Class Wiki - User contributions [en]

Example: Metal Cart

2010-02-16T03:44:27Z

Jimmy.apablaza: /* Comments */

==Problem==
A DC generator is built using a metal cart with metallic wheels that travel around a set of perfectly conducting rails in a large circle. The rails are L m apart and there is a uniform magnetic <math>\vec B</math> field normal to the plane. The cart has a penguin,with mass m, and is driven by a rocket engine having a constant thrust <math> F_1 </math>. A wet polar bear, having stumbled out of a shack where he recently had a bad experience with a battery, lays dying across the tracks acting as a load resistance R over the rails. Find The current as a function of time. What is the current after the generator attains the steady-state condition?

[[Image:Emec_cart_polarBear2.png]]

<blockquote>Problem loosely based on 2.6 from Electric Machinery and Transformers, 3rd ed
<ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001)</ref></blockquote>
==Solution==

For this Problem the large circle will be represented by a pair of parallel wires and the cart as a single wire. This is illustrated below in the top and end view figures.
[[Image:Emec_cart_topview.png]]

[[Image:Emec_cart_endview.png]]

We have two forces, <math> F_1 </math> being the force from the rocket engine and <math> F_2 </math> being the force caused by the current in the conductor and the Magnetic Field.
The resulting Force <math> F_t </math> is simply the sum of <math> F_1 </math> and <math> F_2 </math>

<math> F_2 </math> can be found using Ampere's Law

<math>\vec F=\int\limits_{c} I ~\vec dl\times \vec B~~~~~\Longrightarrow~~~~~ \vec F_2=\int\limits_{0}^{L} I ~\vec dl\times \vec B ~~~~~\Longrightarrow~~~~~ \vec F_2=- I(t) ~B~L ~~ \hat i</math>

We can also say that <math> I(t)=\frac{-e_m(t)}{R} </math>

And <math>~~ {e_m(t)}= \int\limits_{o}^{L} (\vec v \times \vec B)~\vec dl ~~~~~\Longrightarrow~~~~~ {e_m(t)}=-v~L~B </math>

<math> I(t)=\frac{v~L~B}{R} ~~~~~~~~~~~~~~~~~~~~~~\vec F_2=- I(t) ~B~L ~~ \hat i~~~~~\Longrightarrow~~~~~\vec F_2=- \frac{v~L~B}{R} ~B~L ~~ \hat i ~~~~~\Longrightarrow~~~~~\vec F_2=- \frac{v~L^2~B^2}{R} ~~ \hat i</math>

<math> \dot v=\frac{F_1}{m} ~\hat i~+\frac {F_2}{m}~ \hat i ~~~~~\Longrightarrow~~~~~ \dot v= ~\frac{F_1}{m} ~~\hat i~- \frac{v~L^2~B^2}{R m}~~\hat i ~~~~~\Longrightarrow~~~~~ \dot v~ +~ v \left(\frac{L^2~B^2}{R~m}\right) ~-~\frac{F_1}{m}~=~0 </math>

Now we have a lovely differential equation to work with! To attempt to find the current we will take the Laplace transform.

<math> \mathcal{L}\left\{ \dot v ~+~v\left(\frac{L^2~B^2}{R~m}\right)-\frac{F_1}{m} \right\} ~~~~~\Longrightarrow~~~~~ s~V(s)~+~\frac{L^2B^2}{R~m} V(s) ~-~\frac{F_1}{m~s}</math>

Lets title and substitute in the variable <math>~~~~\psi=\frac{L^2~B^2}{R ~m}~~</math> to simplify things

<math> s~V(s)~+~\psi~V(s)~-~\frac{F_1}{m~s}~=~0</math>

<math> V(s)~(s~+~\psi)~=~\frac{F_1}{m~s} ~~~~~\Longrightarrow~~~~~ V(s)~=~\frac {F_1}{m~s~(s~+~\psi)}</math>

Using partial fraction expansion

<math>\frac {F_1}{m~s~(s~+~\psi)}~~~~~~\Longrightarrow~~~~~~\frac{F_1/m\psi}{s}~-~\frac{F_1/m\psi}{s~+~\psi} </math>

<math>V(s)=\frac{F_1/m\psi}{s}~-~\frac{F_1/m\psi}{s~+~\psi}</math>

<math>V(t)= \mathcal{L}\left\{ \frac{F_1/m\psi}{s}~-~\frac{F_1/m\psi}{s~+~\psi} \right\} ~~~~~\Longrightarrow~~~~~ V(t)=\frac {F_0}{m~\psi}\left(u(t)~-~e^{-\psi~t} ~u(t)\right) ~~~~~\Longrightarrow~~~~~ V(t)= \frac {F_0}{m~\psi}\left(1~-~e^{-\psi~t}\right) </math>

<math> V(t)=\frac {F_0}{m~ \frac{L^2~B^2}{R ~m}}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right) ~~~~~\Longrightarrow~~~~~ V(t)=\frac {F_0~R}{L^2~B^2}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)</math>

We know that
<math>~~ e_m(t)~=~V(t)LB </math>

So we can substitute in V(t) to get

<math> e_m(t)= \frac {F_0~R}{L~B}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)</math>

And we know that <math>~~I(t)~=~\frac{e_m(t)}{R}</math>

<math> I(t)=\frac{\frac {F_0~R}{L~B}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)}{R}~~~~~\Longrightarrow~~~~~ I(t)~=~\frac {F_0}{L~B}\left(1~-~e^{-\frac{L^2~B^2}{R ~m}~t}\right)</math>

To find the steady-state current we simply look at the limit of I(t) as <math>t \rightarrow \infty</math>

