https://fweb.wallawalla.edu/class-wiki/api.php?action=feedcontributions&feedformat=atom&user=WilspaClass Wiki - User contributions [en]2026-04-06T12:50:04ZUser contributionsMediaWiki 1.43.0https://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=10226State Space Form2011-03-08T21:49:46Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
Where the bold faced x is a matrix such that:<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}</math><br><br />
This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.<br />
<br />
-Example-<br />
<br />
For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br><br />
[[Image:Circuit1.jpg]]<br />
<br />
Start by using loop (also known as mesh or KVL) analysis.<br />
<br />
Loop 1:<br><br />
<math>-Vin +i_{1}*R + L*\frac{di_{2}}{dt}=0</math><br><br />
Loop 2:<br><br />
<math>-Vin +i_{1}*R + i_{3}*R + L*\frac{di_{3}}{dt} = 0</math><br><br />
<br />
Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br><br />
So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br><br />
<math>-Vin +(x_{1}+x_{2})R + L\dot x_{1}=0</math> and<br><br />
<math>-Vin +(x_{1}+x_{2})R + x_{2}R + L\dot x_{2} = 0</math><br><br />
Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.<br />
<br />
Solving these equations for <math>\dot x_{1}</math> and <math>\dot x_{2}</math> We find:<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
and<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-2\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
Now we can write these equations in stat space matrix form:<br><br />
<math>\mathbf{\dot x}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}<br />
\mathbf{x}+\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}<br />
Vin</math><br><br />
Where<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x_{1} \\<br />
x_{2} \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{\dot x}=<br />
\begin{bmatrix}<br />
\dot x_{1} \\<br />
\dot x_{2} \\<br />
\end{bmatrix}</math><br><br />
So then, since <math>Vin</math> is our forcing function <math>f</math>:<br><br />
<math>\mathbf{A}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}</math><br />
,<math>\mathbf{B}=\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}</math><br><br />
We now have our differential equations in state space form.<br><br />
Now, these equations aren't much use by themselves, we need to use them to find the output of the system y(t). y(t) in state space form is expressed as follows:<br><br />
<math>y(t)=\mathbf{C}\mathbf{x}+\mathbf{D}f(t)</math><br><br />
Lets suppose that y(t) is the voltage across the inductor that has i3 passing through it. That means that:<br><br />
<math>y(t)=L\frac{di_{3}}{dt}</math><br><br />
From this we can see that:<br><br />
<math>y(t)=<br />
\begin{bmatrix}<br />
0 \ L \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}+<br />
\begin{bmatrix}<br />
0 \\<br />
\end{bmatrix}f(t)<br />
</math><br><br />
So we can see that:<br><br />
<math>\mathbf{C}=<br />
\begin{bmatrix}<br />
0 \ L \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{D}=[0]</math>.<br><br />
<br />
<br />
<br />
I'll be adding more on how to use state space form to solve these equations as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Discrete_Fourier_Transforms&diff=10225Discrete Fourier Transforms2011-03-08T21:12:31Z<p>Wilspa: </p>
<hr />
<div>[[User:wilspa|Back to my page]]<br />
<br />
==Paul's DFT Page==<br />
One of the major tools used in signal processing is the DFT, which stands for Discrete Fourier Transform. The reason we need to to a DFT instead of a Fourier Transform is that our computers are limited in their abilites. They use sampling, and they have limited memory, so we have to adapt to the computers.<br />
===What is a DFT?===<br />
A DFT is like doing a Fourier Transform, but instead of doing it with an integral, we do it with discrete values and a sum. A Fourier Transform looks like this:<br><br />
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt<br />
</math><br><br />
Which uses an integral, while the DFT which looks like this:<br><br />
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}<br />
</math><br><br />
Which is using a sum and a discrete series of delta functions x(n) instead of the continuous function x(t).<br />
<br />
===What is it used for?===<br />
The DFT is (as the name suggests) the discrete version of the continuous Fourier Transform. This is important because computers can only truly work in the discrete time domain. You can use it to find out frequency and phase information about a discrete signal.<br />
<br />
===Things to know about a DFT===<br />
A Discrete Fourier transform has some limitations. The first is that it can only test for discrete frequencies. Each frequency in the output of the DFT is a multiple of the fundamental frequency. The fundamental frequency is based on the number of samples used in the transform and the sampling rate. If the sampling rate of a signal is 100 Hz (100 samples per second) and the number of samples stored is 40 samples, that means that the period of the fundamental frequency is<br><br />
