Class Wiki - User contributions [en]

Colby's Asgn

2010-11-08T00:07:40Z

Colby.fullerton: Created page with 'Find the Thevenin equivalent impedance seen looking into the terminals of the following circuit: center To find the Thevenin equivalent…'

Find the Thevenin equivalent impedance seen looking into the terminals of the following circuit:
[[File:Lna hw8 diagram1.jpg|thumb|400px|center]]

To find the Thevenin equivalent impedance I will need to take the Laplace transform of each branch of the circuit and add them in series.

The Laplace transform of a resistor is
<math>\mathcal{L}\{resistor\} = R</math>

The Laplace transform of an inductor is
<math>\mathcal{L}\{inductor\} = sL</math>

The Laplace transform of a capacitor is
<math>\mathcal{L}\{\mbox{capacitor}\} = \frac{1}{sC}</math>

Taking the Laplace transform of the original circuit gives the following.
[[File:Lna hw8 diagram2.jpg|thumb|400px|center]]

The Thevenin equivalent impedance of this circuit is given by the equation
:<math>Z = \frac{1}{\frac{1}{R1+\frac{1}{sC}}+\frac{1}{R2+sL}} \Omega</math>

Substituting the values for R1, R2, C and L into the equation gives
:<math>Z = \frac{1}{\frac{1}{50+\frac{1}{.002s}}+\frac{1}{100+.5s}} \Omega = \frac{1}{\frac{s}{50(s+10)}+.5(s+200)} \Omega = \frac{2(s+10)}{s^2+210.04s+2000} \Omega</math>

File:Lna hw8 diagram2.jpg

2010-11-07T23:54:47Z

Colby.fullerton:

File:Lna hw8 diagram1.jpg

2010-11-07T23:40:09Z

Colby.fullerton:

Fall 2010

2010-11-07T23:34:43Z

Colby.fullerton:

Put links for your reports here.

Links to the posted reports are found under the publishing author's name.

Please number and sort Authors alphabetically.

# [[Banton, Alex]]
#*[[Alex's Octave Assignment]]
#*[[Alex's Assignment #8]]
# [[Bidwell, Kelvin]]
#*[[Kelvin's Octave Assignment]]
# [[Blaire, Matthew]]
#*[[Matthew's Octave Assignment]]
#*[[Matthew's Asgn #8]]
# [[Boyd, Aaron]]
#*[[ Aaron's Octave Assignment]]
#*[[Aaron Boyd's Assignment 8]]
# [[Bryson, David]]
#*[[David's Octave Assignment]]
# [[Colls, David]]
#*[[Colls Octave Assignment]]
#*[[Fourier Series Assignment]]
# [[Fullerton, Colby]]
#*[[Colby's Octave Assignment]]
#*[[Colby's Asgn #8]]
# [[Hildebrand, Kurt]]
#*[[Kurt's Octave Assignment]]
#*[[Kurt's Assignment #8]]
# [[Martinez, Jonathan]]
#*[[Martinez's Octave Assignment]]
#*[[Martinez's Fourier Assignment]]
# [[Morgan, David]]
#*[[David Morgan's Octave Assignment]]
#*[[David Morgan's Fourier Series assignment]]
# [[Roth, Andrew]]
#*[[Andrew Roth's Fourier Series/Laplace Transform Project]]
#*[[Andrew's Octave Assignment]]
#*[[Example: LaTex format]]
#*[[Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency]]
# [[Stirn, Jed]]
#*[[Jed's Octave Assignment]]
#*[[HW#8:Laplace Transforms/Fourier Series]]
# [[Stringer, Robert]]
#*[[Robert's Octave Assignment]]
#*[[Robert's HW #8]]
# [[Wooley, Andy]]
#*[[Andy's Octave Assignment]]
#*[[Andy's Fourier Series Project]]
# [[Zimmerly, Brian]]
#*[[Brian's Octave Assignment]]

Colby's Octave Assignment

2010-09-30T23:22:57Z

Colby.fullerton: Created page with '==Entering Matrices== Matrices can be entered into Octave quite easily. Typing v=[1 2 3] sets the variable v to be a 1x3 matrix. To create a 2x2 matrix you can either type v=[1…'

==Entering Matrices==

Matrices can be entered into Octave quite easily. Typing v=[1 2 3] sets the variable v to be a 1x3 matrix.

To create a 2x2 matrix you can either type v=[1 2; 3 4] or you could type

v=[1 2
3 4]

To create a 3x1 column vector you can type v=[1 2 3]'. The prime ' denotes a Hermitian transpose. Alternatively you could type

v=[1
2
3]

Typing the function eye(n) will create an n x n identity matrix. Typing the function zeros(n) will create an n x n matrix of zeroes. Typing the function zeros(m,n) will create an m x n matrix of zeroes. The function ones(m,n) will create an m x n matrix of ones. If m=1 it is a column vector, if n=1 it is a row vector.

To create a diagonal matrix use the diag function. Typing diag([1 2 3]) will create a 3x3 diagonal matrix with 1, 2 and 3 on the diagonal.

==Matrix Operations==

The inv function inverts matrices. Typing A = inv(B) makes A the inverse of B. In this case A*B and B*A are both the identity matrix.

Say we have entered the matrices v=[1 2 3] and w=[1 2 ; 3 4] and z=[5 6 7]'. We can find the dot product of v and z by typing v*z. We can do matrix-vector multiplication by typing w*[1 1]' or [1 1]*w. These will produce different results.

To find eigenvalues use the eig function. Given a square matrix stored in variable X the command

e = eig(x);

will put the eigenvalues of X into the column vector e. If they are real, they are sorted in ascending order.

To find eigenvectors the format of the output will be changed to

[v,e] = eig(X);

This will put the eigenvectors into the columns of v and the eigenvalues into the diagonal of e. In this case e will be a diagonal square matrix with the eigenvalues on the diagonal.

Fall 2010

2010-09-30T22:50:18Z

Colby.fullerton:

Put links for your reports here.

Links to the posted reports are found under the publishing author's name.

Please number and sort Authors alphabetically.

1. [[Roth, Andrew]]
*[[Andrew's Octave Assignment]]
*[[Example: LaTex format]]
*[[Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency]]

2. [[Wooley, Andy]]
*[[Andy's Octave Assignment]]

3. [[Fullerton, Colby]]
*[[Colby's Octave Assignment]]

Fullerton, Colby

2010-01-17T23:34:11Z

Colby.fullerton:

[[Image:Philosoraptor.jpg‎|300px]]
== Contact info ==

'''Email:''' colby.fullerton@wallawalla.edu

'''Phone/Text:''' (509) 386-7970

File:Philosoraptor.jpg

2010-01-17T23:32:58Z

Colby.fullerton:

Winter 2010

2010-01-17T23:27:53Z

Colby.fullerton:

Put links for your reports here. Someone feel free to edit this better.

Links to the posted reports are found under the publishing author's name. (Like it says above, if you hate it...change it! I promise I won’t cry. Brandon)

1. [[Biesenthal, Dan]]

2. [[Blackley, Ben]]

3. [[Cruz, Jorge]]

4. [[Fullerton, Colby]]
<math>\Rightarrow</math> [[Laplace Transform]]
<math>\Rightarrow</math> [[Class Notes 1-5-2010]]

5. [[Grant, Joshua]]

6. [[Gratias, Ryan]]

7. [[Hawkins, John]]

<math>\Rightarrow</math> [[Exercise: Sawtooth Wave Fourier Transform]]

8. [[Lau, Chris]]
<math>\Rightarrow</math> [[Fourier Series: Explained!]]

9. [[Roath, Brian]]
<math>\Rightarrow</math> [[Laplace Transform]]
<math>\Rightarrow</math> [[Class Notes 1-5-2010]]

10. [[Robbins, David]]

11. [[Roth, Andrew]]
<math>\Rightarrow</math> [[Example: LaTex format (0 points)]]
<math>\Rightarrow</math> [[Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency]]

12. [[Vazquez, Brandon]]

13. [[Vier, Michael]]
<math>\Rightarrow</math> [[Fourier Series: Explained!]]

14. [[Wooley, Thomas]]

15. [[Jaymin, Joseph]]

16. [[Starr, Brielle]]

17. [[Starr, Tyler]]

==Draft Articles==
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.
* [[Gibbs Phenomenon]]

* [[Linear Time Invariant Overview]]

Class Notes 1-5-2010

2010-01-17T23:23:51Z

Colby.fullerton:

[[Image:January_5_graph_1.jpg|200px|thumb|left|Modeling functions as vectors. Using function approximations, the vector path is described.]]
This article covers the notes given in class on January 5, 2010.
==Subjects Covered==
1) Linear Systems

2) Functions as Vectors

[[Image:Figure_1.jpg|200px|thumb|left|Functions graphed in vector form.]]

