Transform Theory MCQ Quiz in मल्याळम - Objective Question with Answer for Transform Theory - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

The Laplace transform of a general exponential signal is given by:

\(L[e^{-at}]\longleftrightarrow \frac{1}{s+a}\)

where 'a' is any positive integer.

Calculation:

Given:

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

\(\frac{1}{{\left( {s + 1} \right)\left( {s - 2} \right)}} = \frac{A}{{\left( {s - 2} \right)}} + \frac{B}{{\left( {s + 1} \right)}} = \frac{1}{3}\left\{ {\frac{1}{{s - 2}} + \frac{{ - 1}}{{s + 1}}} \right\}\)

\({L^{ - 1}}\left( {\frac{1}{{\left( {s + 1} \right)\left( {s - 2} \right)}}} \right) = {L^{ - 1}}\left\{ {\frac{1}{3}\left( {\frac{1}{{s - 2}} - \frac{1}{{s + 1}}} \right)} \right\} = \frac{{{e^{2t}} - {e^{ - t}}}}{3}\)

Additional Information

Some common inverse Laplace transforms are:

Concept:

Laplace transform of f(t):

L {f(t)} = F (s)

Then, Multiplication by tⁿ:           

L {tⁿ f(t)} = (-1)^{n\(\frac{{{d^n}}}{{d{s^n}}}\)} [ F(s)]

Now,

L (cos at) = \(\frac{{\rm{s}}}{{{{\rm{s}}^2} + {{\rm{a}}^2}}}\)

L {t cos at} = (-1) \(\frac{{\rm{d}}}{{{\rm{ds}}}}\left( {\frac{{\rm{s}}}{{{{\rm{s}}^2} + {{\rm{a}}^2}}}} \right)\)

= \(\frac{{{\rm{s\;}}\left( {2{\rm{s}}} \right) - \left( 1 \right)\left( {{{\rm{s}}^2} + {{\rm{a}}^2}} \right)}}{{{{\left( {{{\rm{s}}^2} + {{\rm{a}}^2}} \right)}^2}}}\)

Concept:

This is a definition of Laplace transform

\(L\left\{ {f\left( t \right)} \right\} = \;\mathop \smallint \nolimits_0^\infty {e^{ - st}}f\left( t \right)dt = f\left( s \right)\)

Some important Laplace transforms are:

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

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

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

Concept:

If L{f(t)} = f(s)

To solve the differential equation:

If f(t) is a function of independent variable 't' then

L {f'(t)} = sL{f(t)} - f(0)

L {f''(t)} = s² L{f(t)} - sf(0) - f'(0)

\(L\{sin\;(at)\}=\frac{a}{s^2\;+\;a^2}\)

\(L\{cos\;(at)\}=\frac{s}{s^2\;+\;a^2}\)

Calculation:

Given:

\(f(t)\;+\;\ddot{f}(t)=0\) and f(0) = f'(0) = 1.

We know that:

L{f''(t)} = s2L{f(t)} - sf(0) - f'(0)

∴ s2 f(s) - sf(0) - f'(0) + f(s) = 0

∴ f(s)[s² + 1] - (s + 1) = 0       [∵ f(0) = f'(0) = 1]

\(∴ f(s)=\frac{s\;+\;1}{s^2\;+\;1}\)

\(\frac{s\;+\;1}{s^2\;+\;1}=\frac{s}{s^2\;+\;1}\;+\;\frac{1}{s\;+\;1}\)

∵ L^-1 f(s) = f(t)

\(∴ L^{-1} \left ( \frac{s\;+\;1}{s^2\;+\;1} \right )=L^{-1}\left(\frac{s}{s^2\;+\;1}\;+\;\frac{1}{s\;+\;1} \right)\)

∴ f(t) = cos t + sin t

Concept:

If L{f(t)} = F(s) or L^-1 {f(s)} = f(t)

Then\(L\left\{ {tf\left( t \right)} \right\} = - \frac{d}{{ds}}\left\{ {F\left( s \right)} \right\}\)

\(tf\left( t \right) = {L^{ - 1}}\left\{ { - \frac{d}{{ds}}F\left( s \right)} \right\}\)

\(f\left( t \right) = - \frac{1}{t}{L^{ - 1}}\left\{ {\frac{d}{{ds}}\left( {F\left( s \right)} \right)} \right\}\)

Calculation:

Given:

\(Lf\left( t \right) = \log \left( {\frac{{s + a}}{{s + b}}} \right)\)

\(f\left( t \right) = {L^{ - 1}}\left[ {\log \left( {\frac{{s + a}}{{s + b}}} \right)} \right]={L^{ - 1}}\left\{ {\log \left( {s + a} \right) - \log \left( {s + b} \right)} \right\}\)

\(\Rightarrow - \frac{1}{t}{L^{ - 1}}\left\{ {\frac{d}{{ds}}\left( {\log \left( {s + a} \right) - \log \left( {s + b} \right)} \right)} \right\}\;= - \frac{1}{t}\left\{ {\frac{1}{{s + a}} - \frac{1}{{s + b}}} \right\}\)

\(\Rightarrow - \frac{1}{t}({e^{ - at}} - {e^{ - bt}})\;=\frac{1}{t}\left( {{e^{ - bt}} - {e^{ - at}}} \right)\)

The correct answer is option 1):(Circular convolution)

Concept:

• Circular convolution, also known as cyclic convolution, is a special case of periodic convolution, which is the convolution of two periodic functions that have the same period
• Convolution in the time domain is equal to multiplication in the frequency domain and vice versa.
• If X(ω) and Y(ω) are the Fourier transforms of x(t) and y(t)

then

x(t) * y(t) = X(ω) × Y(ω)

• If X(s) and Y(s) are the Laplace transforms of x(t) and y(t), t

hen x(t) * y(t) = X(s) × Y(s)

• Hence the multiplication of two discrete Fourier transforms (DFTs) is equal to the Circular convolution of two sequences in the time-domain.