<math>\lim_{t\rightarrow \infty} I(t)=\frac{F_0}{L~B} \left( 1- e^{-\infty}\right) ~~~~~\Longrightarrow~~~~~ I(\infty)=\frac{F_0}{L~B} </math>

So the Steady-State Current = <math>\frac{F_0}{L~B}</math>

<math> In~conclusion~it ~can ~be ~seen ~that ~a ~penguin ~driven, ~polar ~bear ~killing ~generator </math>

<math>~would ~be ~a ~viable ~option ~for ~alternative ~energy ~in ~Canada. </math>

==References==

<references />

==Reviewed by==

[[Kirk Betz]] Read and approved 1-26-10

[[Will Griffith]] Approved 1-27-10

==Read By==

==Comments==
*I completely agree with your conclusion (Tim Rasmussen)
*I'm just worried about the ecologists. (J. Apablaza)

Example: Ideal Transformer Exercise

2010-01-27T03:17:40Z

Jimmy.apablaza: /* Reviewed By */

==Author==

John Hawkins

==Problem Statement==

An ideal transformer has a primary winding with 500 turns and a secondary winding with 2000 turns. Given that <math>\ e_1=120\angle{0^\circ}\text{ V, RMS}</math> and <math>\ i_1=(2+3j)\text{ A}</math>, find the load impedance, <math>\ Z_L</math> and the Thevenin equivalent, <math>\ Z_{th}</math>.

==Solution==

We could find the Thevenin impedance directly, but we will save that until the end as a checking mechanism. First, we will find the actual load impedance by finding the current and voltage in the secondary winding and finding their ratio. The equations used are those derived in class by Professor Frohne.

<center>

<math>e_2=\frac{N_2}{N_2}e_1=\frac{2000}{500}(120)=480\text{ V}</math>

<br />

<math>i_2=\frac{N_1}{N_2}i_1=\frac{500}{2000}(2+3j)=\left(\frac{1}{2}+\frac{3}{4}j\right)\text{ A}</math>

<br />

<math>Z_L=\frac{e_2}{i_2}=\frac{480}{\frac{1}{2}+\frac{3}{4}j}=\mathbf{(295.4-443.1j)\ \Omega\ =(532.5\angle{-56.3^\circ})\ \Omega}</math>

<br />

<math>Z_{th}=\left(\frac{N_1}{N_2}\right)^2Z_L=\left(\frac{500}{2000}\right)^2(295.4-443.1j)=\mathbf{(18.5-27.7j)\ \Omega\ =(33.3\angle{-56.3^\circ})\ \Omega}</math>

</center>

<br />

As mentioned at the beginning, this should be the impedance found using the ratio of the primary voltage and current. Using this method, we find that

<br />

<center>

<math>Z_{th}=\frac{e_1}{i_1}=\frac{120}{2+3j}=\mathbf{(18.5-27.7j)\ \Omega\ =(33.3\angle{-56.3^\circ})\ \Omega}</math>

</center>

<br />

This is the same answer as above, which verifies the solutions.

==Reviewed By==

*Tyler Anderson
*Jimmy Apablaza-Lorca

==Read By==

==Comments==
Tyler Anderson: it may be helpful to the readers if you referenced what equations you are using. For example:
<math>e_2=\frac{N_2}{N_2}e_1</math> <math> EQ (5-39)</math>
Otherwise it looks sound to me.

John Hawkins: I didn't use the textbook, so such a reference is not required. I agree that it would be useful for those in the class, but I don't have the same textbook as everyone else, and I doubt anyone would care to know my book's equation numbers. Thanks for reminding me about references, though. I mentioned the class derivation above in the text.

Tyler Anderson: haha fair enough then. props for that. perhaps I could barrow your book sometime? cause ours is absolute crap.

John Hawkins: I hate the book I'm using as well, but if you want to use it sometime that would be fine.

J. Apablaza: Everything looks sound to me. Perhaps, you should include an image so that you can earn some extra points.

Example: Ideal Transformer Exercise

2010-01-27T03:17:16Z

Jimmy.apablaza: /* Reviewed By */

==Author==

John Hawkins

==Problem Statement==

An ideal transformer has a primary winding with 500 turns and a secondary winding with 2000 turns. Given that <math>\ e_1=120\angle{0^\circ}\text{ V, RMS}</math> and <math>\ i_1=(2+3j)\text{ A}</math>, find the load impedance, <math>\ Z_L</math> and the Thevenin equivalent, <math>\ Z_{th}</math>.

==Solution==

We could find the Thevenin impedance directly, but we will save that until the end as a checking mechanism. First, we will find the actual load impedance by finding the current and voltage in the secondary winding and finding their ratio. The equations used are those derived in class by Professor Frohne.

<center>

<math>e_2=\frac{N_2}{N_2}e_1=\frac{2000}{500}(120)=480\text{ V}</math>

<br />

<math>i_2=\frac{N_1}{N_2}i_1=\frac{500}{2000}(2+3j)=\left(\frac{1}{2}+\frac{3}{4}j\right)\text{ A}</math>

<br />

<math>Z_L=\frac{e_2}{i_2}=\frac{480}{\frac{1}{2}+\frac{3}{4}j}=\mathbf{(295.4-443.1j)\ \Omega\ =(532.5\angle{-56.3^\circ})\ \Omega}</math>

<br />

<math>Z_{th}=\left(\frac{N_1}{N_2}\right)^2Z_L=\left(\frac{500}{2000}\right)^2(295.4-443.1j)=\mathbf{(18.5-27.7j)\ \Omega\ =(33.3\angle{-56.3^\circ})\ \Omega}</math>

</center>

<br />

As mentioned at the beginning, this should be the impedance found using the ratio of the primary voltage and current. Using this method, we find that

<br />

<center>

<math>Z_{th}=\frac{e_1}{i_1}=\frac{120}{2+3j}=\mathbf{(18.5-27.7j)\ \Omega\ =(33.3\angle{-56.3^\circ})\ \Omega}</math>

</center>

<br />

This is the same answer as above, which verifies the solutions.

