Laplace Transform MCQ Quiz in मराठी - Objective Question with Answer for Laplace Transform - मोफत PDF डाउनलोड करा

Last updated on Mar 9, 2025

पाईये Laplace Transform उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Laplace Transform एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Laplace Transform MCQ Objective Questions

Top Laplace Transform MCQ Objective Questions

Laplace Transform Question 1:

What is the Laplace transform of v(t) = sin (10t) u(t) ?

  1. \(V(s) = \frac{{10}}{{{s^2} + 10}}\)
  2. \(V(s) = \frac{{1}}{{{s^2} + 10}}\)
  3. \(V(s) = \frac{{1}}{{{s^2} + 100}}\)
  4. \(V(s) = \frac{{10}}{{{s^2} + 100}}\)

Answer (Detailed Solution Below)

Option 4 : \(V(s) = \frac{{10}}{{{s^2} + 100}}\)

Laplace Transform Question 1 Detailed Solution

Concept:

L [f(t)] = F(s)

L [sin(at) u(t)] ↔ \(\frac{a}{{{s^2} + {a^2}}}\)

Application:

With v(t) = sin (10t) u(t), the Laplace transform will be:

\(V(s) = \frac{{10}}{{{s^2} + 100}}\)

Important Points 

Some common Laplace transforms are:

f(t)

F(s)

ROC

δ (t)

1

All s

u(t)

\(\frac{1}{s}\)

Re (s) > 0

t

\(\frac{1}{{{s^2}}}\)

Re (s) > 0

tn

\(\frac{{n!}}{{{s^{n + 1}}}}\)

Re (s) > 0

e-at

\(\frac{1}{{s + a}}\)

Re (s) > -a

t e-at

\(\frac{1}{{{{\left( {s + a} \right)}^2}}}\)

Re (s) > -a

tn e-at

\(\frac{{n!}}{{{{\left( {s + a} \right)}^n}}}\)

Re (s) > -a

sin at

\(\frac{a}{{{s^2} + {a^2}}}\)

Re (s) > 0

cos at

\(\frac{s}{{{s^2} + {a^2}}}\)

Re (s) > 0

Laplace Transform Question 2:

Inverse Laplace transform of \(F\left( s \right) = \frac{{\left( {s + 2} \right)}}{{s\left( {s + 3} \right)\left( {s + 4} \right)}}\) would be

  1. (1/6) + (1/3e-3t) + (1/2e4t)
  2. (1/6) – (1/3e-3t) + (1/2e-4t)
  3. (1-3e-3t) + (1/2e4t)
  4. (1/6) + (1/3e-3t) – (1/2e-4t)

Answer (Detailed Solution Below)

Option 4 : (1/6) + (1/3e-3t) – (1/2e-4t)

Laplace Transform Question 2 Detailed Solution

Concept:

\({e^{ - at}}u\left( t \right) \to \frac{1}{{s + a}}\;\;\;{R_e}\left\{ s \right\} > - a\) 

\(u\left( t \right) \leftrightarrow \frac{1}{s}\;\;\;{R_e}\left\{ s \right\} > - a\)

Calculation:

Given:

\(F\left( s \right) = \frac{{\left( {s + 2} \right)}}{{s\left( {s + 3} \right)\left( {s + 4} \right)}}\) 

\(F\left( s \right) = \frac{{s + 2}}{{s\left( {s + 3} \right)\left( {s + 4} \right)}}\)

\(F\left( s \right) = \frac{A}{s} + \frac{B}{{s + 3}} + \frac{C}{{s + 4}}\) 

After Partial fraction we’ll get;

\(f\left( s \right) = \frac{1}{{6s}} + \frac{1}{{3\left( {s + 3} \right)}} - \frac{1}{{2\left( {s + 4} \right)}}\) 

Now, Taking Laplace inverse of F(s);

\(f\left( t \right) = \frac{1}{6} + \frac{1}{3}{e^{ - 3t}} - \frac{1}{2}{e^{ - 4t}}\) 

Laplace Transform Question 3:

What is Laplace transform of function e-5t cos 4t?