Concept:

The Fourier series for the function f(x) in the interval α < x < α + 2π is given by

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx\)

where

\({a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;\;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;\;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx\)

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

\(f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)

Calculation:

 f(x) = cos2(x)

\(f(x) = \frac{{1 + \cos 2x}}{2} = \frac{1}{2} + \frac{{\cos 2x}}{2}\)

On comparing with: \(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

a₀ = 1

a₁ = 0

a₂ = 1/2

Explanation:

Let F(t) = e^-at – e^-bt

{L{F(t)} = L{e^-at – e^-bt}

\(= \frac{1}{{s + a}} - \frac{1}{{s + b}}\)

Now:

\(L\left[ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right] = \mathop \smallint \limits_s^\infty \left( {\frac{1}{{\left( {s + a} \right)}} - \frac{1}{{\left( {s + b} \right)}}} \right)ds\)

\( = \left[ {\log \left( {s + a} \right) - \log \left( {s + b} \right)} \right]_s^\infty \)

\(= \left[ {\log \left( {\frac{{s + a}}{{s + b}}} \right)} \right]_s^\infty \)

\( = \mathop {\lim }\limits_{s \to \infty } \log \left( {\frac{{s + a}}{{s + b}}} \right) - \log \left( {\frac{{s + a}}{{s + b}}} \right)\)

\( = \mathop {\lim }\limits_{s \to \infty } \log \left( {\frac{{1 + a/s}}{{1 + b/s}}} \right) - \log \left[ {\frac{{s + a}}{{s + b}}} \right]\)

\( = \log \left[ {\frac{{1 + 0}}{{1 + 0}}} \right] + \log {\left( {\frac{{s + a}}{{s + b}}} \right)^{ - 1}} \)

\(L\left[ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right] = \log \left( {\frac{{s + b}}{{s + a}}} \right)\)  ----(1)

∵ By definition

\(L\left\{ {F\left( t \right)} \right\} = \mathop \smallint \limits_o^\infty {e^{ - st}}.F\left( t \right)dt\)

\(L\left[ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right] = \mathop \smallint \limits_0^\infty {e^{ - st}}\left[ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right]dt\)  ----(2)

From (1) and (2) 

\(\mathop \smallint \limits_0^\infty {e^{ - st}}\left[ {\frac{{{e^{ - at}} - {e^{ - bt}}}}{t}} \right]dt= \log \left( {\frac{{s + b}}{{s + a}}} \right)\)

Put s = 0

\(\mathop \smallint \limits_0^\infty \frac{{{e^{ - at}} - {e^{ - bt}}}}{t}dt = \log \frac{b}{a}\)

LT of \(\mathop \smallint \limits_0^\infty \frac{{{e^{ - at}} - {e^{ - bt}}}}{t}dt\) = LT of \(\log \frac{b}{a}\)

\(L~[\log(\frac{b}{a})] = \frac{1}{s}\log \frac{b}{a}\)

Explanation:

Dirac Delta Function

\(\delta \left( {t - a} \right) = \mathop {\lim }\limits_{ \epsilon\to 0} \left\{ {\begin{array}{*{20}{c}} {0,\; - \infty < t < a}\\ {\frac{1}{\epsilon},\;a \le t \le a + \epsilon }\\ {0,\;a + \epsilon < t < \epsilon} \end{array}} \right.\)

L[δ(t – a)] = e^-as

∵\(\;\mathop \smallint \nolimits_0^\infty f\left( t \right)\delta \left( {t - a} \right)dt = f\left( a \right)\)

\(\therefore L\left[ {\delta \left( {t - a} \right)} \right] = \mathop \smallint \nolimits_0^\infty {e^{ - st}}\delta \left( {t - a} \right)dt = {e^{ - as}}\)

\(f\left( t \right) = \frac{1}{{\sqrt t }}\)

\(L\left\{ {\frac{1}{{\sqrt t }}} \right\} = \mathop \smallint \limits_{t = 0}^\infty {e^{ - st}}\frac{1}{{\sqrt t }}dt\)

Put st = u ⇒ t = u/s

du = s dt ⇒ dt = du/s

\(L\left\{ {\frac{1}{{\sqrt t }}} \right\} = \mathop \smallint \limits_{u = 0}^\infty {e^{ - u}}\frac{{\sqrt s }}{{\sqrt u }}\frac{{du}}{s}\)

\(= \frac{1}{{\sqrt s }}\mathop \smallint \limits_{u = 0}^\infty {e^{ - u}}{u^{ - \frac{1}{2}}}\;du\)

\(= \frac{1}{{\sqrt s }}\mathop \smallint \limits_{u = 0}^\infty {e^{ - u}}{u^{\frac{1}{2} - 1}}\;du\)

We know that, \({\rm{\Gamma }}n = \mathop \smallint \limits_0^\infty {e^{ - x}}{x^{n - 1}}dx\)

\(= \frac{1}{{\sqrt s }}{\rm{\Gamma }}\left( {\frac{1}{2}} \right)\)

We know that, \({\rm{\Gamma }}\left( {\frac{1}{2}} \right) = \sqrt \pi\)

\(F\left( s \right) = \sqrt {\frac{\pi }{s}}\)