Laplace transform of the function f(t) is given by $${\text{F}}\left( {\text{s}} \right) = {\text{L}}\left\{ {{\text{f}}\left( {\text{t}} \right)} \right\} = \int_0^\infty {{\text{f}}\left( {\text{t}} \right){{\text{e}}^{ - {\text{st}}}}{\text{dt}}{\text{.}}} $$ Laplace transform of the function shown below is given by <img src="/images/question-image/engineering-maths/transform-theory/1689254973-laplace-transform-of-the-function.png" title="Transform Theory mcq question image" alt="Transform Theory mcq question image">

Let x(t) be a periodic function with period T = 10. The Fourier series coefficients for this series are denoted by $${a_k}$$ , that is
$$x\left( t \right) = \sum\limits_{k = - \infty }^\infty {{a_k}{e^{jk{{2\pi } \over T}t}}} .$$
The same function x(t) can also be considered as a periodic function with period T' = 40. Let b_k be the Fourier series coefficients when period is taken as T'. If $$\sum\limits_{k = - \infty }^\infty {\left| {{a_k}} \right|} = 16,$$ then $$\sum\limits_{k = - \infty }^\infty {\left| {{b_k}} \right|} $$ is equal to

A

256

B

64

C

16

D

4

For a function g(t), it is given that
$$\int\limits_{ - \infty }^{ + \infty } {g\left( t \right){e^{ - j\omega t}}dt = \omega {e^{ - 2{\omega ^2}}}} $$ for any real value $$\omega $$ . If $$y\left( t \right) = \int\limits_{ - \infty }^t {g\left( \tau \right)} d\tau ,\,{\rm{then}}\,\int\limits_{ - \infty }^{ + \infty } {y\left( t \right)} dt$$ is. . . . . . . .

A

0

B

-j

C

$$ - {j \over 2}$$

D

$${j \over 2}$$

The value of $$\int_0^\infty {\frac{1}{{1 + {{\text{x}}^2}}}} {\text{dx}} + \int_0^\infty {\frac{{\sin {\text{x}}}}{{\text{x}}}} {\text{dx}}$$ is

A

$$\frac{\pi }{2}$$

B

$$\pi $$

C

$$\frac{{3\pi }}{2}$$

D

1

The Laplace transform of the causal periodic square wave of period T shown in the figure below is It

A

$$F\left( s \right) = \frac{1}{{1 + {e^{ - \frac{{sT}}{2}}}}}$$

B

$$F\left( s \right) = \frac{1}{{s\left( {1 + {e^{ - \frac{{sT}}{2}}}} \right)}}$$

C

$$F\left( s \right) = \frac{1}{{s\left( {1 - {e^{ - \frac{{sT}}{2}}}} \right)}}$$

D

$$F\left( s \right) = \frac{1}{{1 - {e^{ - sT}}}}$$

Consider a system whose input r and output y are related by the equation $$y\left( t \right) = \int\limits_{ - \infty }^\infty {x\left( {t - \tau } \right)} h\left( {2\tau } \right)d\tau $$

Where h(t) is shown in the graph
Which of the following four properties are possessed by the system? BIBO: Bounded input gives a bounded output
Causal: The system is causal.
LP : The system is low pass.
LTI: The system is linear and time-invariant.

A

Causal, LP

B

BIBO, LTI

C

BIBO, Causal, LTI

D

LP, LTI

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

$$\frac{1}{s}\left$$

If {x} is a continuous, real valued random variable defined over the interval (-$$\infty $$, +$$\infty $$) and its occurrence is defined by the density function given as:
$${\text{f}}\left( {\text{x}} \right) = \frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}$$ where 'a' and 'b' are the statistical attributes of the random variable {x}. The value of the integral $$\int_{ - \infty }^{\text{a}} {\frac{1}{{\sqrt {2\pi } * {\text{b}}}}{{\text{e}}^{\frac{{ - 1}}{2}{{\left( {\frac{{{\text{x}} - {\text{a}}}}{{\text{b}}}} \right)}^2}}}{\text{dx}}} $$

A

1

B

0.5

C

$$\pi $$

D

$$\frac{\pi }{2}$$

The function f(t) has the Fourier transform f(ω)
The Fourier transform of
$$g\left( t \right) = \left( {\int\limits_{ - \infty }^\infty {g\left( t \right){e^{ - j\omega }}dt} } \right)$$ is

A

2πf(-ω)

B

$${1 \over {2\pi }}f\left( \omega \right)$$

C

$${1 \over {2\pi }}f\left( { - \omega } \right)$$

D

None of these

A cylindrical rod of length L and radius r, made of an inhomogeneous dielectric, is placed with its axis along the Z-direction with one end at the origin as shown below.

If the rod carries a polarization $$\overrightarrow {\bf{P}} = \left( {5{z^2} + 7} \right){\bf{\hat k}},$$ the volume bound charge inside the dielectric is

A

zero

B

10πr2L

C

-5πr2L

D

-5πr2L2

A rod of length L with uniform charge density $$\lambda $$ per unit length is in the XY-plane and rotating about Z-axis passing through one of its edge with an angularvelocity $$\overrightarrow \omega $$ as shown in the figure below. $$\left( {{\bf{\hat r}},\,\hat \phi ,\,{\bf{\hat z}}} \right)$$ refer to the unit vectors at Q, $$\overrightarrow {\bf{A}} $$ is the vector potential at a distance d from the origin O along Z-axis for d ≪ L and $$\overrightarrow {\bf{J}} $$ is the current density due to the motion of the rod. Which one of the following statements is correct?

A

$$\overrightarrow {\bf{J}} {\text{ along }}{\bf{\hat r}};\overrightarrow {\bf{A}} {\text{ along }}{\bf{\hat z}};\left| {\overrightarrow {\bf{A}} } \right| \propto \frac{1}{d}$$

B

$$\overrightarrow {\bf{J}} {\text{ along }}\hat \phi ;\overrightarrow {\bf{A}} {\text{ along }}\hat \phi ;\left| {\overrightarrow {\bf{A}} } \right| \propto \frac{1}{{{d^2}}}$$

C

$$\overrightarrow {\bf{J}} {\text{ along }}{\bf{\hat r}};\overrightarrow {\bf{A}} {\text{ along }}{\bf{\hat z}};\left| {\overrightarrow {\bf{A}} } \right| \propto \frac{1}{{{d^2}}}$$

D

$$\overrightarrow {\bf{J}} {\text{ along }}\hat \phi ;\overrightarrow {\bf{A}} {\text{ along }}\hat \phi ;\left| {\overrightarrow {\bf{A}} } \right| \propto \frac{1}{d}$$