==Reviewed By==

Tyler Anderson
Jimmy Apablaza-Lorca

==Read By==

==Comments==
Tyler Anderson: it may be helpful to the readers if you referenced what equations you are using. For example:
<math>e_2=\frac{N_2}{N_2}e_1</math> <math> EQ (5-39)</math>
Otherwise it looks sound to me.

John Hawkins: I didn't use the textbook, so such a reference is not required. I agree that it would be useful for those in the class, but I don't have the same textbook as everyone else, and I doubt anyone would care to know my book's equation numbers. Thanks for reminding me about references, though. I mentioned the class derivation above in the text.

Tyler Anderson: haha fair enough then. props for that. perhaps I could barrow your book sometime? cause ours is absolute crap.

John Hawkins: I hate the book I'm using as well, but if you want to use it sometime that would be fine.

J. Apablaza: Everything looks sound to me. Perhaps, you should include an image so that you can earn some extra points.

Example: Ideal Transformer Exercise

2010-01-27T03:16:53Z

Jimmy.apablaza: /* Comments */

==Author==

John Hawkins

==Problem Statement==

An ideal transformer has a primary winding with 500 turns and a secondary winding with 2000 turns. Given that <math>\ e_1=120\angle{0^\circ}\text{ V, RMS}</math> and <math>\ i_1=(2+3j)\text{ A}</math>, find the load impedance, <math>\ Z_L</math> and the Thevenin equivalent, <math>\ Z_{th}</math>.

==Solution==

We could find the Thevenin impedance directly, but we will save that until the end as a checking mechanism. First, we will find the actual load impedance by finding the current and voltage in the secondary winding and finding their ratio. The equations used are those derived in class by Professor Frohne.

<center>

<math>e_2=\frac{N_2}{N_2}e_1=\frac{2000}{500}(120)=480\text{ V}</math>

<br />

<math>i_2=\frac{N_1}{N_2}i_1=\frac{500}{2000}(2+3j)=\left(\frac{1}{2}+\frac{3}{4}j\right)\text{ A}</math>

<br />

<math>Z_L=\frac{e_2}{i_2}=\frac{480}{\frac{1}{2}+\frac{3}{4}j}=\mathbf{(295.4-443.1j)\ \Omega\ =(532.5\angle{-56.3^\circ})\ \Omega}</math>

<br />

<math>Z_{th}=\left(\frac{N_1}{N_2}\right)^2Z_L=\left(\frac{500}{2000}\right)^2(295.4-443.1j)=\mathbf{(18.5-27.7j)\ \Omega\ =(33.3\angle{-56.3^\circ})\ \Omega}</math>

</center>

<br />

As mentioned at the beginning, this should be the impedance found using the ratio of the primary voltage and current. Using this method, we find that

<br />

<center>

<math>Z_{th}=\frac{e_1}{i_1}=\frac{120}{2+3j}=\mathbf{(18.5-27.7j)\ \Omega\ =(33.3\angle{-56.3^\circ})\ \Omega}</math>

</center>

<br />

This is the same answer as above, which verifies the solutions.

==Reviewed By==

Tyler Anderson

==Read By==

==Comments==
Tyler Anderson: it may be helpful to the readers if you referenced what equations you are using. For example:
<math>e_2=\frac{N_2}{N_2}e_1</math> <math> EQ (5-39)</math>
Otherwise it looks sound to me.

John Hawkins: I didn't use the textbook, so such a reference is not required. I agree that it would be useful for those in the class, but I don't have the same textbook as everyone else, and I doubt anyone would care to know my book's equation numbers. Thanks for reminding me about references, though. I mentioned the class derivation above in the text.

Tyler Anderson: haha fair enough then. props for that. perhaps I could barrow your book sometime? cause ours is absolute crap.

John Hawkins: I hate the book I'm using as well, but if you want to use it sometime that would be fine.

J. Apablaza: Everything looks sound to me. Perhaps, you should include an image so that you can earn some extra points.

Eddy Current

2010-01-20T10:13:45Z

Jimmy.apablaza:

[[Image:Eddy1.png|thumb|Figure 1. Induced magnetic fields in a loaded coil.]]
Eddy current is an electrical phenomenon that occurs when a conductor is exposed to a varying magnetic field which causes a current flow within the conductor.

== Why "Eddy"?==
An Eddy is a term used in fluid mechanics which describes a fluid's swirl and its reverse current after the fluid has overcome an obstacle, like those produced by a dragging oar. A similar phenomenon occurs when a current is induced in a large conductor.

== How does it work? ==
[[Image:Eddy2.png|thumb|Figure 2. Eddy currents.]]
Whenever a current travels in a coil, a magnetic field, or electromagnetic induction, is produced around the coil. This is explained by the Faraday's Law of Induction, which states that a change in the magnetic environment of a coil will induce an electromagnetic force in the coil. The polarity of that emf generated by the change in magnetic flux will also produce a current whose magnetic field is opposed to the change that produces it. This is know as Lenz's Law.

==Applications==

Eddy current is used as a method of measurement, inspection of defects, and correction.

Eddy Current

2010-01-20T10:13:30Z

Jimmy.apablaza:

[[Image:Eddy2.png|thumb|Figure 1. Induced magnetic fields in a loaded coil.]]
Eddy current is an electrical phenomenon that occurs when a conductor is exposed to a varying magnetic field which causes a current flow within the conductor.