  1. \(\frac{{\left( {s + 4} \right)}}{{\left( {{{\left( {s + 4} \right)}^2} + 25} \right)}}\)
  2. \(\frac{5}{{\left( {{{\left( {s + 4} \right)}^2} + 25} \right)}}\)
  3. \(\frac{4}{{\left( {{{\left( {s + 5} \right)}^2} + 16} \right)}}\)
  4. \(\frac{{s + 5}}{{\left( {{{\left( {s + 5} \right)}^2} + 16} \right)}}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{{s + 5}}{{\left( {{{\left( {s + 5} \right)}^2} + 16} \right)}}\)

Laplace Transform Question 3 Detailed Solution

Concept:

Let the Laplace transform of function of f(t) is L [f(t)] = F (s)

By using first shifting rule

 If L [f(t)] = F (s), then

L [eat f(t)] = F (s – a)

\(L\left( \cos at \right)=~\frac{s}{\left( {{s}^{2}}+{{a}^{2}} \right)}\)

Calculation:

Laplace transform of y(t) = e-3t cos 5t

\(L\left( \cos 4t \right)=~\frac{s}{\left( {{s}^{2}}+{{4}^{2}} \right)}\)

By applying the property of shifting,

\(L\left( {{\mathbf{e}}^{-5\mathbf{t}}}~\mathbf{cos}~4\mathbf{t} \right)=~\frac{\left( s+5 \right)}{{{\left( s+5 \right)}^{2}}+{{4}^{2}}}\\=\frac{\left( s+5 \right)}{{{\left( s+5 \right)}^{2}}+{16}}\)

Laplace Transform Question 4:

The Laplace transform of the waveform shown in the figure is \(\frac{1}{s}\left( {\frac{1}{{1 + {e^{ks}}}}} \right)\) such that the value of k is ________.

F1 S.B Madhu 01.07.20 D1

Answer (Detailed Solution Below) -1

Laplace Transform Question 4 Detailed Solution

Solution:

The given periodic function can be expressed as the summation of shifted unit step functions as shown:

F1 S.B Madhu 01.07.20 D2

Similarly x3(t), x4(t), … will be shifted unit step functions.

x1(t) = u(t) - u(t - 1)

x2(t) = u(t - 2) - u(t - 3)

x(t) can be expressed as:

x(t) = x1(t) + x2(t) + x3(t) + ….

x(t) = u(t) - u(t - 1) + u(t - 2) - u(t - 3)+ …      ---(1)

\(u\left( t \right)\mathop \leftrightarrow \limits^{L.T} \frac{1}{s}\) 

Time-shifting affects the frequency as:

\(u\left( {t - {t_0}} \right)\mathop \leftrightarrow \limits^{L.T.} \frac{1}{s} \cdot {e^{ - s{t_0}}}\) 

The Laplace transform of x(t) from equation (1), can be written as:

\(= \frac{1}{s} - \frac{1}{s}{e^{ - s}} + \frac{1}{s}{e^{ - 2s}} - \frac{1}{s}{e^{ - 3s}} + \frac{1}{s}{e^{4s}} - \frac{1}{5}{e^{ - 5s}} + \ldots \) 

\(= \frac{1}{s}\left[ {1 + {e^{ - 2s}} + {e^{ - 4s}} + \ldots } \right] - \frac{1}{s}\left[ {{e^{ - s}} + {e^{ - 3s}} + {e^{ - 5s}} + \ldots } \right]\) 

Simplifying the geometric series, we can write:

\(= \frac{1}{s}\left[ {\frac{1}{{1 - {e^{ - 2s}}}}} \right] - \frac{1}{s}\left[ {\frac{{{e^{ - s}}}}{{1 - {e^{ - 2s}}}}} \right]\) 

\(X\left( s \right) = \frac{1}{s}\left[ {\frac{{1 - {e^{ - s}}}}{{1 - {e^{ - 2s}}}}} \right]\) 

\(= \frac{1}{s}\left[ {\frac{{1 - {e^{ - s}}}}{{\left( {1 + {e^{ - s}}} \right)\left( {1 - {e^{ - s}}} \right)}}} \right]\) 

\(X\left( s \right) = \frac{1}{s}\left[ {\frac{1}{{1 + {e^{ - s}}}}} \right]\)         ---(2)

Given \(X\left( s \right) = \frac{1}{{s\left( {1 + {e^{ks}}} \right)}}\) 

Comparing this with equation (2) we get:

k = -1

Laplace Transform Question 5:

The solution for the differential equation

\(\frac{{{d^2}x}}{{d{t^2}}} = - 9x\)

With initial conditions \(x\left( 0 \right) = 1\ and\ \frac{{dx}}{{dt}}{|_{t = 0}} = 1\), is

