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

f(t) = 2u(t) - 5u(t - 2) + 7u(t - 5)

$\begin{array}{l}\Rightarrow F\left(s\right)=\frac{2}{s}-\frac{5e^{-2s}}{s}+\frac{7e^{-5s}}{s}\\ =\frac{2-5e^{-2s}+7e^{5s}}{s}\end{array}$

Given that,

$F\left(s\right)=\frac{a-5e^{-2s}+b{e}^{-cs}}{s}$

By comparing both equations, a = 2, b = 7, c = 5

⇒ a + b + c = 14

$\mathrm{L}\left({\mathrm{e}}^{-\mathrm{a}\mathrm{t}}-{\mathrm{e}}^{-\mathrm{b}\mathrm{t}}\right)=\frac{1}{\mathrm{s}+\mathrm{a}}-\frac{1}{\mathrm{s}+\mathrm{b}}$

We know that $\mathrm{F}\left\{\frac{\mathrm{f}\left(\mathrm{t}\right)}{\mathrm{t}}\right\}=\underset{\mathrm{s}}{\overset{\mathrm{\infty }}{\int }}\mathrm{F}\left(\mathrm{s}\right)\mathrm{d}\mathrm{s}$

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

$\begin{array}{l}=\log {{\left(\frac{s+a}{s+b}\right)}^{\mathrm{\infty }}}_s\\ =\log 1-\log \left(\frac{s+a}{s+b}\right)\\ =\log \left(\frac{s+b}{s+a}\right)\end{array}$

$Y\left(s\right)=L\left(y\left(t\right)\right)=\frac{5}{2\left(s-1\right)}-\frac{2}{s-2}+\frac{1}{2\left(s-3\right)}$

Apply inverse Laplace transform,

$\Rightarrow y\left(t\right)=\frac{5}{2}{e}^t-2e^{2t}+\frac{1}{2}{e}^{3t}$

Differentiate with respect to ‘t’.

$\Rightarrow y^{\prime }\left(t\right)=\frac{5}{2}{e}^t-4e^{2t}+\frac{3}{2}{e}^{3t}$

$\begin{array}{l}y\left(0\right)=\frac{5}{2}-2+\frac{1}{2}=1\\ y^{\prime }\left(0\right)=\frac{5}{2}-4+\frac{3}{2}=0\end{array}$

⇒ y(0) + y'(0) = 1

Y^’’ + 25 y = a sin 5t

By applying Laplace transform to the given differential equation, we get

$\left[s^{2}y\left(s\right)-sy\left(0\right)-y^{\prime }\left(0\right)\right]+25y\left(s\right)=\frac{5a}{s^2+25}$

$\Rightarrow \left[s^{2}y\left(s\right)-4s-0\right]+25y\left(s\right)=\frac{5a}{s^2+25}$

$\Rightarrow y\left(s\right)\left[s^2+25\right]=4s+\frac{5a}{s^2+25}$

$\Rightarrow y\left(s\right)=\frac{4s\left(s^2+25\right)+5a}{{\left(s^2+25\right)}^2}$

$\Rightarrow y\left(s\right)=\frac{4s^3+100s+5a}{{\left(s^2+25\right)}^2}$

Given that,

$y\left(s\right)=\frac{b{s}^3+100s+20}{{\left(s^2+25\right)}^2}$

By comparing both the equations a = 4, b = 4

⇒ a + b = 8

If f(t) is a periodic function with period T,

i.e. f(t + T) = f (t), then

$\mathrm{L}\left\{\mathrm{f}\left(\mathrm{t}\right)\right\}=\frac{{\int }_{\mathrm{o}}^{\mathrm{T}}{\mathrm{e}}^{-\mathrm{s}\mathrm{t}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{d}\mathrm{t}}{1-{\mathrm{e}}^{-\mathrm{S}\mathrm{T}}}$

Here the period of f(t) = 2π 

$\mathrm{L}\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{1-{\mathrm{e}}^{-2\pi \mathrm{s}}}\left\{\underset{\mathrm{o}}{\overset{\pi }{\int }}{\mathrm{e}}^{-\mathrm{s}\mathrm{t}}\mathrm{t}\mathrm{d}\mathrm{t}+\underset{\pi }{\overset{2\pi }{\int }}{\mathrm{e}}^{-\mathrm{s}\mathrm{t}}\left(\pi -\mathrm{t}\right)\mathrm{d}\mathrm{t}\right\}$

