- #1
PiRho31416
- 19
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[Solved] Laplace Transform
[tex]\frac{d^{2}y}{dt^{2}}+4y=sin(t),\quad y(0)=0,\quad\frac{dy}{dt}(0)=0[/tex]
Laplace transform is defined as:
[tex]\mathcal{L}\{f(t)\} = \int_{-\infty}^{\infty}f(t)e^{st}dt[/tex]
[tex]
\frac{d^{2}y}{dt^{2}}+4y=sin(t),\quad y(0)=0,\quad\frac{dy}{dt}(0)=0 [/tex]
[tex]s^{2}\mathcal{L}\{y\}-sy'(0)-y(0)-4\mathcal{L}\{y\}=\frac{1}{s^{2}+1}[/tex]
[tex] \mathcal{L}\{y\}(s^{2}-4)=\frac{1}{s^{2}+1} [/tex]
[tex] \Rightarrow \frac{1}{s^{2}+1}\cdot \frac{1}{s^{2}-4}=\frac{As+B}{s^2+1}\cdot\frac{Cs+D}{s^2-4} [/tex]
[tex] 1 = (As+B)(s^2-4)\cdot(Cs+D)(s^2+1)[/tex]
[tex] 1 = As^3-4As+Bs^2-4B+Cs^3+Cs+Ds^2+D[/tex]
[tex] 1 = s^3(A+C)+s^2(B+D)+s(C-4A)+D-4B[/tex]
[tex]\begin{bmatrix}
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
-4 & 0 & 1 & 0\\
0 & -4 &0 & 1
\end{bmatrix}
\begin{bmatrix}
A\\
B\\
C\\
D
\end{bmatrix}
=\begin{bmatrix}
0\\
0\\
1\\
0
\end{bmatrix}
\\\
\begin{bmatrix}
A\\
B\\
C\\
D
\end{bmatrix} =
\begin{bmatrix}
-1/5 \\
0 \\
1/5 \\
0
\end{bmatrix}[/tex]
[tex]\mathcal{L}\{Y\}=\frac{1}{5}\frac{1}{s^2-1}-\frac{1}{5}\frac{1}{s^2+1} [/tex]
[tex]\mathcal{L}\{Y\}=\frac{1}{5}\frac{1}{s^2-1}-\frac{1}{5}\frac{1}{s^2+1}[/tex]
[tex]\Rightarrow\mathcal{L}\{\frac{1}{5}\frac{1}{s^2+1}\}=\frac{1}{5}sin(t) [/tex]
[tex]\Rightarrow\mathcal{L}\{\frac{-1}{5}\frac{1}{s^2-1}\}=\mathcal{L}\{\frac{-1}{5}\frac{1}{(s+1)(s-1)}\} [/tex]
So this is where I get stuck using convolution. Since we know
[tex]\Rightarrow \mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t}[/tex]
[tex]\Rightarrow \mathcal{L}^{-1}\{\frac{1}{s-1}\} = e^{t}[/tex]
[tex]f(t)=e^{t}[/tex]
[tex]g(t)=e^{t}[/tex]
[tex]f(t) \ast g(t) = \int_{0}^{t}f(\tau)g(\tau-t)\,\,d\tau[/tex]
[tex]\frac{1}{5}\int_{0}^{t}e^{-\tau}e^{\tau-t}\,\,d\tau
= \frac{1}{5}\int_{0}^{t}e^{-\tau+\tau-t},\,\,d\tau
= \frac{1}{5}\int_{0}^{t}e^{-t}\,\,d\tau \\
=\frac{1}{5}\tau e^{-t} \bigg|_{\tau=0}^{\tau=t} = \frac{1}{5}te^{-t}
[/tex]
Homework Statement
[tex]\frac{d^{2}y}{dt^{2}}+4y=sin(t),\quad y(0)=0,\quad\frac{dy}{dt}(0)=0[/tex]
Homework Equations
Laplace transform is defined as:
[tex]\mathcal{L}\{f(t)\} = \int_{-\infty}^{\infty}f(t)e^{st}dt[/tex]
The Attempt at a Solution
[tex]
\frac{d^{2}y}{dt^{2}}+4y=sin(t),\quad y(0)=0,\quad\frac{dy}{dt}(0)=0 [/tex]
[tex]s^{2}\mathcal{L}\{y\}-sy'(0)-y(0)-4\mathcal{L}\{y\}=\frac{1}{s^{2}+1}[/tex]
[tex] \mathcal{L}\{y\}(s^{2}-4)=\frac{1}{s^{2}+1} [/tex]
[tex] \Rightarrow \frac{1}{s^{2}+1}\cdot \frac{1}{s^{2}-4}=\frac{As+B}{s^2+1}\cdot\frac{Cs+D}{s^2-4} [/tex]
[tex] 1 = (As+B)(s^2-4)\cdot(Cs+D)(s^2+1)[/tex]
[tex] 1 = As^3-4As+Bs^2-4B+Cs^3+Cs+Ds^2+D[/tex]
[tex] 1 = s^3(A+C)+s^2(B+D)+s(C-4A)+D-4B[/tex]
[tex]\begin{bmatrix}
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
-4 & 0 & 1 & 0\\
0 & -4 &0 & 1
\end{bmatrix}
\begin{bmatrix}
A\\
B\\
C\\
D
\end{bmatrix}
=\begin{bmatrix}
0\\
0\\
1\\
0
\end{bmatrix}
\\\
\begin{bmatrix}
A\\
B\\
C\\
D
\end{bmatrix} =
\begin{bmatrix}
-1/5 \\
0 \\
1/5 \\
0
\end{bmatrix}[/tex]
[tex]\mathcal{L}\{Y\}=\frac{1}{5}\frac{1}{s^2-1}-\frac{1}{5}\frac{1}{s^2+1} [/tex]
[tex]\mathcal{L}\{Y\}=\frac{1}{5}\frac{1}{s^2-1}-\frac{1}{5}\frac{1}{s^2+1}[/tex]
[tex]\Rightarrow\mathcal{L}\{\frac{1}{5}\frac{1}{s^2+1}\}=\frac{1}{5}sin(t) [/tex]
[tex]\Rightarrow\mathcal{L}\{\frac{-1}{5}\frac{1}{s^2-1}\}=\mathcal{L}\{\frac{-1}{5}\frac{1}{(s+1)(s-1)}\} [/tex]
So this is where I get stuck using convolution. Since we know
[tex]\Rightarrow \mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t}[/tex]
[tex]\Rightarrow \mathcal{L}^{-1}\{\frac{1}{s-1}\} = e^{t}[/tex]
[tex]f(t)=e^{t}[/tex]
[tex]g(t)=e^{t}[/tex]
[tex]f(t) \ast g(t) = \int_{0}^{t}f(\tau)g(\tau-t)\,\,d\tau[/tex]
[tex]\frac{1}{5}\int_{0}^{t}e^{-\tau}e^{\tau-t}\,\,d\tau
= \frac{1}{5}\int_{0}^{t}e^{-\tau+\tau-t},\,\,d\tau
= \frac{1}{5}\int_{0}^{t}e^{-t}\,\,d\tau \\
=\frac{1}{5}\tau e^{-t} \bigg|_{\tau=0}^{\tau=t} = \frac{1}{5}te^{-t}
[/tex]
Last edited: