Inverse Laplace Transform MCQ Quiz - Objective Question with Answer for Inverse Laplace Transform - Download Free PDF

Explanation:

$\text{Given},\text{ f}\left(\text{t}\right)={\text{e}}^{\text{at}}$

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

$\text{L}\left[{\text{e}}^{\text{at}}\right]={\int }_0^{\mathrm{\infty }}{\text{e}}^{-\text{st}}{\text{e}}^{\text{at}}\text{dt}={\int }_0^{\mathrm{\infty }}{\text{e}}^{-(\text{s}-\text{a})\text{t}}\text{dt}$

$\therefore \text{L}\left[{\text{e}}^{\text{at}}\right]={\left(\frac{{\text{e}}^{-(\text{s}-\text{a})\text{t}}}{-(\text{s}-\text{a})}\right)}_0^{\mathrm{\infty }}=-\frac{1}{\text{s}-\text{a}}(0-1)=\frac{1}{\text{s}-\text{a}}$

∴ Laplace transform of L[eat] is $\frac{1}{\mathbf{s}-\mathbf{a}}$

The correct answer is _____.

Detailed Solution:

Apply Laplace Transform of the circuit:

Apply KVL in the loop:

$\frac{-V_0}{s}+LsI(s)+RI(s)+\frac{V_0{e}^{-t_{0}s}}{s}=0$

$I(s)=\frac{V_0(1-e^{t_{0}s})}{s(Ls+R)}$

Apply Inverse Laplace Transform:

i(t) = $\frac{V_0}{R}(1-e^{\frac{R}{L}t})u(t)-\frac{V_0}{R}(1-e^{\frac{R}{L}(t-t_0)})u(t-t_0))A$

Concept:

If L-1{F(s)} = f(t)

then

L-1F(s – a) = eat.f(t) and L-1{F(s + a)} = e-at.f(t)

Calculation:

Given, 

$\mathrm{F}(\mathrm{s})=\frac{1}{{\mathrm{s}}^3-\mathrm{s}}=\frac{1}{\mathrm{s}({\mathrm{s}}^2-1)}=\frac{1}{\mathrm{s}(\mathrm{s}-1)(\mathrm{s}+1)}$

On partial fraction decomposition,

$\mathrm{F}(\mathrm{s})=-\frac{1}{\mathrm{s}}+\frac{\frac{1}{2}}{\mathrm{s}-1}+\frac{\frac{1}{2}}{\mathrm{s}+1}$

Thus, ${\mathrm{L}}^{-1}(\mathrm{F}(\mathrm{s}))={\mathrm{L}}^{-1}(-\frac{1}{\mathrm{s}})+{\mathrm{L}}^{-1}\left(\frac{\frac{1}{2}}{\mathrm{s}-1}\right)+{\mathrm{L}}^{-1}\left(\frac{\frac{1}{2}}{\mathrm{s}+1}\right)$

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

= $-1+\frac{1}{2}{\mathrm{e}}^{\mathrm{t}}+\frac{1}{2}{\mathrm{e}}^{-\mathrm{t}}$

Concept Used:-

Laplace transform is the particular integral formulas and very useful to solve the equations of linear differential in some certain conditions.

Some useful formulas of Laplace transform which we will use in the given problem are, 

$\begin{array}{rl}& L^{-1}\left\{\frac{s-a}{(s-a{)}^2+b^2}\right\}=e^{at}\cos (bt)\\ & L^{-1}\left\{\frac{1}{(s-a{)}^2+b^2}\right\}=e^{at}\sin (bt)\end{array}$

Explanation:-

Given function is, 

$\frac{6\mathrm{s}-4}{{\mathrm{s}}^2-4\mathrm{s}+20}$

Here, the parameter s is real. Its Laplace transform will be $L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}$.

Rewrite the expression to use the direct result of Laplace transform as,

$$\begin{gathered} \Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=L^{-1}\left\{{\displaystyle \frac{6s-12+8}{s^2-4s+4+16}}\right\} \\ \Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=L^{-1}\left\{{\displaystyle \frac{6(s-2)+8}{(s-2{)}^2+16}}\right\} \\ \Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=L^{-1}\{{\displaystyle \frac{6(s-2)}{(s-2{)}^2+16}}+{\displaystyle \frac{8}{(s-2{)}^2+16}}\} \\ \Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=6L^{-1}\left\{{\displaystyle \frac{(s-2)}{(s-2{)}^2+16}}\right\}+2L^{-1}\left\{{\displaystyle \frac{4}{(s-2{)}^2+16}}\right\} \\ \Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=6L^{-1}\left\{{\displaystyle \frac{(s-2)}{(s-2{)}^2+4^2}}\right\}+2L^{-1}\left\{{\displaystyle \frac{4}{(s-2{)}^2+4^2}}\right\} \end{gathered}$$Now, using the above Laplace transform values, we get,

$\Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=$ 6e2t cos 4t + 2e2t sin 4t

So, the inverse Laplace transform of the given function is 6e2t cos 4t + 2e2t sin 4t.

Hence, the correct option is 2.