<math>T_0=\frac{40 samples}{100 samples/second}=0.4 seconds</math>.<br><br />
Therefore the fundamental frequency will be:<br><br />
<math>f_0=\frac{1}{T_0}=2.5 Hz</math>.<br><br />
This is important because if your signal contains a frequency that is not an exact multiple of the fundamental frequency, it will not show up as a single spike like it would in the continuous Fourier Transform.<br><br />
<br />
Another important thing to know is the range of frequencies that the DFT shows. A DFT shows frequencies in the range <math>[0,2\pi)</math>. However, only 1/2 the frequencies are positive frequencies. In the range <math>[0,\pi]</math> the frequencies are positive. The negative frequencies are in the range <math>(\pi,2\pi)</math>. Why are the negative frequencies above the positive frequencies? That is a good question, and the answer has to do with the unit circle. I need to do more explanation here.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Discrete_Fourier_Transforms&diff=10224Discrete Fourier Transforms2011-03-08T21:07:33Z<p>Wilspa: </p>
<hr />
<div>[[User:wilspa|Back to my page]]<br />
<br />
==Paul's DFT Page==<br />
One of the major tools used in signal processing is the DFT, which stands for Discrete Fourier Transform. The reason we need to to a DFT instead of a Fourier Transform is that our computers are limited in their abilites. They use sampling, and they have limited memory, so we have to adapt to the computers.<br />
===What is a DFT?===<br />
A DFT is like doing a Fourier Transform, but instead of doing it with an integral, we do it with discrete values and a sum. A Fourier Transform looks like this:<br><br />
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt<br />
</math><br><br />
Which uses an integral, while the DFT which looks like this:<br><br />
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}<br />
</math><br><br />
Which is using a sum and a discrete series of delta functions x(n) instead of the continuous function x(t).<br />
<br />
===What is it used for?===<br />
The DFT is (as the name suggests) the discrete version of the continuous Fourier Transform. This is important because computers can only truly work in the discrete time domain. You can use it to find out frequency and phase information about a discrete signal.<br />
<br />
===Things to know about a DFT===<br />
A Discrete Fourier transform has some limitations. The first is that it can only test for discrete frequencies. Each frequency in the output of the DFT is a multiple of the fundamental frequency. The fundamental frequency is based on the number of samples used in the transform and the sampling rate. If the sampling rate of a signal is 100 Hz (100 samples per second) and the number of samples stored is 40 samples, that means that the period of the fundamental frequency is<br><br />
<math>T_0=\frac{40 samples}{100 samples/second}=0.4 seconds</math>.<br><br />
Therefore the fundamental frequency will be:<br><br />
<math>f_0=1/T_0=2.5 Hz</math>.<br><br />
This is important because if your signal contains a frequency that is not an exact multiple of the fundamental frequency, it will not show up as a single spike like it would in the continuous Fourier Transform.<br><br />
<br />
Another important thing to know is the range of frequencies that the DFT shows. A DFT shows frequencies in the range <math>[0,2\pi)</math>. However, only 1/2 the frequencies are positive frequencies. In the range <math>[0,\pi]</math> the frequencies are positive. The negative frequencies are in the range <math>(\pi,2\pi)</math>. Why are the negative frequencies above the positive frequencies? That is a good question, and the answer has to do with the unit circle. I need to do more explanation here.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=10217State Space Form2011-03-07T23:20:35Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
Where the bold faced x is a matrix such that:<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}</math><br><br />
This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.<br />
<br />
-Example-<br />
<br />
For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br><br />
[[Image:Circuit1.jpg]]<br />
<br />
Start by using loop (also known as mesh or KVL) analysis.<br />
<br />
Loop 1:<br><br />
<math>-Vin +i_{1}*R + L*\frac{di_{2}}{dt}=0</math><br><br />
Loop 2:<br><br />
<math>-Vin +i_{1}*R + i_{3}*R + L*\frac{di_{3}}{dt} = 0</math><br><br />
<br />
Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br><br />
So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br><br />
<math>-Vin +(x_{1}+x_{2})R + L\dot x_{1}=0</math> and<br><br />
<math>-Vin +(x_{1}+x_{2})R + x_{2}R + L\dot x_{2} = 0</math><br><br />
Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.<br />
<br />
Solving these equations for <math>\dot x_{1}</math> and <math>\dot x_{2}</math> We find:<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
and<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-2\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
Now we can write these equations in stat space matrix form:<br><br />
<math>\mathbf{\dot x}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}<br />
\mathbf{x}+\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}<br />
Vin</math><br><br />
Where<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x_{1} \\<br />
x_{2} \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{\dot x}=<br />
\begin{bmatrix}<br />
\dot x_{1} \\<br />
\dot x_{2} \\<br />
\end{bmatrix}</math><br><br />
So then, since <math>Vin</math> is our forcing function <math>f</math>:<br><br />
<math>\mathbf{A}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}</math><br />
,<math>\mathbf{B}=\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}</math><br><br />
We now have our differential equations in state space form.<br><br />
Now, these equations aren't much use by themselves, we need to use them to find the output of the system y(t). y(t) in state space form is expressed as follows:<br><br />
<math>y(t)=\mathbf{C}\mathbf{x}+\mathbf{D}f(t)</math><br><br />
Lets suppose that y(t) is the voltage across the inductor that has i3 passing through it. That means that:<br><br />
<math>y(t)=L\frac{di_{3}}{dt}</math><br><br />
From this we can see that:<br><br />
<math>y(t)=<br />
\begin{bmatrix}<br />
0 \ 1 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}+<br />
\begin{bmatrix}<br />
0 \\<br />
\end{bmatrix}f(t)<br />
</math><br><br />
So we can see that:<br><br />
<math>\mathbf{C}=<br />
\begin{bmatrix}<br />
0 \ 1 \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{D}=[0]</math>.<br><br />
<br />
<br />
<br />
I'll be adding more on how to use state space form to solve these equations as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=10216State Space Form2011-03-07T23:19:50Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
Where the bold faced x is a matrix such that:<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}</math><br><br />
This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.<br />
<br />
-Example-<br />
<br />
For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br><br />
[[Image:Circuit1.jpg]]<br />
<br />
Start by using loop (also known as mesh or KVL) analysis.<br />
<br />
Loop 1:<br><br />
<math>-Vin +i_{1}*R + L*\frac{di_{2}}{dt}=0</math><br><br />
Loop 2:<br><br />
<math>-Vin +i_{1}*R + i_{3}*R + L*\frac{di_{3}}{dt} = 0</math><br><br />
<br />
Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br><br />
So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br><br />
<math>-Vin +(x_{1}+x_{2})R + L\dot x_{1}=0</math> and<br><br />
<math>-Vin +(x_{1}+x_{2})R + x_{2}R + L\dot x_{2} = 0</math><br><br />
Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.<br />
<br />
Solving these equations for <math>\dot x_{1}</math> and <math>\dot x_{2}</math> We find:<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
and<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-2\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
Now we can write these equations in stat space matrix form:<br><br />
<math>\mathbf{\dot x}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}<br />
\mathbf{x}+\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}<br />
Vin</math><br><br />
Where<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x_{1} \\<br />
x_{2} \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{\dot x}=<br />
\begin{bmatrix}<br />
\dot x_{1} \\<br />
\dot x_{2} \\<br />
\end{bmatrix}</math><br><br />
So then, since <math>Vin</math> is our forcing function <math>f</math>:<br><br />
<math>\mathbf{A}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}</math><br />
,<math>\mathbf{B}=\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}</math><br><br />
We now have our differential equations in state space form.<br><br />
Now, these equations aren't much use by themselves, we need to use them to find the output of the system y(t). y(t) in state space form is expressed as follows:<br><br />
<math>y(t)=\mathbf{C}\mathbf{x}+\mathbf{D}f(t)</math><br><br />
Lets suppose that y(t) is the voltage across the inductor that has i3 passing through it. That means that:<br><br />
<math>y(t)=L\frac{di_{3}}{dt}</math><br><br />
From this we can see that:<br><br />
<math>y(t)=<br />
\begin{bmatrix}<br />
0 \ 1 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}+<br />
\begin{bmatrix}<br />
0 \\<br />
\end{bmatrix}f(t)<br />
</math><br><br />
So we can see that:<br><br />
<math>C=<br />
\begin{bmatrix}<br />
0 \ 1 \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>D=[0]</math>.<br><br />
<br />
<br />
<br />
