Class Wiki - User contributions [en]

Derivative Operator Picture Code

2010-02-05T06:19:08Z

John.hawkins: Main Content

The figure in [[Derivative Matrix for a Function Vector]] was generated with the following code:

 

<source lang="matlab">

%

% _-_,, ,, _-_- ,,

% ( // || /, _ ; || '

% _|| /'\\ ||/\\ \\/\\ || __ < \, \\/\/\ ||/\ \\ \\/\\ _-_,

% _|| || || || || || || ~||- - /-|| || | | ||_< || || || ||_.

% || || || || || || || ||===|| (( || || | | || | || || || ~ ||

% -__-, \\,/ \\ |/ \\ \\ ( \_, | \/\\ \\/\\/ \\,\ \\ \\ \\ ,-_-

% _/ `

%

% John Hawkins

% LNA

% Derivative Matrix Example

% 2 Feb 2010

function DerivativeExample

x=linspace(0,4*pi,500)';

y=sin(x);

dydx=cos(x);

h.fig=figure('color','w','units','normalized','position',[.2 .2 .6 .6]);

plot(x,y,'color',[.8 .8 .8]);

hold on;

plot(x,dydx,'color',[.8 .8 .8]);

numStartPoints=10;

x=linspace(0,4*pi,numStartPoints)';

y=sin(x);

dydx=DERIV(x,y);

h.y=plot(x,y,'b.');

h.construct=plot(0,0,'r');

h.dydx=plot(0,0,'r.');

legend([h.y,h.construct,h.dydx],'y','Construction Line','dy/dx');

xlabel('x'); ylabel('y');

axis equal;

title('Taking the Derivative of a Finite Function Vector for y=sin x',...

'fontname','times','fontsize',20);

set([h.construct,h.dydx],'visible','off');

fig = getframe(h.fig);

[im,map] = rgb2ind(fig.cdata,256,'nodither');

set(h.construct,'visible','on','xdata',x([1,2]),'ydata',y([1,2]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'visible','on','xdata',x(1),'ydata',dydx(1));

im=CAPTUREFIG(im,map,h);

for n=2:numStartPoints-1

set(h.construct,'xdata',x([n-1,n+1]),'ydata',y([n-1,n+1]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'xdata',x(1:n),'ydata',dydx(1:n));

im=CAPTUREFIG(im,map,h);

end

set(h.construct,'xdata',x([numStartPoints-1,numStartPoints]),...

'ydata',y([numStartPoints-1,numStartPoints]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'xdata',x,'ydata',dydx);

im=CAPTUREFIG(im,map,h);

set(h.construct,'visible','off');

im=CAPTUREFIG(im,map,h);

for k=1:5

im=CAPTUREFIG(im,map,h);

end

for n=[15 20 30 50 100 200 500]

x=linspace(0,4*pi,n)';

y=sin(x);

dydx=DERIV(x,y);

set(h.y,'xdata',x,'ydata',y);

set(h.dydx,'xdata',x,'ydata',dydx);

im=CAPTUREFIG(im,map,h);

end

for k=1:5

im=CAPTUREFIG(im,map,h);

end

imwrite(im,map,'Derivative.gif','DelayTime',.5,'LoopCount',inf)

function dydx=DERIV(x,y)

num=numel(y);

deltax=x(2)-x(1);

operator=zeros(num);

operator(1,1)=-2;

operator(1,2)=2;

for k=2:num-1

operator(k,k+1)=1;

operator(k,k-1)=-1;

end

operator(num,num-1)=-2;

operator(num,num)=2;

operator=operator/(2*deltax);

dydx=operator*y;

function im=CAPTUREFIG(im,map,h)

fig = getframe(h.fig);

im(:,:,1,size(im,4)+1) = rgb2ind(fig.cdata,map,'nodither');

</source>

Derivative Matrix for a Function Vector

2010-02-05T06:18:57Z

John.hawkins: Moving code to its own page

Problem 5 Exam 1

2010-02-05T06:11:17Z

John.hawkins:

==General Problem Statement==
Set up and solve a simple spring mass problem that models a car's shock absorber system.

==Designed Problem Statement==

A car of mass (<math> m \ </math>) 1000 kg rests on a shock absorber that has a damper coefficient (<math> \beta \ </math>) of 2,000 N<math>\cdot</math>s/m and a spring coefficient (<math> k \ </math>) of 400,000 N/m. A tree lands on the car, delivering a force of <math>f_t</math> to the car and consequently the shock absorber.

Find the transfer function of the shock absorber and create the Bode plot of it.

[[Image:SD_FBD.jpg|thumb|800px|right|The shock absorber in its uncompressed state and compressed state.]]

==Equations==

The force of the tree and the force of the car's weight due to gravity can be combined into one equation. 
<math> f(t) = f_t + mg \ </math> 
The damping force of the damper in the shock absorber has an equation similar to a spring. 
<math> f_D = \beta \frac {dy} {dt} </math> 
The spring on the shock absorber has the following equation to define its reactive force against the tree and car. 
<math> f_s = ky(t) \ </math> 

Doing the sum of the forces in the y-direction gives us the following equation: 
<math> \Sigma F_y = f(t) - f_D - f_s = ma_y \ </math> 

The acceleration in the y-direction can be substituted out as <math> \frac {d^2y} {dt^2} </math>. And we can then substitute in the remaining definitions of two of the three force variables in the summation equation above. The resulting force of the tree falling on the car and the car's weight need not be substituted in. This will give us the equation here: 
<math> \Sigma F_y = f(t) - \beta \frac {dy} {dt} - ky(t) = m\frac {d^2y} {dt^2} \ </math> 

Now that we have this equation, we can take the Laplace transform of it: 
<math> \mathcal{L} \left\{\Sigma F_y = f(t) - \beta \frac {dy} {dt} - ky(t) = m\frac {d^2y} {dt^2} \right\} </math> 

We can now progress through and take the Laplace of each of the terms to find the output function <math>Y(s) \ </math>. 

<math> F(s) - \beta (sY(s) - y(0)) - kY(s) = m(s^2Y(s) - sy(0) - y'(0)) \ </math> 

Since this was a second order differential equation before we took the Laplace Transform of it, we need two initial conditions. For this example, we can say that: 
<math> y(0) = y'(0) = 0 \ </math> 

Using these initial conditions, we can simplify the transformed equation. 
<math> F(s) - \beta (sY(s)) - kY(s) = m(s^2Y(s)) \ </math> 

Next we can move the terms around so that we solve for <math>Y(s) \ </math>, our output function. 
<math> Y(s) = \frac {F(s)} {ms^2 + \beta s + k} = \frac {\frac {F(s)} {m}} {s^2 + \frac {\beta s} {m} + \frac {k} {m}} \ </math> 

Since our output function is: 
<math> Y(s) = \frac {\frac {F(s)} {m}} {s^2 + \frac {\beta s} {m} + \frac {k} {m}} \ </math> 

We can now plug it into the transfer function equation, which is written as: 
<math> H(s) = \frac {Y(s)} {F(s)} \ </math> The input function is <math> F(s) \ </math>, which is the Laplaced version of <math> f(t) \ </math>, because it includes the force of the tree falling on the car, which is the factor that disturbs the system. 

Plugging in our output function into the transfer function equation, we get the following: 
<math> H(s) = \frac {\frac {\frac {F(s)} {m}} {s^2 + \frac {\beta s} {m} + \frac {k} {m}}} {F(s)} \ </math> 

This can simplify to: 
<math> H(s) = \frac {\frac {1} {m}} {s^2 + \frac {\beta s} {m} + \frac {k} {m}} \ </math> 

Plugging in the given values for <math> m \ </math>, <math> \beta \ </math>, and <math> k \ </math>, the simplified transfer function becomes: 
<math> H(s) = \frac {\frac {1} {1000}} {s^2 + \frac {2000s} {1000} + \frac {400000} {1000}} \ </math> 
<math> H(s) = \frac {\frac {1} {1000}} {s^2 + 2s + 400} \ </math> 

This final transfer function generates the MATLAB-created figure below are the final solutions.

[[Image:Bode_Plot_2.JPG|thumb|560px|center|The Bode plot of the transfer function.]] 

==Author==
Brian Roath

==Read By==
John Hawkins

Derivative Matrix for a Function Vector

2010-02-03T06:04:32Z

John.hawkins: Added final equation.

==Author==

John Hawkins

==Problem==

The following problem statement is that proposed by Prof. Frohne on the main class page:

# Explore how a linear operator, like for example <math>d \over dt</math> can be represented as some kind of a matrix multiply (with perhaps an infinite number of dimensions).

==Solution==

The solution presented here is limited to presenting a matrix representation of the operator <math>d / dx</math>.

First, let us consider the finite case. Let <math>x</math> be the independent variable with <math>n</math> components and let the components of <math>x</math> be defined such that <math>x_{m+1}=x_m+\Delta x</math> for all <math>m<n</math> and for some constant <math>\Delta x</math>. Also let <math>y=f(x)</math>. We wish to find the slope of the function at <math>y(x_m)</math> for any <math>m</math>. To do this for the finite case, we will approximate the slope at a point <math>y(x_m)</math> as being the slope of the line between the two points <math>y(x_{m-1})</math> and <math>y(x_{m+1})</math> given that <math>1<m<n</math>. For the case when <math>m=1</math> or <math>m=n</math>, we approximate <math>dy/dx</math> as the slope the lines between <math>y(x_1)</math> and <math>y(x_n)</math> respectively. These approximations are accomplished by the following matrix:

<center><math>

\frac{d}{dx} = \dfrac{1}{2(\Delta x)}\begin{bmatrix}

-2&2&0&0&&&&&\\

-1&0&1&0&&&\cdots&&\\

0&-1&0&1&&&&&\\

\\

&&&&\ddots&&&&\\

\\

&\vdots&&&&-1&0&1&0\\

&&&&&0&-1&0&1\\

&&&&&0&0&-2&2

\end{bmatrix}

</math></center>

Note that we need to divide by <math>1/2(\Delta x)</math> because the distance between <math>x_{m-1}</math> and <math>x_{m+1}</math> is <math>2\Delta x</math>. Similarly, the nonzero entries in the first and last rows are 2 to cancel out the 2 in the denominator. Using this matrix, we find that

<center><math>

\left(\frac{d}{dx}\right)y=\dfrac{1}{2(\Delta x)}\begin{bmatrix}

-2&2&0&0&&&&&\\

-1&0&1&0&&&\cdots&&\\

0&-1&0&1&&&&&\\

\\

&&&&\ddots&&&&\\

\\

&\vdots&&&&-1&0&1&0\\

&&&&&0&-1&0&1\\

&&&&&0&0&-2&2

\end{bmatrix}
\begin{bmatrix}y(x_1)\\y(x_2)\\y(x_3)\\ \\ \vdots \\ \\ y(x_{n-2})\\y(x_{n-1})\\y(x_n)\end{bmatrix}
=

\begin{bmatrix}

\left[y(x_2)-y(x_1)\right]/\Delta x\\

\left[y(x_3)-y(x_1)\right]/2(\Delta x)\\

\left[y(x_4)-y(x_2)\right]/2(\Delta x)\\

\\

\vdots \\

\\

\left[y(x_{n-1})-y(x_{n-3})\right]/2(\Delta x)\\

\left[y(x_n)-y(x_{n-2})\right]/2(\Delta x)\\

\left[y(x_n)-y(x_{n-1})\right]/\Delta x

\end{bmatrix}

</math></center>

which corresponds with the slopes we wished to obtain.

==Solution Picture==

The figure below was created to demonstrate how the above matrix works. First, 10 points of the function are displayed. Then for each point the line between the two adjacent points is drawn and the slope of that line is shown to be <math>dy/dx</math> for the original point, calculated using the matrix shown above. After this has been done in the case of 10 points, the function and derivative vectors are shown for 15, 20, 30, 50, 100, 200, and 500 points.

[[Image:Derivative.gif]]

==Code==

The figure above was generated with the following code:

 

<source lang="matlab">

%

% _-_,, ,, _-_- ,,

% ( // || /, _ ; || '

% _|| /'\\ ||/\\ \\/\\ || __ < \, \\/\/\ ||/\ \\ \\/\\ _-_,

% _|| || || || || || || ~||- - /-|| || | | ||_< || || || ||_.

% || || || || || || || ||===|| (( || || | | || | || || || ~ ||

% -__-, \\,/ \\ |/ \\ \\ ( \_, | \/\\ \\/\\/ \\,\ \\ \\ \\ ,-_-

% _/ `

%

% John Hawkins

% LNA

% Derivative Matrix Example

% 2 Feb 2010

function DerivativeExample

x=linspace(0,4*pi,500)';

y=sin(x);

dydx=cos(x);

h.fig=figure('color','w','units','normalized','position',[.2 .2 .6 .6]);

plot(x,y,'color',[.8 .8 .8]);

hold on;

plot(x,dydx,'color',[.8 .8 .8]);

numStartPoints=10;

x=linspace(0,4*pi,numStartPoints)';

y=sin(x);

dydx=DERIV(x,y);

h.y=plot(x,y,'b.');

h.construct=plot(0,0,'r');

h.dydx=plot(0,0,'r.');

legend([h.y,h.construct,h.dydx],'y','Construction Line','dy/dx');

xlabel('x'); ylabel('y');

axis equal;

title('Taking the Derivative of a Finite Function Vector for y=sin x',...

'fontname','times','fontsize',20);

set([h.construct,h.dydx],'visible','off');

fig = getframe(h.fig);

[im,map] = rgb2ind(fig.cdata,256,'nodither');

set(h.construct,'visible','on','xdata',x([1,2]),'ydata',y([1,2]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'visible','on','xdata',x(1),'ydata',dydx(1));

im=CAPTUREFIG(im,map,h);

for n=2:numStartPoints-1

set(h.construct,'xdata',x([n-1,n+1]),'ydata',y([n-1,n+1]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'xdata',x(1:n),'ydata',dydx(1:n));

im=CAPTUREFIG(im,map,h);

end

set(h.construct,'xdata',x([numStartPoints-1,numStartPoints]),...

'ydata',y([numStartPoints-1,numStartPoints]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'xdata',x,'ydata',dydx);

im=CAPTUREFIG(im,map,h);

set(h.construct,'visible','off');

im=CAPTUREFIG(im,map,h);

for k=1:5

im=CAPTUREFIG(im,map,h);

end

for n=[15 20 30 50 100 200 500]

x=linspace(0,4*pi,n)';

y=sin(x);

dydx=DERIV(x,y);

set(h.y,'xdata',x,'ydata',y);

set(h.dydx,'xdata',x,'ydata',dydx);

im=CAPTUREFIG(im,map,h);

end

for k=1:5

im=CAPTUREFIG(im,map,h);

end

imwrite(im,map,'Derivative.gif','DelayTime',.5,'LoopCount',inf)

function dydx=DERIV(x,y)

num=numel(y);

deltax=x(2)-x(1);

operator=zeros(num);

operator(1,1)=-2;

operator(1,2)=2;

for k=2:num-1

operator(k,k+1)=1;

operator(k,k-1)=-1;

end

operator(num,num-1)=-2;

operator(num,num)=2;

operator=operator/(2*deltax);

dydx=operator*y;

function im=CAPTUREFIG(im,map,h)

fig = getframe(h.fig);

im(:,:,1,size(im,4)+1) = rgb2ind(fig.cdata,map,'nodither');

</source>

==Reviewed By==

==Read By==

==Comments==

Derivative Matrix for a Function Vector

2010-02-03T05:47:38Z

John.hawkins:

==Author==

John Hawkins

==Problem==

The following problem statement is that proposed by Prof. Frohne on the main class page:

# Explore how a linear operator, like for example <math>d \over dt</math> can be represented as some kind of a matrix multiply (with perhaps an infinite number of dimensions).

==Solution==

The solution presented here is limited to presenting a matrix representation of the operator <math>d / dx</math>.

First, let us consider the finite case. Let <math>x</math> be the independent variable with <math>n</math> components and let the components of <math>x</math> be defined such that <math>x_{m+1}=x_m+\Delta x</math> for all <math>m<n</math> and for some constant <math>\Delta x</math>. Also let <math>y=f(x)</math>. We wish to find the slope of the function at <math>y(x_m)</math> for any <math>m</math>. To do this for the finite case, we will approximate the slope at a point <math>y(x_m)</math> as being the slope of the line between the two points <math>y(x_{m-1})</math> and <math>y(x_{m+1})</math> given that <math>1<m<n</math>. For the case when <math>m=1</math> or <math>m=n</math>, we approximate <math>dy/dx</math> as the slope the lines between <math>y(x_1)</math> and <math>y(x_n)</math> respectively. These approximations are accomplished by the following matrix:

<center><math>

\frac{d}{dx} = \dfrac{1}{2(\Delta x)}\begin{bmatrix}

-2&2&0&0&&&&&\\

-1&0&1&0&&&\cdots&&\\

0&-1&0&1&&&&&\\

\\

&&&&\ddots&&&&\\

\\

&\vdots&&&&-1&0&1&0\\

&&&&&0&-1&0&1\\

&&&&&0&0&-2&2

\end{bmatrix}

</math></center>

Note that we need to divide by <math>1/2(\Delta x)</math> because the distance between <math>x_{m-1}</math> and <math>x_{m+1}</math> is <math>2\Delta x</math>. Similarly, the nonzero entries in the first and last rows are 2 to cancel out the 2 in the denominator.

==Solution Picture==

The figure below was created to demonstrate how the above matrix works. First, 10 points of the function are displayed. Then for each point the line between the two adjacent points is drawn and the slope of that line is shown to be <math>dy/dx</math> for the original point, calculated using the matrix shown above. After this has been done in the case of 10 points, the function and derivative vectors are shown for 15, 20, 30, 50, 100, 200, and 500 points.

[[Image:Derivative.gif]]

==Code==

The figure above was generated with the following code:

 

<source lang="matlab">

%

% _-_,, ,, _-_- ,,

% ( // || /, _ ; || '

% _|| /'\\ ||/\\ \\/\\ || __ < \, \\/\/\ ||/\ \\ \\/\\ _-_,

% _|| || || || || || || ~||- - /-|| || | | ||_< || || || ||_.

% || || || || || || || ||===|| (( || || | | || | || || || ~ ||

% -__-, \\,/ \\ |/ \\ \\ ( \_, | \/\\ \\/\\/ \\,\ \\ \\ \\ ,-_-

% _/ `

%

% John Hawkins

% LNA

% Derivative Matrix Example

% 2 Feb 2010

function DerivativeExample

x=linspace(0,4*pi,500)';

y=sin(x);

dydx=cos(x);

h.fig=figure('color','w','units','normalized','position',[.2 .2 .6 .6]);

plot(x,y,'color',[.8 .8 .8]);

hold on;

plot(x,dydx,'color',[.8 .8 .8]);

numStartPoints=10;

x=linspace(0,4*pi,numStartPoints)';

y=sin(x);

dydx=DERIV(x,y);

h.y=plot(x,y,'b.');

h.construct=plot(0,0,'r');

h.dydx=plot(0,0,'r.');

legend([h.y,h.construct,h.dydx],'y','Construction Line','dy/dx');

xlabel('x'); ylabel('y');

axis equal;

title('Taking the Derivative of a Finite Function Vector for y=sin x',...

'fontname','times','fontsize',20);

set([h.construct,h.dydx],'visible','off');

fig = getframe(h.fig);

[im,map] = rgb2ind(fig.cdata,256,'nodither');

set(h.construct,'visible','on','xdata',x([1,2]),'ydata',y([1,2]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'visible','on','xdata',x(1),'ydata',dydx(1));

im=CAPTUREFIG(im,map,h);

for n=2:numStartPoints-1

set(h.construct,'xdata',x([n-1,n+1]),'ydata',y([n-1,n+1]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'xdata',x(1:n),'ydata',dydx(1:n));

im=CAPTUREFIG(im,map,h);

end

set(h.construct,'xdata',x([numStartPoints-1,numStartPoints]),...

'ydata',y([numStartPoints-1,numStartPoints]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'xdata',x,'ydata',dydx);

im=CAPTUREFIG(im,map,h);

set(h.construct,'visible','off');

im=CAPTUREFIG(im,map,h);

for k=1:5

im=CAPTUREFIG(im,map,h);

end

for n=[15 20 30 50 100 200 500]

x=linspace(0,4*pi,n)';

y=sin(x);

dydx=DERIV(x,y);

set(h.y,'xdata',x,'ydata',y);

set(h.dydx,'xdata',x,'ydata',dydx);

im=CAPTUREFIG(im,map,h);

end

for k=1:5

im=CAPTUREFIG(im,map,h);

end

imwrite(im,map,'Derivative.gif','DelayTime',.5,'LoopCount',inf)

function dydx=DERIV(x,y)

num=numel(y);

deltax=x(2)-x(1);

operator=zeros(num);

operator(1,1)=-2;

operator(1,2)=2;

for k=2:num-1

operator(k,k+1)=1;

operator(k,k-1)=-1;

end

operator(num,num-1)=-2;

operator(num,num)=2;

operator=operator/(2*deltax);

dydx=operator*y;

function im=CAPTUREFIG(im,map,h)

fig = getframe(h.fig);

im(:,:,1,size(im,4)+1) = rgb2ind(fig.cdata,map,'nodither');

</source>

==Reviewed By==

==Read By==

==Comments==

Winter 2010

2010-02-03T05:46:34Z

John.hawkins:

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]]
*[[Fourier Example]](**Check out the bonus video, really helpful**)

4. [[Fullerton, Colby]]
*[[Laplace Transform]]
*[[Class Notes 1-5-2010]]

5. [[Grant, Joshua]]

6. [[Gratias, Ryan]]

7. [[Hawkins, John]]

*[[Exercise: Sawtooth Wave Fourier Transform]]
*[[Exercise: Sawtooth Redone With Exponential Basis Functions]]

*[[Exercise: Solving an IVP Problem with Laplace Transforms]]

*[[Derivative Matrix for a Function Vector]]

 
8. [[Lau, Chris]]
*[[Fourier Series: Explained!]]

9. [[Roath, Brian]]
*[[Laplace Transform]]
*[[Class Notes 1-5-2010]]

10. [[Robbins, David]]

11. [[Roth, Andrew]]
*[[Example: LaTex format (0 points)]]
*[[Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency]]

12. [[Vazquez, Brandon]]

13. [[Vier, Michael]]
* [[Fourier Series: Explained!]]

14. [[Wooley, Thomas]]

15. [[Jaymin, Joseph]]
*[[Basic_Laplace_Transforms]]

16. [[Starr, Brielle]]

17. [[Starr, Tyler]]

==Article Suggestions or Homework==
(Please put a note when these are published.)

# Explore how a linear operator, like for example <math>d \over dt</math> can be represented as some kind of a matrix multiply (with perhaps an infinite number of dimensions).
## One solution: [[Derivative Matrix for a Function Vector]] (John Hawkins).
# Suppose you had to approximate a vector by using the first few dimensions. Show that if you wish to minimize the error, defined as the length squared of the difference of your approximate vector and the real vector, that the coefficients (or components) of the approximate vector would still be the same as the ones in the same dimensions of the exact vector. Now, apply this to the Fourier series.
# Describe the e Gram-Schmidt Orthogonalization process for taking a set of non orthogonal vectors and using them to find an orthogonal set. How does this apply to functions?
# Solve a circuit using Laplace Transforms.
# Set up and solve a simple spring mass problem that models a car's shock absorber system.
# Find the steady state response of a simple circuit (with at least one capacitor or inductor) to a triangle wave using Fourier series, and again with Laplace transforms. Compare and contrast the solutions.
# Find the Laplace transform of <math>cos(\omega_0 t) x(t)</math>. What does this mean if the function <math>x(t) = cos(\omega_1 t)</math>?

===More Specific Elementary Problems===
# Solve the following differential equation using Laplace transforms.<math>\dot y + 10y ~=~ u(t)</math>, <math>y(0) = 4</math>.
## If the input is considered to be <math>u(t)</math> and the output <math>y(t)</math>, what is the transfer function?
## What is the output, <math>y(t)</math>, in sinusoidal steady state, if <math>u(t)</math> is replaced with <math>cos(\omega t)</math>?
# A series RLC circuit with <math>R=10 \Omega</math>, <math>L=1 H</math>, and <math>C = 1 F</math> is driven by <math>tcos(t)u(t) V</math>. What is the current, <math>i(t)</math> if the initial current is 1 A and the initial capactor voltage is 2 volts?
# If a linear time invariant system has a transfer function H(s), what is the steady state response of that system to the the the [http://fweb/class-wiki/index.php/Exercise:_Sawtooth_Wave_Fourier_Transform triangle wave]?
# Write <math>x(t)</math> as a linear combination of time shifted impulse functions.
# Find the Laplace transform of <math>x(t-3)(t-3)e^{4(t-3)} u(t-3)</math>.

==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]]

* [[Solving Series RLC Circuit Using Laplace Transforms]]

* [[Laplace Transform of a Triangle Wave]]

* [[ProblemCh14-22]]

* [[Problem 5 Exam 1]]

Derivative Matrix for a Function Vector

2010-02-03T05:37:46Z

John.hawkins: Main Content

==Author==

John Hawkins

==Problem==

The following problem statement is that proposed by Prof. Frohne on the main class page:

# Explore how a linear operator, like for example <math>d \over dt</math> can be represented as some kind of a matrix multiply (with perhaps an infinite number of dimensions).

==Solution==

The solution presented here is limited to presenting a matrix representation of the operator <math>d / dx</math>.

First, let us consider the finite case. Let <math>x</math> be the independent variable with <math>n</math> components and let the components of <math>x</math> be defined such that <math>x_{m+1}=x_m+\Delta x</math> for all <math>m<n</math> and for some constant <math>\Delta x</math>. Also let <math>y=f(x)</math>. We wish to find the slope of the function at <math>y(x_m)</math> for any <math>m</math>. To do this for the finite case, we will approximate the slope at a point <math>y(x_m)</math> as being the slope of the line between the two points <math>y(x_{m-1})</math> and <math>y(x_{m+1})</math> given that <math>1<m<n</math>. For the case when <math>m=1</math> or <math>m=n</math>, we approximate <math>dy/dx</math> as the slope the lines between <math>y(x_1)</math> and <math>y(x_n)</math> respectively. These approximations are accomplished by the following matrix:

<center><math>

\frac{d}{dx} = \dfrac{1}{2(\Delta x)}\begin{bmatrix}

-2&2&0&0&&&&&\\

-1&0&1&0&&&\cdots&&\\

0&-1&0&1&&&&&\\

\\

&&&&\ddots&&&&\\

\\

&\vdots&&&&-1&0&1&0\\

&&&&&0&-1&0&1\\

&&&&&0&0&-2&2

\end{bmatrix}

</math></center>

Note that we need to divide by <math>1/2(\Delta x)</math> because the distance between <math>x_{m-1}</math> and <math>x_{m+1}</math> is <math>2\Delta x</math>. Similarly, the nonzero entries in the first and last rows are 2 to cancel out the 2 in the denominator.

==Solution Picture==

The figure below was created to demonstrate how the above matrix works. First, 10 points of the function are displayed. Then for each point the line between the two adjacent points is drawn and the slope of that line is shown to be <math>dy/dx</math> for the original point, calculated using the matrix shown above. After this has been done in the case of 10 points, the function and derivative vectors are shown for 15, 20, 30, 50, 100, 200, and 500 points.

[[Image:Derivative.gif]]

==Code==

The figure above was generated with the following code:

 

<source lang="matlab">

%

% _-_,, ,, _-_- ,,

% ( // || /, _ ; || '

% _|| /'\\ ||/\\ \\/\\ || __ < \, \\/\/\ ||/\ \\ \\/\\ _-_,

% _|| || || || || || || ~||- - /-|| || | | ||_< || || || ||_.

% || || || || || || || ||===|| (( || || | | || | || || || ~ ||

% -__-, \\,/ \\ |/ \\ \\ ( \_, | \/\\ \\/\\/ \\,\ \\ \\ \\ ,-_-

% _/ `

%

% John Hawkins

% LNA

% Derivative Matrix Example

% 2 Feb 2010

function DerivativeExample

x=linspace(0,4*pi,500)';

y=sin(x);

dydx=cos(x);

h.fig=figure('color','w','units','normalized','position',[.2 .2 .6 .6]);

plot(x,y,'color',[.8 .8 .8]);

hold on;

plot(x,dydx,'color',[.8 .8 .8]);

numStartPoints=10;

x=linspace(0,4*pi,numStartPoints)';

y=sin(x);

dydx=DERIV(x,y);

h.y=plot(x,y,'b.');

h.construct=plot(0,0,'r');

h.dydx=plot(0,0,'r.');

legend([h.y,h.construct,h.dydx],'y','Construction Line','dy/dx');

xlabel('x'); ylabel('y');

axis equal;

title('Taking the Derivative of a Finite Function Vector for y=sin x',...

'fontname','times','fontsize',20);

set([h.construct,h.dydx],'visible','off');

fig = getframe(h.fig);

[im,map] = rgb2ind(fig.cdata,256,'nodither');

set(h.construct,'visible','on','xdata',x([1,2]),'ydata',y([1,2]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'visible','on','xdata',x(1),'ydata',dydx(1));

im=CAPTUREFIG(im,map,h);

for n=2:numStartPoints-1

set(h.construct,'xdata',x([n-1,n+1]),'ydata',y([n-1,n+1]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'xdata',x(1:n),'ydata',dydx(1:n));

im=CAPTUREFIG(im,map,h);

end

set(h.construct,'xdata',x([numStartPoints-1,numStartPoints]),...

'ydata',y([numStartPoints-1,numStartPoints]));

im=CAPTUREFIG(im,map,h);

set(h.dydx,'xdata',x,'ydata',dydx);

im=CAPTUREFIG(im,map,h);

set(h.construct,'visible','off');

im=CAPTUREFIG(im,map,h);

for k=1:5

im=CAPTUREFIG(im,map,h);

end

for n=[15 20 30 50 100 200 500]

x=linspace(0,4*pi,n)';

y=sin(x);

dydx=DERIV(x,y);

set(h.y,'xdata',x,'ydata',y);

set(h.dydx,'xdata',x,'ydata',dydx);

im=CAPTUREFIG(im,map,h);

end

for k=1:5

im=CAPTUREFIG(im,map,h);

end

imwrite(im,map,'Derivative.gif','DelayTime',.5,'LoopCount',inf)

function dydx=DERIV(x,y)

num=numel(y);

deltax=x(2)-x(1);

operator=zeros(num);

operator(1,1)=-2;

operator(1,2)=2;

for k=2:num-1

operator(k,k+1)=1;

operator(k,k-1)=-1;

end

operator(num,num-1)=-2;

operator(num,num)=2

operator=operator/(2*deltax);

dydx=operator*y;

function im=CAPTUREFIG(im,map,h)

fig = getframe(h.fig);

im(:,:,1,size(im,4)+1) = rgb2ind(fig.cdata,map,'nodither');

</source>

==Reviewed By==

==Read By==

==Comments==

File:Derivative.gif

2010-02-03T04:58:13Z

John.hawkins:

Winter 2010

2010-02-03T04:56:07Z

John.hawkins:

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]]
*[[Fourier Example]](**Check out the bonus video, really helpful**)

4. [[Fullerton, Colby]]
*[[Laplace Transform]]
*[[Class Notes 1-5-2010]]

5. [[Grant, Joshua]]

6. [[Gratias, Ryan]]

7. [[Hawkins, John]]

*[[Exercise: Sawtooth Wave Fourier Transform]]
*[[Exercise: Sawtooth Redone With Exponential Basis Functions]]

*[[Exercise: Solving an IVP Problem with Laplace Transforms]]

 
8. [[Lau, Chris]]
*[[Fourier Series: Explained!]]

9. [[Roath, Brian]]
*[[Laplace Transform]]
*[[Class Notes 1-5-2010]]

10. [[Robbins, David]]

11. [[Roth, Andrew]]
*[[Example: LaTex format (0 points)]]
*[[Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency]]

12. [[Vazquez, Brandon]]

13. [[Vier, Michael]]
* [[Fourier Series: Explained!]]

14. [[Wooley, Thomas]]

15. [[Jaymin, Joseph]]
*[[Basic_Laplace_Transforms]]

16. [[Starr, Brielle]]

17. [[Starr, Tyler]]

==Article Suggestions or Homework==
(Please put a note when these are published.)

# Explore how a linear operator, like for example <math>d \over dt</math> can be represented as some kind of a matrix multiply (with perhaps an infinite number of dimensions).
## One solution: [[Derivative Matrix for a Function Vector]] (John Hawkins).
# Suppose you had to approximate a vector by using the first few dimensions. Show that if you wish to minimize the error, defined as the length squared of the difference of your approximate vector and the real vector, that the coefficients (or components) of the approximate vector would still be the same as the ones in the same dimensions of the exact vector. Now, apply this to the Fourier series.
# Describe the e Gram-Schmidt Orthogonalization process for taking a set of non orthogonal vectors and using them to find an orthogonal set. How does this apply to functions?
# Solve a circuit using Laplace Transforms.
# Set up and solve a simple spring mass problem that models a car's shock absorber system.
# Find the steady state response of a simple circuit (with at least one capacitor or inductor) to a triangle wave using Fourier series, and again with Laplace transforms. Compare and contrast the solutions.
# Find the Laplace transform of <math>cos(\omega_0 t) x(t)</math>. What does this mean if the function <math>x(t) = cos(\omega_1 t)</math>?

===More Specific Elementary Problems===
# Solve the following differential equation using Laplace transforms.<math>\dot y + 10y ~=~ u(t)</math>, <math>y(0) = 4</math>.
## If the input is considered to be <math>u(t)</math> and the output <math>y(t)</math>, what is the transfer function?
## What is the output, <math>y(t)</math>, in sinusoidal steady state, if <math>u(t)</math> is replaced with <math>cos(\omega t)</math>?
# A series RLC circuit with <math>R=10 \Omega</math>, <math>L=1 H</math>, and <math>C = 1 F</math> is driven by <math>tcos(t)u(t) V</math>. What is the current, <math>i(t)</math> if the initial current is 1 A and the initial capactor voltage is 2 volts?
# If a linear time invariant system has a transfer function H(s), what is the steady state response of that system to the the the [http://fweb/class-wiki/index.php/Exercise:_Sawtooth_Wave_Fourier_Transform triangle wave]?
# Write <math>x(t)</math> as a linear combination of time shifted impulse functions.
# Find the Laplace transform of <math>x(t-3)(t-3)e^{4(t-3)} u(t-3)</math>.

==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]]

* [[Solving Series RLC Circuit Using Laplace Transforms]]

* [[Laplace Transform of a Triangle Wave]]

* [[ProblemCh14-22]]

* [[Problem 5 Exam 1]]

Example: Ideal Transformer Exercise

2010-02-01T07:13:53Z

John.hawkins:

==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_1}e_1=\frac{2000}{500}(120)=480\text{ V}</math>

 

<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>

 

<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>

 

<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>

 

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

 

<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>

 

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.

Exercise: Solving an IVP Problem with Laplace Transforms

2010-01-29T07:36:57Z

John.hawkins: Main content

==Author==

John Hawkins

==Problem Statement==

Solve the following initial value problem using Laplace Transforms:
 
<center>

<math>y^{\prime\prime}-4y=12x,\ \ y(0)=4,\ \ y^\prime(0)=1.</math>

</center>
 
Note: This problem was solved by Zill without the use of Laplace Transforms.<ref>Dennis G. Zill, ''A first course in Differential Equations'', 8th ed., Int. ed (Belmont, CA: Thomson Learning, 2005), 128.</ref>
==Solution==
Given the initial ODE
 

<center><math>y^{\prime\prime}-4y=12x</math></center>

 
we take the Laplace transform of both sides
 

<center><math>\mathcal{L}\left\{y^{\prime\prime}-4y\right\}=\mathcal{L}\left\{12x\right\}</math></center>

 
Using the transforms displayed in [[Laplace Transform]], we find this to be
 

<center><math>[s^2Y(s)-sy(0)-y^\prime(0)]-4Y(s)=\frac{12}{s^2}</math></center>

 
which, with initial values substituted, gives
 

<center><math>\Rightarrow (s^2-4)Y(s)-4s-1=\frac{12}{s^2}</math></center>

 
Hence,
 

<center>

<math>Y(s)=\frac{\frac{12}{s^2}+4s+1}{s^2-4}</math>

 

::<math>=\frac{12+4s^3+s^2}{s^4-4s^2}</math>

 

:::<math>=\frac{12+s^2+4s^3}{s^2(s-2)(s+2)}</math>

</center>

 
Using a calculator to expand this, we have

 

<center><math>Y(s)=\frac{1}{s+2}+\frac{3}{s-2}-\frac{3}{s^2}</math></center>

 
And therefore, using the equations on [[Laplace Transform]] to perform an inverse Laplace transform, we have our solution:
 

<center><math>\ y(x)=e^{-2x}+3e^{2x}-3x</math></center>

 
This equation matches that found by Zill, providing confirmation of a correct solution.

==References==
<references />
==Reviewed By==
==Read By==
==Comments==

Electromechanical Energy Conversion

2010-01-29T06:59:25Z

John.hawkins:

[[Rules]]

[[Class Roster]]

[[Points]]

[http://people.wallawalla.edu/~rob.frohne/ClassNotes/engr431index.htm Class Notes]

==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 anybody know of a way to put a box around text or math equations? Would I just have to create a 1x1 table?

==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.

==Article Suggestions==
(Please link these to the article when you complete it.)
# Rewrite the notes for the wiki. (Really, this will help get ready for the exam. The professor who did them is the one who is writing the exam questions.)
#*[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/engr431index.htm Click here for the Class Notes.]
* Effects of the Non-Linear B-H Curve
# Draw and explain the effect of the non-linear B-H curve on current waveforms for a voltage excited inductor. (This is mostly done in the notes.)
# Explain how to measure the B-H curve experimentally.
# If the B-H curve was traced out more quickly in the experiment above, would the curve look different? If so why?
# Show how to calculate the core losses of a nonlinear inductor using its i-v curve.
* Transformers
# Explore transformers with more than one secondary winding.
#* What is the input impedance of an idea transformer with two secondaries, one with N2 turns and one with N3 turns, each with a different load resistor on attached.
#* How do the mutual impedances relate to the turns ratios in transformers with more than one secondary?
#* Develop a circuit model for a non-ideal transformer with multiple secondaries. (Follow the way we did it in class. You will have more mutual inductances. You may or may not be able to do it by using ideal transformers, but you should be able to do it with mutual impedances for sure.)
# Develop the theory of auto-transformers.
#* Come up with an ideal transformer model for auto-transformers.
#* Which part of the windings carries more current?
#* Is there any isolation with auto-transformers?
#* What if you have more than one tap on the auto-transformer? Can you come up with a model for that?
#* Develop a circuit model for a single tap auto-transformer.
# Explore how leakage flux affects the inductance of an inductor. What if that flux is then recovered and the effect accounted for by mutual inductance? Does the result agree with the simple calculation of inductance without leakage?
# Describe the coupling factor, k, used in Spice simulators and other circuit simulators. Relate it to the leakage, magnetizing, and mutual inductances.
# Derive the <math>Y/\Delta</math> transformations.
# Explore the voltage regulation <math>(V_{full ~load} - V_{no ~load} ) \over {V_{full ~load} } </math>x 100% as a function of the power factor angle on the load of a transformer. (You will note some surprising results in some cases.)
# Describe the open circuit and short circuit test as applied to transformers.
# Explore how much flux is in the core of a loaded ideal transformer.
* Three Phase Circuits
# Calculate and compare how much power can be delivered with three phase circuits as compared to a single phase circuits. Assume that the same amount of copper is available for the transmission lines of both systems.
# Compare and contrast the types of three phase transformers. (See the class handout.) Explain why each is better or worse in each situation.
#* Explain what the big deal is with third harmonic currents in three phase transformers. Where do they come from? Why does the delta winding minimize their effect on voltage distortion?

And if you don't understand any of the above, please see the teacher! He is willing to sit down with you and work any and/or all of the above.

==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]]
* [[Example: Ampere's Law]] (Tyler Anderson)
* [[Ampere's Law]]
* [[DC Motor]]
* [[Fringing]]
* [[Electrostatics]]
* [[Faraday's Law]]
* [[Eddy Current]]
* [[Example Problems of Magnetic Circuits]]
* [[Ohm's Law and Reluctance]]
* [[Magnetic Flux]]

* [[Reference Terms and Units]] (Amy Crosby)
* [[Ideal Transformer Example|Example: Ideal Transformer]]
* [[Problem Set 1]](Jodi Hodge)
* [[ANOTHER IDEAL TRANSFORMER!!!!!!!!!]]
* [[Example: Magnetic Field]] (Amy Crosby)
* [[Class Notes]](Tyler Anderson)
* [[The Class Notes]] ([[Kirk Betz]])
* [[Transformer Example Problem]] (Aric Vyhmeister, Kevin Starkey, Nick Christman)

==Reviewed Articles==
These articles have been reviewed and submitted.
* [[Gauss Meters]] (Tyler Anderson)
* [[Nick_ENGR431_P1|Magnetostatics]] (Nick Christman)
* [[Magnetic Flux]] (Jason Osborne)
*[[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]] (Chris Lau)
* [[An Ideal Transformer Example]] (Chris Lau)
* [[Magnetic Circuit]] (John Hawkins)
* [[Example: Ideal Transformer Exercise]] (John Hawkins)
* [[AC Motors]]
* [[Example Problem - Toroid]] ([[Kirk Betz]])
* [[Transformer_example_problem|Ideal Transformer Example]] (Tim Rasmussen)
* [[AC vs. DC]] (Wesley Brown)
* [[Example: Metal Cart]] (Amy Crosby)

Winter 2010

2010-01-29T06:52:29Z

John.hawkins:

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]]
*[[Fourier Example]](**Check out the bonus video, really helpful**)

4. [[Fullerton, Colby]]
*[[Laplace Transform]]
*[[Class Notes 1-5-2010]]

5. [[Grant, Joshua]]

6. [[Gratias, Ryan]]

7. [[Hawkins, John]]

*[[Exercise: Sawtooth Wave Fourier Transform]]
*[[Exercise: Sawtooth Redone With Exponential Basis Functions]]

*[[Exercise: Solving an IVP Problem with Laplace Transforms]]

 
8. [[Lau, Chris]]
*[[Fourier Series: Explained!]]

9. [[Roath, Brian]]
*[[Laplace Transform]]
*[[Class Notes 1-5-2010]]

10. [[Robbins, David]]

11. [[Roth, Andrew]]
*[[Example: LaTex format (0 points)]]
*[[Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency]]

12. [[Vazquez, Brandon]]

13. [[Vier, Michael]]
* [[Fourier Series: Explained!]]

14. [[Wooley, Thomas]]

15. [[Jaymin, Joseph]]
*[[Basic_Laplace_Transforms]]

16. [[Starr, Brielle]]

17. [[Starr, Tyler]]

==Article Suggestions or Homework==
(Please put a note when these are published.)

# Explore how a linear operator, like for example <math>d \over dt</math> can be represented as some kind of a matrix multiply (with perhaps an infinite number of dimensions).
# Suppose you had to approximate a vector by using the first few dimensions. Show that if you wish to minimize the error, defined as the length squared of the difference of your approximate vector and the real vector, that the coefficients (or components) of the approximate vector would still be the same as the ones in the same dimensions of the exact vector. Now, apply this to the Fourier series.
# Describe the e Gram-Schmidt Orthogonalization process for taking a set of non orthogonal vectors and using them to find an orthogonal set. How does this apply to functions?
# Solve a circuit using Laplace Transforms.
# Set up and solve a simple spring mass problem that models a car's shock absorber system.
# Find the steady state response of a simple circuit (with at least one capacitor or inductor) to a triangle wave using Fourier series, and again with Laplace transforms. Compare and contrast the solutions.
# Find the Laplace transform of <math>cos(\omega_0 t) x(t)</math>. What does this mean if the function <math>x(t) = cos(\omega_1 t)</math>?

==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]]

* [[Solving Series RLC Circuit Using Laplace Transforms]]

* [[Laplace Transform of a Triangle Wave]]

* [[ProblemCh14-22]]

Example: Ideal Transformer Exercise

2010-01-22T09:45:54Z

John.hawkins: Made Comments Section

==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>

 

<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>

 

<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>

 

<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>

 

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

 

<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>

 

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.

Laplace Transform

2010-01-22T09:30:51Z

John.hawkins:

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(at)\right\}=\int_0^{\infty} e^{-st} g(at) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right), \mbox{ for} ~a > 0.</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==

Jaymin Joseph
 

John Hawkins

==Comments==
John Hawkins:

*Nice list of transforms! Where did you find it? Inside the back cover of the textbook has a good list, but none including transforms of <math>\ g(t)</math>. I see your reference to the textbook. What page?

*I believe that

:<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>

should be

:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \frac {A} {s}+ \frac {B} {s+3}+ \frac {C} {s-5}</math>

*Also, the solution <math> y(t)= -\frac {2} {5} - \frac {1} {4} e^{-3t} - \frac {7} {20} e^{5t} </math> does not match the initial condition of <math>\ y(0)=1</math>.

Laplace Transform

2010-01-22T07:53:32Z

John.hawkins: Comments

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(at)\right\}=\int_0^{\infty} e^{-st} g(at) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right), \mbox{ for} ~a > 0.</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==

Jaymin Joseph
 

John Hawkins

==Comments==

*John Hawkins: Nice list of transforms! Where did you find it? Inside the back cover of the textbook has a good list, but none including transforms of <math>\ g(t)</math>. I see your reference to the textbook. What page?

*John Hawkins: I believe that

:<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>

should be

:<math> \frac {6+s-s^2} {s(s+3)(s-5)} = \frac {A} {s}+ \frac {B} {s+3}+ \frac {C} {s-5}</math>

*John Hawkins: Also, the solution <math> y(t)= -\frac {2} {5} - \frac {1} {4} e^{-3t} - \frac {7} {20} e^{5t} </math> does not match the initial condition of <math>\ y(0)=1</math>.

Laplace Transform

2010-01-22T07:39:37Z

John.hawkins: Comment

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(at)\right\}=\int_0^{\infty} e^{-st} g(at) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right), \mbox{ for} ~a > 0.</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==

Jaymin Joseph
 

John Hawkins

==Comments==

*John Hawkins: I believe that instead of

:<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>

you wanted to write

:<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>

Electromechanical Energy Conversion

2010-01-21T19:49:55Z

John.hawkins:

[[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]]
* [[Example: Ampere's Law]] (Tyler Anderson)
* [[Ampere's Law]]
* [[DC Motor]]
* [[Fringing]]
* [[Electrostatics]]
* [[Faraday's Law]]
* [[Eddy Current]]
* [[Example Problems of Magnetic Circuits]]
* [[Ohm's Law and Reluctance]]
* [[Magnetic Flux]]
* [[An Ideal Transformer Example]]
* [[Example: Ideal Transformer Exercise]] (John Hawkins)
* [[Reference Terms and Units]] (Amy Crosby)
* [[Ideal Transformer Example|Example: Ideal Transformer]]
* [[Problem Set 1]](Jodi Hodge)
* [[ANOTHER IDEAL TRANSFORMER!!!!!!!!!]]
* [[Example: Magnetic Field]] (Amy Crosby)
* [[Example: Metal Cart]] (Amy Crosby)

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

* [[Magnetic Circuit]] (John Hawkins)
* [[AC Motors]]
* [[Example Problem - Toroid]] ([[Kirk Betz]])
* [[Transformer_example_problem|Ideal Transformer Example]] (Tim Rasmussen)
* [[AC vs. DC]] (Wesley Brown)

Class Roster

2010-01-21T19:08:17Z

John.hawkins:

===Class of 2010===
#[[Eric Clay]]
#[[Jason Osborne]]
#[[Tim Van Arsdale]]
#[[Kirk Betz]]
#[[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]]

Example: Ideal Transformer Exercise

2010-01-21T19:02:29Z

John.hawkins:

==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>

 

<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>

 

<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>

 

<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>

 

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

 

<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>

 

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

==Reviewed By==

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.

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

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

==Read By==

Ideal Transformer Example

2010-01-21T19:00:27Z

John.hawkins:

An idea transformer has a 275-turn primary and 825-turn secondary. The primary is connected to a 200-V, 60-Hz source. The secondary supplies a load of 5 A at a lagging power factor of 0.5. Find the turns-ratio, the current in the primary, the power supplied to the load, and the flux in the core.

===Solution===
(A) <math>\ {turns-ratio}=\frac{N_{1}}{N_{2}}</math>
<math>\ =\frac{275}{825}</math>
<math>\ =0.333</math>

(B) Because <math>\ {I_{2}}=5 A</math>, the current in the primary is...

<math>\ {I_{1}}=\frac{I_{2}}{turns-ratio}</math>
<math>\ =\frac{5}{0.333}</math>
<math>\ =15 A</math>

(C) <math>\ {V_{2}}=\frac{V_{1}}{turns-ratio}</math>
<math>\ =\frac{200}{0.333}</math>
<math>\ =600 V</math>

Therefore, the power supplied to the load is...

<math>\ {P_{L}}=V_{2} I_{2}\cos(\theta)</math>
<math>\ =600 * 5 * 0.5</math>
<math>\ =1500 W</math>

(D) <math>\ {\phi_{m}}=\frac{E_{1}}{4.44 f N_{1}}</math>
<math>\ =\frac{V_{1}}{4.44 f N_{1}}</math>
<math>\ =\frac{200}{4.44 * 60 * 275}</math>
<math>\ =2.73 mWb</math>

==Author==
[[Kyle Lafferty]]

==Reviewers==
Aric Vyhmeister

Erik Biesenthal

==Readers==
Aric Vyhmeister

Erik Biesenthal
John Hawkins

==Comments==

The way to do capital phi, <math>\ \Phi</math>, for the flux is to capitalize the first letter of the word, i.e. \Phi instead of \phi. And if you want a space between the number and the units, space is \(space), as in "\ ".

Ideal Transformer Example

2010-01-21T18:57:55Z

John.hawkins:

Magnetic Circuit

2010-01-21T18:51:42Z

John.hawkins:

=Author: John Hawkins=
==Problem Statement==

[[Image:Img001.jpg|thumb|500px|right]]
Problem 2.16 from ''Electric Machinery and Transformers'', 3rd ed:
<blockquote>A magnetic circuit is given in Figure P2.16. What must be the current in the 1600-turn coil to set up a flux density of 0.1 T in the air-gap? All dimensions are in centimeters. Assume that magnetic flux density varies as <math>B=[1.5H/(750+H)]</math>.<ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 129.</ref></blockquote>
==Solution==

First, we note that the problem statement is incomplete. Assume that the core has a relative permeability of 500. Hence, for all magnetic sections excluding the air gap,

<center><math>\mu=\mu_r\mu_0=(500)(4\pi\times10^{-7})=6.2832\times10^{-4}</math></center>

Also, as recommended in the text, we will neglect fringing.

The lengths and areas of each of the sections to be evaluated are given in the following table.

{| class="wikitable" border="1"
|+ ''Table 1: Lengths and Areas for the pertinent secions of the magnetic circuit.''
! '''Section''' !! fg !! def !! ghc !! dc !! dabc
|-
| Length <math>l</math> (m)|| 0.01 || 0.555|| 0.555 || 0.48 || 1.38
|-
| Area <math>A</math> (m2) || 0.0032 || 0.0032 || 0.0032 || 0.0096 || 8.0e-4
|}

We must now work backward from the air-gap, since the value of the flux-density is given there. We need only employ the analagous equations to Ohm's Law, KVL, and KCL. All units are standard units.
 
'''Air Gap:'''

 

<center>
<math>\mathcal{R}_{fg}=\frac{l_{fg}}{\mu A_{fg}} = 4973.6</math>
 
<math>\Phi_{fg}=B_{fg}A_{fg}=3.20\times10^{-4}</math>
 
<math>\mathcal{F}_{fg}=\mathcal{R}_{fg}\Phi_{fg}=1.5915</math>
</center>

 
 
'''Right Arms:'''
 

<center>
<math>\Phi_{def}=\Phi_{ghc} = \Phi_{fg} = 3.20\times10^{-4}</math>
 
<math>\mathcal{R}_{def}=\mathcal{R}_{ghc}=\frac{l_{def}}{\mu A_{def}}=2.7603\times10^5</math>
 
<math>\mathcal{F}_{def}=\mathcal{F}_{ghc}=\mathcal{R}_{def}\Phi_{def}=88.331</math>
</center>

 
 
'''Center Column:'''
 
<center>

<math>\mathcal{F}_{dc}=\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=178.25</math>
 
<math>\mathcal{R}_{dc}=\frac{l_{dc}}{\mu A_{dc}}=79,577</math>
 
<math>\Phi_{dc}=\frac{\mathcal{F}_{dc}}{\mathcal{R}_{dc}}=0.0022</math>
</center>

 
 
'''Left Arm:'''
 
<center>

<math>\ \Phi_{dabc}=\Phi_{dc}-\Phi_{def}=0.0019</math>
 
<math>\mathcal{R}_{dabc}=\frac{l_{dabc}}{\mu A_{dabc}}=2.745\times 10^6</math>
 
<math>\mathcal{F}_{dabc}=\mathcal{F}_{dabc}\Phi_{dabc}=5,271.2</math>
</center>

 
 
'''Conclusions:'''
 
<center>

<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+

\mathcal{F}_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+

\mathcal{F}_{ghc}=5,627.7</math>
 
<math>\mathbf{i=\frac{\mathcal{F}_{Total}}{N}=3.52 A}</math>
</center>

 

Which is the quantity we were looking for.

 

Calculations were performed using the following [[Magnetic Circuit Matlab Script]].

==References==

<references />

==Reviewed By==
Amy Crosby
==Read By==

==Comments==

Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency

2010-01-20T10:30:22Z

John.hawkins:

22 lines (currently)

1 reference

1 figure

123 points

==Period, Frequency, and Angular Frequency==
[[Image:Sinewave.png|450px|thumb|right|Picture of a Sine Wave where f(x)=sin(x)<ref>http://en.wikipedia.org/wiki/File:Sine.svg</ref>]]

===Period===

Long long ago, in a high school class called trigonometry, we leaned about '''periodic functions'''. A periodic function is a function that repeats itself over and over for infinity. The '''period''' of the function is the distance of one iteration that is infinitely repeating.

<center>A signal <math>f(t)</math> is periodic if, for some <math>T > 0</math> and all ''t'',</center>

<center><math>f(t+T) = f(t)</math><ref>DeCarlo/Lin, Linear Circuit Analysis--Time Domain, Phasor, and Laplace Transform Approaches, Second Edition. Figure 22.1</ref></center>

Where T is the period

The picture to the right shows the plot of the standard sine function whose period is <math>2\pi</math>. What the plot does not show is that the line keeps extending and repeating the bumps and valleys over the whole x axis, or <math>(-\infty,\infty)</math>. But wait! Can't the period also be <math>4\pi</math> or <math>6\pi</math>? In fact it can. Because the graph of sin(x) repeats itself every <math>2\pi</math> units, the period of the function is actually <math>2\pi n</math> where n is any whole number from zero to <math>\infty</math>

===Frequency and Angular Frequency===

The '''Frequency''' is the number of periods per second and is defined mathematically as

<math>f = \frac{1}{T}</math>

The standard unit of measurement for frequency is Hz (Hertz). 1 Hz = 1 cycle/second

The '''Angular Frequency''' is defined as

<math>\omega = 2\pi f = \frac{2\pi}{T}</math>

The standard unit of measurement for angular frequency is in radians/second.

===Fundamental Period, Frequency, and Angular Frequency===

The '''fundamental period''' is the smallest positive real number <math>T_0</math> for which the periodic equation <math>f(t+T) = f(t)</math> holds true.

The '''fundamental frequency''' is defined as <math>f_0 = \frac{1}{T_0}</math>.

The '''fundamental angular frequency''' is defined as <math>\omega_0 = 2\pi f_0 = \frac{2\pi}{T_0}</math>.

== References ==

<references />

==Author==
Andrew Roth

==Reviewers==
Brandon Vazquez

Ben Blackley

==Readers==
Thomas Wooley

Jaymin Joseph

John Hawkins

Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency

2010-01-20T10:30:02Z

John.hawkins:

Exercise: Sawtooth Redone With Exponential Basis Functions

2010-01-20T04:48:48Z

John.hawkins: Finished with primary content.

==Author==

John Hawkins

==Problem Statement==

Find the Fourier Tranform with exponential basis functions of the sawtooth wave given by the equation

 

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

 

Note that this is the same function solved in [[Exercise: Sawtooth Wave Fourier Transform]], but solved differently to compare the two methods.
==Solution==
The goal of this method is to find the coefficients <math>a_n</math> such that

 

<center><math>x(t)=\sum_{n=-\infty}^\infty a_n e^{j2\pi nt/T}</math></center>

 

In class we showed not only that this was possible, but also that

 

<center><math>a_n=\frac{1}{T}\int_c^{c+T}x(t) e^{-j2\pi nt/T}dt</math></center>

 

Noting that our period for this function is <math>T=1</math> and that an obvious choice for <math>c</math> is zero, we proceed:

 
<center>

<math>a_n=\frac{1}{1}\int_0^1te^{-j2\pi nt/1}dt</math>

</center>
 

For <math>n\neq 0</math>, the above integral is solved easiest using integration by parts. When <math>\ n=0</math>,however, IBP does not work, so the case when <math>\ n=0</math> needs to be considered separately. In this case,

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

For <math>n\neq 0</math>, we continue with IBP, letting
 

<center>

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

 

<math>dv=e^{-j2\pi nt}dt \quad\Rightarrow\quad v=\frac{1}{-j2\pi n}e^{-j2\pi nt}</math>

</center>

 
This gives
 

<center>

<math>a_n=t\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}\Bigg|_0^1-\int_0^1\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}dt</math>
 

<math>=\left[\frac{1}{-j2\pi n}e^{-j2\pi n}-0\right]-\left(\frac{1}{-j2\pi n}\right)^2e^{-j2\pi nt}\Bigg|_0^1</math>

 

<math>=\frac{1}{-j2\pi n}e^{-j2\pi n}-\left(\frac{1}{-j2\pi n}\right)^2\left(e^{-j2\pi n}-1\right)</math>

</center>
 

But

 
<center>

<math>\ e^{-j2\pi n}=\cos(-2\pi n)+j\sin(-2\pi n) = 1+j0=1</math>

</center>
 

So

 
<center>

<math>a_n=\frac{1}{-j2\pi n}(1)-\left(\frac{1}{-j2\pi n}\right)^2(1-1)=\frac{1}{-j2\pi n}</math>

</center>
 

Therefore,

 
<center><math>x(t)=\frac{1}{2}-\sum_{n=\pm 1}^{\pm \infty}\frac{1}{j2\pi n}e^{j2\pi nt}</math></center>
 
==Solution Graphs==
I modified the Matlab code used in the [[Exercise: Sawtooth Wave Fourier Transform]] to generate the solution graphs using the equation found above instead of the previously found solution. This code can be found here: [[Sawtooth2 Matlab Code]]. It generates the following analagous three graphs, which as hoped appear exactly identical to those found using the other method. Note that the "terms" mentioned in the titles of the graphs should now be interpreted as the sum of the <math>n</math>th and <math>-n</math>th terms.
 