Explanation:

$F\left(s\right)=\frac{3\left(s+3\right)}{s^2+6s+8}$

By applying partial fraction method:

$F\left(s\right)=\frac{3\left(s+3\right)}{\left(s+2\right)\left(s+4\right)}=\frac{A}{\left(s+2\right)}+\frac{B}{\left(s+4\right)}$

35 + 9 = A (s + 4) + B (s + 2)

Comparing both the sides:

A + B = 3 and 4A + 2B = 9

A = B = 1.5

$F\left(s\right)=\frac{1.5}{\left(s+2\right)}+\frac{1.5}{\left(s+4\right)}$

By applying inverse Laplace transform, we get:

f(t) = 1.5e^-2t + 1.5e^-4t

At t = 0.5

$f\left(0.5\right)=1.5\left(\frac{1}{e}+\frac{1}{e^2}\right)=1.5\left(\frac{1+e}{e^2}\right)$

∴ the value of K is 1.5

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:

If L^-1{F(s)} = f(t)

then

L^-1F(s – a) = e^at.f(t) =  e^at. L^-1{F(s)}

Additional Information

L^-1{F(s + a)} = e^-at.f(t)

From the basic definition of Laplace transform:

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

f(t) = eat

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

$\mathrm{F}\left(\mathrm{s}\right)=\frac{1}{\mathrm{s}-\mathrm{a}}$

Concept Used:-

Laplace transform is the particular integral formulas and very useful to solve the equations of linear differential in some certain conditions.

Some useful formulas of Laplace transform which we will use in the given problem are, 

$\begin{array}{rl}& L^{-1}\left\{\frac{s-a}{(s-a{)}^2+b^2}\right\}=e^{at}\cos (bt)\\ & L^{-1}\left\{\frac{1}{(s-a{)}^2+b^2}\right\}=e^{at}\sin (bt)\end{array}$

Explanation:-

Given function is, 

$\frac{6\mathrm{s}-4}{{\mathrm{s}}^2-4\mathrm{s}+20}$

Here, the parameter s is real. Its Laplace transform will be $L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}$.

Rewrite the expression to use the direct result of Laplace transform as,

$$\begin{gathered} \Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=L^{-1}\left\{{\displaystyle \frac{6s-12+8}{s^2-4s+4+16}}\right\} \\ \Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=L^{-1}\left\{{\displaystyle \frac{6(s-2)+8}{(s-2{)}^2+16}}\right\} \\ \Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=L^{-1}\{{\displaystyle \frac{6(s-2)}{(s-2{)}^2+16}}+{\displaystyle \frac{8}{(s-2{)}^2+16}}\} \\ \Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=6L^{-1}\left\{{\displaystyle \frac{(s-2)}{(s-2{)}^2+16}}\right\}+2L^{-1}\left\{{\displaystyle \frac{4}{(s-2{)}^2+16}}\right\} \\ \Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=6L^{-1}\left\{{\displaystyle \frac{(s-2)}{(s-2{)}^2+4^2}}\right\}+2L^{-1}\left\{{\displaystyle \frac{4}{(s-2{)}^2+4^2}}\right\} \end{gathered}$$Now, using the above Laplace transform values, we get,

$\Rightarrow L^{-1}\left\{\frac{6s-4}{s^2-4s+20}\right\}=$ 6e2t cos 4t + 2e2t sin 4t

So, the inverse Laplace transform of the given function is 6e2t cos 4t + 2e2t sin 4t.

Hence, the correct option is 2.

Explanation:

$\text{Given},\text{ f}\left(\text{t}\right)={\text{e}}^{\text{at}}$

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

$\text{L}\left[{\text{e}}^{\text{at}}\right]={\int }_0^{\mathrm{\infty }}{\text{e}}^{-\text{st}}{\text{e}}^{\text{at}}\text{dt}={\int }_0^{\mathrm{\infty }}{\text{e}}^{-(\text{s}-\text{a})\text{t}}\text{dt}$

$\therefore \text{L}\left[{\text{e}}^{\text{at}}\right]={\left(\frac{{\text{e}}^{-(\text{s}-\text{a})\text{t}}}{-(\text{s}-\text{a})}\right)}_0^{\mathrm{\infty }}=-\frac{1}{\text{s}-\text{a}}(0-1)=\frac{1}{\text{s}-\text{a}}$

∴ Laplace transform of L[eat] is $\frac{1}{\mathbf{s}-\mathbf{a}}$

Concept:

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

$L[\cos at]⟷\frac{s}{s^2+a^2}$

Laplace Inverse of $L^{-1}\left[\frac{s}{s^2+{\alpha }^2}\right]=\cos \alpha t$

where 'a' is any positive integer.

Explanation:

$F\left(s\right)=\frac{3\left(s+3\right)}{s^2+6s+8}$

By applying partial fraction method:

$F\left(s\right)=\frac{3\left(s+3\right)}{\left(s+2\right)\left(s+4\right)}=\frac{A}{\left(s+2\right)}+\frac{B}{\left(s+4\right)}$

35 + 9 = A (s + 4) + B (s + 2)

Comparing both the sides:

A + B = 3 and 4A + 2B = 9

A = B = 1.5

$F\left(s\right)=\frac{1.5}{\left(s+2\right)}+\frac{1.5}{\left(s+4\right)}$

By applying inverse Laplace transform, we get:

f(t) = 1.5e^-2t + 1.5e^-4t

At t = 0.5

$f\left(0.5\right)=1.5\left(\frac{1}{e}+\frac{1}{e^2}\right)=1.5\left(\frac{1+e}{e^2}\right)$

∴ the value of K is 1.5

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:

If L^-1{F(s)} = f(t)

then

L^-1F(s – a) = e^at.f(t) =  e^at. L^-1{F(s)}

Additional Information

L^-1{F(s + a)} = e^-at.f(t)

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:

$F\left(s\right)=\frac{2s^2-4}{\left(s-3\right)\left(s^2-s-2\right)}$

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

$=\frac{7}{2\left(s-3\right)}-\frac{1}{6\left(s+1\right)}-\frac{4}{3\left(s-2\right)}$

Apply Inverse Laplace transform on both sides,