I'll be adding more on how to use state space form to solve these equations as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=10215State Space Form2011-03-07T23:19:16Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
Where the bold faced x is a matrix such that:<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}</math><br><br />
This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.<br />
<br />
-Example-<br />
<br />
For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br><br />
[[Image:Circuit1.jpg]]<br />
<br />
Start by using loop (also known as mesh or KVL) analysis.<br />
<br />
Loop 1:<br><br />
<math>-Vin +i_{1}*R + L*\frac{di_{2}}{dt}=0</math><br><br />
Loop 2:<br><br />
<math>-Vin +i_{1}*R + i_{3}*R + L*\frac{di_{3}}{dt} = 0</math><br><br />
<br />
Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br><br />
So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br><br />
<math>-Vin +(x_{1}+x_{2})R + L\dot x_{1}=0</math> and<br><br />
<math>-Vin +(x_{1}+x_{2})R + x_{2}R + L\dot x_{2} = 0</math><br><br />
Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.<br />
<br />
Solving these equations for <math>\dot x_{1}</math> and <math>\dot x_{2}</math> We find:<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
and<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-2\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
Now we can write these equations in stat space matrix form:<br><br />
<math>\mathbf{\dot x}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}<br />
\mathbf{x}+\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}<br />
Vin</math><br><br />
Where<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x_{1} \\<br />
x_{2} \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{\dot x}=<br />
\begin{bmatrix}<br />
\dot x_{1} \\<br />
\dot x_{2} \\<br />
\end{bmatrix}</math><br><br />
So then, since <math>Vin</math> is our forcing function <math>f</math>:<br><br />
<math>\mathbf{A}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}</math><br />
,<math>\mathbf{B}=\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}</math><br><br />
We now have our differential equations in state space form.<br><br />
Now, these equations aren't much use by themselves, we need to use them to find the output of the system y(t). y(t) in state space form is expressed as follows:<br><br />
<math>y(t)=\mathbf{C}\mathbf{x}+\mathbf{D}f(t)</math><br><br />
Lets suppose that y(t) is the voltage across the inductor that has i3 passing through it. That means that:<br><br />
<math>y(t)=L\frac{di_{3}}{dt}</math><br><br />
From this we can see that:<br><br />
<math>y(t)=<br />
\begin{bmatrix}<br />
0 \ 1 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}+<br />
\begin{bmatrix}<br />
0 \\<br />
\end{bmatrix}f(t)<br />
</math><br><br />
So we can see that:<br><br />
</math>C=<br />
\begin{bmatrix}<br />
0 \ 1 \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>D=[0]</math>.<br><br />
<br />
<br />
<br />
I'll be adding more on how to use state space form to solve these equations as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=10214State Space Form2011-03-07T23:15:44Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
Where the bold faced x is a matrix such that:<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}</math><br><br />
This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.<br />
<br />
-Example-<br />
<br />
For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br><br />
[[Image:Circuit1.jpg]]<br />
<br />
Start by using loop (also known as mesh or KVL) analysis.<br />
<br />
Loop 1:<br><br />
<math>-Vin +i_{1}*R + L*\frac{di_{2}}{dt}=0</math><br><br />
Loop 2:<br><br />
<math>-Vin +i_{1}*R + i_{3}*R + L*\frac{di_{3}}{dt} = 0</math><br><br />
<br />
Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br><br />
So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br><br />
<math>-Vin +(x_{1}+x_{2})R + L\dot x_{1}=0</math> and<br><br />
<math>-Vin +(x_{1}+x_{2})R + x_{2}R + L\dot x_{2} = 0</math><br><br />
Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.<br />
<br />
Solving these equations for <math>\dot x_{1}</math> and <math>\dot x_{2}</math> We find:<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
and<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-2\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
Now we can write these equations in stat space matrix form:<br><br />
<math>\mathbf{\dot x}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}<br />
\mathbf{x}+\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}<br />
Vin</math><br><br />
Where<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x_{1} \\<br />
x_{2} \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{\dot x}=<br />
\begin{bmatrix}<br />
\dot x_{1} \\<br />
\dot x_{2} \\<br />
\end{bmatrix}</math><br><br />
So then, since <math>Vin</math> is our forcing function <math>f</math>:<br><br />
