Transform Theory MCQ

Test your knowledge with important Transform Theory MCQ and their applications. These MCQs are beneficial for competitive exams too. Explore 30 + more Transform Theory MCQs at Bissoy

• The inverse Laplace transform of the function $$\frac{{s + 5}}{{\left( {s + 1} \right)\left( {s + 3} \right)}}$$   is
  ক

  $$2{e^{ - t}} - {e^{ - 3t}}$$

  খ

  $$2{e^{ - t}} - 2{e^{ - 3t}}$$

  গ

  $${e^{ - t}} - 2{e^{ - 3t}}$$

  ঘ

  $${e^{ - t}} + {e^{ - 3t}}$$
• The trigonometric Fourier series of an even function does not have the
  ক

  DC term

  খ

  cosine terms

  গ

  sine terms

  ঘ

  odd harmonic terms
• The Laplace transform of $$x\left( t \right)$$  is $$X\left( s \right) = \frac{{2{s^2} + 3}}{{{s^2} + 5s + 1}}$$<br>The initial value of $$x\left( t \right)$$  is
  ক

  0

  খ

  2

  গ

  3

  ঘ

  Does not exist
• The Laplace transform of $$x\left( t \right)$$  is $$X\left( s \right) = \frac{{{e^{ - 3s}}\left( {2{s^2} + 1} \right)}}{{s\left( {s + 1} \right)\left( {s + 4} \right)}}$$<br>The final value of $$x\left( t \right)$$  is
  ক

  2

  খ

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

  গ

  -3

  ঘ

  Does not exist
• The bilateral Laplace transform of $${e^{\left( {3t + 6} \right)}}u\left( {t + 3} \right)$$   is
  ক

  $$\frac{{{e^{3s}}}}{{s - 3}},\,\,\,\,\,\operatorname{Re} \left( s \right) &gt; 3$$

  খ

  $$\frac{{{e^{3s}}}}{{s - 3}},\,\,\,\,\,\operatorname{Re} \left( s \right)

  গ

  $$\frac{{{e^{3\left( {s - 1} \right)}}}}{{s - 3}},\,\,\,\,\,\operatorname{Re} \left( s \right) &gt; 3$$

  ঘ

  $$\frac{{{e^{3\left( {s - 1} \right)}}}}{{s - 3}},\,\,\,\,\,\operatorname{Re} \left( s \right)
• The Laplace transform of $$x\left( t \right)$$  is $$X\left( s \right) = {e^{ - 2s}}\frac{{6{s^2} + s}}{{{s^2} + 2s - 2}}$$<br>The initial value of $$x\left( t \right)$$  is
  ক

  6

  খ

  2

  গ

  3

  ঘ

  0
• The bilateral Laplace transform of u(-t + 3) is
  ক

  $$\frac{{1 - {e^{ - 3s}}}}{s},\,\,\,\,\,\operatorname{Re} \left( s \right) &gt; 0$$

  খ

  $$\frac{{ - {e^{ - 3s}}}}{s},\,\,\,\,\,\operatorname{Re} \left( s \right)

  গ

  $$1 - \frac{{{e^{ - 3s}}}}{s},\,\,\,\,\,\operatorname{Re} \left( s \right)

  ঘ

  $$\frac{{ - {e^{ - 3s}}}}{s},\,\,\,\,\,\operatorname{Re} \left( s \right) &gt; 0$$
• The bilateral Laplace transform of $${e^t}\cos 2tu\left( { - t} \right) + {e^{ - t}}u\left( t \right) + {e^{\frac{t}{2}}}u\left( t \right)$$      is
  ক

  [ <p><span>A. $$\frac{{1 - s}}{{{{\left( {s - 1} \right)}^2} + 4}} + \frac{1}{{s + 1}} + \frac{1}{{s - 0.5}},\,0.5 <p><span>B. $$\frac{{1 - s}}{{{{\left( {s - 1} \right)}^2} + 4}} + \frac{1}{{s + 1}} + \frac{1}{{s - 0.5}},\, - 1

