Laplace Transform

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[Solved] Laplace Transform

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:

Answers and Replies

  • #2
Ray Vickson
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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(\tau)=e^{-\tau}[/tex]
[tex]g(\tau-t)=e^{\tau-t][/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]

Why not use 1/[(s+1)(s-1)] = (1/2)[1/(s-1) - 1/(s+1)]?

RGV
 
  • #3
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0
Uhm. The Laplace transform of +4y is not -4Y.
 
  • #5
19
0
Uhm. The Laplace transform of +4y is not -4Y.

Pretty sure it is.
 
  • #6
HallsofIvy
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Pretty sure it's NOT!
If
[tex]\mathcal{L}(y)= \int_0^\infty e^{-st}y(t)dt[/tex]
then
[tex]\mathcal{L}(4y)= \int_0^\infty e^{-st}4y(t)dt= 4\int_0^\infty e^{-st}y(t)dt= 4\mathcal{L}(y)[/tex]
not [itex]-4\mathcal{L}(y)[/itex].
 
  • #7
19
0
Oh crap. The negative sign! Sorry it was late last night and somehow I thought he was saying [tex]\mathcal{L}\{y(t)\}\neq Y[/tex]

Sorry about that. He even says it's +4y not -4y. Geez :-(

Pretty sure it's NOT!
If
[tex]\mathcal{L}(y)= \int_0^\infty e^{-st}y(t)dt[/tex]
then
[tex]\mathcal{L}(4y)= \int_0^\infty e^{-st}4y(t)dt= 4\int_0^\infty e^{-st}y(t)dt= 4\mathcal{L}(y)[/tex]
not [itex]-4\mathcal{L}(y)[/itex].
 
  • #8
Ray Vickson
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For what it's worth: the convolution of exp(t) and exp(-t) is sinh(t), not c*t*exp(-t) or c*t*exp(t). You made an error in your convolution integral.

RGV
 
  • #9
vela
Staff Emeritus
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Why not use 1/[(s+1)(s-1)] = (1/2)[1/(s-1) - 1/(s+1)]
For what it's worth: the convolution of exp(t) and exp(-t) is sinh(t), not c*t*exp(-t) or c*t*exp(t). You made an error in your convolution integral.
The OP made several mistakes. There should be no 1/(s2-1) term in the problem in the first place.
 

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