State Space Form - Revision history

Wilspa at 21:49, 8 March 2011

2011-03-08T21:49:46Z

2011-03-07T23:20:35Z

← Older revision		Revision as of 16:20, 7 March 2011
Line 126:		Line 126:
	</math><br>		</math><br>
	So we can see that:<br>		So we can see that:<br>
	<math>C=		<math>\mathbf{C}=
	\begin{bmatrix}		\begin{bmatrix}
	0 \ 1 \\		0 \ 1 \\
	\end{bmatrix}</math><br>		\end{bmatrix}</math><br>
	and<br>		and<br>
	<math>D=[0]</math>.<br>		<math>\mathbf{D}=[0]</math>.<br>



	I'll be adding more on how to use state space form to solve these equations as I have time.		I'll be adding more on how to use state space form to solve these equations as I have time.

2011-03-07T23:19:50Z

2011-03-07T23:19:16Z

← Older revision		Revision as of 16:19, 7 March 2011
Line 114:		Line 114:
	From this we can see that:<br>		From this we can see that:<br>
	<math>y(t)=		<math>y(t)=
			\begin{bmatrix}
			0 \ 1 \\
			\end{bmatrix}
			\begin{bmatrix}
			x1 \\
			x2 \\
			\end{bmatrix}+
	\begin{bmatrix}		\begin{bmatrix}
	0 \\		0 \\
	1 \\		\end{bmatrix}f(t)
	\end{bmatrix}~~\mathbf{x}+~~[0]~~f(t)~~		</math><br>
	</math>		So we can see that:<br>
			</math>C=
			\begin{bmatrix}
			0 \ 1 \\
			\end{bmatrix}</math><br>
			and<br>
			<math>D=[0]</math>.<br>



	I'll be adding more on how to use state space form to solve these equations as I have time.		I'll be adding more on how to use state space form to solve these equations as I have time.

2011-03-07T23:15:44Z

@@ Line 42: / Line 42: @@
 Or, more generally,<br>
 <math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br>
 This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.
@@ Line 102: / Line 108: @@
 \end{bmatrix}</math><br>
 We now have our differential equations in state space form.<br>
 I'll be adding more on how to use state space form to solve these equations as I have time.

2011-03-07T22:50:30Z

← Older revision		Revision as of 15:50, 7 March 2011
Line 91:		Line 91:
	\end{bmatrix}</math><br>		\end{bmatrix}</math><br>
	So then, since <math>Vin</math> is our forcing function <math>f</math>:<br>		So then, since <math>Vin</math> is our forcing function <math>f</math>:<br>
	<math>\mathbf{A}=		<math>\mathbf{A}=-\frac{R}{L}
	\begin{bmatrix}		\begin{bmatrix}
	1 & 1 \\		1 & 1 \\
	1& 2 \\		1& 2 \\
	\end{bmatrix}</math>		\end{bmatrix}</math>
	,<math>\mathbf{B}=		,<math>\mathbf{B}=\frac{1}{L}
	\begin{bmatrix}		\begin{bmatrix}
	1 \\		1 \\

2011-03-07T22:10:19Z

← Older revision		Revision as of 15:10, 7 March 2011
Line 101:		Line 101:
	1 \\		1 \\
	\end{bmatrix}</math><br>		\end{bmatrix}</math><br>
	We now have our differential equations in state space form.		We now have our differential equations in state space form.<br>
			I'll be adding more on how to use state space form to solve these equations as I have time.

2010-09-10T19:18:21Z

← Older revision		Revision as of 12:18, 10 September 2010
Line 46:		Line 46:
	-Example-		-Example-

	For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)		For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br>
	[[Image:Circuit1.jpg]]		[[Image:Circuit1.jpg]]

2010-09-10T19:18:04Z

@@ Line 58: / Line 58: @@
 Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br>
 So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br>
-<math>-Vin +(x_{1}+x_{2})*R + L*\dot x_{1}=0</math> and<br>
+<math>-Vin +(x_{1}+x_{2})R + L\dot x_{1}=0</math> and<br>
-<math>-Vin +(x_{1}+x_{2})*R + x_{2}*R + L*\dot x_{2} = 0</math><br>
+<math>-Vin +(x_{1}+x_{2})R + x_{2}R + L\dot x_{2} = 0</math><br>
 Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.

2010-09-10T19:03:29Z

@@ Line 42: / Line 42: @@
 Or, more generally,<br>
 <math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br>
-This is called the state space representation of the differential equation.
+This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.
-Example coming
+-Example-