<math>\mathbf{A}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}</math><br />
,<math>\mathbf{B}=\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}</math><br><br />
We now have our differential equations in state space form.<br><br />
Now, these equations aren't much use by themselves, we need to use them to find the output of the system y(t). y(t) in state space form is expressed as follows:<br><br />
<math>y(t)=\mathbf{C}\mathbf{x}+\mathbf{D}f(t)</math><br><br />
Lets suppose that y(t) is the voltage across the inductor that has i3 passing through it. That means that:<br><br />
<math>y(t)=L\frac{di_{3}}{dt}</math><br><br />
From this we can see that:<br><br />
<math>y(t)=<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}\mathbf{x}+[0]f(t)<br />
</math><br />
<br />
I'll be adding more on how to use state space form to solve these equations as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=10213State Space Form2011-03-07T22:50:30Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.<br />
<br />
-Example-<br />
<br />
For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br><br />
[[Image:Circuit1.jpg]]<br />
<br />
Start by using loop (also known as mesh or KVL) analysis.<br />
<br />
Loop 1:<br><br />
<math>-Vin +i_{1}*R + L*\frac{di_{2}}{dt}=0</math><br><br />
Loop 2:<br><br />
<math>-Vin +i_{1}*R + i_{3}*R + L*\frac{di_{3}}{dt} = 0</math><br><br />
<br />
Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br><br />
So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br><br />
<math>-Vin +(x_{1}+x_{2})R + L\dot x_{1}=0</math> and<br><br />
<math>-Vin +(x_{1}+x_{2})R + x_{2}R + L\dot x_{2} = 0</math><br><br />
Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.<br />
<br />
Solving these equations for <math>\dot x_{1}</math> and <math>\dot x_{2}</math> We find:<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
and<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-2\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
Now we can write these equations in stat space matrix form:<br><br />
<math>\mathbf{\dot x}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}<br />
\mathbf{x}+\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}<br />
Vin</math><br><br />
Where<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x_{1} \\<br />
x_{2} \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{\dot x}=<br />
\begin{bmatrix}<br />
\dot x_{1} \\<br />
\dot x_{2} \\<br />
\end{bmatrix}</math><br><br />
So then, since <math>Vin</math> is our forcing function <math>f</math>:<br><br />
<math>\mathbf{A}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}</math><br />
,<math>\mathbf{B}=\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}</math><br><br />
We now have our differential equations in state space form.<br><br />
I'll be adding more on how to use state space form to solve these equations as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=10212State Space Form2011-03-07T22:10:19Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.<br />
<br />
-Example-<br />
<br />
For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br><br />
[[Image:Circuit1.jpg]]<br />
<br />
Start by using loop (also known as mesh or KVL) analysis.<br />
<br />
Loop 1:<br><br />
<math>-Vin +i_{1}*R + L*\frac{di_{2}}{dt}=0</math><br><br />
Loop 2:<br><br />
<math>-Vin +i_{1}*R + i_{3}*R + L*\frac{di_{3}}{dt} = 0</math><br><br />
<br />
Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br><br />
So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br><br />
<math>-Vin +(x_{1}+x_{2})R + L\dot x_{1}=0</math> and<br><br />
<math>-Vin +(x_{1}+x_{2})R + x_{2}R + L\dot x_{2} = 0</math><br><br />
Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.<br />
<br />
Solving these equations for <math>\dot x_{1}</math> and <math>\dot x_{2}</math> We find:<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
and<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-2\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
Now we can write these equations in stat space matrix form:<br><br />
<math>\mathbf{\dot x}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}<br />
\mathbf{x}+\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}<br />
Vin</math><br><br />
Where<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x_{1} \\<br />
x_{2} \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{\dot x}=<br />
\begin{bmatrix}<br />
\dot x_{1} \\<br />
\dot x_{2} \\<br />
\end{bmatrix}</math><br><br />
So then, since <math>Vin</math> is our forcing function <math>f</math>:<br><br />
<math>\mathbf{A}=<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}</math><br />
,<math>\mathbf{B}=<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}</math><br><br />
We now have our differential equations in state space form.<br><br />