  খ

  $$\frac{{s - 1}}{{{{\left( {s - 1} \right)}^2} + 4}} + \frac{1}{{s + 1}} + \frac{1}{{s - 0.5}},\,0.5

  গ

  $$\frac{{s - 1}}{{{{\left( {s - 1} \right)}^2} + 4}} + \frac{1}{{s + 1}} + \frac{1}{{s - 0.5}},\, - 1
• The bilateral Laplace transform of $${e^{ - 1}}u\left( {t + 2} \right)$$   is
  ক

  $$\frac{{{e^{2\left( {s + 1} \right)}}}}{{s + 1}},\,\,\,\,\operatorname{Re} \left( s \right) &gt; - 1$$

  খ

  $$\frac{1}{{1 + s}},\,\,\,\,\operatorname{Re} \left( s \right)

  গ

  $$\frac{{{e^{2\left( {s + 1} \right)}}}}{{s + 1}},\,\,\,\,\operatorname{Re} \left( s \right)

  ঘ

  $$\frac{1}{{1 + s}},\,\,\,\,\operatorname{Re} \left( s \right) &gt; - 1$$
• The Laplace transform of $$i\left( t \right)$$  is given by $$I\left( s \right) = \frac{2}{{s\left( {1 + s} \right)}}.$$    As $$t \to \infty $$  the value of $$i\left( t \right)$$  tends to
  ক

  0

  খ

  1

  গ

  2

  ঘ

  $$\infty $$
• 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
  ক

  $$sF\left( s \right) - f\left( 0 \right)$$

  খ

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

  গ

  $$\int_0^5 {F\left( \tau \right)d\tau } $$

  ঘ

  $$\frac{1}{s}\left$$
• 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
  ক

  0

  খ

  1

  গ

  $$\frac{5}{2}$$

  ঘ

  $$\infty $$
• Which of the following correctly defines Laplace transform of a function in the time domain?
  ক

  $$L\left\{ {f\left( t \right)} \right\} = \int_{{0^ - }}^\infty {f\left( t \right){e^{ - st}}dt} $$

  খ

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

  গ

  $$L\left\{ {f\left( t \right)} \right\} = \int_{{0^ - }}^\infty {f{{\left( t \right)}^{ - st}}{e^{ - st}}dt} $$

  ঘ

  $$L\left\{ {f\left( t \right)} \right\} = \int_{{0^ - }}^\infty {f\left( s \right){e^{ - st}}dt} $$
• 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
  ক

  45°

  খ

  0°

  গ

  -45°

  ঘ

  -90°
• Find the inverse Laplace transform of $$F\left( s \right) = \frac{{{s^2} + 2s - 2}}{{s\left( {s + 4} \right)\left( {s - 5} \right)}}$$
  ক

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

  খ

  $$\left( {\frac{1}{{10}} + \frac{1}{6}{e^{6t}} + \frac{{10}}{{15}}{e^{5t}}} \right)u\left( t \right)$$

  গ

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

  ঘ

  $$\left( {\frac{1}{{10}} - \frac{1}{6}{e^{4t}} + \frac{{10}}{{15}}{e^{ - 5t}}} \right)u\left( t \right)$$
• The initial value theorem does not hold good for which of the following functions?
  ক

  Ramp function

  খ

  Delta function

  গ

  Step function

  ঘ

  Hyperbolic function
• Which of the following is the advantage of using Laplace transform techniques?
  ক