[[Image:Sawtooth2_First_100_Terms.jpg|thumb|800px|center]]

 
[[Image:Sawtooth2_First_n_Terms.jpg|thumb|800px|center]]

 
[[Image:Sawtooth2_Error.jpg|thumb|800px|center]]

 

==Analytical Comparison of Two Solutions==
To convince myself that the two solutions are actually the same, I performed the following analysis. Let <math>t_n</math> be the nth term of the solution found on this page. Then for <math>n\neq 0</math>,

 
<center>

<math>t_n+t_{-n}=\frac{1}{-j2\pi n}e^{j2\pi nt}+\frac{1}{-j2\pi (-n)}e^{j2\pi (-n)t}</math>

 

<math>=\frac{1}{-j2\pi n} (\cos 2\pi nt+j\sin 2\pi nt)-\frac{1}{-j2\pi n}\left(\cos(-2\pi nt)+j\sin(-2\pi nt)\right)</math>

 

<math>=\frac{1}{-j2\pi n}\left(\cos 2\pi nt+j\sin 2\pi nt-\cos 2\pi nt+j\sin 2\pi nt \right)</math>

 

<math>=\frac{1}{-j2\pi n}\left(2j\sin 2\pi nt\right)</math>

 

<math>=-\frac{1}{\pi n}\sin 2\pi nt</math>

</center>

 

Therefore,

 
<center><math>x(t)=\frac{1}{2}-\sum_{n=\pm 1}^{\pm \infty}\frac{1}{j2\pi n}e^{j2\pi nt}=\frac{1}{2}-\sum_{n=1}^\infty\frac{1}{\pi n}\sin 2\pi nt</math></center>

 

As we had hoped and expected.

 

==Reviewed By==
==Read By==
==Comments==

Exercise: Sawtooth Redone With Exponential Basis Functions

2010-01-20T01:01:03Z

John.hawkins:

==Author==

John Hawkins

==Problem Statement==

Find the Fourier Tranform with exponential basis functions of the sawtooth wave given by the equation

 

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

 

Note that this is the same function solved in [[Exercise: Sawtooth Wave Fourier Transform]], but solved differently to compare the two methods.
==Solution==
The goal of this method is to find the coefficients <math>a_n</math> such that

 

<center><math>x(t)=\sum_{n=-\infty}^\infty a_n e^{j2\pi nt/T}</math></center>

 

In class we showed not only that this was possible, but also that

 

<center><math>a_n=\frac{1}{T}\int_c^{c+T}x(t) e^{-j2\pi nt/T}dt</math></center>

 

Noting again that our period for this function is <math>T=1</math> and that an obvious choice for <math>c</math> is zero, we proceed:

 
<center>

<math>a_n=\frac{1}{1}\int_0^1te^{-j2\pi nt/1}dt</math>

</center>
 

Again, the case when <math>\ n=0</math> needs to be considered separately. In this case,

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

For <math>n\neq 0</math>, the above integral is solved easiest using integration by parts. So letting
 

<center>

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

 

<math>dv=e^{-j2\pi nt}dt \quad\Rightarrow\quad v=\frac{1}{-j2\pi n}e^{-j2\pi nt}</math>

</center>

 
we have
 

<center>

<math>a_n=t\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}\Bigg|_0^1-\int_0^1\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}dt</math>
 

<math>=\left[\frac{1}{-j2\pi n}e^{-j2\pi n}-0\right]-\left(\frac{1}{-j2\pi n}\right)^2e^{-j2\pi nt}\Bigg|_0^1</math>

 

<math>=\frac{1}{-j2\pi n}e^{-j2\pi n}-\left(\frac{1}{-j2\pi n}\right)^2\left(e^{-j2\pi n}-1\right)</math>

</center>
 

But

 
<center>

<math>\ e^{-j2\pi n}=\cos(-2\pi n)+j\sin(-2\pi n) = 1+j0=1</math>

</center>
 

So

 
<center>

<math>a_n=\frac{1}{-j2\pi n}(1)-\left(\frac{1}{-j2\pi n}\right)^2(1-1)=\frac{1}{-j2\pi n}</math>

</center>
 

Therefore,

 
<center><math>x(t)=\frac{1}{2}-\sum_{n=\pm 1}^{\pm \infty}\frac{1}{j2\pi n}e^{j2\pi nt}</math></center>
 
==Solution Graphs==
I modified the Matlab code used in the [[Exercise: Sawtooth Wave Fourier Transform]] to generate the solution graphs using the equation found above instead of the previously found solution. This code can be found here: [[Sawtooth2 Matlab Code]]. It generates the following analagous three graphs, which as hoped appear exactly identical to those found using the other method. Note that the "terms" mentioned in the titles of the graphs should now be interpreted as the sum of the nth and -nth terms.
 

[[Image:Sawtooth2_First_100_Terms.jpg|thumb|800px|center]]

 
[[Image:Sawtooth2_First_n_Terms.jpg|thumb|800px|center]]

 
[[Image:Sawtooth2_Error.jpg|thumb|800px|center]]

 

==Analytical Comparison of Two Solutions==
To convince myself that the two solutions are actually the same, I performed the following analysis. Let <math>t_n</math> be the nth term of the solution found on this page. Then for <math>n\neq 0</math>,

 
<center>

<math>t_n+t_{-n}=\frac{1}{-j2\pi n}e^{j2\pi nt}+\frac{1}{-j2\pi (-n)}e^{j2\pi (-n)t}</math>

 

<math>=\frac{1}{-j2\pi n} (\cos 2\pi nt+j\sin 2\pi nt)-\frac{1}{-j2\i n}\left(\cos(-2\pi nt)+j\sin(-2\pi nt)\right)</math>

 

<math>=\frac{1}{-j2\pi n}\left(\cos 2\pi nt+j\sin 2\pi nt-\cos 2\pi nt+j\sin 2\pi nt \right)</math>

 

<math>=\frac{1}{-j2\pi n}\left(2j\sin 2\pi nt\right)</math>

 

<math>=-\frac{1}{\pi n}\sin 2\pi nt</math>

</center>

Sawtooth2 Matlab Code

2010-01-20T00:49:12Z

John.hawkins:

Sawtooth2 Matlab Code

2010-01-20T00:48:31Z

John.hawkins: New page: <source lang="matlab"> % % _-_,, ,, _-_- ,, % ( // ...

Exercise: Sawtooth Redone With Exponential Basis Functions

2010-01-20T00:45:42Z

John.hawkins: Intermediate progress

File:Sawtooth2 Error.jpg

2010-01-20T00:41:18Z

John.hawkins:

File:Sawtooth2 First n Terms.jpg

2010-01-20T00:40:59Z

John.hawkins:

File:Sawtooth2 First 100 Terms.jpg

2010-01-20T00:39:12Z

John.hawkins:

Exercise: Sawtooth Redone With Exponential Basis Functions

2010-01-20T00:17:54Z

John.hawkins: First bit of work

Exercise: Sawtooth Redone With Exponential Basis Functions

2010-01-19T23:57:19Z

John.hawkins: New page: ==Problem Statement== Find the Fourier Tranform with exponential basis functions of the sawtooth wave given by the equation <center><math>x(t)=t-\lfloor t \rfloor</math></cente...