:<math>v_\mathrm{x} = \vec{v} \cdot \mathbf{\hat{i}}</math>
:<math> \vec{v} = v_\mathrm{x} \mathbf{\hat{i}} + v_\mathrm{y} \mathbf{\hat{j}} </math>
:<math> \vec{v} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} </math>
:<math> \langle v_x, v_y\rangle</math>
:<math> \vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos\theta </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} (\mathbf{\hat{i}} \cdot \mathbf{\hat{i}}) + v_\mathrm{y} \mathbf{\hat{j}} \cdot \mathbf{\hat{i}} </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} </math>
:<math> \vec{v} \cdot \mathbf{\hat{a}}_\mathrm{m} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} \cdot \mathbf{\hat{a}}_\mathrm{m} = v_\mathrm{m} </math>
:<math> \delta_\mathrm{i,m} \equiv \begin{cases} 1 & \mbox{if } i = m, \\ 0 & \mbox{else} \end{cases}</math>

==Example==
[[Image:January_5_graph_2.jpg|200px|thumb|left|Function waves with varying periods based on the function x(t) = x(t+T)]]
Given function: <math> x(t) = x(t+T) \,</math>
:<math> x(t) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \right] </math>
1) Use vector analogy
:<math> x(t) \cdot \sin \left( \frac {2\pi mt} {T} \right) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \cdot \sin \left( \frac {2\pi mt} {T} \right) \right] </math>

:<math> \int_{\frac {-T} {2}}^{\frac {T} {2}} x(t) \sin \left( \frac {2\pi mt} {T} \right) \,dt = v_m</math>

==External Links==
*[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/ENGR351/2010w/Keystone/index.php Class Notes].

==Authors==
Colby Fullerton

Brian Roath

==Read By==

==Reviewed By==

Class Notes 1-5-2010

2010-01-17T23:22:02Z

Colby.fullerton: /* External Links */

[[Image:January_5_graph_1.jpg|200px|thumb|left|Modeling functions as vectors. Using function approximations, the vector path is described.]]
This article covers the notes given in class on January 5, 2010.
==Subjects Covered==
1) Linear Systems

2) Functions as Vectors

[[Image:Figure_1.jpg|200px|thumb|left|Functions graphed in vector form.]]

:<math>v_\mathrm{x} = \vec{v} \cdot \mathbf{\hat{i}}</math>
:<math> \vec{v} = v_\mathrm{x} \mathbf{\hat{i}} + v_\mathrm{y} \mathbf{\hat{j}} </math>
:<math> \vec{v} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} </math>
:<math> \langle v_x, v_y\rangle</math>
:<math> \mathbf{\hat{u}} \cdot \mathbf{\hat{v}} = |\mathbf{\hat{u}}| |\mathbf{\hat{v}}| \cos\theta </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} (\mathbf{\hat{i}} \cdot \mathbf{\hat{i}}) + v_\mathrm{y} \mathbf{\hat{j}} \cdot \mathbf{\hat{i}} </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} </math>
:<math> \vec{v} \cdot \mathbf{\hat{a}}_\mathrm{m} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} \cdot \mathbf{\hat{a}}_\mathrm{m} = v_\mathrm{m} </math>
:<math> \delta_\mathrm{i,m} \equiv \begin{cases} 1 & \mbox{if } i = m, \\ 0 & \mbox{else} \end{cases}</math>

==Example==
[[Image:January_5_graph_2.jpg|200px|thumb|left|Function waves with varying periods based on the function x(t) = x(t+T)]]
Given function: <math> x(t) = x(t+T) \,</math>
:<math> x(t) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \right] </math>
1) Use vector analogy
:<math> x(t) \cdot \sin \left( \frac {2\pi mt} {T} \right) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \cdot \sin \left( \frac {2\pi mt} {T} \right) \right] </math>

:<math> \int_{\frac {-T} {2}}^{\frac {T} {2}} x(t) \sin \left( \frac {2\pi mt} {T} \right) \,dt = v_m</math>

==External Links==
*[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/ENGR351/2010w/Keystone/index.php Class Notes].

==Authors==
Colby Fullerton

Brian Roath

Class Notes 1-5-2010

2010-01-17T23:17:04Z

Colby.fullerton:

[[Image:January_5_graph_1.jpg|200px|thumb|left|Modeling functions as vectors. Using function approximations, the vector path is described.]]
This article covers the notes given in class on January 5, 2010.
==Subjects Covered==
1) Linear Systems

2) Functions as Vectors

[[Image:Figure_1.jpg|200px|thumb|left|Functions graphed in vector form.]]

:<math>v_\mathrm{x} = \vec{v} \cdot \mathbf{\hat{i}}</math>
:<math> \mathbf{\hat{v}} = v_\mathrm{x} \mathbf{\hat{i}} + v_\mathrm{y} \mathbf{\hat{j}} </math>
:<math> \mathbf{\hat{v}} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} </math>
:<math> \langle v_x, v_y\rangle</math>
:<math> \mathbf{\hat{u}} \cdot \mathbf{\hat{v}} = |\mathbf{\hat{u}}| |\mathbf{\hat{v}}| \cos\theta </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} (\mathbf{\hat{i}} \cdot \mathbf{\hat{i}}) + v_\mathrm{y} \mathbf{\hat{j}} \cdot \mathbf{\hat{i}} </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} </math>
:<math> \vec{v} \cdot \mathbf{\hat{a}}_\mathrm{m} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} \cdot \mathbf{\hat{a}}_\mathrm{m} = v_\mathrm{m} </math>
:<math> \delta_\mathrm{i,m} \equiv \begin{cases} 1 & \mbox{if } i = m, \\ 0 & \mbox{else} \end{cases}</math>
==Example==
[[Image:January_5_graph_2.jpg|200px|thumb|left|Function waves with varying periods based on the function x(t) = x(t+T)]]
:<math> x(t) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \right] </math>
1) Use vector analogy
:<math> x(t) \cdot \sin \left( \frac {2\pi mt} {T} \right) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \cdot \sin \left( \frac {2\pi mt} {T} \right) \right] </math>

:<math> \int_{\frac {-T} {2}}^{\frac {T} {2}} x(t) \sin \left( \frac {2\pi mt} {T} \right) \,dt = v_m</math>

==External Links==
*[[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/ENGR351/2010w/Keystone/index.php Class Notes].].

==Authors==
Colby Fullerton

Brian Roath

Class Notes 1-5-2010

2010-01-17T23:15:33Z

Colby.fullerton:

[[Image:January_5_graph_1.jpg|200px|thumb|left|Modeling functions as vectors. Using function approximations, the vector path is described.]]
This article covers the notes given in class on January 5, 2010.
==Subjects Covered==
1) Linear Systems

2) Functions as Vectors

[[Image:Figure_1.jpg|200px|thumb|left|Functions graphed in vector form.]]

:<math>v_\mathrm{x} = \vec{v} \cdot \mathbf{\hat{i}}</math>
:<math> \mathbf{\hat{v}} = v_\mathrm{x} \mathbf{\hat{i}} + v_\mathrm{y} \mathbf{\hat{j}} </math>
:<math> \mathbf{\hat{v}} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} </math>
:<math> \langle v_x, v_y\rangle</math>
:<math> \mathbf{\hat{u}} \cdot \mathbf{\hat{v}} = |\mathbf{\hat{u}}| |\mathbf{\hat{v}}| \cos\theta </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} (\mathbf{\hat{i}} \cdot \mathbf{\hat{i}}) + v_\mathrm{y} \mathbf{\hat{j}} \cdot \mathbf{\hat{i}} </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} </math>
:<math> \vec{v} \cdot \mathbf{\hat{a}}_\mathrm{m} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} \cdot \mathbf{\hat{a}}_\mathrm{m} = v_\mathrm{m} </math>
:<math> \delta_\mathrm{i,m} \equiv \begin{cases} 1 & \mbox{if } i = m, \\ 0 & \mbox{else} \end{cases}</math>
:<math> x(t) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \right] </math>

==Example==
[[Image:January_5_graph_2.jpg|200px|thumb|left|Function waves with varying periods based on the function x(t) = x(t+T)]]
:<math> x(t) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \right] </math>
1) Use vector analogy
:<math> x(t) \cdot \sin \left( \frac {2\pi mt} {T} \right) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \cdot \sin \left( \frac {2\pi mt} {T} \right) \right] </math>

:<math> \int_{\frac {-T} {2}}^{\frac {T} {2}} x(t) \sin \left( \frac {2\pi mt} {T} \right) \,dt = v_m</math>

==External Links==
*[[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/ENGR351/2010w/Keystone/index.php Class Notes].].

