6.4
- Page ID
- 155602
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Find the Laplace transform for \( g(t)= \begin{cases}0 & t<4 \\
2 & 4 \leq t<10 \\
t & t \geq 10\end{cases} \\\)
Solution
\[ \notag
\begin{aligned}
& \text { Hence } g(t)=2 u_4(t)+(t-2) u_{10}(t) \\
& \text { Solve } 3 y^{\prime \prime}+y^{\prime}+y=2 u_4(t)+(t-2) u_{10}(t) \text {, } \\
& y(0)=0, y^{\prime}(0)=0 . \\
& 3 \mathcal{L}\left(y^{\prime \prime}\right)+\mathcal{L}\left(y^{\prime}\right)+\mathcal{L}(y)=\mathcal{L}\left(2 u_4(t)\right)+\mathcal{L}\left((t-2) u_{10}(t)\right) \\
& 3\left[s^2 \mathcal{L}(y)-s y(0)-y^{\prime}(0)\right]+s \mathcal{L}(y)-y(0)+\mathcal{L}(y) \\
& =\mathcal{L}\left(2 u_4(t)\right)+\mathcal{L}\left((t-10+8) u_{10}(t)\right) \\
&
\end{aligned}
\]
Thm: \(\mathcal{L}\left(u_c(t) f(t-c)\right)=e^{-c s} \mathcal{L}(f(t))\).
Thus \(\mathcal{L}\left(u_c(t) f(t)\right)=e^{-c s} \mathcal{L}(f(t+c))\)
\[
\begin{aligned}
& 3\left[s^2 \mathcal{L}(y)\right]+s \mathcal{L}(y)+\mathcal{L}(y)=e^{-4 s} \mathcal{L}(2)+e^{-10 s} \mathcal{L}((t+8)) \\
& \mathcal{L}(y)\left[3 s^2+s+1\right]=2 e^{-4 s} \mathcal{L}(1)+e^{-10 s} \mathcal{L}(t)+8 e^{-10 s} \mathcal{L}(1) \\
& \mathcal{L}(y)\left[3 s^2+s+1\right]=e^{-4 s} \frac{2}{s}+e^{-10 s} \frac{1}{s^2}+e^{-10 s} \frac{8}{s} \\
& \mathcal{L}(y)=e^{-4 s} \frac{2}{s\left[3 s^2+s+1\right]}+e^{-10 s} \frac{1}{s^2\left[3 s^2+s+1\right]}+8 e^{-10 s} \frac{1}{s\left[3 s^3+s+1\right]} \\
& y=2 \mathcal{L}^{-1}\left(e^{-4 s} \frac{1}{s\left[3 s^2+s+1\right]}\right)+\mathcal{L}^{-1}\left(e^{-10 s} \frac{1}{s^2\left[3 s^2+s+1\right]}\right) \\
& +8 \mathcal{L}^{-1}\left(e^{-10 s} \frac{1}{s\left[3 s^2+s+1\right]}\right)
\end{aligned}
\]
\[ \notag
y=u_4(t) f(t-4)+u_{10} h(t-10)+8 u_{10} f(t-10)
\]
where \(f(t)=\mathcal{L}^{-1}\left(\frac{1}{s\left[3 s^2+s+1\right]}\right)\) and \(h(t)=\mathcal{L}^{-1}\left(\frac{1}{s^2\left[3 s^2+s+1\right]}\right)\)
\[
\begin{aligned}
& \frac{1}{s\left[3 s^2+s+1\right]}=\frac{A}{s}+\frac{B s+C}{3 s^2+s+2} \\
& 1=A\left(3 s^2+s+1\right)+(B s+C) s \\
& 0 s^2+0 s+1=(3 A+B) s^2+(A+C) s+A \\
& 0=3 A+B, 0=A+C, 1=A
\end{aligned}
\]
Hence \(A=1, B=-3 A=-3, C=-A=-1\)
\[\notag \begin{align}
f(t) & =\mathcal{L}^{-1}\left(\frac{1}{s\left[3 s^2+s+1\right]}\right) \notag \\
& =\mathcal{L}^{-1}\left(\frac{1}{s}+\frac{-3 s-1}{3 s^2+s+1}\right) \notag\\
& =\mathcal{L}^{-1}\left(\frac{1}{s}+\frac{-3 s-1}{3 s^2+s+1}\right) \notag\\
& =1+\mathcal{L}^{-1}\left(\frac{-3 s-1}{3\left[s^2+\frac{1}{3} s+ \frac{1}{36} - \frac{1}{36}+\frac{1}{3}\right]}\right)\notag\\
& =1+\mathcal{L}^{-1}\left(\frac{-3 s-1}{3\left[\left(s+\frac{1}{6}\right)^2-\frac{1}{36}+\frac{1}{3}\right]}\right) \notag\\
& =1+\mathcal{L}^{-1}\left(\frac{-3 s-1}{3\left[\left(s+\frac{1}{6}\right)^2-\frac{1}{36}+\frac{1}{3}\right]}\right)\notag\\
& =1+\mathcal{L}^{-1}\left(\frac{-3\left(s+\frac{1}{3}\right)}{3\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}\right)\notag \\
& =1+\mathcal{L}^{-1}\left(\frac{-\left(s+\frac{1}{6}-\frac{1}{6}+\frac{1}{3}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}\right)\notag\\
& =1+\mathcal{L}^{-1}\left(\frac{-\left(s+\frac{1}{6}+\frac{1}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}\right) \notag\\
& =1+\mathcal{L}^{-1}\left(\frac{-\left(s+\frac{1}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}+\frac{-\frac{1}{6}}\notag\\{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}\right) \notag\\
