f (t) is?

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Explanation: The Laplace transform is a powerful analytical technique that is widely used to study the behavior of linear, lumped parameter circuits. L(f(t)) = F (s)

2. Laplace transform changes the ____ domain function to the _____ domain function.

a) time, time

b) time, frequency

c) frequency, time

d) frequency, frequency

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Explanation: Laplace transform changes the time domain function f (t) to the frequency domain function F(s). Similarly Laplace transformation converts frequency domain function F(s) to the time domain function f(t).

3. In the bilateral Laplace transform, the lower limit is?

a) 0

b) 1

c) ∞

d) – ∞

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Explanation: If the lower limit is 0, then the transform is referred to as one-sided or unilateral Laplace transform. In the two-sided or bilateral Laplace transform, the lower limit is – ∞.

4. The unit step is not defined at t =?

a) 0

b) 1

c) 2

d) 3

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Explanation: If k is 1, the function is defined as unit step function. And the unit step is not defined at t =0.

5. The total period of the function shown in the figure is 4 sec and the amplitude is 10. Find the function f_{1} (t) from t = 0 to 1 in terms of unit step function.

a) 10t [u (t) – u (t + 1)]

b) 10t [u (t) + u (t – 1)]

c) 10t [u (t) + u (t + 1)]

d) 10t [u (t) – u (t – 1)].

View Answer

Explanation: The function shown in the figure is made up of linear segments with break points at 0, 1,3 and 4 seconds. From the graph, f

_{1}(t) = 10t for 0 < t < 1. In terms of unit step function, f

_{1}(t) = 10t [u (t) – u (t – 1)].

6. Find the function f_{2} (t) from the time t = 1 to 3 sec.

a) (-10t+20) [u (t-1) +u (t-3)]

b) (-10t+20) [u (t-1) – u (t-3)]

c) (-10t-20) [u (t-1) + u (t-3)]

d) (-10t-20) [u (t-1) – u (t-3)].

View Answer

Explanation: From the graph, f

_{2}(t) = -10t + 20 for 1 < t < 3. In terms of unit step function, f

_{2}(t) = (-10t+20) [u (t-1) – u (t-3)]. This function turn off at t = 1, turn off at t = 3.

7. Find the function f_{3} (t) from the time t = 3 to 4 sec.

a) (20t – 40) [u (t-3) – u (t-4)]

b) (20t + 40) [u (t-3) – u (t-4)]

c) (20t + 40) [u (t-3) + u (t-4)]

d) (20t – 40) [u (t-3) + u (t-4)].

View Answer

Explanation: From the graph, f

_{3}(t) = 20t – 40 for 3 < t < 4. In terms of unit step function, f

_{3}(t) = (20t – 40) [u (t-3) – u (t-4)]. This function turn off at t = 3, turn off at t = 4.

8. Find the expression of f (t) in the graph shown in question 5.

a) 10t [u (t) – u (t – 1)] – (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)]

b) 10t [u (t) – u (t – 1)] – (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)]

c) 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)]

d) 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] – (20t – 40) [u (t-3) – u (t-4)] View Answer

Explanation: We use the step function to initiate and terminate these linear segments at the proper times. The expression of f (t) is f (t) = 10t [u (t) – u (t – 1)] + (-10t+20) [u (t-1) – u (t-3)] + (20t – 40) [u (t-3) – u (t-4)].

9. In the graph shown below, find the expression f (t).

a) 2t

b) 3t

c) 4t

d) 5t

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Explanation: The waveform shown in the figure starts at t = 0 and ends at t = 5 sec. The equation for the above waveform is f (t) = 4t.

10. Find the function f (t) in terms of unit step function in the graph shown in question 9.

a) 4t [u (t) – u (t + 5)]

b) 4t [u (t) + u (t + 5)]

c) 4t [u (t) – u (t – 5)]

d) 4t [u (t) + u (t – 5)].

View Answer

Explanation: The waveform shown in the figure starts at t = 0 and ends at t = 5 sec. In terms of unit step function the waveform can be expressed as f (t) = 4t [u (t) – u (t – 5)].

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