  1. t2 + t + 1
  2. \(sin\ 3t + \frac{1}{3}\cos 3t + \frac{2}{3}\)
  3. \(\frac{1}{{3}}\sin 3t + \cos 3t\)
  4. cos 3t + t

Answer (Detailed Solution Below)

Option 3 : \(\frac{1}{{3}}\sin 3t + \cos 3t\)

Laplace Transform Question 5 Detailed Solution

Given the differential equation,

\(\frac{{{d^2}x}}{{d{t^2}}} = - 9x\)

Taking the Laplace transform,

s2X(s) – sX(0) – X(0) = -9X(s)

or s2X(s) – s(1) – (1) = - 9X(s)

or X(s) (s2+9) = s + 1

\(X\left( s \right) = \frac{{s + 1}}{{{s^2} + 9}} = \frac{s}{{{s^2} + 9}} + \frac{1}{{{s^2} + 9}}\)

\(x\left( t \right) = cos\ 3t + \frac{1}{3}\sin 3t\) (Taking  inverse Laplace.)

Laplace Transform Question 6:

Match the following and choose the correct alternative from List - I and List - II.

List - I List - II
(Time function) (Laplace transform)
A. 1 1. 1/s
B. t 2. 1/s2
C. sin ωt 3. \(\frac{\text{s}}{\left(\text{s}^{2}+\omega^{2}\right)}\)
D. cos ωt 4. \(\frac{\omega}{\left(\text{s}^{2}+\omega^{2}\right)}\)

  1. A - 2, B - 1, C - 3, D - 4
  2. A - 1, B - 2, C - 4, D - 3
  3. A - 1, B - 2, C - 3, D - 4
  4. A - 2, B - 1, C - 4, D - 3

Answer (Detailed Solution Below)

Option 2 : A - 1, B - 2, C - 4, D - 3

Laplace Transform Question 6 Detailed Solution

Laplace Transform:

The Laplace transform of \({t^n} u(t)={n\over s^{n+1}}\)

1.)  \( L[{t^0}]={1 \over s}\)

2.) \( L[{t}]={1 \over s^2}\)

3.) \( L[{sin \space\omega t}]={\omega \over s^2+\omega^2}\)

4.) \( L[{cos \space\omega t}]={s \over s^2+\omega^2}\)

Hence, option 2 is correct.

Laplace Transform Question 7:

Fourier transform and Laplace transform are related through 

  1. Time domain
  2. Frequency domain
  3. Both time and frequency domains
  4. None of these

Answer (Detailed Solution Below)

Option 2 : Frequency domain

Laplace Transform Question 7 Detailed Solution

The Fourier transform of a function is equal to its two-sided Laplace transform evaluated on the imaginary axis of the s-plane.

Explanation:

The Fourier transform of x(t) is, denoted by X(jω), is defined as:

\(X\left( {j\omega } \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - j\omega t}}dt\)

The Laplace transform of x(t), denoted by X(s), is defined as:

\(X\left( s \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - st}}dt\)

Where s is a continuous complex variable.

We can also express s as: s = σ + jω

Where σ and ω are the real and imaginary parts of s, respectively

The Laplace transform can be written as:

\(X\left( {\sigma + j\omega } \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - \left( {\sigma + j\omega } \right)t}}dt = \mathop \smallint \limits_{ - \infty }^\infty \left( {x\left( t \right){e^{ - \sigma t}}} \right){e^{ - j\omega t}}dt\)

By comparing the above Laplace and Fourier transform equations, it is clear that Laplace transform of x(t) is equal to the Fourier transform of \(x\left( t \right){e^{ - \sigma t}}\).

When σ = 0 or s = jω, both are identical.

\({\left. {X\left( s \right)} \right|_{s = j\omega }} = X\left( {j\omega } \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - j\omega t}}dt\)

That is, Laplace transform generalizes Fourier transform.

Laplace Transform Question 8:

If F(s) is the Laplace transform of f(t), then how will F(s) be written in partial fraction method?