== Why "Eddy"?==
An Eddy is a term used in fluid mechanics which describes a fluid's swirl and its reverse current after the fluid has overcome an obstacle, like those produced by a dragging oar. A similar phenomenon occurs when a current is induced in a large conductor.

== How does it work? ==
[[Image:Eddy1.png|thumb|Figure 2. Eddy currents.]]
Whenever a current travels in a coil, a magnetic field, or electromagnetic induction, is produced around the coil. This is explained by the Faraday's Law of Induction, which states that a change in the magnetic environment of a coil will induce an electromagnetic force in the coil. The polarity of that emf generated by the change in magnetic flux will also produce a current whose magnetic field is opposed to the change that produces it. This is know as Lenz's Law.

==Applications==

Eddy current is used as a method of measurement, inspection of defects, and correction.

File:Eddy2.png

2010-01-20T10:10:41Z

Jimmy.apablaza: Eddy Currents

Eddy Currents

File:Eddy1.png

2010-01-20T10:10:16Z

Jimmy.apablaza: Diagram - Induced magnetic field in a coil

Diagram - Induced magnetic field in a coil

Eddy Current

2010-01-20T10:09:05Z

Jimmy.apablaza:

Eddy current is an electrical phenomenon that occurs when a conductor is exposed to a varying magnetic field which causes a current flow within the conductor.

== Why "Eddy"?==
An Eddy is a term used in fluid mechanics which describes a fluid's swirl and its reverse current after the fluid has overcome an obstacle, like those produced by a dragging oar. A similar phenomenon occurs when a current is induced in a large conductor.

== How does it work? ==

Whenever a current travels in a coil, a magnetic field, or electromagnetic induction, is produced around the coil. This is explained by the Faraday's Law of Induction, which states that a change in the magnetic environment of a coil will induce an electromagnetic force in the coil. The polarity of that emf generated by the change in magnetic flux will also produce a current whose magnetic field is opposed to the change that produces it. This is know as Lenz's Law.

==Applications==

Eddy current is used as a method of measurement, inspection of defects, and correction.

Eddy Current

2010-01-20T08:07:10Z

Jimmy.apablaza: /* How does it work? */

Eddy Current

2010-01-20T08:05:28Z

Jimmy.apablaza: /* How does it work? */

Eddy current is an electrical phenomenon that occurs when a conductor is exposed to a varying magnetic field which causes a current flow within the conductor.

== Why "Eddy"?==
An Eddy is a term used in fluid mechanics which describes a fluid's swirl and its reverse current after the fluid has overcome an obstacle, like those produced by a dragging oar. A similar phenomenon occurs when a current is induced in a large conductor.

== How does it work? ==

Whenever a current travels in a coil, a magnetic field, or electromagnetic induction, around the coil. This is explained by the Faraday's Law of Induction, which states that a change in the magnetic environment of a coil will induce an electromagnetic force in the coil. The polarity of that emf generated by the change in magnetic flux will also produce a current whose magnetic field is opposed to the change that produces it. This is know as Lenz's Law

Eddy Current

2010-01-20T07:57:58Z

Jimmy.apablaza:

Eddy Current

2010-01-20T07:48:27Z

Jimmy.apablaza: /* Why "Eddy"? */

Eddy Current

2010-01-20T07:19:58Z

Jimmy.apablaza:

Eddy Current

2010-01-18T23:47:12Z

Jimmy.apablaza: New page: Eddy current is an electrical phenomenon that occurs when a conductor is exposed to a varying magnetic field which causes a current flow within the conductor. ==Why "Eddy"?==

Eddy current is an electrical phenomenon that occurs when a conductor is exposed to a varying magnetic field which causes a current flow within the conductor.

==Why "Eddy"?==

Jimmy Apablaza

2010-01-18T23:43:46Z

Jimmy.apablaza:

== My Articles ==

* [[Electrostatics]]
* [[Faraday's Law]]
* [[Eddy Current]]

Jimmy Apablaza

2010-01-18T23:42:42Z

Jimmy.apablaza: New page: == My Articles ==

== My Articles ==

Class Roster

2010-01-18T23:42:22Z

Jimmy.apablaza:

===Class of 2010===
#[[Eric Clay]]
#[[Jason Osborne]]
#Tim Van Arsdale
#[[Kirk Betz]]
#Corneliu Turturica
#[[Jimmy Apablaza]]
#[[Will Griffith]]
#[[Greg Fong]]
#[[Tyler Anderson]]
#[[Andrew Sell]]
#[[Lau, Chris]]
#[[Kyle Lafferty]]
#[[Matthew Fetke]]
#[[Wesley Brown]]
#[[Erik Biesenthal]]
#[[Jodi Hodge]]
#[[David Robbins]]
#[[Amy Crosby]]
#[[Tim Rasmussen]]
#[[Kevin Starkey EMEC]]
#[[John Hawkins]]
#[[Alex Roddy]]
#[[Aric Vyhmeister]]
#[[Nick Christman]]

Electromechanical Energy Conversion

2010-01-18T23:41:47Z

Jimmy.apablaza:

[[Rules]]

[[Class Roster]]

[[Points]]

==Questions==

What do we do when we are finished with the draft and ready to publish?
* If it's been approved by the reviewers, move it to the articles section

Does anyone know why my LaTEX stuff is changing sizes throughout my article? [[An Ideal Transformer Example]]

*(John Hawkins) As I understand it, the text is made full size (larger) if there is ever a function call, i.e. something starting with a backslash, excluding some things like greek letters. I have just put "\ " (the function call for a space) at the beginning of an equation and had it work. If you don't want to change anything about your equation but just want it displayed full size, type "\,\!" (small forward space and small backward space) somewhere in your equation.
*Thanks John!

==Announcements==

If anyone wants to write the derivation of Ampere's Law you can put it on my (Wesley Brown) [[Ampere's Law]] page and be a co-author.

