8.2.1: The Inverse Laplace Transform (Exercises)

Last updated
Save as PDF

Page ID: 30765

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

Q8.2.1

1. Use the table of Laplace transforms to find the inverse Laplace transform.

\( \dfrac{3}{(s-7)^4}\)
\( \dfrac{2s-4}{s^2-4s+13}\)
\( \dfrac{1}{s^2+4s+20}\)
\( \dfrac{2}{s^2+9}\)
\( \dfrac{s^2-1}{(s^2+1)^2}\)
\( \dfrac{1}{(s-2)^2-4}\)
\( \dfrac{12s-24}{(s^2-4s+85)^2}\)
\( \dfrac{2}{(s-3)^2-9}\)
\( \dfrac{s^2-4s+3}{(s^2-4s+5)^2}\)

2. Use Theorem 8.2.1 and the table of Laplace transforms to find the inverse Laplace transform.

\( \dfrac{2s+3}{(s-7)^4}\)
\( \dfrac{s^2-1}{(s-2)^6}\)
\( \dfrac{s+5}{ s^2+6s+18}\)
\( \dfrac{2s+1}{ s^2+9}\)
\( \dfrac{s}{ s^2+2s+1}\)
\( \dfrac{s+1}{ s^2-9}\)
\( \dfrac{s^3+2s^2-s-3}{(s+1)^4}\)
\( \dfrac{2s+3}{(s-1)^2+4}\)
\( \dfrac{1}{ s}-\dfrac{s}{ s^2+1}\)
\( \dfrac{3s+4}{ s^2-1}\)
\( \dfrac{3}{ s-1}+\dfrac{4s+1}{ s^2+9}\)
\( \dfrac{3}{(s+2)^2}-\dfrac{2s+6}{ s^2+4}\)

3. Use Heaviside’s method to find the inverse Laplace transform.

\( \dfrac{3-(s+1)(s-2)}{(s+1)(s+2)(s-2)}\)
\( \dfrac{7+(s+4)(18-3s)}{(s-3)(s-1)(s+4)}\)
\( \dfrac{2+(s-2)(3-2s)}{(s-2)(s+2)(s-3)}\)
\( \dfrac{3-(s-1)(s+1)}{(s+4)(s-2)(s-1)}\)
\( \dfrac{3+(s-2)(10-2s-s^2)}{(s-2)(s+2)(s-1)(s+3)}\)
\( \dfrac{3+(s-3)(2s^2+s-21)}{(s-3)(s-1)(s+4)(s-2)}\)

4. Find the inverse Laplace transform.

\( \dfrac{2+3s}{(s^2+1)(s+2)(s+1)}\)
\( \dfrac{3s^2+2s+1}{(s^2+1)(s^2+2s+2)}\)
\( \dfrac{3s+2}{(s-2)(s^2+2s+5)}\)
\( \dfrac{3s^2+2s+1}{(s-1)^2(s+2)(s+3)}\)
\( \dfrac{2s^2+s+3}{(s-1)^2(s+2)^2}\)
\( \dfrac{3s+2}{(s^2+1)(s-1)^2}\)

5. Use the method of Example 8.2.9 to find the inverse Laplace transform.

\( \dfrac{3s+2}{(s^2+4)(s^2+9)}\)
\( \dfrac{-4s+1}{(s^2+1)(s^2+16)}\)
\( \dfrac{5s+3}{(s^2+1)(s^2+4)}\)
\( \dfrac{-s+1}{(4s^2+1)(s^2+1)}\)
\( \dfrac{17s-34}{(s^2+16)(16s^2+1)}\)
\( \dfrac{2s-1}{(4s^2+1)(9s^2+1)}\)

6. Find the inverse Laplace transform.

\( \dfrac{17 s-15}{(s^2-2s+5)(s^2+2s+10)}\)
\( \dfrac{8s+56}{(s^2-6s+13)(s^2+2s+5)}\)
\( \dfrac{s+9}{(s^2+4s+5)(s^2-4s+13)}\)
\( \dfrac{3s-2}{(s^2-4s+5)(s^2-6s+13)}\)
\( \dfrac{3s-1}{(s^2-2s+2)(s^2+2s+5)}\)
\( \dfrac{20s+40}{(4s^2-4s+5)(4s^2+4s+5)}\)

7. Find the inverse Laplace transform.

\( \dfrac{1}{ s(s^2+1)}\)
\( \dfrac{1}{(s-1)(s^2-2s+17)}\)
\( \dfrac{3s+2}{(s-2)(s^2+2s+10)}\)
\( \dfrac{34-17s}{(2s-1)(s^2-2s+5)}\)
\( \dfrac{s+2}{(s-3)(s^2+2s+5)}\)
\( \dfrac{2s-2}{(s-2)(s^2+2s+10)}\)

8. Find the inverse Laplace transform.

\( \dfrac{2s+1}{(s^2+1)(s-1)(s-3)}\)
\( \dfrac{s+2}{(s^2+2s+2)(s^2-1)}\)
\( \dfrac{2s-1}{(s^2-2s+2)(s+1)(s-2)}\)
\( \dfrac{s-6}{(s^2-1)(s^2+4)}\)
\( \dfrac{2s-3}{ s(s-2)(s^2-2s+5)}\)
\( \dfrac{5s-15}{(s^2-4s+13)(s-2)(s-1)}\)

9. Given that \(f(t)\leftrightarrow F(s)\), find the inverse Laplace transform of \(F(as-b)\), where \(a>0\).

1. If \(s_1\), \(s_2\), …, \(s_n\) are distinct and \(P\) is a polynomial of degree less than \(n\), then \[{P(s)}{(s-s_1)(s-s_2)\cdots(s-s_n)}= {A_1}{ s-s_1}+{A_2}{ s-s_2}+\cdots+{A_n}{ s-s_n}.\nonumber \] Multiply through by \(s-s_i\) to show that \(A_i\) can be obtained by ignoring the factor \(s-s_i\) on the left and setting \(s=s_i\) elsewhere.
2. Suppose \(P\) and \(Q_1\) are polynomials such that \(\mbox{degree}(P)\le\mbox{degree}(Q_1)\) and \(Q_1(s_1)\ne0\). Show that the coefficient of \(1/(s-s_1)\) in the partial fraction expansion of \[F(s)={P(s)}{(s-s_1)Q_1(s)}\nonumber \] is \(P(s_1)/Q_1(s_1)\).
3. Explain how the results of (a) and (b) are related.

Search

Text Color

Text Size

Margin Size

Font Type

Q8.2.1