==Authors==
Colby Fullerton

Brian Roath

Class Notes 1-5-2010

2010-01-17T23:09:15Z

Colby.fullerton:

==Subjects Covered==
1) Linear Systems

2) Functions as Vectors

[[Image:Figure_1.jpg|400px|thumb|left|Functions graphed in vector form.]]

[[Image:January_5_graph_1.jpg|500px|thumb|left|Modeling functions as vectors. Using function approximations, the vector path is described.]]

[[Image:January_5_graph_2.jpg|400px|thumb|left|Function waves with varying periods based on the function x(t) = x(t+T)]]

:<math>v_\mathrm{x} = \vec{v} \cdot \mathbf{\hat{i}}</math>
:<math> \mathbf{\hat{v}} = v_\mathrm{x} \mathbf{\hat{i}} + v_\mathrm{y} \mathbf{\hat{j}} </math>
:<math> \mathbf{\hat{v}} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} </math>
:<math> \langle v_x, v_y\rangle</math>
:<math> \mathbf{\hat{u}} \cdot \mathbf{\hat{v}} = |\mathbf{\hat{u}}| |\mathbf{\hat{v}}| \cos\theta </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} (\mathbf{\hat{i}} \cdot \mathbf{\hat{i}}) + v_\mathrm{y} \mathbf{\hat{j}} \cdot \mathbf{\hat{i}} </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} </math>
:<math> \vec{v} \cdot \mathbf{\hat{a}}_\mathrm{m} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} \cdot \mathbf{\hat{a}}_\mathrm{m} = v_\mathrm{m} </math>
:<math> \delta_\mathrm{i,m} \equiv \begin{cases} 1 & \mbox{if } i = m, \\ 0 & \mbox{else} \end{cases}</math>
:<math> x(t) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \right] </math>

1) Use vector analogy
:<math> x(t) \cdot \sin \left( \frac {2\pi mt} {T} \right) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \cdot \sin \left( \frac {2\pi mt} {T} \right) \right] </math>

:<math> \int_{\frac {-T} {2}}^{\frac {T} {2}} x(t) \sin \left( \frac {2\pi mt} {T} \right) \,dt = v_m</math>

==External Links==
*[[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/ENGR351/2010w/Keystone/index.php Class Notes].].

==Authors==
Colby Fullerton

Brian Roath

Class Notes 1-5-2010

2010-01-17T23:04:25Z

Colby.fullerton:

==Subjects Covered==
1) Linear Systems

2) Functions as Vectors

[[Image:Figure_1.jpg|400px|thumb|left|Functions graphed in vector form.]]

[[Image:January_5_graph_1.jpg|500px|thumb|left|Modeling functions as vectors. Using function approximations, the vector path is described.]]

[[Image:January_5_graph_2.jpg|400px|thumb|left|Function waves with varying periods based on the function x(t) = x(t+T)]]

:<math>v_\mathrm{x} = \vec{v} \cdot \mathbf{\hat{i}}</math>
:<math> \mathbf{\hat{v}} = v_\mathrm{x} \mathbf{\hat{i}} + v_\mathrm{y} \mathbf{\hat{j}} </math>
:<math> \mathbf{\hat{v}} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} </math>
:<math> \langle v_x, v_y\rangle</math>
:<math> \mathbf{\hat{u}} \cdot \mathbf{\hat{v}} = |\mathbf{\hat{u}}| |\mathbf{\hat{v}}| \cos\theta </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} (\mathbf{\hat{i}} \cdot \mathbf{\hat{i}}) + v_\mathrm{y} \mathbf{\hat{j}} \cdot \mathbf{\hat{i}} </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} </math>
:<math> \vec{v} \cdot \mathbf{\hat{a}}_\mathrm{m} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} \cdot \mathbf{\hat{a}}_\mathrm{m} = v_\mathrm{m} </math>
:<math> \delta_\mathrm{i,m} = \begin{cases} 1 & \mbox{if } i = m, \\ 0 & \mbox{else} \end{cases}</math>
:<math> x(t) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \right] </math>

1) Use vector analogy
:<math> x(t) \cdot \sin \left( \frac {2\pi mt} {T} \right) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \cdot \sin \left( \frac {2\pi mt} {T} \right) \right] </math>

:<math> \int_{\frac {-T} {2}}^{\frac {T} {2}} x(t) \sin \left( \frac {2\pi mt} {T} \right) \,dt = v_m</math>

==External Links==
*[[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/ENGR351/2010w/Keystone/index.php Class Notes].].

==Authors==
Colby Fullerton

Brian Roath

Class Notes 1-5-2010

2010-01-17T22:54:36Z

Colby.fullerton:

==Subjects Covered==
1) Linear Systems

2) Functions as Vectors

[[Image:Figure_1.jpg|400px|thumb|left|Functions graphed in vector form.]]

[[Image:January_5_graph_1.jpg|500px|thumb|left|Modeling functions as vectors. Using function approximations, the vector path is described.]]

[[Image:January_5_graph_2.jpg|400px|thumb|left|Function waves with varying periods based on the function x(t) = x(t+T)]]

:<math>v_\mathrm{x} = \vec{v} \cdot \mathbf{\hat{i}}</math>
:<math> \mathbf{\hat{v}} = v_\mathrm{x} \mathbf{\hat{i}} + v_\mathrm{y} \mathbf{\hat{j}} </math>
:<math> \mathbf{\hat{v}} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} </math>
:<math> \langle v_x, v_y\rangle</math>
:<math> \mathbf{\hat{u}} \cdot \mathbf{\hat{v}} = |\mathbf{\hat{u}}| |\mathbf{\hat{v}}| \cos\theta </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} (\mathbf{\hat{i}} \cdot \mathbf{\hat{i}}) + v_\mathrm{y} \mathbf{\hat{j}} \cdot \mathbf{\hat{i}} </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} </math>
:<math> \vec{v} \cdot \mathbf{\hat{a}}_\mathrm{m} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} \cdot \mathbf{\hat{a}}_\mathrm{m} = v_\mathrm{m} </math>
:<math> x(t) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \right] </math>

1) Use vector analogy
:<math> x(t) \cdot \sin \left( \frac {2\pi mt} {T} \right) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \cdot \sin \left( \frac {2\pi mt} {T} \right) \right] </math>

:<math> \int_{\frac {-T} {2}}^{\frac {T} {2}} x(t) \sin \left( \frac {2\pi mt} {T} \right) \,dt = v_m</math>

==External Links==
*[[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/ENGR351/2010w/Keystone/index.php Class Notes].].

==Authors==
Colby Fullerton

Brian Roath

Class Notes 1-5-2010

2010-01-17T22:54:02Z

Colby.fullerton:

==Subjects Covered==
1) Linear Systems

2) Functions as Vectors

[[Image:Figure_1.jpg|400px|thumb|left|Functions graphed in vector form.]]

[[Image:January_5_graph_1.jpg|500px|thumb|left|Modeling functions as vectors. Using function approximations, the vector path is described.]]

[[Image:January_5_graph_2.jpg|400px|thumb|left|Function waves with varying periods based on the function x(t) = x(t+T)]]

:<math>v_\mathrm{x} = \vec{v} \cdot \mathbf{\hat{i}}</math>
:<math> \mathbf{\hat{v}} = v_\mathrm{x} \mathbf{\hat{i}} + v_\mathrm{y} \mathbf{\hat{j}} </math>
:<math> \mathbf{\hat{v}} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} </math>
:<math> \langle v_x, v_y\rangle</math>
:<math> \mathbf{\hat{u}} \cdot \mathbf{\hat{v}} = |\mathbf{\hat{u}}| |\mathbf{\hat{v}}| \cos\theta </math>

:<math> x(t) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \right] </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} (\mathbf{\hat{i}} \cdot \mathbf{\hat{i}}) + v_\mathrm{y} \mathbf{\hat{j}} \cdot \mathbf{\hat{i}} </math>
:<math> \vec{v} \cdot \mathbf{\hat{i}} = v_\mathrm{x} </math>
:<math> \vec{v} \cdot \mathbf{\hat{a}}_\mathrm{m} = \sum_{i} v_\mathrm{i} \mathbf{\hat{a}}_\mathrm{i} \cdot \mathbf{\hat{a}}_\mathrm{m} = v_\mathrm{m} </math>

1) Use vector analogy
:<math> x(t) \cdot \sin \left( \frac {2\pi mt} {T} \right) = \sum^\infty_{n=1} \left[ b_n \sin \left( \left( \frac {2\pi n} {T} \right) t \right) \cdot \sin \left( \frac {2\pi mt} {T} \right) \right] </math>

:<math> \int_{\frac {-T} {2}}^{\frac {T} {2}} x(t) \sin \left( \frac {2\pi mt} {T} \right) \,dt = v_m</math>

==External Links==
*[[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/ENGR351/2010w/Keystone/index.php Class Notes].].