$=\frac{1}{1-{\mathrm{e}}^{-2\pi \mathrm{s}}}\{{\left[t\left(\frac{e^{-st}}{-s}\right)-1\left(\frac{e^{-st}}{s^2}\right)\right]}_0^{\pi }+$

${\left[\left(\pi -\mathrm{t}\right)\left(\frac{{\mathrm{e}}^{-\mathrm{s}\mathrm{t}}}{-\mathrm{s}}\right)-\left(-1\right)\left(\frac{{\mathrm{e}}^{2\pi \mathrm{s}}}{{\mathrm{s}}^2}\right)\right]}_{\pi }^{2\pi }\}$

$=\frac{1}{1-e^{-2\pi s}}\left\{\frac{-\pi e^{-\pi s}}{s}-\frac{e^{-\pi s}}{s^2}+\frac{1}{s^2}+\frac{\pi e^{-2\pi s}}{s}+\frac{e^{-2\pi s}}{s^2}-\frac{e^{-\pi s}}{s^2}\right\}$

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

We know that $2{\sin }^{2}2\mathrm{t}=\left(1-\cos 4\mathrm{t}\right)$

Apply Laplace transform on both sides then we get,

$\begin{array}{l}\mathrm{L}\mathrm{T}\left\{2{\sin }^{2}2\mathrm{t}\right\}=\mathrm{L}\mathrm{T}\left\{1\right\}-\mathrm{L}\mathrm{T}\left\{\cos 4\mathrm{t}\right\}\\ =\frac{1}{\mathrm{s}}-\frac{\mathrm{s}}{{\mathrm{s}}^2+16}\\ =\frac{16}{\mathrm{s}\left({\mathrm{s}}^2+16\right)}\end{array}$

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:

$L\left\{f\left(t\right)\right\}=\underset{0}{\overset{T}{\int }}{e}^{-st}\left(\frac{t}{T}\right)dt+\underset{T}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}dt$

$=\frac{1}{T}\underset{0}{\overset{T}{\int }}t{e}^{-st}dt+\underset{T}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}dt$

$=\frac{1}{T}{\left[t\int e^{-st}dt-\int 1.\int e^{-st}dtdt\right]}_0^T+\underset{T}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}dt$

$=\frac{1}{T}{\left[\frac{t{e}^{-st}}{-s}-\int \frac{e^{-st}}{-s}dt\right]}_0^T+\underset{T}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}dt$

$=\frac{1}{T}{\left[\frac{t{e}^{-st}}{-s}-\frac{e^{-st}}{s^2}\right]}_0^T+{\left[\frac{e^{-st}}{-s}\right]}_T^{\mathrm{\infty }}$

$=\frac{1}{T}\left[\frac{T{e}^{-sT}}{-s}-\frac{e^{-sT}}{s^2}+\frac{1}{s^2}\right]+\left[0+\frac{e^{-sT}}{s}\right]$

$=\frac{\left(1-e^{-sT}\right)}{s^{2}T}$

Concept:

L{sin at} = $\frac{a}{s^2+a^2}$

Division by t:

$L\{\frac{1}{t}f(t)\}={\int }_s^{\mathrm{\infty }}F(s)ds$

Laplace transform of Integral:

$L\{{\int }_0^{t}f(t)dt=\frac{1}{s}F(s)$

Shifting property:

L{eat⋅f(t)} = F(s - a)

Calculation:

Given:

$\left\{e^{-t}\underset{0}{\overset{t}{\int }}\frac{\sin t}{t}dt\right\}$

Let f(t) = sin t

$L\left\{\sin t\right\}=\frac{1}{s^2+1}$

$L\left\{\frac{\sin t}{t}\right\}=\underset{s}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\left(s^2+1\right)}ds={\left[ta{n}^{-1}s\right]}_s^{\mathrm{\infty }}$

$\therefore \frac{\pi }{2}-{\tan }^{-1}s={\cot }^{-1}s$

$L\left\{\underset{0}{\overset{t}{\int }}\frac{\sin t}{t}dt\right\}=\frac{1}{s}{\cot }^{-1}s$

$\therefore L\left\{e^{-t}\underset{0}{\overset{t}{\int }}\frac{\sin t}{t}dt\right\}=\frac{1}{\left(s+1\right)}{\cot }^{-1}\left(s+1\right)$