The unilateral Laplace transform of f(t) is $${1 \over {{s^2} + s + 1}}.$$ Which one of the following is the unilateral Laplace transform of g(t) = t.f(t)?

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
Transform Theory mcq question image

A

$$\frac{{1 - {{\text{e}}^{ - 2{\text{s}}}}}}{{\text{s}}}$$

B

$$\frac{{1 - {{\text{e}}^{ - {\text{s}}}}}}{{2{\text{s}}}}$$

C

$$\frac{{2 - 2{{\text{e}}^{ - {\text{s}}}}}}{{\text{s}}}$$

D

$$\frac{{1 - 2{{\text{e}}^{ - {\text{s}}}}}}{{\text{s}}}$$

In which of the following useful signals, is the bilateral Laplace Transform different from the unilateral Laplace Transform?

A

d(t)

B

s(t)

C

u(t)

D

all of the mentioned

Simplify : $${{{5 \over 3} \times {7 \over {51}}{\text{ of }}{{17} \over 5} - {1 \over 3}} \over {{2 \over 9} \times {5 \over 7}{\text{ of }}{{28} \over 5} - {2 \over 3}}}$$

A

$$\frac{1}{2}$$

B

4

C

2

D

$$\frac{1}{4}$$

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

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

Given that F(s) is the one-sided Laplace transform of f(t), the Laplace transform of $$\int\limits_0^t {f\left( \tau \right)} d\tau $$ is

A

sF(s) - f(0)

B

$${1 \over s}F\left( s \right)$$

C

$$\int\limits_0^s {F\left( \tau \right)} d\tau $$

D

$${1 \over s}\left$$

Laplace transform function f (t) is F(S), then how will you represent Laplace transform for differential of f (t)?

A

S⁄F(S)

B

F(S)

C

S.F(S)

D

F’(S)

Function f (t) has Laplace transform F(S). How will you represent Laplace transform of integral of f(t)?

A

S.F(S)

B

F(S)

C

F(S2)

D

F(S)⁄S

If the Laplace transform of $${{\text{e}}^{\omega {\text{t}}}}$$ is $$\frac{1}{{{\text{s}} - \omega }},$$ the Laplace transform of tcosh t is

A

$$\frac{{1 + {{\text{s}}^2}}}{{{{\left( {{{\text{s}}^2} - 1} \right)}^2}}}$$

B

$$\frac{{{\text{st}}}}{{\left( {{{\text{s}}^2} - 1} \right)}}$$

C

$$\frac{{1 - {{\text{s}}^2}}}{{{{\left( {{{\text{s}}^2} - 1} \right)}^2}}}$$

D

$$\frac{{1 + {{\text{s}}^2}}}{{1 - {{\text{s}}^2}}}$$

Laplace transform of cos (ωt) is $$\frac{{\text{s}}}{{{{\text{s}}^2} + {\omega ^2}}}.$$ The Laplace transform of e^-2t cos(4t) is

A

$$\frac{{{\text{s}} - 2}}{{{{\left( {{\text{s}} - 2} \right)}^2} + 16}}$$

B

$$\frac{{{\text{s}} + 2}}{{{{\left( {{\text{s}} - 2} \right)}^2} + 16}}$$

C

$$\frac{{{\text{s}} - 2}}{{{{\left( {{\text{s}} + 2} \right)}^2} + 16}}$$

D

$$\frac{{{\text{s}} + 2}}{{{{\left( {{\text{s}} + 2} \right)}^2} + 16}}$$

The Laplace transform of a function f(t) u(t), where f(t) is periodic with period T, is A(s) times the Laplace transform of its first period. Then

A

A(s) = s

B

$$A\left( s \right) = {1 \over {\left( {1 - \exp \left( { - {T_s}} \right)} \right)}}$$

C

$$A\left( s \right) = {1 \over {\left( {1 + \exp \left( { - {T_s}} \right)} \right)}}$$

D

A(s) = exp(T_s)

The unilateral Laplace transform of f(t) is $${1 \over {{s^2} + s + 1}}.$$ Which one of the following is the unilateral Laplace transform of g(t) = t.f(t)?

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