==Authors==
Colby Fullerton

Brian Roath

Class Notes 1-5-2010

2010-01-17T22:39:54Z

Colby.fullerton:

Class Notes 1-5-2010

2010-01-17T22:29:39Z

Colby.fullerton:

Class Notes 1-5-2010

2010-01-17T22:21:52Z

Colby.fullerton:

Class Notes 1-5-2010

2010-01-17T22:06:30Z

Colby.fullerton:

Class Notes 1-5-2010

2010-01-17T22:06:11Z

Colby.fullerton:

Laplace Transform

2010-01-17T22:05:22Z

Colby.fullerton: Undo revision 8078 by Colby.fullerton (Talk)

Laplace transforms are an adapted integral form of a differential equation (created and introduced by the French mathematician Pierre-Simon Laplace (1749-1827)) used to describe electrical circuits and physical processes. Adapted from previous notions given by other notable mathematicians and engineers like Joseph-Louis Lagrange (1736-1812) and Leonhard Euler (1707-1783), Laplace transforms are used to be a more efficient and easy-to-recognize form of a mathematical equation.

==Standard Form==
This is the standard form of a Laplace transform that a function will undergo.
:<math>F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>

==Sample Functions==
The following is a list of commonly seen functions of which the Laplace transform is taken. The start function is noted within the Laplace symbol <math> \mathcal{L} \left\{ \right\} </math>.
:<math>F(s) = \mathcal{L} \left\{1\right\}=\int_0^{\infty} e^{-st} \,dt = </math> <math> \frac {1}{s}</math>

:<math>F(s) = \mathcal{L} \left\{t^n\right\}=\int_0^{\infty} e^{-st} t^n \,dt = </math> <math> \frac {n!}{s^{n+1}}</math>

:<math>F(s) = \mathcal{L} \left\{e^{at}\right\}=\int_0^{\infty} e^{-st} e^{at} \,dt = </math> <math> \frac {1}{s-a}</math>

:<math>F(s) = \mathcal{L} \left\{sin(\omega t)\right\}=\int_0^{\infty} e^{-st} sin(\omega t) \,dt = </math> <math> \frac {\omega}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{t^n g(t)\right\}=\int_0^{\infty} e^{-st} t^n g(t) \,dt = </math> <math> \frac {(-1)^n d^n G(s)} {ds^n} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{t cos(\omega t)\right\}=\int_0^{\infty} e^{-st} t cos(\omega t) \,dt = </math> <math> \frac {s^2-\omega^2} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>

:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = G(s-a) </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{te^{-t}\right\}=\int_0^{\infty} e^{-st} te^{-t} \,dt = </math> <math> \frac {1} {(s+1)^2} </math>

:<math>F(s) = \mathcal{L} \left\{1-e^{-t/T}\right\}=\int_0^{\infty} e^{-st} (1-e^{-t/T}) \,dt = </math> <math> \frac {1} {s(1+Ts)} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} sin(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} sin(\omega t) \,dt = </math> <math> \frac {\omega} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} cos(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} cos(\omega t) \,dt = </math> <math> \frac {s-a} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{u(t)\right\}=\int_0^{\infty} e^{-st} u(t) \,dt = </math> <math> \frac {1} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = e^{-as} G(s) </math>

:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} e^{-st} g'(t) \,dt = sG(s) - g(0) </math>

:<math>F(s) = \mathcal{L} \left\{g''(t)\right\}=\int_0^{\infty} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) </math>

:<math>F(s) = \mathcal{L} \left\{g^{(n)}(t)\right\}=\int_0^{\infty} e^{-st} g^{(n)}(t) \,dt = s^n \cdot G(s) - s^{n-1} \cdot g(0) - s^{n-2} \cdot g'(0) - ... - g^{(n-1)}(0) </math>

==Transfer Function==
The Laplace transform of the impulse response of a circuit with no initial conditions is called the transfer function. If a single-input, single-output circuit has no internal stored energy and all the independent internal sources are zero, the transfer function is

: <math>H(s) = \frac{ \mathcal{L} (response~signal)} { \mathcal{L} (input~signal)}. </math>

Impedances and admittances are special cases of transfer functions.

==Example==
Solve the differential equation:
:<math>y''-2y'-15y=6 \qquad y(0)=1 \qquad y'(0)=3</math>

We start by taking the Laplace transform of each term.
:<math>\mathcal{L} \left\{y''\right\}-2\mathcal{L} \left\{y'\right\}-15\mathcal{L} \left\{y\right\}=\mathcal{L} \left\{6\right\}</math>

The next step is to perform the respective Laplace transforms, using the information given above.
:<math> (s^2\mathcal{L} \left\{y\right\}-s-3)-2(s\mathcal{L} \left\{y\right\}-1)-15\mathcal{L} \left\{y\right\} = \mathcal{L} \left\{6\right\}</math>

Using association, the equation is rearranged:
:<math> (s^2-2s-15)\mathcal{L} \left\{y\right\}= \frac {6} {s} -s+3-2</math>

Continuing on using the method of partial fractions, the equation is progressed:
:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \left(\frac {A} {s}\right) \left(\frac {B} {s+3}\right) \left(\frac {C} {s-5}\right)</math>

:<math> A(s+3)(s-5)+Bs(s-5)+Cs(s+3)=-s^2+s+6 \,</math>

:<math> A+B+C=-1 \,</math>
:<math> -2A-5B+3C=1 \,</math>
:<math> -15A=6 \,</math>

:<math> A=\frac {-2} {5} \qquad B=\frac {-1} {4} \qquad C=\frac {-7} {20} </math>

Plugging the above values back into the equation further up, we get:
:<math> \mathcal{L} \left\{y\right\} = \frac {\frac {-2} {5}} {s} + \frac {\frac {-1} {4}} {s+3} + \frac {\frac {-7} {20}} {s-5} </math>

Applying anti-Laplace transforms, we get the equation:
:<math> y= \mathcal{L}^{-1} \left\{\frac {\frac {-2} {5}} {s}\right\} + \mathcal{L}^{-1} \left\{\frac {\frac {-1} {4}} {s+3}\right\} + \mathcal{L}^{-1} \left\{\frac {\frac {-7} {20}} {s-5}\right\} </math>

Applying the Laplace transforms in reverse (as the above equation utilizes inverse Laplace transforms) for the above equation, we get the solution:
:<math> y(t)= \frac {-2} {5} + \frac {-1} {4} e^{-3t} + \frac {-7} {20} e^{5t} </math>

==References==
DeCarlo, Raymond A.; Lin, Pen-Min (2001), Linear Circuit Analysis, Oxford University Press, ISBN 0-19-513666-7 .
== External links ==
*[http://www.intmath.com/Laplace-transformation/Intro.php The Laplace Transform].

==Authors==
Colby Fullerton

Brian Roath

==Reviewed By==
David Robbins

Thomas Wooley

==Read By==

Laplace Transform

2010-01-17T22:04:51Z

Colby.fullerton: /* External links */

Laplace transforms are an adapted integral form of a differential equation (created and introduced by the French mathematician Pierre-Simon Laplace (1749-1827)) used to describe electrical circuits and physical processes. Adapted from previous notions given by other notable mathematicians and engineers like Joseph-Louis Lagrange (1736-1812) and Leonhard Euler (1707-1783), Laplace transforms are used to be a more efficient and easy-to-recognize form of a mathematical equation.

==Standard Form==
This is the standard form of a Laplace transform that a function will undergo.
:<math>F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>

==Sample Functions==
The following is a list of commonly seen functions of which the Laplace transform is taken. The start function is noted within the Laplace symbol <math> \mathcal{L} \left\{ \right\} </math>.
:<math>F(s) = \mathcal{L} \left\{1\right\}=\int_0^{\infty} e^{-st} \,dt = </math> <math> \frac {1}{s}</math>

:<math>F(s) = \mathcal{L} \left\{t^n\right\}=\int_0^{\infty} e^{-st} t^n \,dt = </math> <math> \frac {n!}{s^{n+1}}</math>

:<math>F(s) = \mathcal{L} \left\{e^{at}\right\}=\int_0^{\infty} e^{-st} e^{at} \,dt = </math> <math> \frac {1}{s-a}</math>

:<math>F(s) = \mathcal{L} \left\{sin(\omega t)\right\}=\int_0^{\infty} e^{-st} sin(\omega t) \,dt = </math> <math> \frac {\omega}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{t^n g(t)\right\}=\int_0^{\infty} e^{-st} t^n g(t) \,dt = </math> <math> \frac {(-1)^n d^n G(s)} {ds^n} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{t cos(\omega t)\right\}=\int_0^{\infty} e^{-st} t cos(\omega t) \,dt = </math> <math> \frac {s^2-\omega^2} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>

:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = G(s-a) </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{te^{-t}\right\}=\int_0^{\infty} e^{-st} te^{-t} \,dt = </math> <math> \frac {1} {(s+1)^2} </math>

:<math>F(s) = \mathcal{L} \left\{1-e^{-t/T}\right\}=\int_0^{\infty} e^{-st} (1-e^{-t/T}) \,dt = </math> <math> \frac {1} {s(1+Ts)} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} sin(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} sin(\omega t) \,dt = </math> <math> \frac {\omega} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} cos(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} cos(\omega t) \,dt = </math> <math> \frac {s-a} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{u(t)\right\}=\int_0^{\infty} e^{-st} u(t) \,dt = </math> <math> \frac {1} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = e^{-as} G(s) </math>

:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} e^{-st} g'(t) \,dt = sG(s) - g(0) </math>

:<math>F(s) = \mathcal{L} \left\{g''(t)\right\}=\int_0^{\infty} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) </math>

:<math>F(s) = \mathcal{L} \left\{g^{(n)}(t)\right\}=\int_0^{\infty} e^{-st} g^{(n)}(t) \,dt = s^n \cdot G(s) - s^{n-1} \cdot g(0) - s^{n-2} \cdot g'(0) - ... - g^{(n-1)}(0) </math>

==Transfer Function==
The Laplace transform of the impulse response of a circuit with no initial conditions is called the transfer function. If a single-input, single-output circuit has no internal stored energy and all the independent internal sources are zero, the transfer function is

: <math>H(s) = \frac{ \mathcal{L} (response~signal)} { \mathcal{L} (input~signal)}. </math>

Impedances and admittances are special cases of transfer functions.

==Example==
Solve the differential equation:
:<math>y''-2y'-15y=6 \qquad y(0)=1 \qquad y'(0)=3</math>

We start by taking the Laplace transform of each term.
:<math>\mathcal{L} \left\{y''\right\}-2\mathcal{L} \left\{y'\right\}-15\mathcal{L} \left\{y\right\}=\mathcal{L} \left\{6\right\}</math>

The next step is to perform the respective Laplace transforms, using the information given above.
:<math> (s^2\mathcal{L} \left\{y\right\}-s-3)-2(s\mathcal{L} \left\{y\right\}-1)-15\mathcal{L} \left\{y\right\} = \mathcal{L} \left\{6\right\}</math>

Using association, the equation is rearranged:
:<math> (s^2-2s-15)\mathcal{L} \left\{y\right\}= \frac {6} {s} -s+3-2</math>

Continuing on using the method of partial fractions, the equation is progressed:
:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \left(\frac {A} {s}\right) \left(\frac {B} {s+3}\right) \left(\frac {C} {s-5}\right)</math>

:<math> A(s+3)(s-5)+Bs(s-5)+Cs(s+3)=-s^2+s+6 \,</math>

:<math> A+B+C=-1 \,</math>
:<math> -2A-5B+3C=1 \,</math>
:<math> -15A=6 \,</math>

:<math> A=\frac {-2} {5} \qquad B=\frac {-1} {4} \qquad C=\frac {-7} {20} </math>

Plugging the above values back into the equation further up, we get:
:<math> \mathcal{L} \left\{y\right\} = \frac {\frac {-2} {5}} {s} + \frac {\frac {-1} {4}} {s+3} + \frac {\frac {-7} {20}} {s-5} </math>

Applying anti-Laplace transforms, we get the equation:
:<math> y= \mathcal{L}^{-1} \left\{\frac {\frac {-2} {5}} {s}\right\} + \mathcal{L}^{-1} \left\{\frac {\frac {-1} {4}} {s+3}\right\} + \mathcal{L}^{-1} \left\{\frac {\frac {-7} {20}} {s-5}\right\} </math>

Applying the Laplace transforms in reverse (as the above equation utilizes inverse Laplace transforms) for the above equation, we get the solution:
:<math> y(t)= \frac {-2} {5} + \frac {-1} {4} e^{-3t} + \frac {-7} {20} e^{5t} </math>

==References==
DeCarlo, Raymond A.; Lin, Pen-Min (2001), Linear Circuit Analysis, Oxford University Press, ISBN 0-19-513666-7 .
== External links ==
*[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/ENGR351/2010w/Keystone/index.php Class Notes].

==Authors==
Colby Fullerton

Brian Roath

==Reviewed By==
David Robbins

Thomas Wooley

==Read By==

Class Notes 1-5-2010

2010-01-17T22:03:01Z

Colby.fullerton:

Class Notes 1-5-2010

2010-01-17T22:00:43Z

Colby.fullerton:

[[Image:January_5_graph_1.jpg‎]]

==Authors==
Colby Fullerton

Brian Roath

File:January 5 graph 1.jpg

2010-01-17T21:57:46Z

Colby.fullerton:

Exercise: Sawtooth Wave Fourier Transform

2010-01-13T05:02:22Z

Colby.fullerton: /* Reviewed By */

==Problem Statement==

Find the Fourier Tranform of the sawtooth wave given by the equation

 

<center><math>x(t)=t-\lfloor t \rfloor</math></center>

 

 

==Solution==

As shown in class, the general equation for the Fourier Transform for a periodic function with period <math>T</math> is given by

 

<center><math> x(t)=\frac{a_0}{2}+\sum^\infty_{n=1} \left[a_n\cos\frac{2\pi nt}{T}+b_n\sin\frac{2\pi nt}{T}\right]</math></center>

 

where

 

<center><math>\begin{cases}

a_n=\frac{2}{T}\int_c^{c+T}x(t)\cos\frac{2\pi nt}{T}dt\\

b_n=\frac{2}{T}\int_c^{c+T}x(t)\sin\frac{2\pi nt}{T}dt

\end{cases} \ \ \ \ n=0,1,2,3\dots</math></center>

 

For the sawtooth function given, we note that <math>T=1</math>, and an obvious choice for <math>c</math> is 0 since this allows us to reduce the equation to <math>x(t)=t</math>. It remains, then, only to find the expression for <math>a_n</math> and <math>b_n</math>. We proceed first to find <math>b_n</math>.

 

<center><math>b_n=\frac{2}{1}\int_0^1t\sin 2\pi nt\ dt</math></center>

 

which is solved easiest with integration by parts, letting

 

<center>

<math>u=t\qquad\Rightarrow\qquad du=dt</math>

 

<math>dv=\sin2\pi nt\ dt\qquad\Rightarrow\qquad v=-\frac{1}{2\pi n}\cos 2\pi nt</math>

</center>

 

so

 

<center>

<math>b_n=2\left[t\left(-\frac{1}{2\pi n}\right)\cos 2\pi nt\bigg|_0^1+\frac{1}{2\pi n}\int_0^1\cos 2\pi nt\ dt \right]</math>

 

<math>=2\left[\left(-\frac{1}{2\pi n}\cos 2\pi n - 0\right)+\left(\frac{1}{2\pi n}\right)^2\sin 2\pi nt \bigg|_0^1\right]</math>

 

<math>=2\left[-\frac{1}{2\pi n}(1)+0\right]</math>

 

<math>=-\frac{1}{\pi n}</math>

</center>

Now, for <math>a_n</math> we must consider the case when <math>n=0</math> separately.

 

<center>

<math>a_0=\frac{2}{1}\int_0^1t\ dt=t^2\bigg|_0^1=1</math>

</center>

 

For <math>n=1,2,3\dots</math>, we have

 

<center>

<math>a_n=\frac{2}{1}\int_0^1t\cos 2\pi nt\ dt</math>

</center>

 

which again is best solved using integration by parts, this time with

 

<center>

<math>u=t\qquad\Rightarrow\qquad du=dt</math>

 

<math>dv=\cos 2\pi nt\ dt\qquad \Rightarrow\qquad v=\frac{1}{2\pi n}\sin 2\pi nt</math>

 

</center>

so

 

<center>

<math>a_n=2\left[t\left(\frac{1}{2\pi n}\right)\sin 2\pi nt\bigg|_0^1-\int_0^1\frac{1}{2\pi n}\sin 2\pi nt\ dt\right]</math>

 

<math>=2\left[\left(\frac{1}{2\pi n}\sin 2\pi n-0\right)-\left[-\left(\frac{1}{2\pi n}\right)^2\cos 2\pi nt\right]_0^1\right]</math>

 

<math>=2\left[0+\left(\frac{1}{2\pi n}\right)^2\left(\cos 2\pi n-\cos 0\right)\right]</math>

 

<math>\ =0</math>

</center>

 

Therefore, the Fourier Transform representation of the sawtooth wave given is:

 

<center><math>x(t)=\frac{1}{2}-\sum_{n=1}^\infty \frac{1}{\pi n}\sin 2\pi nt</math></center>

 

==Solution Graphs==
The figures below graph the first few iterations of the above solution. The first graph shows the solution truncated after the first 100 terms of the infinite sum, as well as each of the contributing sine waves with offset. The second figure shows the function truncated after 1, 3, 5, 10, 50, and 100 terms. The last figure shows the Error between the Fourier Series truncated after the first 100 terms and the function itself. These figures were constructed using the following matlab code: [[SawToothFourier]].

 

[[Image:First_100_Terms.jpg|thumb|800px|center]]

 
[[Image:First_n_Terms.jpg|thumb|800px|center]]

 
[[Image:Error.jpg|thumb|800px|center]]

==Author==

John Hawkins

==Read By==

==Reviewed By==
Colby Fullerton

Winter 2010

2010-01-12T04:24:33Z

Colby.fullerton:

Put links for your reports here. Someone feel free to edit this better.

Links to the posted reports are found under the publishing author's name. (Like it says above, if you hate it...change it! I promise I won’t cry. Brandon)

1. [[Biesenthal, Dan]]

2. [[Blackley, Ben]]

3. [[Cruz, Jorge]]

4. [[Fullerton, Colby]]
<math>\Rightarrow</math> [[Laplace Transform]]

5. [[Grant, Joshua]]

6. [[Gratias, Ryan]]

7. [[Hawkins, John]]

8. [[Lau, Chris]]

9. [[Roath, Brian]]
<math>\Rightarrow</math> [[Laplace Transform]]

10. [[Robbins, David]]

11. [[Roth, Andrew]]
<math>\Rightarrow</math> [[Example: LaTex format (0 points)]]
<math>\Rightarrow</math> [[Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency]]

12. [[Vazquez, Brandon]]

13. [[Vier, Michael]]

14. [[Wooley, Thomas]]

15. [[Jaymin, Joseph]]

==Draft Articles==
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.
* [[Fourier Series: Explained!]]

Laplace Transform

2010-01-12T03:56:12Z

Colby.fullerton: /* External links */

Laplace transforms are an adapted integral form of a differential equation (created and introduced by the French mathematician Pierre-Simon Laplace (1749-1827)) used to describe electrical circuits and physical processes. Adapted from previous notions given by other notable mathematicians and engineers like Joseph-Louis Lagrange (1736-1812) and Leonhard Euler (1707-1783), Laplace transforms are used to be a more efficient and easy-to-recognize form of a mathematical equation.

==Standard Form==
This is the standard form of a Laplace transform that a function will undergo.
:<math>F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>

==Sample Functions==
The following is a list of commonly seen functions of which the Laplace transform is taken. The start function is noted within the Laplace symbol <math> \mathcal{L} \left\{ \right\} </math>.
:<math>F(s) = \mathcal{L} \left\{1\right\}=\int_0^{\infty} e^{-st} \,dt = </math> <math> \frac {1}{s}</math>

:<math>F(s) = \mathcal{L} \left\{t^n\right\}=\int_0^{\infty} e^{-st} t^n \,dt = </math> <math> \frac {n!}{s^{n+1}}</math>

:<math>F(s) = \mathcal{L} \left\{e^{at}\right\}=\int_0^{\infty} e^{-st} e^{at} \,dt = </math> <math> \frac {1}{s-a}</math>

:<math>F(s) = \mathcal{L} \left\{sin(\omega t)\right\}=\int_0^{\infty} e^{-st} sin(\omega t) \,dt = </math> <math> \frac {\omega}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{t^n g(t)\right\}=\int_0^{\infty} e^{-st} t^n g(t) \,dt = </math> <math> \frac {(-1)^n d^n G(s)} {ds^n} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{t cos(\omega t)\right\}=\int_0^{\infty} e^{-st} t cos(\omega t) \,dt = </math> <math> \frac {s^2-\omega^2} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>

:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = G(s-a) </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{te^{-t}\right\}=\int_0^{\infty} e^{-st} te^{-t} \,dt = </math> <math> \frac {1} {(s+1)^2} </math>

:<math>F(s) = \mathcal{L} \left\{1-e^{-t/T}\right\}=\int_0^{\infty} e^{-st} (1-e^{-t/T}) \,dt = </math> <math> \frac {1} {s(1+Ts)} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} sin(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} sin(\omega t) \,dt = </math> <math> \frac {\omega} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} cos(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} cos(\omega t) \,dt = </math> <math> \frac {s-a} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{u(t)\right\}=\int_0^{\infty} e^{-st} u(t) \,dt = </math> <math> \frac {1} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = e^{-as} G(s) </math>

:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} e^{-st} g'(t) \,dt = sG(s) - g(0) </math>

:<math>F(s) = \mathcal{L} \left\{g''(t)\right\}=\int_0^{\infty} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) </math>

:<math>F(s) = \mathcal{L} \left\{g^{(n)}(t)\right\}=\int_0^{\infty} e^{-st} g^{(n)}(t) \,dt = s^n \cdot G(s) - s^{n-1} \cdot g(0) - s^{n-2} \cdot g'(0) - ... - g^{(n-1)}(0) </math>

==Transfer Function==
The Laplace transform of the impulse response of a circuit with no initial conditions is called the transfer function. If a single-input, single-output circuit has no internal stored energy and all the independent internal sources are zero, the transfer function is

: <math>H(s) = \frac{ \mathcal{L} (response~signal)} { \mathcal{L} (input~signal)}. </math>

Impedances and admittances are special cases of transfer functions.

==References==
DeCarlo, Raymond A.; Lin, Pen-Min (2001), Linear Circuit Analysis, Oxford University Press, ISBN 0-19-513666-7 .
== External links ==
*[http://www.intmath.com/Laplace-transformation/Intro.php The Laplace Transform].

==Authors==
Colby Fullerton

Brian Roath

==Reviewed By==

==Read By==

Laplace Transform

2010-01-12T03:55:54Z

Colby.fullerton:

Laplace transforms are an adapted integral form of a differential equation (created and introduced by the French mathematician Pierre-Simon Laplace (1749-1827)) used to describe electrical circuits and physical processes. Adapted from previous notions given by other notable mathematicians and engineers like Joseph-Louis Lagrange (1736-1812) and Leonhard Euler (1707-1783), Laplace transforms are used to be a more efficient and easy-to-recognize form of a mathematical equation.

==Standard Form==
This is the standard form of a Laplace transform that a function will undergo.
:<math>F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>

==Sample Functions==
The following is a list of commonly seen functions of which the Laplace transform is taken. The start function is noted within the Laplace symbol <math> \mathcal{L} \left\{ \right\} </math>.
:<math>F(s) = \mathcal{L} \left\{1\right\}=\int_0^{\infty} e^{-st} \,dt = </math> <math> \frac {1}{s}</math>

:<math>F(s) = \mathcal{L} \left\{t^n\right\}=\int_0^{\infty} e^{-st} t^n \,dt = </math> <math> \frac {n!}{s^{n+1}}</math>

:<math>F(s) = \mathcal{L} \left\{e^{at}\right\}=\int_0^{\infty} e^{-st} e^{at} \,dt = </math> <math> \frac {1}{s-a}</math>

:<math>F(s) = \mathcal{L} \left\{sin(\omega t)\right\}=\int_0^{\infty} e^{-st} sin(\omega t) \,dt = </math> <math> \frac {\omega}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{t^n g(t)\right\}=\int_0^{\infty} e^{-st} t^n g(t) \,dt = </math> <math> \frac {(-1)^n d^n G(s)} {ds^n} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{t cos(\omega t)\right\}=\int_0^{\infty} e^{-st} t cos(\omega t) \,dt = </math> <math> \frac {s^2-\omega^2} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>

:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = G(s-a) </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{te^{-t}\right\}=\int_0^{\infty} e^{-st} te^{-t} \,dt = </math> <math> \frac {1} {(s+1)^2} </math>

:<math>F(s) = \mathcal{L} \left\{1-e^{-t/T}\right\}=\int_0^{\infty} e^{-st} (1-e^{-t/T}) \,dt = </math> <math> \frac {1} {s(1+Ts)} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} sin(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} sin(\omega t) \,dt = </math> <math> \frac {\omega} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} cos(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} cos(\omega t) \,dt = </math> <math> \frac {s-a} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{u(t)\right\}=\int_0^{\infty} e^{-st} u(t) \,dt = </math> <math> \frac {1} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = e^{-as} G(s) </math>

:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} e^{-st} g'(t) \,dt = sG(s) - g(0) </math>

:<math>F(s) = \mathcal{L} \left\{g''(t)\right\}=\int_0^{\infty} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) </math>

:<math>F(s) = \mathcal{L} \left\{g^{(n)}(t)\right\}=\int_0^{\infty} e^{-st} g^{(n)}(t) \,dt = s^n \cdot G(s) - s^{n-1} \cdot g(0) - s^{n-2} \cdot g'(0) - ... - g^{(n-1)}(0) </math>

==Transfer Function==
The Laplace transform of the impulse response of a circuit with no initial conditions is called the transfer function. If a single-input, single-output circuit has no internal stored energy and all the independent internal sources are zero, the transfer function is

: <math>H(s) = \frac{ \mathcal{L} (response~signal)} { \mathcal{L} (input~signal)}. </math>

Impedances and admittances are special cases of transfer functions.

==References==
DeCarlo, Raymond A.; Lin, Pen-Min (2001), Linear Circuit Analysis, Oxford University Press, ISBN 0-19-513666-7 .
== External links ==
*[http://www.intmath.com/Laplace-transformation/Intro.php The Laplace Transforms].
==Authors==
Colby Fullerton

Brian Roath

==Reviewed By==

==Read By==

Laplace Transform

2010-01-12T03:54:51Z

Colby.fullerton:

Laplace transforms are an adapted integral form of a differential equation (created and introduced by the French mathematician Pierre-Simon Laplace (1749-1827)) used to describe electrical circuits and physical processes. Adapted from previous notions given by other notable mathematicians and engineers like Joseph-Louis Lagrange (1736-1812) and Leonhard Euler (1707-1783), Laplace transforms are used to be a more efficient and easy-to-recognize form of a mathematical equation.

==Standard Form==
This is the standard form of a Laplace transform that a function will undergo.
:<math>F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>

==Sample Functions==
The following is a list of commonly seen functions of which the Laplace transform is taken. The start function is noted within the Laplace symbol <math> \mathcal{L} \left\{ \right\} </math>.
:<math>F(s) = \mathcal{L} \left\{1\right\}=\int_0^{\infty} e^{-st} \,dt = </math> <math> \frac {1}{s}</math>

:<math>F(s) = \mathcal{L} \left\{t^n\right\}=\int_0^{\infty} e^{-st} t^n \,dt = </math> <math> \frac {n!}{s^{n+1}}</math>

:<math>F(s) = \mathcal{L} \left\{e^{at}\right\}=\int_0^{\infty} e^{-st} e^{at} \,dt = </math> <math> \frac {1}{s-a}</math>

:<math>F(s) = \mathcal{L} \left\{sin(\omega t)\right\}=\int_0^{\infty} e^{-st} sin(\omega t) \,dt = </math> <math> \frac {\omega}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{t^n g(t)\right\}=\int_0^{\infty} e^{-st} t^n g(t) \,dt = </math> <math> \frac {(-1)^n d^n G(s)} {ds^n} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{t cos(\omega t)\right\}=\int_0^{\infty} e^{-st} t cos(\omega t) \,dt = </math> <math> \frac {s^2-\omega^2} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>

:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = G(s-a) </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{te^{-t}\right\}=\int_0^{\infty} e^{-st} te^{-t} \,dt = </math> <math> \frac {1} {(s+1)^2} </math>

:<math>F(s) = \mathcal{L} \left\{1-e^{-t/T}\right\}=\int_0^{\infty} e^{-st} (1-e^{-t/T}) \,dt = </math> <math> \frac {1} {s(1+Ts)} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} sin(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} sin(\omega t) \,dt = </math> <math> \frac {\omega} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} cos(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} cos(\omega t) \,dt = </math> <math> \frac {s-a} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{u(t)\right\}=\int_0^{\infty} e^{-st} u(t) \,dt = </math> <math> \frac {1} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = e^{-as} G(s) </math>

:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} e^{-st} g'(t) \,dt = sG(s) - g(0) </math>

:<math>F(s) = \mathcal{L} \left\{g''(t)\right\}=\int_0^{\infty} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) </math>

:<math>F(s) = \mathcal{L} \left\{g^{(n)}(t)\right\}=\int_0^{\infty} e^{-st} g^{(n)}(t) \,dt = s^n \cdot G(s) - s^{n-1} \cdot g(0) - s^{n-2} \cdot g'(0) - ... - g^{(n-1)}(0) </math>

==Transfer Function==
The Laplace transform of the impulse response of a circuit with no initial conditions is called the transfer function. If a single-input, single-output circuit has no internal stored energy and all the independent internal sources are zero, the transfer function is

: <math>H(s) = \frac{ \mathcal{L} (response~signal)} { \mathcal{L} (input~signal)}. </math>

Impedances and admittances are special cases of transfer functions.

==References==
DeCarlo, Raymond A.; Lin, Pen-Min (2001), Linear Circuit Analysis, Oxford University Press, ISBN 0-19-513666-7 .
== External links ==
*[http://www.intmath.com/Laplace-transformation/Intro.php Help:Laplace Transforms].
==Authors==
Colby Fullerton

Brian Roath

==Reviewed By==

==Read By==

Laplace Transform

2010-01-12T03:08:03Z

Colby.fullerton:

Laplace Transform

2010-01-12T03:06:24Z

Colby.fullerton:

Template:Citation

2010-01-12T03:04:27Z

Colby.fullerton: New page: * {{citation|first1=Raymond A.|last1=DeCarlo|first2=Pen-Min|last2=Lin|title=Linear Circuit Analysis|publisher=Oxford University Press|year=2001|isbn=0-19-513666-7}}.

* {{citation|first1=Raymond A.|last1=DeCarlo|first2=Pen-Min|last2=Lin|title=Linear Circuit Analysis|publisher=Oxford University Press|year=2001|isbn=0-19-513666-7}}.

Laplace Transform

2010-01-12T03:04:01Z

Colby.fullerton:

Laplace Transform

2010-01-12T02:49:19Z

Colby.fullerton:

==Standard Form==
:<math>F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>

==Sample Functions==
:<math>F(s) = \mathcal{L} \left\{1\right\}=\int_0^{\infty} e^{-st} \,dt = </math> <math> \frac {1}{s}</math>

:<math>F(s) = \mathcal{L} \left\{t^n\right\}=\int_0^{\infty} e^{-st} t^n \,dt = </math> <math> \frac {n!}{s^{n+1}}</math>

:<math>F(s) = \mathcal{L} \left\{e^{at}\right\}=\int_0^{\infty} e^{-st} e^{at} \,dt = </math> <math> \frac {1}{s-a}</math>

:<math>F(s) = \mathcal{L} \left\{sin(\omega t)\right\}=\int_0^{\infty} e^{-st} sin(\omega t) \,dt = </math> <math> \frac {\omega}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{t^n g(t)\right\}=\int_0^{\infty} e^{-st} t^n g(t) \,dt = </math> <math> \frac {(-1)^n d^n G(s)} {ds^n} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{t cos(\omega t)\right\}=\int_0^{\infty} e^{-st} t cos(\omega t) \,dt = </math> <math> \frac {s^2-\omega^2} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>

:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = G(s-a) </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{te^{-t}\right\}=\int_0^{\infty} e^{-st} te^{-t} \,dt = </math> <math> \frac {1} {(s+1)^2} </math>

:<math>F(s) = \mathcal{L} \left\{1-e^{-t/T}\right\}=\int_0^{\infty} e^{-st} (1-e^{-t/T}) \,dt = </math> <math> \frac {1} {s(1+Ts)} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} sin(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} sin(\omega t) \,dt = </math> <math> \frac {\omega} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} cos(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} cos(\omega t) \,dt = </math> <math> \frac {s-a} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{u(t)\right\}=\int_0^{\infty} e^{-st} u(t) \,dt = </math> <math> \frac {1} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = e^{-as} G(s) </math>

:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} e^{-st} g'(t) \,dt = sG(s) - g(0) </math>

:<math>F(s) = \mathcal{L} \left\{g''(t)\right\}=\int_0^{\infty} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) </math>

:<math>F(s) = \mathcal{L} \left\{g^{(n)}(t)\right\}=\int_0^{\infty} e^{-st} g^{(n)}(t) \,dt = s^n \cdot G(s) - s^{n-1} \cdot g(0) - s^{n-2} \cdot g'(0) - ... - g^{(n-1)}(0) </math>

== External links ==
*[http://www.intmath.com/Laplace-transformation/Intro.php Help:Laplace Transforms].
==Authors==
Colby Fullerton

Brian Roath

==Reviewed By==

==Read By==

Laplace Transform

2010-01-12T02:47:23Z

Colby.fullerton:

Laplace Transform

2010-01-12T02:42:38Z

Colby.fullerton: /* Sample Functions */

==Standard Form==
:<math>F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>

==Sample Functions==
:<math>F(s) = \mathcal{L} \left\{1\right\}=\int_0^{\infty} e^{-st} \,dt = </math> <math> \frac {1}{s}</math>

:<math>F(s) = \mathcal{L} \left\{t^n\right\}=\int_0^{\infty} e^{-st} t^n \,dt = </math> <math> \frac {n!}{s^{n+1}}</math>

:<math>F(s) = \mathcal{L} \left\{e^{at}\right\}=\int_0^{\infty} e^{-st} e^{at} \,dt = </math> <math> \frac {1}{s-a}</math>

:<math>F(s) = \mathcal{L} \left\{sin(\omega t)\right\}=\int_0^{\infty} e^{-st} sin(\omega t) \,dt = </math> <math> \frac {\omega}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{t^n g(t)\right\}=\int_0^{\infty} e^{-st} t^n g(t) \,dt = </math> <math> \frac {(-1)^n d^n G(s)} {ds^n} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{t cos(\omega t)\right\}=\int_0^{\infty} e^{-st} t cos(\omega t) \,dt = </math> <math> \frac {s^2-\omega^2} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>

:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = G(s-a) </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{te^{-t}\right\}=\int_0^{\infty} e^{-st} te^{-t} \,dt = </math> <math> \frac {1} {(s+1)^2} </math>

:<math>F(s) = \mathcal{L} \left\{1-e^{-t/T}\right\}=\int_0^{\infty} e^{-st} (1-e^{-t/T}) \,dt = </math> <math> \frac {1} {s(1+Ts)} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} sin(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} sin(\omega t) \,dt = </math> <math> \frac {\omega} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} cos(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} cos(\omega t) \,dt = </math> <math> \frac {s-a} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{u(t)\right\}=\int_0^{\infty} e^{-st} u(t) \,dt = </math> <math> \frac {1} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = e^{-as} G(s) </math>

:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} e^{-st} g'(t) \,dt = sG(s) - g(0) </math>

:<math>F(s) = \mathcal{L} \left\{g''(t)\right\}=\int_0^{\infty} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) </math>

:<math>F(s) = \mathcal{L} \left\{g^{(n)}(t)\right\}=\int_0^{\infty} e^{-st} g^{(n)}(t) \,dt = s^n \cdot G(s) - s^{n-1} \cdot g(0) - s^{n-2} \cdot g'(0) - ... - g^{(n-1)}(0) </math>

:<math>F(s) = \mathcal{L} \left\{\int_0^{\t} g(t) \,dt \right\}=\int_0^{\t} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) </math>

==External Links==

==Authors==

==Reviewed By==

==Read By==

Laplace Transform

2010-01-12T02:38:46Z

Colby.fullerton:

==Standard Form==
:<math>F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>

==Sample Functions==
:<math>F(s) = \mathcal{L} \left\{1\right\}=\int_0^{\infty} e^{-st} \,dt = </math> <math> \frac {1}{s}</math>

:<math>F(s) = \mathcal{L} \left\{t^n\right\}=\int_0^{\infty} e^{-st} t^n \,dt = </math> <math> \frac {n!}{s^{n+1}}</math>

:<math>F(s) = \mathcal{L} \left\{e^{at}\right\}=\int_0^{\infty} e^{-st} e^{at} \,dt = </math> <math> \frac {1}{s-a}</math>

:<math>F(s) = \mathcal{L} \left\{sin(\omega t)\right\}=\int_0^{\infty} e^{-st} sin(\omega t) \,dt = </math> <math> \frac {\omega}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{t^n g(t)\right\}=\int_0^{\infty} e^{-st} t^n g(t) \,dt = </math> <math> \frac {(-1)^n d^n G(s)} {ds^n} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{t cos(\omega t)\right\}=\int_0^{\infty} e^{-st} t cos(\omega t) \,dt = </math> <math> \frac {s^2-\omega^2} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>

:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = G(s-a) </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{te^{-t}\right\}=\int_0^{\infty} e^{-st} te^{-t} \,dt = </math> <math> \frac {1} {(s+1)^2} </math>

:<math>F(s) = \mathcal{L} \left\{1-e^{-t/T}\right\}=\int_0^{\infty} e^{-st} (1-e^{-t/T}) \,dt = </math> <math> \frac {1} {s(1+Ts)} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} sin(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} sin(\omega t) \,dt = </math> <math> \frac {\omega} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} cos(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} cos(\omega t) \,dt = </math> <math> \frac {s-a} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{u(t)\right\}=\int_0^{\infty} e^{-st} u(t) \,dt = </math> <math> \frac {1} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = e^{-as} G(s) </math>

:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} e^{-st} g'(t) \,dt = sG(s) - g(0) </math>

:<math>F(s) = \mathcal{L} \left\{g''(t)\right\}=\int_0^{\infty} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) </math>

:<math>F(s) = \mathcal{L} \left\{g^{(n)}(t)\right\}=\int_0^{\infty} e^{-st} g^{(n)}(t) \,dt = s^n \cdot G(s) - s^{n-1} \cdot g(0) - s^{n-2} \cdot g'(0) - ... - g^{(n-1)}(0) </math>

==External Links==

==Authors==

==Reviewed By==

==Read By==

Laplace Transform

2010-01-12T02:28:51Z

Colby.fullerton:

Standard Form:
:<math>F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt </math>

Sample Functions:
:<math>F(s) = \mathcal{L} \left\{1\right\}=\int_0^{\infty} e^{-st} \,dt = </math> <math> \frac {1}{s}</math>

:<math>F(s) = \mathcal{L} \left\{t^n\right\}=\int_0^{\infty} e^{-st} t^n \,dt = </math> <math> \frac {n!}{s^{n+1}}</math>

:<math>F(s) = \mathcal{L} \left\{e^{at}\right\}=\int_0^{\infty} e^{-st} e^{at} \,dt = </math> <math> \frac {1}{s-a}</math>

:<math>F(s) = \mathcal{L} \left\{sin(\omega t)\right\}=\int_0^{\infty} e^{-st} sin(\omega t) \,dt = </math> <math> \frac {\omega}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>

:<math>F(s) = \mathcal{L} \left\{t^n g(t)\right\}=\int_0^{\infty} e^{-st} t^n g(t) \,dt = </math> <math> \frac {(-1)^n d^n G(s)} {ds^n} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{t cos(\omega t)\right\}=\int_0^{\infty} e^{-st} t cos(\omega t) \,dt = </math> <math> \frac {s^2-\omega^2} {(s^2+\omega^2)^2} </math>

:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>

:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = G(s-a) </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4

:<math>F(s) = \mathcal{L} \left\{te^{-t}\right\}=\int_0^{\infty} e^{-st} te^{-t} \,dt = </math> <math> \frac {1} {(s+1)^2} </math>

:<math>F(s) = \mathcal{L} \left\{1-e^{-t/T}\right\}=\int_0^{\infty} e^{-st} (1-e^{-t/T}) \,dt = </math> <math> \frac {1} {s(1+Ts)} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} sin(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} sin(\omega t) \,dt = </math> <math> \frac {\omega} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{e^{at} cos(\omega t)\right\}=\int_0^{\infty} e^{-st} e^{at} cos(\omega t) \,dt = </math> <math> \frac {s-a} {(s-a)^2 + \omega^2} </math>

:<math>F(s) = \mathcal{L} \left\{u(t)\right\}=\int_0^{\infty} e^{-st} u(t) \,dt = </math> <math> \frac {1} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math>

:<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = e^{-as} G(s) </math>

:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} g'(t) \,dt = sG(s) - g(0) </math>

Laplace Transform

2010-01-12T02:26:03Z

Colby.fullerton:

Laplace Transform

2010-01-12T02:23:51Z

Colby.fullerton:

Fullerton, Colby

2010-01-12T00:57:17Z

Colby.fullerton: New page: == Contact info == '''Email:''' colby.fullerton@wallawalla.edu '''Phone/Text:''' (509) 386-7970

== Contact info ==

'''Email:''' colby.fullerton@wallawalla.edu

'''Phone/Text:''' (509) 386-7970