Properties MCQ Quiz - Objective Question with Answer for Properties - Download Free PDF

The correct answer is ∫₀^∞ f(t) e^-st dt.

Key Points

• The **Laplace transformation** of a function **f(t)** is used to transform a time-domain function into a complex frequency-domain function.
• The correct definition of the Laplace transformation is given by the integral **∫₀^∞ f(t) e^-st dt**. This is the correct mathematical expression for the Laplace transform.
• In this integral, **f(t)** is the original function of time **t**, and **s** is a complex number frequency parameter **(s = σ + jω)**.
• The exponential term **e^-st** serves to weight the function **f(t)** and transform it into the complex frequency domain.

 Additional Information

• Incorrect Options Explanation
• **∫₀^∞ (1)/(f(t) e^st) dt**: This expression is incorrect as it involves an inverse function and an incorrect exponential term.
• **∫₀^∞ f(t) e^st dt**: This is incorrect because the exponential term should be **e^-st** for the Laplace transformation.
• **∫₀^∞ (1)/(f(t) e^-st) dt**: This is incorrect due to the inverse function and incorrect formulation.
• Key Topics Related to Laplace Transform
• The **Laplace transform** is widely used in engineering and physics to solve differential equations, control systems, and signal processing.
• It helps in converting differential equations into algebraic equations which are easier to handle.
• **Inverse Laplace Transform**: This is used to convert the function back from the frequency domain to the time domain.
• **Properties of Laplace Transform**: Linearity, time shifting, frequency shifting, and convolution theorem are some key properties.
• **Applications**: Commonly used in electrical engineering for circuit analysis, in mechanical engineering for system dynamics, and in control systems for stability analysis.

$\frac{dt}{dt}+ay=e^{-bt}$

Integrating factor $=e^{\int pdt}=e^{at}$

Solution of y is,

$y\left(e^{at}\right)=\int e^{at}.e^{-bt}dt+c$

$\Rightarrow y{e}^{at}=\frac{e^{\left(a-b\right)t}}{a-b}+c$

Given that, y(0) = 0

$\Rightarrow 0=\frac{1}{a-b}+c\Rightarrow c=-\frac{1}{\left(a-b\right)}$

Now, the solution becomes,

$y{e}^{at}=\frac{e^{\left(a-b\right)t}}{a-b}-\frac{1}{a-b}$

$\Rightarrow y\left(t\right)=\frac{e^{-bt}}{a-b}-\frac{e^{-at}}{a-b}$

Apply Laplace transform

$\Rightarrow y\left(s\right)=\frac{1}{a-b}\left[\frac{1}{s+b}-\frac{1}{s+a}\right]$

Concept

A second-order differential equation is represented as:

y" + ay' + by = f(t)

The Laplace transform of the above equation with the initial condition is:

$s^{2}Y(s)-sY(0)-Y^{\prime }(0)+a[sY(s)-Y(0)]+bY(s)=F(s)$

Calculation

Given, y(0) = 5, y'(0) = 10

$s^{2}Y(s)-5s-10+a(sY(s)-5)+bY(s)=F(s)$

Hence, the correct answer is option 2.

Concept:

$L^{-1}\{cosh\omega t\}=\frac{s}{s^2-{\omega }^2}$

${L}^{-1}\{cosh2t\}=\frac{s}{s^2-2^2}$

$=\frac{s}{s^2-4}$

Additional Information Some common inverse Laplace transforms are:

Concept:

$L\left\{f\left(t\right)\right\}=F\left(s\right)=\underset{0}{\overset{\mathrm{\infty }}{\int }}f\left(t\right).e^{-st}dt$

According to the property of Laplace

$L\left\{t^n.f\left(t\right)\right\}={\left(-1\right)}^n\times \frac{d^n}{d{s}^n}\left\{F\left(s\right)\right\}$

Calculation:

Given:

f(t) = 3t⁴ = t⁴.3

$L\left(3\right)=3\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}dt=\frac{3}{s}[e^{-st}{]}_0^{\mathrm{\infty }}=\frac{3}{s}$

Now

$L\{t^4.3\}={\left(-1\right)}^4\times \frac{d^4}{d{s}^4}\left(\frac{3}{s}\right)$

$L\{t^4.3\}=3\times \frac{4!}{s^5}=\frac{72}{s^5}$

Note: You can also solve it directly by using:

$L\left(t^n\right)=\frac{n!}{s^{n+1}}$

Concept:

Some pairs of Laplace transforms are given below.

$e^{-at}\leftrightarrow \frac{1}{s+a}$

$t^n{e}^{-at}\leftrightarrow \frac{n!}{{\left(s+a\right)}^{n+1}}$

Calculation:

Given:

$H\left(s\right)=\frac{s+3}{s^2+2s+1}$

$\Rightarrow \frac{s+3}{{\left(s+1\right)}^2}=\frac{s+1}{{\left(s+1\right)}^2}+\frac{2}{{\left(s+1\right)}^2}$

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

By applying inverse Laplace transform

Given that,

z(t) = x(t) * y(t)

Apply Laplace transform,

Z(s) = X(s) . Y(s)

Now apply time scaling property,

$z\left(ct\right)\leftrightarrow \frac{1}{c}Z\left(\frac{s}{c}\right)$

$z\left(ct\right)\leftrightarrow \frac{1}{c}X\left(\frac{s}{c}\right)Y\left(\frac{s}{c}\right)$