I'll be adding more on how to use state space form to solve these equations as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=9613State Space Form2010-09-10T19:18:21Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.<br />
<br />
-Example-<br />
<br />
For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br><br />
[[Image:Circuit1.jpg]]<br />
<br />
Start by using loop (also known as mesh or KVL) analysis.<br />
<br />
Loop 1:<br><br />
<math>-Vin +i_{1}*R + L*\frac{di_{2}}{dt}=0</math><br><br />
Loop 2:<br><br />
<math>-Vin +i_{1}*R + i_{3}*R + L*\frac{di_{3}}{dt} = 0</math><br><br />
<br />
Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br><br />
So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br><br />
<math>-Vin +(x_{1}+x_{2})R + L\dot x_{1}=0</math> and<br><br />
<math>-Vin +(x_{1}+x_{2})R + x_{2}R + L\dot x_{2} = 0</math><br><br />
Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.<br />
<br />
Solving these equations for <math>\dot x_{1}</math> and <math>\dot x_{2}</math> We find:<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
and<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-2\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
Now we can write these equations in stat space matrix form:<br><br />
<math>\mathbf{\dot x}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}<br />
\mathbf{x}+\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}<br />
Vin</math><br><br />
Where<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x_{1} \\<br />
x_{2} \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{\dot x}=<br />
\begin{bmatrix}<br />
\dot x_{1} \\<br />
\dot x_{2} \\<br />
\end{bmatrix}</math><br><br />
So then, since <math>Vin</math> is our forcing function <math>f</math>:<br><br />
<math>\mathbf{A}=<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}</math><br />
,<math>\mathbf{B}=<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}</math><br><br />
We now have our differential equations in state space form.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=9612State Space Form2010-09-10T19:18:04Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.<br />
<br />
-Example-<br />
<br />
For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br />
[[Image:Circuit1.jpg]]<br />
<br />
Start by using loop (also known as mesh or KVL) analysis.<br />
<br />
Loop 1:<br><br />
<math>-Vin +i_{1}*R + L*\frac{di_{2}}{dt}=0</math><br><br />
Loop 2:<br><br />
<math>-Vin +i_{1}*R + i_{3}*R + L*\frac{di_{3}}{dt} = 0</math><br><br />
<br />
Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br><br />
So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br><br />
<math>-Vin +(x_{1}+x_{2})R + L\dot x_{1}=0</math> and<br><br />
<math>-Vin +(x_{1}+x_{2})R + x_{2}R + L\dot x_{2} = 0</math><br><br />
Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.<br />
<br />
Solving these equations for <math>\dot x_{1}</math> and <math>\dot x_{2}</math> We find:<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
and<br><br />
<math>\dot x_{1}=-\frac{R}{L}x_{1}-2\frac{R}{L}x_{2}+\frac{1}{L}Vin</math><br><br />
Now we can write these equations in stat space matrix form:<br><br />
<math>\mathbf{\dot x}=-\frac{R}{L}<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}<br />
\mathbf{x}+\frac{1}{L}<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}<br />
Vin</math><br><br />
Where<br><br />
<math>\mathbf{x}=<br />
\begin{bmatrix}<br />
x_{1} \\<br />
x_{2} \\<br />
\end{bmatrix}</math><br><br />
and<br><br />
<math>\mathbf{\dot x}=<br />
\begin{bmatrix}<br />
\dot x_{1} \\<br />
\dot x_{2} \\<br />
\end{bmatrix}</math><br><br />
So then, since <math>Vin</math> is our forcing function <math>f</math>:<br><br />
<math>\mathbf{A}=<br />
\begin{bmatrix}<br />
1 & 1 \\<br />
1& 2 \\<br />
\end{bmatrix}</math><br />
,<math>\mathbf{B}=<br />
\begin{bmatrix}<br />
1 \\<br />
1 \\<br />
\end{bmatrix}</math><br><br />
We now have our differential equations in state space form.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=9611State Space Form2010-09-10T19:03:29Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
This is called the state space representation of the differential equation. This can be quite useful because the entire description of the differential equation is available in the matrix, and is easily manipulated using linear algebra.<br />
<br />
-Example-<br />
<br />
For the circuit below, find a set of state variable equations (there are several ways to do this, I will choose the one I feel is most intuitive.)<br />
[[Image:Circuit1.jpg]]<br />
<br />
Start by using loop (also known as mesh or KVL) analysis.<br />
<br />
Loop 1:<br><br />
<math>-Vin +i_{1}*R + L*\frac{di_{2}}{dt}=0</math><br><br />
Loop 2:<br><br />
<math>-Vin +i_{1}*R + i_{3}*R + L*\frac{di_{3}}{dt} = 0</math><br><br />
<br />
Lets let:<br> <math>x_{1}=i_{2}</math> and<br> <math> x_{2}=i_{3}</math><br><br />
So, <math>i_{1}=x_{1}+x_{2}</math>. Substituting into the equations above, we get:<br><br />