==Draft Articles==
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.
* [[Ferromagnetism]]
* [[Magnetic Circuits]]
* [[Gauss Meters]]
* [[Ampere's Law]]
* [[DC Motor]]
* [[AC vs. DC]]
* [[AC Motors]]
* [[Fringing]]
* [[Electrostatics]]
* [[Faraday's Law]]
* [[Eddy Current]]
* [[Example Problems of Magnetic Circuits]]
* [[Magnetic Circuit]] (John Hawkins)
* [[Ohm's Law and Reluctance]]
* [[Magnetic Flux]]
* [[An Ideal Transformer Example]]
* [[Example: Ideal Transformer Exercise]] (John Hawkins)
* [[Reference Terms and Units]] (Amy Crosby)

==Reviewed Articles==
These articles have been reviewed and submitted.
* [[Nick_ENGR431_P1|Magnetostatics]] (Nick Christman)
* [[Magnetic Flux]] (Jason Osborne)
*[[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]] (Chris Lau)

Electrostatics

2010-01-11T08:00:47Z

Jimmy.apablaza:

Electrostatics is the study of the electric phenomena under statics conditions. (Under construction...)

Nick ENGR431 P1

2010-01-11T07:55:36Z

Jimmy.apablaza:

== Nick Christman: Paper 1 - Magnetostatics ==

Back to [[Nick ENGR431 | Nick's EMEC Wiki]]

Back to [[Electromechanical Energy Conversion | Class EMEC Wiki]]
<br/>
<br/>

<div style="text-align: center; font-size:15pt;">'''Brief Introduction to Magnetostatics'''</div>
<div style="text-align: center; font-size:13pt;">'''Nick Christman'''</div>

<br/>
<p>
According to several sources, there is an ancient story about a Cretan shepherd named Magnes who accidentally discovered a naturally occurring magnetic material known as lodestone (or loadstone). As the legend goes, Magnes was herding his sheep one day when the iron nails in his shoes and the iron tip on his staff became unusually attracted to a large, dark stone – lodestones contain a mineral, now known as magnetite (<math>Fe_{3}O_{4}</math>), that consists of naturally occurring magnetic properties<ref>Jezek, How Magnets Work</ref>. The first record of using magnets is somewhat of a debatable topic because magnetic properties were first recorded by Greek philosophers possibly as early as the 7th century BC and the Chinese have written records dating circa 4th century BC. In either case, the awareness of magnetic properties has been around for quite some time and the study of magnetism continues to be important and popular portion of research topics.
</p>
<br/>
<p>
Magnetostatics, the study of static magnetic fields, is only a small portion of the overall study of magnetic properties. As just stated, magnetostatics implies that the magnetic field is static – that is, the flow of current that creates the magnetic field is steady or direct current (DC)<ref>Wikipedia, Magnetostatics</ref> – and this allows scientists to make very accurate approximations of how magnetic fields act. In this document, the theory and application behind magnetostatics will be addressed.
</p>
<br/>
<p>
It is said that currents in opposite directions repel while currents in the same directions attract. To demonstrate this, take a common example that is illustrated in David Griffiths ''Introduction to Electrodynamics'', 3rd edition: “[Imagine] two wires hang from the ceiling, a few centimeters apart; when I turn on a current so that it passes up one wire and back down the other, the wires jump apart. Moreover, I could hook up my demonstration so as to make the current flow up both wires; in this case they are found to attract” <ref>Griffiths, p. 203</ref>(Griffiths 203). This demonstration is illustrated in Figure 1. Through experimentation it has been proven that the forces of attraction in this system are not due to electrostatics, but instead are the result of magnetic forces – ''in addition to an electric field, a moving charge also generates a magnetic field''<ref>Griffiths, p. 203</ref> (Griffiths 203). In fact, if one were to allow a current to flow down a straight wire, then a magnetic field circling the wire will be formed – recall the right-hand-rule which states that if you grasp the wire with your right hand (thumb pointing in the direction of current) then your fingers curl in the direction of the magnetic field. Place two wires in close proximity to each other, such as those mentioned in the demonstration above, and there will be a magnetic force between the two wires – hence, in one case the two wires repel and in the other they attract.
</p>
[[Image:Paper1_img1.png |thumb|center|upright=2.5|Figure 1: Illustration of (A) currents in opposite directions repel and (B) currents in the same directions attract. Image created by Nick Christman.]]
<p>
This concept of magnetic force is an interesting, yet complicating topic that is often covered in series of undergraduate and graduate level courses. Generally speaking, however, there is one law that encapsulates all of magnetostatics, which is Lorentz Force Law

<div style="text-align:center;">
<math>\textstyle F = Q(v \times B) </math><ref>Griffiths, p. 204</ref>
</div>

where Q is the charge for which the magnetic field is acting upon, B is the magnetic field, and v is the velocity of the charge Q. (Note that this force, F, is due to the magnetic field acting on Q and that there are no external electric fields present. In the presence of an external electric field, there would be an additional term.) Furthermore, Lorentz Force Law can be expanded in order to account for a current carrying wire rather than a point charge. In the case of a current carrying wire, Lorentz Force Law becomes

<div style="text-align:center;">
<math> \textstyle F = \int (I \times B)dl = \int I(dl \times B) </math><ref>Griffiths, p. 204</ref>
</div>
where <math>I</math>, the current, is represented by a vector pointing in the same direction as <math>dl</math>. In most cases, however, current is treated as constant; thus, Lorentz Force Law can be written as

<div style="text-align:center;">
<math> \textstyle F = I \int (dl \times B) </math><ref>Griffiths, p. 209</ref>
</div>
</p>
<br/>
<p>
Overall, magnetostatics has been derived from a compilation of different sources and it is used to more simply model the properties of magnetism. In both the constant and non-constant current cases presented above, it is clear that the Lorentz Force Law is an important aspect of magnetostatics that can be expanded to account for nearly every study of magnetism. This short overview of magnetostatics is important when one desires to develop and understand magnetic circuits and, ultimately, transformers.
</p>
<br/>
----

<div style="text-align: center; font-size:13pt;">'''References''':</div>

Griffiths, David J. "Introduction to Electrodynamics." Upper Saddle River, N.J: Prentice Hall, 1999.

