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:

$L[e^{-at}]⟷\frac{1}{s+a}$

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) = $\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{a}}^2}$

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

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

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:

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

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

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

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]

$\therefore f(s)=\frac{s+1}{s^2+1}$

$\frac{s+1}{s^2+1}=\frac{s}{s^2+1}+\frac{1}{s+1}$

∵ L^-1 f(s) = f(t)

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

∴ 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

$f\left(x\right)=\frac{a_o}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos nx+\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin nx$

where

$a_o=\frac{1}{\pi }\underset{\alpha }{\overset{\alpha +2\pi }{\int }}f\left(x\right)dx;a_n=\frac{1}{\pi }\underset{\alpha }{\overset{\alpha +2\pi }{\int }}f\left(x\right)\cos nxdx;b_n=\frac{1}{\pi }\underset{\alpha }{\overset{\alpha +2\pi }{\int }}f\left(x\right)\sin nxdx$

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

$f\left(x\right)=\frac{a_o}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\frac{\cos n\pi x}{L}$

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

$f\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin \frac{n\pi x}{L}$

Calculation:

 f(x) = cos2(x)

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

On comparing with: $f\left(x\right)=\frac{a_o}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\frac{\cos n\pi x}{L}$

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:

$L\left[\frac{e^{-at}-e^{-bt}}{t}\right]=\underset{s}{\overset{\mathrm{\infty }}{\int }}\left(\frac{1}{\left(s+a\right)}-\frac{1}{\left(s+b\right)}\right)ds$

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

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

$=\underset{s\to \mathrm{\infty }}{lim}\log \left(\frac{s+a}{s+b}\right)-\log \left(\frac{s+a}{s+b}\right)$

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

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

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

∵ By definition

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

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

From (1) and (2) 

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}\left[\frac{e^{-at}-e^{-bt}}{t}\right]dt=\log \left(\frac{s+b}{s+a}\right)$

Put s = 0

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-at}-e^{-bt}}{t}dt=\log \frac{b}{a}$

LT of $\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-at}-e^{-bt}}{t}dt$ = LT of $\log \frac{b}{a}$

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

Explanation:

Dirac Delta Function

$\delta \left(t-a\right)=\underset{ϵ\to 0}{lim}\{\begin{array}{c}0,-\mathrm{\infty }<t<a\\ \frac{1}{ϵ},a\le t\le a+ϵ\\ 0,a+ϵ<t<ϵ\end{array}$

L[δ(t – a)] = e^-as

∵${\int }_0^{\mathrm{\infty }}f\left(t\right)\delta \left(t-a\right)dt=f\left(a\right)$

$\therefore L\left[\delta \left(t-a\right)\right]={\int }_0^{\mathrm{\infty }}{e}^{-st}\delta \left(t-a\right)dt=e^{-as}$

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

$L\left\{\frac{1}{\sqrt{t}}\right\}=\underset{t=0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}\frac{1}{\sqrt{t}}dt$

Put st = u ⇒ t = u/s

du = s dt ⇒ dt = du/s

$L\left\{\frac{1}{\sqrt{t}}\right\}=\underset{u=0}{\overset{\mathrm{\infty }}{\int }}{e}^{-u}\frac{\sqrt{s}}{\sqrt{u}}\frac{du}{s}$

$=\frac{1}{\sqrt{s}}\underset{u=0}{\overset{\mathrm{\infty }}{\int }}{e}^{-u}{u}^{-\frac{1}{2}}du$

$=\frac{1}{\sqrt{s}}\underset{u=0}{\overset{\mathrm{\infty }}{\int }}{e}^{-u}{u}^{\frac{1}{2}-1}du$

We know that, $\mathrm{\Gamma }n=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-x}{x}^{n-1}dx$

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

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

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