$z\left(ct\right)\leftrightarrow \frac{1}{c}X\left(\frac{s}{c}\right)\times c.\frac{1}{c}Y\left(\frac{s}{c}\right)$

Apply inverse Laplace

z(ct) = c x(ct) * y(ct)

Concept:

Bilateral Laplace transform:

$L\left[x\left(t\right)\right]=x\left(s\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)e^{-st}dt$

Unilateral Laplace transform:

$L\left[x\left(t\right)\right]=x\left(s\right)=\underset{0}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)e^{-st}dt$

Some important Laplace transforms:

Frequency shifting property:

If X(s) is the Laplace transform of x(t), then

$e^{at}X\left(s\right)\leftrightarrow X\left(s-a\right)$

Calculation:

f(t) = e^2t sin (5t) u(t)

$L\left(\sin \left(5t\right)u\left(t\right)\right)=\frac{5}{s^2+25}$

By using frequency shifting property,

 The Laplace transform of f(t) is

$F\left(s\right)=\frac{5}{{\left(s-2\right)}^2+25}=\frac{5}{s^2-4s+29}$

Concept:

Shifting property:

L{e^atf(t)} = F(s - a) 

Calculation:

Given:

eat cos ωt, ∴ f(t) = cos ωt

Laplace transformation of $\cos \omega t=\frac{s}{s^2+{\omega }^2}$

From first shifting property:

L{eatf(t)} = F(s - a)

Concept:

$x\left(t\right)\stackrel{LT}{\leftrightarrow }X\left(s\right)$

$\frac{dx\left(t\right)}{dt}\stackrel{L.T}{\leftrightarrow }sX\left(s\right)$

$sX\left(s\right)\stackrel{I.L.T}{\leftrightarrow }\frac{dx\left(t\right)}{dt}$

$L\left\{\frac{x\left(t\right)}{t}\right\}=\frac{}{}\underset{s}{\overset{0}{\int }}\left\{x\left(t\right)\right\}ds$

Calculation:

Given that Laplace transform $f\left(t\right)=2\sqrt{\frac{t}{\pi }}$ is $s^{-\frac{3}{2}}$

Given as $g\left(t\right)=\frac{1}{\sqrt{\pi t}}$

$\begin{array}{l}\Rightarrow g\left(t\right)=\frac{2\sqrt{\frac{t}{\pi }}}{2t}=\frac{f\left(t\right)}{2t}\\ L\left\{g\left(t\right)\right\}=L\left\{\frac{f\left(t\right)}{2t}\right\}=\frac{1}{2}\underset{s}{\overset{0}{\int }}\left\{f\left(t\right)\right\}ds\\ =\frac{1}{2}\underset{s}{\overset{\mathrm{\infty }}{\int }}{s}^{-\frac{3}{2}}ds=\frac{1}{2}{\left(\frac{s^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}\right)}_s^{\mathrm{\infty }}\\ =\frac{1}{2}\left(-2\right)\left[0-s^{-\frac{1}{2}}\right]=s^{-\frac{1}{2}}=\frac{1}{\sqrt{s}}\end{array}$

Concept:

$L\left(t^n\right)=\frac{n!}{s^{n+1}}$

$L\left(e^{at}\right)=\frac{1}{s-a}$

$L\left(t^n{e}^{at}\right)=\frac{n!}{{\left(s-a\right)}^{n+1}}$

Calculation:

f(t) = 2t²e^-t

$F\left(s\right)=L\left(f\left(t\right)\right)=2L\left(t^2{e}^{-t}\right)=2\times \frac{2!}{{\left(s+1\right)}^{2+1}}=\frac{4}{{\left(s+1\right)}^3}$

$F\left(1\right)=\frac{4}{{\left(1+1\right)}^3}=\frac{1}{2}=0.5$

$f\left(t\right)=\begin{array}{cc}2;& 0<t<1\\ 0;& otherwise\end{array}$

Concept:

Some important Laplace transforms are:

$L\left(t^n\right)=\frac{n!}{s^{n+1}}$

$L\left(t^n\right)=\frac{n!}{s^{n+1}}$

$L\left(t^n{e}^{at}\right)=\frac{n!}{{\left(s-a\right)}^{n+1}}$

$L\left\{\cos \omega t\right\}=\frac{s}{s^2+{\omega }^2}$

$L\left\{\sin \left(\omega t\right)\right\}=\frac{\omega }{s^2+{\omega }^2}$

Concept:

Laplace transform is defined by:

$F\left(s\right)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}f\left(t\right)dt$

Where, 

f(t) is the function varying with time t

F(s) is the Laplace transform of f(t)

Analysis:

$sinh(at)=\frac{e^{at}-e^{-at}}{2}$​​

$F\left(s\right)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}[\frac{e^{at}-e^{-at}}{2}]dt$

Solving the above, we get:

​$\frac{a}{s^2-a^2}$​

Concept:

This is a definition of Laplace transform

$L\left\{f\left(t\right)\right\}={\int }_0^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt=f\left(s\right)$

Some important Laplace transforms are:

$L\left(t^n\right)=\frac{n!}{s^{n+1}}$

$L\left(t^n\right)=\frac{n!}{s^{n+1}}$

$L\left(t^n{e}^{at}\right)=\frac{n!}{{\left(s-a\right)}^{n+1}}$