& =1+\mathcal{L}^{-1}\left(\frac{-\left(s+\frac{1}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}+\frac{-\frac{1}{6} \frac{6}{\sqrt{11}} \frac{\sqrt{11}}{6}}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}\right) \notag\\
& =1+\mathcal{L}^{-1}\left(\frac{-\left(s+\frac{1}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}+\frac{-\frac{1}{\sqrt{11}} \frac{\sqrt{11}}{6}}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}\right) \notag\\
& =1-e^{-\frac{1}{6} t} \cos \frac{\sqrt{11}}{6} t-\frac{1}{\sqrt{11}} e^{-\frac{1}{6} t} \sin \frac{\sqrt{11}}{6} t \notag
\end{align}\]
\[ \notag \begin{aligned}
& h(t)=\mathcal{L}^{-1}\left(\frac{1}{s^2\left[3 s^2+s+1\right]}\right) \\
& \frac{1}{s^2\left[3 s^2+s+1\right]}=\frac{A s+D}{s^2}+\frac{B s+C}{3 s^2+s+2} \\
& 1=(A s+D)\left(3 s^2+s+1\right)+(B s+C) s^2 \\
& 0 s^3+0 s^2+0 s+1=(3 A+B) s^3+(A+3 D+C) s^2+(A+D) s+D \\
& 0=3 A+B, 0=A+3 D+C, 0=A+D, 1=D . \\
& \text { Hence } D=1, A=-D=-1, C=-A-3 D=1-3=-2, \\
& B=-3 A=3 .
\end{aligned} \]
\[ \notag
\begin{aligned}
& \frac{1}{s^2\left[3 s^2+s+1\right]}=\frac{-s+1}{s^2}+\frac{3 s-2}{3 s^2+s+1}=\frac{-s}{s^2}+\frac{1}{s^2}+\frac{3\left(s-\frac{2}{3}\right)}{3\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]} \\
& =\frac{-1}{s}+\frac{1}{s^2}+\frac{\left(s+\frac{1}{6}-\frac{1}{6}-\frac{2}{3}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}=\frac{-1}{s}+\frac{1}{s^2}+\frac{\left(s+\frac{1}{6}-\frac{5}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]} \\
& =\frac{-1}{s}+\frac{1}{s^2}+\frac{\left(s+\frac{1}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}-\frac{\left(\frac{5}{6}\right)\left(\frac{6}{\sqrt{11}} \frac{\sqrt{11}}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}\\
& =\frac{-1}{s}+\frac{1}{s^2}+\frac{\left(s+\frac{1}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}-\frac{\left(\frac{5}{6}\right)\left(\frac{6}{\sqrt{11}} \frac{\sqrt{11}}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}\\
& =\frac{-1}{s}+\frac{1}{s^2}+\frac{\left(s+\frac{1}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}-\frac{\frac{5}{\sqrt{11}}\left(\frac{\sqrt{11}}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}\\
\end{aligned}
\]
\[ \notag
\begin{aligned}
h(t) & =\mathcal{L}^{-1}\left(\frac{1}{s^2\left[3 s^2+s+1\right]}\right) \\
& =\mathcal{L}^{-1}\left(\frac{-1}{s}+\frac{1}{s^2}+\frac{\left(s+\frac{1}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}-\frac{\frac{5}{\sqrt{11}}\left(\frac{\sqrt{11}}{6}\right)}{\left[\left(s+\frac{1}{6}\right)^2+\frac{11}{36}\right]}\right) \\
& =-1+t+e^{-\frac{1}{6} t} \cos \frac{\sqrt{11}}{6} t-\frac{5}{\sqrt{11}} e^{-\frac{1}{6} t} \cos \frac{\sqrt{11}}{6} t
\end{aligned}
\]
Hence the final answer is
\[ \notag
\begin{aligned}
& y=u_4(t) f(t-4)+u_{10} h(t-10)+8 u_{10} f(t-10) \\
& =u_4(t)\left[1-e^{-\frac{1}{6}(t-4)} \cos \frac{\sqrt{11}}{6}(t-4)-\frac{1}{\sqrt{11}} e^{-\frac{1}{6}(t-4)} \sin \frac{\sqrt{11}}{6}(t-4)\right] \\
& +u_{10}\left[-1+t+e^{-\frac{1}{6}(t-10)} \cos \frac{\sqrt{11}}{6}(t-10)-\frac{5}{\sqrt{11}} e^{-\frac{1}{6}(t-10)} \cos \frac{\sqrt{11}}{6}(t-10)\right] \\
& +8\left[1-e^{-\frac{1}{6}(t-10)} \cos \frac{\sqrt{11}}{6}(t-10)-\frac{1}{\sqrt{11}} e^{-\frac{1}{6}(t-10)} \sin \frac{\sqrt{11}}{6}(t-10)\right]
\end{aligned}
\]