  Permits use of simple algebra

  খ

  Converts functions in the $$t$$-domain into $$s$$-domain

  গ

  Initial conditions are automatically taken care of

  ঘ

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

  $$ - {e^{ - t}} + 2{e^{ - 2t}}$$

  খ

  $${e^{ - t}} - 2{e^{ - 2t}}$$

  গ

  $${e^{ - t}} + 2{e^{ - 2t}}$$

  ঘ

  $$2{e^{ - t}} + {e^{ - 2t}}$$
• Laplace transform analysis gives
  ক

  Time domain response only

  খ

  Frequency domain response only

  গ

  Both A and B

  ঘ

  None of the above
• 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
  ক

  f(t) = sin t

  খ

  f(t) = e^-t sin t

  গ

  f(t) = e^-t

  ঘ

  f(t) = 1 - e^-t
• Laplace transform of 8t³
  ক

  $$\frac{8}{{{{\text{S}}^4}}}$$

  খ

  $$\frac{{16}}{{{{\text{S}}^4}}}$$

  গ

  $$\frac{{24}}{{{{\text{S}}^4}}}$$

  ঘ

  $$\frac{{48}}{{{{\text{S}}^4}}}$$
• The Fourier series expansion of the saw-toothed waveform f(x) = x in (-π, π) of period 2π gives the series, $$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \,....$$<br>The sum is equal to
  ক

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

  খ

  $$\frac{{{\pi ^2}}}{4}$$

  গ

  $$\frac{{{\pi ^2}}}{{16}}$$

  ঘ

  $$\frac{\pi }{4}$$
• 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<br><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">
  ক

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

  খ

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

  গ

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

  ঘ

  $$\frac{{1 - 2{{\text{e}}^{ - {\text{s}}}}}}{{\text{s}}}$$
• Given the Fourier series in (-π, π) for f(x) = x cosx, the value of a₀ will be
  ক

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

  খ

  0

  গ

  2

  ঘ

  $$\frac{{{{\left( { - 1} \right)}^{\text{n}}}2{\text{n}}}}{{{{\text{n}}^2} - 1}}$$
• Let \[{\text{f}}\left( {\text{x}} \right) = \left\{ {\begin{array}{*{20}{c}} { - \pi ,\,\,{\text{if}}}&amp;{ - \pi
  ক

  $$\frac{4}{5}$$

  খ

  $$\frac{5}{4}$$

  গ

  $$\frac{4}{3}$$

  ঘ

  $$\frac{3}{4}$$
• The inverse Laplace transform of $${\text{H}}\left( {\text{s}} \right) = \frac{{{\text{s}} + 3}}{{{{\text{s}}^2} + 2{\text{s}} + 1}}$$    for t ≥ 0 is
  ক

  2t e^-t + e^-t

  খ

  3e^-t

  গ

  3t e^-t + e^-t

  ঘ

  4t e^-t + e^-t
• The Laplace transform of a function f(t) is $$\frac{1}{{{{\text{s}}^2}\left( {{\text{s}} + 1} \right)}}.$$  The function f(t) is
  ক

  t - 1 + e^-1

  খ

  t + 1 + e^-1

  গ

  -1 + e^-1

  ঘ

  2t + e^t
• Laplace transform of cos (ωt) is
  ক

  $$\frac{{\text{s}}}{{{{\text{s}}^2} + {\omega ^2}}}$$

  খ

  $$\frac{\omega }{{{{\text{s}}^2} + {\omega ^2}}}$$

  গ

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

  ঘ

  $$\frac{\omega }{{{{\text{s}}^2} - {\omega ^2}}}$$
• Laplace transform of sin ht is
  ক

  $$\frac{1}{{{{\text{S}}^2} - 1}}$$

  খ

  $$\frac{1}{{1 - {{\text{S}}^4}}}$$

  গ

  $$\frac{{\text{S}}}{{{{\text{S}}^4} - 1}}$$

  ঘ

  $$\frac{{\text{S}}}{{1 - {{\text{S}}^4}}}$$
• The Fourier series of the function,<br>\[\begin{array}{*{20}{c}} {{\text{f}}\left( {\text{x}} \right) = 0,}&amp;{ - \pi The convergence of the above Fourier series at x = 0 gives
  ক

  $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{1}{{{{\text{n}}^2}}} = \frac{{{\pi ^2}}}{6}} $$

  খ

  $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{{\text{n}} + 1}}}}{{{{\text{n}}^2}}} = \frac{{{\pi ^2}}}{{12}}} $$

  গ

  $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{1}{{{{\left( {{\text{2n}} - 1} \right)}^2}}} = \frac{{{\pi ^2}}}{8}} $$

  ঘ

  $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{{\text{n}} + 1}}}}{{{\text{2n}} - 1}} = \frac{\pi }{4}} $$