==Problem Statement==

Find the Fourier Tranform with exponential basis functions of the sawtooth wave given by the equation

 

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

Winter 2010

2010-01-19T23:54:33Z

John.hawkins:

Reference Terms and Units

2010-01-19T06:33:45Z

John.hawkins:

Magnetic Circuit

2010-01-18T18:55:13Z

John.hawkins: Error correction

Magnetic Circuit Matlab Script

2010-01-18T18:48:16Z

John.hawkins: Error correction

% John Hawkins

% EMEC

% Magnetic Circuit Calculations

% Professor Rob Frohne

% January 10, 2009
 
% Problem Statement Constraints

N=1600;

Bfg=.1;

 % Assumed relative permeability

mu0=4e-7*pi;

mur=500;

mu=mur*mu0

 % Geometry

lfg=1e-2

ldef=31e-2+24.5e-2

lghc=ldef

ldc=48e-2

ldabc=45e-2+48e-2+45e-2

Afg=4e-2*8e-2

Adef=Afg

Aghc=Afg

Adc=8e-2*12e-2

Adabc=2e-2*4e-2

 % Air gap

Rfg=lfg/(mu*Afg)

Ifg=Bfg*Afg

Ffg=Rfg*Ifg

 % Right arms

Idef=Ifg

Rdef=ldef/(mu*Adef)

Fdef=Rdef*Idef

Ighc=Idef;

Rghc=Rdef;

Fghc=Fdef;

 % Center Column

Fdc=Fdef+Ffg+Fghc

Rdc=ldc/(mu*Adc)

Idc=Fdc/Rdc

 % Left arm

Idabc=Idc-Idef

Rdabc=ldabc/(mu*Adabc)

Fdabc=Rdabc*Idabc

 % Results

Ftotal = Fdabc+Fdc+Fdef+Ffg+Fghc

i=Ftotal/N

Reference Terms and Units

2010-01-18T18:41:48Z

John.hawkins:

This article is for the conference on Monday January 11

{| border="1" cellpadding="5" cellspacing="0"

|-

|Variable

|Description

|Unit variables
|Units

|-

|<math>\vec E</math>

|Electric Field Intensity

|<math> V/m\,\!</math> or <math> N/C\,\!</math>
|Volts per Meter or Newtons per Coulomb
|-

|<math>\vec D</math>

|Electric Flux Density

|<math> C/m^2\,\!</math>
|Coulombs per Meter squared
|-

|<math>\vec H</math>

|Magnetic Flux Intensity

|<math> A/m \,\!</math>
|Amps per Meter
|-

|<math>\vec B</math>

|Magnetic Flux Density or Magnetic Induction

|<math> T \,\!</math> or <math>W/m^2\,\!</math> or <math> G \,\!</math>
|Tessla or Watts per Meter squared or Gauss'
|}

Relationships

<math>\vec D=\in \vec E</math>

*In free space:

*<math>\in_0=1/(36*\pi)*10^{-9} \,\!</math> F/m

<math>\vec B=\mu \vec H</math>

*In free space:

*<math>\mu_0=4*\pi*10^{-7} \,\!</math> H/m

==Reviewed By==

==Read By==
John Hawkins

Comments and questions:

*I love the article on units. I hate these units where I don't know what applies to what or what comes from what. Good idea. I'm also interested in the derivation of them. Would you be interested for me to contribute with the derivation of some of these?

*I think Tesla are equivalent to <math>Wb/m^2\ </math>, Webers per meter squared.

*(It doesn't really matter, but the character you're using for epsilon is actually the set inclusion operator. \epsilon or \varepsilon is technically correct.)

An Ideal Transformer Example

2010-01-18T13:30:04Z

John.hawkins:

Consider a simple, transformer with two windings. Find the current provided by the voltage source.
* Winding 1 has a sinusoidal voltage of <math>120\sqrt{2}\angle{0}</math>° applied to it at a frequency of 60Hz.
* <math>\frac{N_{1}}{N_{2}}=3</math>
* The combined load on winding 2 is <math>\ {Z_{L}}=(5+j3)\Omega</math>
===Solution===
<math>\ {e_{1}}(t)={V_{1}}\cos(\omega t)</math>

<math>\ \omega=2\pi f</math>, so <math>\ \omega=120\pi</math>

Therefore, <math>\ {e_{1}}(t)={V_{1}}\cos(120\pi t)</math>

Now the Thevenin equivalent impedance, <math>\ {Z_{th}}</math>, is found through the following steps:

<math>{Z_{th}} = \frac{e_{1}}{i_{1}}</math>

<math>=\frac{\frac{N_{1}}{N_{2}}{e_{2}}}{\frac{N_{2}}{N_{1}}{i_{2}}}</math>

<math>=(\frac{N_{1}}{N_{2}})^2{R_{L}}</math>

Now, substituting:

<math>\ {Z_{th}} = 3^2(5+j3)</math>

<math>\ =(45+j27)\Omega</math>

Since <math>{i_{1}}=\frac{e_{1}}{R_{th}}</math>,

<math>{i_{1}}=\frac{120\sqrt{2}}{45+j27}</math>
===Contributors===

[[Lau, Chris|Christopher Garrison Lau I]]

===Read By===

John Hawkins

An Ideal Transformer Example

2010-01-18T13:29:43Z

John.hawkins:

Example: Ideal Transformer Exercise

2010-01-18T13:29:03Z

John.hawkins:

==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>

 

<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>

 

<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>

 

<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>

 

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

 

<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>

 

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

==Reviewed By==

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.

* 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.

==Read By==

Electromechanical Energy Conversion

2010-01-18T03:03:00Z

John.hawkins:

[[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]]
* [[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)

==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)

Magnetic Circuit

2010-01-18T02:59:58Z

John.hawkins:

=Author: John Hawkins=
==Problem Statement==

[[Image:Img001.jpg|thumb|500px|right]]
Problem 2.16 from ''Electric Machinery and Transformers'', 3rd ed:
<blockquote>A magnetic circuit is given in Figure P2.16. What must be the current in the 1600-turn coil to set up a flux density of 0.1 T in the air-gap? All dimensions are in centimeters. Assume that magnetic flux density varies as <math>B=[1.5H/(750+H)]</math>.<ref>Guru and Huseyin, ''Electric Machinery and Transformers'', 3rd ed. (New York: Oxford University Press, 2001), 129.</ref></blockquote>
==Solution==

First, we note that the problem statement is incomplete. Assume that the core has a relative permeability of 500. Hence, for all magnetic sections excluding the air gap,

<center><math>\mu=\mu_r\mu_0=(500)(4\pi\times10^{-7})=6.2832\times10^{-4}</math></center>

Also, as recommended in the text, we will neglect fringing.

The lengths and areas of each of the sections to be evaluated are given in the following table.

{| class="wikitable" border="1"
|+ ''Table 1: Lengths and Areas for the pertinent secions of the magnetic circuit.''
! '''Section''' !! fg !! def !! ghc !! dc !! dabc
|-
| Length <math>l</math> (m)|| 0.01 || 0.555|| 0.555 || 0.48 || 1.4
|-
| Area <math>A</math> (m2) || 0.0032 || 0.0032 || 0.0032 || 0.0096 || 8.0e-4
|}

We must now work backward from the air-gap, since the value of the flux-density is given there. We need only employ the analagous equations to Ohm's Law, KVL, and KCL. All units are standard units.
 
'''Air Gap:'''

 

<center>
<math>\mathcal{R}_{fg}=\frac{l_{fg}}{\mu A_fg} = 4973.6</math>
 
<math>\Phi_{fg}=B_{fg}A_{fg}=3.20\times10^{-4}</math>
 
<math>\mathcal{F}_{fg}=\mathcal{R}_{fg}\Phi_{fg}=1.5915</math>
</center>

 
 
'''Right Arms:'''
 

<center>
<math>\Phi_{def}=\Phi_{ghc} = \Phi_{fg} = 3.20\times10^{-4}</math>
 
<math>\mathcal{R}_{def}=\mathcal{R}_{ghc}=\frac{l_{def}}{\mu A_{def}}=2.7603\times10^5</math>
 
<math>\mathcal{F}_{def}=\mathcal{F}_{ghc}=\mathcal{R}_{def}\Phi_{def}=88.331</math>
</center>

 
 
'''Center Column:'''
 
<center>

<math>\mathcal{F}_{dc}=\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=178.25</math>
 
<math>\mathcal{R}_{dc}=\frac{l_{dc}}{\mu A_{dc}}=79,577</math>
 
<math>\Phi_{dc}=\frac{\mathcal{F}_{dc}}{\mathcal{R}_{dc}}=0.0022</math>
</center>

 
 
'''Left Arm:'''
 
<center>

<math>\Phi_{dabc}=\Phi_{dc}-\Phi_{def}=0.0019</math>
 
<math>\mathcal{R}_{dabc}=\frac{l_{dabc}}{\mu A_{dabc}}=2.785\times 10^6</math>
 
<math>\mathcal{F}_{dabc}=\mathcal{F}_{dabc}\Phi_{dabc}=5347.6</math>
</center>

 
 
'''Conclusions:'''
 
<center>

<math>\mathcal{F}_{Total}=\mathcal{F}_{dabc}+\mathcal{F}_{dc}+\mathcal{F}_{def}+\mathcal{F}_{fg}+\mathcal{F}_{ghc}=57,041</math>
 
<math>\mathbf{i=\frac{\mathcal{F}_{Total}}{N}=3.57 A}</math>
</center>

 

Which is the quantity we were looking for.

 

Calculations were performed using the following [[Magnetic Circuit Matlab Script]].

==References==

<references />

==Reviewed By==
==Read By==

Electromechanical Energy Conversion

2010-01-18T02:36:04Z

John.hawkins:

[[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]]
* [[Example problems of magnetic circuits]]
* [[Magnetic Circuit]]
* [[Ohm's Law and Reluctance]]
* [[Magnetic Flux]]
* [[An Ideal Transformer Example]]
* [[Example: Ideal Transformer Exercise]] (John Hawkins)

==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)

Example: Ideal Transformer Exercise

2010-01-18T02:33:03Z

John.hawkins: Main Content

Electromechanical Energy Conversion

2010-01-18T01:51:02Z

John.hawkins:

[[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.

==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]]
* [[Example problems of magnetic circuits]]
* [[Magnetic Circuit]]
* [[Ohm's Law and Reluctance]]
* [[Magnetic Flux]]
* [[An Ideal Transformer Example]]

* [[Example: Ideal Transformer Exercise]]

==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]]

Electromechanical Energy Conversion

2010-01-18T01:47:07Z

John.hawkins:

[[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. Even "\ " (the function call for a space) works for me. 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.

==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]]
* [[Example problems of magnetic circuits]]
* [[Magnetic Circuit]]
* [[Ohm's Law and Reluctance]]
* [[Magnetic Flux]]
* [[An Ideal Transformer Example]]

* [[Example: Ideal Transformer Exercise]]

==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]]

Electromechanical Energy Conversion

2010-01-18T01:42:06Z

John.hawkins:

[[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. Even "\ " (the function call for a space) works for me. 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.

==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]]
* [[Example problems of magnetic circuits]]
* [[Magnetic Circuit]]
* [[Ohm's Law and Reluctance]]
* [[Magnetic Flux]]
* [[An Ideal Transformer Example]]

==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]]