  1. F(s) = D(s)/N(s)
  2. L-1 [F(s)]
  3. F(s) = N(s)/D(s)
  4. F(s) = D(s) x N(s).

Answer (Detailed Solution Below)

Option 3 : F(s) = N(s)/D(s)

Laplace Transform Question 8 Detailed Solution

  • In the partial fraction method of Laplace transforms, the function F(s) is generally represented as a rational function, which is the ratio of two polynomials:
     \(F(S) = \frac{N(S)}{D(S)} = \frac{Numerator\ polynomial}{Denominator \ polynomial}\)
  • The goal of partial fraction decomposition is to express as a sum of simpler fractions to facilitate the inverse Laplace transform.
  • These simpler fractions correspond to known Laplace transform pairs, making it easier to find f(t)" id="MathJax-Element-8-Frame" role="presentation" style="position: relative;" tabindex="0"> f(t)" id="MathJax-Element-15-Frame" role="presentation" style="position: relative;" tabindex="0"> f(t)" id="MathJax-Element-13-Frame" role="presentation" style="position: relative;" tabindex="0"> f(t)" id="MathJax-Element-327-Frame" role="presentation" style="position: relative;" tabindex="0"> f(t)" id="MathJax-Element-10-Frame" role="presentation" style="position: relative;" tabindex="0"> f(t)" id="MathJax-Element-102-Frame" role="presentation" style="position: relative;" tabindex="0"> f(t) from " id="MathJax-Element-9-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-16-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-14-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-328-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-11-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-103-Frame" role="presentation" style="position: relative;" tabindex="0">
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Laplace Transform Question 9:

Determine transfer function if the impulse response is e-2t.

  1. 1 / (s + 2)
  2. 1 / (s - 2)
  3. 1 / (s + 2)2
  4. 1 / (s - 2)2

Answer (Detailed Solution Below)

Option 1 : 1 / (s + 2)

Laplace Transform Question 9 Detailed Solution

Concept:

Transfer function:

It is the ratio of the Laplace transform of the output variable to the input variable with initial conditions zero.

F1 Uday.S 23-12-20 Savita D1

TF = \(\left. {\frac{{L\left[ {output} \right]}}{{L\left[ {input} \right]}}} \right|\)L[output]L[input]|" id="MathJax-Element-28-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">L[output]L[input]|" id="MathJax-Element-71-Frame" role="presentation" style="position: relative;" tabindex="0">L[output]L[input]|" id="MathJax-Element-19-Frame" role="presentation" style="position: relative;" tabindex="0">L[output]L[input]|" id="MathJax-Element-21-Frame" role="presentation" style="position: relative;" tabindex="0">L[output]L[input]|  

Initial conditions = 0

TF = C(s) / R(s)

If R(s) = 1 i.e impulse input

TF = C(s)

Impulse response (IR) = c(t) = L-1 [C(s)] = L-1[TF]

TF = L-1[IR]

Hence, TF may be also called the impulse response of the system.

Explanation:

From the property of Laplace transform

\(LT[e^{-at}]→\frac{1}{s+a}\)

Put a = 2, 

\(LT[e^{-2t}]→\frac{1}{s+2}\)

Hence, T.F = 1 / (s+2)

Additional Information

f(t) L{f(t)} = F(s)
\(\delta(t)\) 1
u(t) \(\frac{1}{s}\)
\(e^{at}u(t)\) \(\frac{1}{s-a}\)
\(e^{-at}u(t)\) \(\frac{1}{s+a}\)
sin (at) \(\frac{a}{s^2+a^2}\)
cos (at) \(\frac{s}{s^2+a^2}\)
\(e^{at}x(t)\) X(s - a)

Laplace Transform Question 10:

Time shifting in Laplace transform if L{x(t)} = X(s), then L[x(t - t0)] is:

  1. \({e^{ - s\left( {{t_0} - t} \right)}}X\left( {s + {s_0}} \right)\)
  2. \({e^{ - s\left( {{t_0} - t} \right)}}X\left( {s - {s_0}} \right)\)
  3. \({e^{ - s\left( {t - {t_0}} \right)}}X\left( {\frac{s}{{{s_0}}}} \right)\)
  4. \({e^{ - s{t_0}}}X\left( s \right)\)

Answer (Detailed Solution Below)

Option 4 : \({e^{ - s{t_0}}}X\left( s \right)\)

Laplace Transform Question 10 Detailed Solution

Concept:

The Laplace transform F(s) of a function f(t) is defined by:

\(L\text{(}f\left( t \right)\text{ }\!\!\}\!\!\text{ }=F\left( s \right)=\underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{-st}}f\left( t \right)dt\)

From the time-shifting property of Laplace transform:

\(L\left\{ f\left( t-a \right) \right\}={{e}^{-as}}F\left( s \right)\)

Application:

L{x(t)} = X(s)

L(x(t-t0)) = \({e^{ - s{t_0}}}X\left( s \right)\)

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