Jezek, Geno. "History of Magnets." How Magnets Work. Web. 9 Jan. 2010. <http://www.howmagnetswork.com/history.html>.

Mohan, Ned. Electric Drives An Integrative Approach. Minneapolis: Mnpere, 2004.

Wikepdia. "Magnetostatics." Wikipedia, the free encyclopedia. Web. 10 Jan. 2010. <http://en.wikipedia.org/wiki/Magnetostatics>.

<br/>

----

<div style="text-align: center; font-size:13pt;">'''In-text Citations''':</div>
<references/>

<br/>

----

<div style="text-align: center; font-size:13pt;">'''Statistics''':</div>

Word Count: 845 (content, equations, work cited, and title)

Image Count: 1

Potential Points: TBD

----

<div style="text-align: center; font-size:13pt;">'''Reviewers''':</div>

* Kevin Starkey

<br/>

----

<div style="text-align: center; font-size:13pt;">'''Readers''':</div>

* Jimmy Apablaza

Electrostatics

2010-01-11T07:50:56Z

Jimmy.apablaza: /* Electrostatics */

Under construction...

Electrostatics

2010-01-11T07:48:35Z

Jimmy.apablaza: New page: = Electrostatics =

= Electrostatics =

Electromechanical Energy Conversion

2010-01-11T07:48:17Z

Jimmy.apablaza: /* Draft Articles */

[[Rules]]

[[Class Roster]]

[[Points]]

==Articles==
None published to date

==Questions==

What do we do when we are finished with the draft and ready to publish?

==Draft Articles==
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.
* [[Ferromagnetism]]
* [[Magnetic Circuits]]
* [[Gauss Meters]]
* [[Ampere's Law]]
* [[DC Motor]]
* [[AC vs. DC]]
* [[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]]
* [[AC Motors]]
* [[Fringing]]
* [[Nick_ENGR431_P1|Magnetostatics]]
* [[Electrostatics]]
* [[Example problems of magnetic circuits]]
* [[Magnetic Circuit]]
* [[Ohm's Law and Reluctance]]

An Interesting Application for Magnets

2010-01-10T22:48:32Z

Jimmy.apablaza:

===An Interesting Application for Magnets===
By: Will Griffith

Reviewed By: Jimmy Apablaza

Science fiction has sparked many great ideas. Satellites for example were first thought of in some sicence fiction stories. Specificaly communication satalites were first theorized by Arthur C. Clarke and later were found to truly be possible. Not everything theorized in science fiction can become a reality. A good example is Jack Williamson's story "The Metal Man." In this story a scientist studies a location riddled with radiation. The cave is riddled with strange crystalline creatures and structures. He tries to escape but before he can reach home the affects of the radiation turn his body into metal. As of now no known form of radiation could cause this, and it is unlikely that any would. Still at the time radiation was not understood and so this author came up with a possible idea of something radiation can do. still can give use good stepping stones in the right direction, and often expand our thinking to relms that we would have never dreamed of going. In our discussion of magnetic circuits I am reminded of an idea I derived from a science fiction book. My idea comes from Ray Bradbury's book Fahrenheit 451. In this book there are firemen and in there fire station they have a fire-poll. This is no ordinary poll though. This fire-poll can go not only down up it can take it's rider up. This got me thinking one day about how someone could do this. I started thinking about Maglev trains and how they work. For propulsion the rail changes part of it's magnetic field to pull the front of the train forward and push the back of the train the same direction. It also levitates the train above the rail so there is hopefully no friction involved. Well this concept could be applied to a fire-poll. You could just make the main poll magnetic and then hook an apparatus too it. This could either just be something you gripped, or better something you could grip and a place to put one or both feet. Realistically this would be impractical energy wise, it would be better to use conventional elevators or just the stairs. But still this would be amazing to see.
<br>
[[Image:Asm0001.jpg]]

This is a concept design I made in Pro-E.
(This article was submitted for review 1-7-10 at 10:30pm).

I am making a video too but it wont be up untell later and so I will just submit it individually then.

Electromechanical Energy Conversion

2010-01-04T20:45:54Z

Jimmy.apablaza:

Class 2010
#Eric Clay
#Jason Osborne
#Tim Van Arsdale
#Kirk Betz
#Corneliu Turturica
#Jimmy Apablaza

Coupled Oscillator: Double Pendulum

2009-12-13T19:46:41Z

Jimmy.apablaza: /* Matrix Exponential */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are negligible
* no damping forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>\underline{x} = c_1 \underline{k_1} e^{\lambda_1 t} + c_2 \underline{k}_2 e^{\lambda_2 t} + c_3 \underline{k_3} e^{\lambda_3 t} + c_4 \underline{k_4} e^{\lambda_4 t}</math>

: <math>

\underline{x}
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
A
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

So, the exponential matrix becomes,

:<math>
\dot{\underline{x}}=\widehat{A}\underline{x}=T^{-1}e^{\hat{A}t}T\underline{x}=e^{At}\underline{x}
</math>

where

:<math>
e^{\hat{A}t}
=
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
</math>

Hence,

: <math>
\dot{\underline{x}}
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
\underline{x}
</math>

so, the matrix exponential can be solved using matrix multiplication.