<math>-Vin +(x_{1}+x_{2})*R + L*\dot x_{1}=0</math> and<br><br />
<math>-Vin +(x_{1}+x_{2})*R + x_{2}*R + L*\dot x_{2} = 0</math><br><br />
Where <math>\dot x_{1}</math> is the derivative of <math>x</math>.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:Circuit1.jpg&diff=9610File:Circuit1.jpg2010-09-10T18:46:05Z<p>Wilspa: </p>
<hr />
<div></div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=9602User:Wilspa2010-09-02T05:08:14Z<p>Wilspa: </p>
<hr />
<div><center><br />
==Paul Wilson==<br />
Working on my masters in EE<br><br />
</center><br />
<br><br />
===My Wiki Contributions===<br />
[[Laplace Transform]]<br />
<br><br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
<br><br />
[[Discrete Fourier Transforms]]<br />
<br><br />
[[Relationship between e, sin and cos]]<br />
<br><br />
[[State Space Form]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=9601State Space Form2010-09-01T15:51:26Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
This is called the state space representation of the differential equation.<br />
<br />
Example coming</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=9600State Space Form2010-08-30T19:08:10Z<p>Wilspa: </p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.<br />
Consider the differential equation <br />
<math>\frac{d^2 y}{d t^2}+2\frac{xy}{dt}+3y=f(t)</math> <br><br />
or <br><br />
<math>\ddot y +2\dot y + 3y =f(t) </math> <br><br />
The state of this equation can be described using what is called state space form. State space form gives the blah blah more here.<br />
<br />
let <math>x1 = y</math> <br><br />
let <math>x2=\dot y </math> <br><br />
so <math>\dot x2 = \ddot y </math> <br><br />
<br />
<br />
We can now re-write the equation above to be:<br><br />
<math>\dot x2 + 2x2 + 3 x1 = f(t)</math><br><br />
so <br><math>\dot x2 = -3x1 -2x2 + f(t)</math><br><br />
and from the definition above <br><math>\dot x1 = x2</math><br><br />
<br />
<br />
<br />
We can take this and put it into matrix form:<br />
<math><br />
\begin{bmatrix}<br />
\dot x1 \\<br />
\dot x2 \\<br />
\end{bmatrix}<br />
=<br />
\begin{bmatrix}<br />
0 & 1 \\<br />
-3 & -2 \\<br />
\end{bmatrix}<br />
\begin{bmatrix}<br />
x1 \\<br />
x2 \\<br />
\end{bmatrix}<br />
+<br />
\begin{bmatrix}<br />
0 \\<br />
1 \\<br />
\end{bmatrix}<br />
f(t)<br />
</math><br><br />
Or, more generally,<br><br />
<math>\mathbf{\dot x} = \mathbf{Ax+B}f</math><br><br />
This is called the state space representation of the differential equation.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=9599State Space Form2010-08-30T18:41:47Z<p>Wilspa: </p>
<hr />
<div>== State Space Form ==<br />
<br />
<br />
In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=9598User:Wilspa2010-08-30T18:39:01Z<p>Wilspa: /* My Wiki Contributions */</p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
===My Wiki Contributions===<br />
[[Laplace Transform]]<br />
<br><br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
<br><br />
[[Discrete Fourier Transforms]]<br />
<br><br />
[[Relationship between e, sin and cos]]<br />
<br><br />
[[State Space Form]]<br />
<br />
===My Senior Project===<br />
For my senior project I'm working on making a pink noise generator. I'm planning to put the details here on this wiki as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=State_Space_Form&diff=9597State Space Form2010-08-30T18:38:41Z<p>Wilspa: Created page with 'In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more her…'</p>
<hr />
<div>In my Signals and Systems II class at University of Idaho, we are learning about the state space form of representing a solution to a LTI differential equation. I'll add more here soon.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=9596User:Wilspa2010-08-30T18:37:49Z<p>Wilspa: /* My Wiki Contributions */</p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
===My Wiki Contributions===<br />
[[Laplace Transform]]<br />
<br><br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
<br><br />
[[Discrete Fourier Transforms]]<br />
<br><br />
[[Relationship between e, sin and cos]]<br />
<br />
<br><br />
[[State Space Form]]<br />
<br />
===My Senior Project===<br />
For my senior project I'm working on making a pink noise generator. I'm planning to put the details here on this wiki as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=9595User:Wilspa2010-08-30T18:36:18Z<p>Wilspa: /* My Wiki Contributions */</p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
===My Wiki Contributions===<br />
[[Laplace Transform]]<br />
<br><br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
<br><br />
[[Discrete Fourier Transforms]]<br />
<br><br />
[[Relationship between e, sin and cos]]<br />
<br />
===My Senior Project===<br />
For my senior project I'm working on making a pink noise generator. I'm planning to put the details here on this wiki as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=FIR_Filter_Example&diff=4386FIR Filter Example2008-10-01T16:36:21Z<p>Wilspa: </p>
