Laplace Transform MCQ Quiz in मल्याळम - Objective Question with Answer for Laplace Transform - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 18, 2025
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) ?
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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 ________.
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:
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
Answer (Detailed Solution Below)
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. | \(\frac{\omega}{\left(\text{s}^{2}+\omega^{2}\right)}\) |
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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) f(t) f(t) f(t) f(t) f(t)
F(s) .
Laplace Transform Question 9:
Determine transfer function if the impulse response is e-2t.
Answer (Detailed Solution Below)
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.
TF = \(\left. {\frac{{L\left[ {output} \right]}}{{L\left[ {input} \right]}}} \right|\)
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:
Answer (Detailed Solution Below)
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)) = \({e^{ - s{t_0}}}X\left( s \right)\)