Coupled Oscillator: Double Pendulum

2009-12-13T19:38:13Z

Jimmy.apablaza: /* Matrix Exponential */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are negligible
* no damping forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>\underline{x} = c_1 \underline{k_1} e^{\lambda_1 t} + c_2 \underline{k}_2 e^{\lambda_2 t} + c_3 \underline{k_3} e^{\lambda_3 t} + c_4 \underline{k_4} e^{\lambda_4 t}</math>

: <math>

\underline{x}
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
A
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

So, the exponential matrix becomes,

:<math>
\dot{\underline{x}}=\widehat{A}\underline{x}=T^{-1}e^{\hat{A}t}T\underline{x}=e^{At}\underline{x}
</math>

where

:<math>
e^{\hat{A}t}
=
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
</math>

Hence,
:<math>
\dot{\underline{x}}
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
\underline{x}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T18:47:21Z

Jimmy.apablaza: /* Matrix Exponential */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are negligible
* no damping forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>\underline{x} = c_1 \underline{k_1} e^{\lambda_1 t} + c_2 \underline{k}_2 e^{\lambda_2 t} + c_3 \underline{k_3} e^{\lambda_3 t} + c_4 \underline{k_4} e^{\lambda_4 t}</math>

: <math>

\underline{x}
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
A
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

So, the exponential matrix becomes,

:<math>
\dot{\underline{x}}=\widehat{A}\underline{x}=T^{-1}e^{\hat{A}t}T\underline{x}=e^{At}\underline{x}
</math>

where

:<math>
e^{\hat{A}t}
=
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
</math>

Hence,
:<math>
\dot{\underline{x}}
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
\underline{x}
</math>

:<math>
\dot{\underline{x}}
=
\begin{bmatrix}
-0.25 e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & 0.25 e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & 0.25 e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & 0.25 e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
\underline{x}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T18:40:45Z

Jimmy.apablaza: /* Matrix Exponential */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are negligible
* no damping forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>\underline{x} = c_1 \underline{k_1} e^{\lambda_1 t} + c_2 \underline{k}_2 e^{\lambda_2 t} + c_3 \underline{k_3} e^{\lambda_3 t} + c_4 \underline{k_4} e^{\lambda_4 t}</math>

: <math>

\underline{x}
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
A
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

So, the exponential matrix becomes,

:<math>
\dot{\underline{x}}=\widehat{A}\underline{x}=T^{-1}e^{\hat{A}t}T\underline{x}=e^{At}\underline{x}
</math>

where

:<math>
e^{\hat{A}t}
=
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
</math>

Hence,
:<math>
\dot{\underline{x}}
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
\underline{x}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T18:40:04Z

Jimmy.apablaza:

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are negligible
* no damping forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>\underline{x} = c_1 \underline{k_1} e^{\lambda_1 t} + c_2 \underline{k}_2 e^{\lambda_2 t} + c_3 \underline{k_3} e^{\lambda_3 t} + c_4 \underline{k_4} e^{\lambda_4 t}</math>

: <math>

\underline{x}
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
A
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

So, the exponential matrix becomes,

:<math>
\dot{\underline{x}}=\widehat{A}\underline{x}=T^{-1}e^{\hat{A}t}T\underline{x}=e^{At}\underline{x}
</math>

where

:<math>
e^{\hat{A}t}
=
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
</math>

Hence,
:<math>
\dot{\underline{x}}
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T18:38:26Z

Jimmy.apablaza: /* Matrix Exponential */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are negligible
* no damping forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>\underline{x} = c_1 \underline{k_1} e^{\lambda_1 t} + c_2 \underline{k}_2 e^{\lambda_2 t} + c_3 \underline{k_3} e^{\lambda_3 t} + c_4 \underline{k_4} e^{\lambda_4 t}</math>

: <math>

\underline{x}
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
A
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

So, the exponential matrix becomes,

:<math>
\dot{\underline{x}}=\widehat{A}\underline{x}=T^{-1}e^{\hat{A}t}T\underline{x}=e^{At}\underline{x}
</math>

where

:<math>
e^{\hat{A}t}
=
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
</math>

Hence,
:<math>
\dot{\underline{x}}
=
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T18:22:00Z

Jimmy.apablaza: /* Standard Equation */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are negligible
* no damping forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>\underline{x} = c_1 \underline{k_1} e^{\lambda_1 t} + c_2 \underline{k}_2 e^{\lambda_2 t} + c_3 \underline{k_3} e^{\lambda_3 t} + c_4 \underline{k_4} e^{\lambda_4 t}</math>

: <math>

\underline{x}
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
\hat{A}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T18:19:11Z

Jimmy.apablaza: /* Standard Equation */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are negligible
* no damping forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>\underline{x} = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>

: <math>

\underline{x}
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
\hat{A}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T00:22:09Z

Jimmy.apablaza: /* Problem Statement */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are negligible
* no damping forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>x = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>

: <math>

x
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
\hat{A}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T00:21:49Z

Jimmy.apablaza: /* Problem Statement */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are negligible
* no dumping forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>x = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>

: <math>

x
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
\hat{A}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T00:20:39Z

Jimmy.apablaza: /* State Space */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are neligible
* no dumpung forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>x = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>

: <math>

x
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
\hat{A}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T00:16:44Z

Jimmy.apablaza: /* Eigenvalues & Eigenvectors */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are neligible
* no dumpung forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. Knowing <math>g=32</math>, and assuming that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>x = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>

: <math>

x
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
\hat{A}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T00:16:19Z

Jimmy.apablaza:

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are neligible
* no dumpung forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. Knowing <math>g=32</math>, and assuming that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues & Eigenvectors ===
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVl()''''' function,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>x = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>

