# Network Theory Questions and Answers – Definition of the Laplace Transform

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?
a) $$\int_0^{\infty}$$ f(t) e-st
b) $$\int_{-\infty}^0$$ f(t) e-st
c) $$\int_0^{\infty}$$ f(t) est
d) $$\int_{-\infty}^0$$ f(t) est

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

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) – ∞

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

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 f1 (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)]

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, f1 (t) = 10t for 0 < t < 1. In terms of unit step function, f1 (t) = 10t [u (t) – u (t – 1)].

6. Find the function f2 (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)]

Explanation: From the graph, f2 (t) = -10t + 20 for 1 < t < 3. In terms of unit step function, f2 (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 f3 (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)]

Explanation: From the graph, f3 (t) = 20t – 40 for 3 < t < 4. In terms of unit step function, f3 (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 below.

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)]

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

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 below.

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)]

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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