Coupled Oscillator: Jonathan Schreven - Revision history

Jonathan.schreven: /* Matrix Exponential */

2009-12-11T22:41:56Z

Matrix Exponential

← Older revision		Revision as of 15:41, 11 December 2009
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	\end{bmatrix}		\end{bmatrix}
	</math>		</math>
			You can solve this with a computer program or your calculator and plug it into the equation for '''A'''. I have not listed the answer for this problem here because it is very messy and extremely long. I did calculate it to make sure it is solvable. But if your numbers are easier to work with you would finish by plugging this value into the equation below.
			:<math>\bar{x}=e^{\bold{A}t}\bar{x}(0)</math>

2009-12-11T00:45:17Z

@@ Line 249: / Line 249: @@
 :<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bar{z}(0)</math>
 :<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}\bar{x}(0)</math>
-:<math>\bar{x}=e^{\bold{A}t}\bar{x}(0)</math>
+Notice here that
-where
+:<math>e^{\bold{A}t}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}</math>
-:<math>e^{\bold{A}t}=</math>
+:<math>e^{\bold{A}t}=\begin{bmatrix}

2009-12-11T00:20:12Z

Matrix Exponential

← Older revision		Revision as of 17:20, 10 December 2009
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	:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}\bar{x}(0)</math>		:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}\bar{x}(0)</math>
	:<math>\bar{x}=e^{\bold{A}t}\bar{x}(0)</math>		:<math>\bar{x}=e^{\bold{A}t}\bar{x}(0)</math>
			where
			:<math>e^{\bold{A}t}=</math>

2009-12-11T00:01:32Z

Solving

← Older revision		Revision as of 17:01, 10 December 2009
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	We can now plug these eigen vectors and eigen values into the standard equation		We can now plug these eigen vectors and eigen values into the standard equation
	:<math>x=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>		:<math>\bar{x}=c_1k_1e^{\lambda_1 t}+c_2k_2e^{\lambda_2 t}+c_3k_3e^{\lambda_3 t}+c_4k_4e^{\lambda_4 t}</math>

	And our final answer is		And our final answer is
	:<math>x=c_1\begin{bmatrix}		:<math>\bar{x}=c_1\begin{bmatrix}
	0.2149i \\		0.2149i \\
	-0.5722 \\		-0.5722 \\

2009-12-11T00:00:42Z

Solving

← Older revision		Revision as of 17:00, 10 December 2009
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	-0.5407i \\		-0.5407i \\
	0.6421		0.6421
	\end{bmatrix}e^{1.~~18766it~~}+c_4\begin{bmatrix}		\end{bmatrix}e^{1.1877it}+c_4\begin{bmatrix}
	0.3500i \\		0.3500i \\
	0.4157 \\		0.4157 \\
	0.5407i \\		0.5407i \\
	0.6421		0.6421
	\end{bmatrix}e^{-1.~~18766it~~}</math>		\end{bmatrix}e^{-1.1877it}</math>


	== Matrix Exponential ==		== Matrix Exponential ==

2009-12-11T00:00:16Z

Eigen Values

2009-12-10T23:59:12Z

Matrix Exponential

← Older revision		Revision as of 16:59, 10 December 2009
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	and get		and get
	:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bar{z}(0)</math>		:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bar{z}(0)</math>
			:<math>\bar{x}=\bold{T^{-1}}e^{\bold{\hat{A}}t}\bold{T}\bar{x}(0)</math>
			:<math>\bar{x}=e^{\bold{A}t}\bar{x}(0)</math>

2009-12-10T23:49:23Z

Matrix Exponential

@@ Line 205: / Line 205: @@
 :<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
 :<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}</math>
 where
 :<math>\bold{\hat{A}}=\bold{TAT^{-1}}</math>
 :<math>\bold{\hat{A}}=
 \begin{bmatrix}
@@ Line 233: / Line 225: @@
 -0.2783i & 0.2783i & -0.5407i & 0.5407i \\
 .7409 & 0.7409 & 0.6421 & 0.6421
-\end{bmatrix}
+\end{bmatrix}</math>
+:<math>=
 \begin{bmatrix}
 .6626i & 0 & 0 & 0 \\
@@ Line 241: / Line 233: @@
 & 0 & 0 & 1.1877i
 \end{bmatrix}</math>

2009-12-10T23:21:24Z

@@ Line 165: / Line 165: @@
 == Matrix Exponential ==
-In this section we will use matrix exponentials to solve the same problem. First we start with this identity.
+We already know what the matrix A is from our state space equation
 :<math>\bar{z}=\bold{T}\bar{x}\,</math>
-This can be rearranged by multiplying the inverse of '''T''' to the left side of the equation.
+This can be rearranged by multiplying '''T-inverse''' to the left side of the equations.
 :<math>\bold{T^{-1}}\bar{z}=\bar{x}\,</math>
-Now we can use another identity that we already know
+Now we can bring in the standard form of a state space equation
 :<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>
@@ Line 187: / Line 212: @@
 If we take the Laplace transform of this equation we can come up with the following
-:<math>\bar{z}=e^{\bold{A}t}\bar{z}(0)</math>
+:<math>\bar{z}=e^{\bold{\hat{A}}t}\bar{z}(0)</math>
-If we calculate the value of <math>\bold{\hat{A}}</math> we will find that it is <math>\lambda\bold{I}</math>
+We know the values of T, A, and T^{-1}. If we calculate the value of <math>\bold{\hat{A}}</math> we will find that it is <math>\hat{\lambda}\bold{I}</math>
-:<math>\bold{\hat{A}}=\begin{bmatrix}
+:<math>\bold{\hat{A}}=
 .6626i & 0 & 0 & 0 \\
 & -2.6626i & 0 & 0 \\
 & 0 & 1.1877i & 0 \\
 & 0 & 0 & 1.1877i
 \end{bmatrix}</math>

2009-12-10T22:54:52Z

Matrix Exponential

← Older revision		Revision as of 15:54, 10 December 2009
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	If we calculate the value of <math>\bold{\hat{A}}</math> we will find that it is <math>\lambda\bold{I}</math>		If we calculate the value of <math>\bold{\hat{A}}</math> we will find that it is <math>\lambda\bold{I}</math>
	:<math>\bold{\hat{A}}=\begin{bmatrix}		:<math>\bold{\hat{A}}=\begin{bmatrix}
	& 0 & 0 & 0 \\		2.6626i & 0 & 0 & 0 \\
	0 & & 0 & 0 \\		0 & -2.6626i & 0 & 0 \\
	0 & 0 & & 0 \\		0 & 0 & 1.1877i & 0 \\
	0 & 0 & 0 &		0 & 0 & 0 & 1.1877i
	\end{bmatrix}</math>		\end{bmatrix}</math>