Transform Theory MCQs Question and Answer in Engineering Maths

31

The inverse Laplace transform of the function $${\text{F}}\left( {\text{s}} \right) = \frac{1}{{{\text{s}}\left( {{\text{s}} + 1} \right)}}$$ is given by

A) f(t) = sin t B) f(t) = e^-t sin t C) f(t) = e^-t D) f(t) = 1 - e^-t

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Answer & Solution

Answer: Option D

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32

Laplace transform analysis gives

A) Time domain response only B) Frequency domain response only C) Both A and B D) None of the above

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Answer & Solution

Answer: Option C

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33

Inverse Laplace transform of the function $$\frac{s}{{{s^2} + 3s + 2}},$$ is

A) $$ - {e^{ - t}} + 2{e^{ - 2t}}$$ B) $${e^{ - t}} - 2{e^{ - 2t}}$$ C) $${e^{ - t}} + 2{e^{ - 2t}}$$ D) $$2{e^{ - t}} + {e^{ - 2t}}$$

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Answer & Solution

Answer: Option A

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34

Which of the following is the advantage of using Laplace transform techniques?

A) Permits use of simple algebra B) Converts functions in the $$t$$-domain into $$s$$-domain C) Initial conditions are automatically taken care of D) All of the above

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Answer & Solution

Answer: Option D

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35

The initial value theorem does not hold good for which of the following functions?

A) Ramp function B) Delta function C) Step function D) Hyperbolic function

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Answer & Solution

Answer: Option B

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36

Find the inverse Laplace transform of $$F\left( s \right) = \frac{{{s^2} + 2s - 2}}{{s\left( {s + 4} \right)\left( {s - 5} \right)}}$$

A) $$\left( {\frac{1}{{10}} + \frac{1}{6}{e^{ - 4t}} + \frac{{11}}{{15}}{e^{5t}}} \right)u\left( t \right)$$ B) $$\left( {\frac{1}{{10}} + \frac{1}{6}{e^{6t}} + \frac{{10}}{{15}}{e^{5t}}} \right)u\left( t \right)$$ C) $$\left( {1 + \frac{1}{6}{e^{ - 4t}} + \frac{{10}}{{15}}{e^{5t}}} \right)u\left( t \right)$$ D) $$\left( {\frac{1}{{10}} - \frac{1}{6}{e^{4t}} + \frac{{10}}{{15}}{e^{ - 5t}}} \right)u\left( t \right)$$

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Answer & Solution

Answer: Option A

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37

Give transfer function $$H\left( s \right) = \frac{{s + 2}}{{{s^2} + s + 4}},$$ under steady state condition, the sinusoidal input and output are, respectively x(t) = cos 2t, y(t) = cos(2t + $$\phi $$), then angle $$\phi $$ will be

A) 45° B) 0° C) -45° D) -90°

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Answer & Solution

Answer: Option C

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38

Which of the following correctly defines Laplace transform of a function in the time domain?

A) $$L\left\{ {f\left( t \right)} \right\} = \int_{{0^ - }}^\infty {f\left( t \right){e^{ - st}}dt} $$ B) $$L\left\{ {f\left( t \right)} \right\} = \int_{{0^ - }}^\infty {f\left( t \right){e^{ + st}}dt} $$ C) $$L\left\{ {f\left( t \right)} \right\} = \int_{{0^ - }}^\infty {f{{\left( t \right)}^{ - st}}{e^{ - st}}dt} $$ D) $$L\left\{ {f\left( t \right)} \right\} = \int_{{0^ - }}^\infty {f\left( s \right){e^{ - st}}dt} $$

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Answer & Solution

Answer: Option A

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39

The Laplace transform of $$I\left( t \right)$$ is given by $$I\left( s \right) = \frac{5}{{s\left( {{s^2} + 2} \right)}}.$$ As $$t \to \infty $$ the value of $$I\left( t \right)$$ tends to

A) 0 B) 1 C) $$\frac{5}{2}$$ D) $$\infty $$

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Answer & Solution

Answer: Option C

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40

Given that $$F\left( s \right)$$ is the one-side Laplace transform of $$f\left( t \right),$$ the Laplace transform of $$\int_0^t {f\left( \tau \right)d\tau } $$ is

A) $$sF\left( s \right) - f\left( 0 \right)$$ B) $$\frac{1}{s}F\left( s \right)$$ C) $$\int_0^5 {F\left( \tau \right)d\tau } $$ D) $$\frac{1}{s}\left[ {F\left( s \right) - f\left( 0 \right)} \right]$$

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Answer & Solution

Answer: Option B

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