: <math>

x
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
\hat{A}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T00:02:26Z

Jimmy.apablaza: /* Matrix Exponential */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are neligible
* no dumpung forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. Knowing <math>g=32</math>, and assuming that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues ===
The eigenvalues are obtained from <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

According to my TI-89, the eigenvalues are,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

and the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>x = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>

: <math>

x
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
\hat{A}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

Coupled Oscillator: Double Pendulum

2009-12-13T00:01:20Z

Jimmy.apablaza:

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are neligible
* no dumpung forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. Knowing <math>g=32</math>, and assuming that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues ===
The eigenvalues are obtained from <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

According to my TI-89, the eigenvalues are,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

and the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>x = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>

: <math>

x
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
\hat{A}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
</math>

Coupled Oscillator: Double Pendulum

2009-12-12T23:34:20Z

Jimmy.apablaza:

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are neligible
* no dumpung forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. Knowing <math>g=32</math>, and assuming that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues ===
The eigenvalues are obtained from <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

According to my TI-89, the eigenvalues are,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

and the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>x = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>

: <math>

x
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
\hat{A}
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Let's consider the space state equation, The matrix exponential is defined as,

: <math>\ddot{x} = \widehat{A} \, \underline{x}(t)</math>

where

:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}=\bold{TAT^{-1}}\bar{z}</math>

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

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 bring in the standard form of a state space equation
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
where
:<math>\bold{\hat{A}}=</math>

Coupled Oscillator: Double Pendulum

2009-12-12T23:08:54Z

Jimmy.apablaza: /* = Matrix Exponential */

By '''Jimmy Apablaza'''

This problem is described in Page 321-322, Section 7.6 of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3).

__TOC__

[[Image:Fig1_Double_Pendulum.png|thumb|Figure 1. Coupled Pendulum.‎]]
= Problem Statement =
Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are neligible
* no dumpung forces act on the system
* positive direction to the right.

The system of differential equations describing the motion is nonlinear

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0</math>

In order to linearize these equations, we assume that the displacements <math>\theta_1</math> and <math>\theta_2</math> are small enough so that <math>cos(\theta_1-\theta_2)\approx1</math> and <math>sin(\theta_1-\theta_2)\approx0</math>. Thus,

: <math>(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0</math>
: <math>m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0</math>

== Solution ==

Since our concern is about the motion functions, we will assign the masses <math>m_1</math> and <math>m_2</math>, the rod lenghts <math>l_1</math> and <math>l_1</math>, and gravitational force <math>g</math> constants to different variables as follows,

: <math>A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g</math>

Hence,

: <math>A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0</math>
: <math>D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0</math>

Solving for <math>\theta_1^{''}</math> and <math>\theta_2^{''}</math> we obtain,

: <math>\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1</math>
: <math>\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2</math>

Therefore,

: <math>\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2 </math>
: <math>\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2</math>

=== State Space ===

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
& & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2} & 0 \\
& & & \\
0 & 0 & 0 & 1 \\
& & & \\
\dfrac{BC}{AD-B^2} & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

</math>

Let's plug some numbers. Knowing <math>g=32</math>, and assuming that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, the constants defined previously become,

: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>

Hence, the state space matrix is,

: <math>
\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

</math>

=== Eigenvalues ===
The eigenvalues are obtained from <math>\widehat{A}</math>'s identity matrix,

: <math>
[\lambda I-A]

=

\begin{bmatrix}
\lambda & 1 & 0 & 0 \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3} & 0 \\
0 & 0 & \lambda & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

</math>

According to my TI-89, the eigenvalues are,

: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_2= -2 \mathbf{i}</math>
: <math>\lambda_3= 1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>

and the eigenvectors,

: <math>
k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}

</math>

=== Standard Equation ===
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

: <math>x = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>

: <math>

x
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4 \\
0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4 \\
-0.4 \mathbf{i} \\
-0.8 \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
-0.67621 \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806 \\
0.58554 \mathbf{i} \\
0.67621 \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}
</math>

=== Matrix Exponential ===
Let's consider the space state equation, The matrix exponential is defined as,

:<math>\dot{\bar{z}}=\bold{\hat{A}}\bar{z}=\bold{TAT^{-1}}\bar{z}</math>

Now we can use the equation for a transfer function to help us solve through the use of matrix exponentials.
:<math>\bar{z}=\bold{T}\bar{x}\,</math>

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 bring in the standard form of a state space equation
:<math>\dot{\bar{x}}=\bold{A}\bar{x}</math>

Combining the two equations we then get
:<math>\bold{T^{-1}}\dot{\bar{z}}=\bold{AT^{-1}}\bar{z}</math>

Multiplying both sides of the equation on the left by '''T''' we get
:<math>\dot{\bar{z}}=\bold{TAT^{-1}}\bar{z}</math>
where
:<math>\bold{\hat{A}}=</math>

Coupled Oscillator: Double Pendulum

2009-12-12T23:02:19Z

Jimmy.apablaza:

Coupled Oscillator: Double Pendulum

2009-12-12T07:00:02Z

Jimmy.apablaza:

Coupled Oscillator: Double Pendulum

2009-12-12T06:53:38Z

Jimmy.apablaza: /* Standard Equation */

Coupled Oscillator: Double Pendulum

2009-12-12T06:47:55Z

Jimmy.apablaza:

Coupled Oscillator: Double Pendulum

2009-12-12T06:20:54Z

Jimmy.apablaza: /* Eigenvalues */

Coupled Oscillator: Double Pendulum

2009-12-12T06:08:41Z

Jimmy.apablaza: /* Eigenvalues */

Coupled Oscillator: Double Pendulum

2009-12-12T05:57:37Z

Jimmy.apablaza: /* Eigenvalues */