<hr />
<div>*[[Signals and systems|Signals and Systems]]<br />
The filter coefficients are given by:<br />
<math><br />
h_m = { T } \int_{-{1\over 4T}}^{{1\over 4T}} H(f)e^{j2\pi m f T}\,df<br />
</math><br />
where H(f) is the desired frequency response. See the notes from October 31, 2005. As an example of this filter we will make a Matlab script that will computer the frequency response for a low pass filter having a cutoff frequency of 1/(4T), and using 2M+1 coefficients. Note that it is periodic with period 1/T. This is the case with all digital filters.<br />
<br />
Here is the plot the Matlab code produces:<br />
<br />
[[Image:Frequency_Response_Pict.png]]<br />
<br />
The Matlab code to see the frequency response is given below:<br />
<br />
% This shows how to find the frequency response for an FIR filter with cutoff 1/4/T and 2M+1 coefficients.<br />
<br />
clf;<br />
<br />
sum=0;<br />
<br />
T=1;<br />
<br />
fs=1/1000/T;<br />
<br />
f=-2/T:fs:2/T<br />
;<br />
M=100;<br />
<br />
for m=-M:M;<br />
<br />
: if m==0<br />
<br />
:: h=1/2;<br />
<br />
: else<br />
<br />
:: h=sin(pi*m/2)/(pi*m);<br />
<br />
: end<br />
<br />
: sum=sum+h*exp(-i*2*pi*f*m*T);<br />
<br />
end<br />
<br />
plot(f,20*log10(abs(sum)))<br />
<br />
title('Frequency Response of Our FIR Filter')<br />
<br />
xlabel('Frequency (1/T)')<br />
<br />
ylabel('Response (db)')<br />
<br />
text(-1.5,-5,'M = 100')<br />
<br />
% End of Matlab code.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=FIR_Filter_Example&diff=4385FIR Filter Example2008-10-01T16:29:21Z<p>Wilspa: </p>
<hr />
<div>*[[Signals and systems|Signals and Systems]]<br />
The filter coefficients are given by:<br />
<math><br />
h_m = { T } \int_{-{1\over 4T}}^{{1\over 4T}} H(f)e^{j2\pi m f T}\,df<br />
</math><br />
where H(f) is the desired frequency response. See the notes from October 31, 2005. As an example of this filter we will make a Matlab script that will computer the frequency response for a low pass filter having a cutoff frequency of 1/(4T), and using 2M+1 coefficients. Note that it is periodic with period 1/T. This is the case with all digital filters.<br />
<br />
Here is the plot the Matlab code produces:<br />
<br />
[[Image:Frequency_Response_Pict.png]]<br />
<br />
The Matlab code to see the frequency response is given below:<br />
<br />
% This shows how to find the frequency response for an FIR filter with cutoff 1/4/T and 2M+1 coefficients.<br />
<br />
clf;<br />
<br />
sum=0;<br />
<br />
T=1;<br />
<br />
fs=1/1000/T;<br />
<br />
f=-2/T:fs:2/T<br />
;<br />
M=100;<br />
<br />
for m=-M:M;<br />
<br />
if m==0<br />
<br />
h=1/2;<br />
<br />
else<br />
<br />
h=sin(pi*m/2)/(pi*m);<br />
<br />
end<br />
<br />
sum=sum+h*exp(-i*2*pi*f*m*T);<br />
<br />
end<br />
<br />
plot(f,20*log10(abs(sum)))<br />
<br />
title('Frequency Response of Our FIR Filter')<br />
<br />
xlabel('Frequency (1/T)')<br />
<br />
ylabel('Response (db)')<br />
<br />
text(-1.5,-5,'M = 100')<br />
<br />
% End of Matlab code.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=4384User:Wilspa2008-10-01T04:58:03Z<p>Wilspa: /* My Wiki Contributions */</p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
===My Wiki Contributions===<br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
<br><br />
[[Discrete Fourier Transforms]]<br />
<br><br />
[[Relationship between e, sin and cos]]<br />
<br />
===My Senior Project===<br />
For my senior project I'm working on making a pink noise generator. I'm planning to put the details here on this wiki as I have time.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Signals_and_Systems&diff=4383Signals and Systems2008-10-01T04:56:25Z<p>Wilspa: /* Course Pages */</p>
<hr />
<div>== Topics ==<br />
[[Fourier series - by Ray Betz|Overview of Signals and Systems]]<br />
<br />
===Individual Subjects===<br />
*[[Linear Time Invarient System|Linear Time Invarient Systems]]<br />
**[[The Game|"The Game"]]<br />
*[[Orthogonal functions|Orthogonal Functions]]<br />
*[[Energy in a signal|Finding the Energy in a Signal]]<br />
**[[Rayleigh's Theorem]]<br />
*[[Fourier series|Fourier Series]]<br />
*[[Fourier transform|Fourier Transforms]]<br />
**[[Discrete Fourier transform]]<br />
*[[Sampling]]<br />
*[[FIR Filter Example]]<br />
*[[Relationship between e, sin and cos]]<br />
<br />
===Course Pages===<br />
[[2005-2006 Assignments]]<br />
<br />
[[2006-2007 Assignments]]<br />
<br />
[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/engr455index.htm Class notes for Signals & Systems]<br />
<br />
<br />
<br />
[[User:Frohro|Instructor: Rob Frohne]]<br />
<br />
==2004-2005 contributors==<br />
<br />
[[User:Barnsa|Sam Barnes]]<br />
<br />
[[User:Santsh|Shawn Santana]]<br />
<br />
[[User:Goeari|Aric Goe]]<br />
<br />
[[User:Caswto|Todd Caswell]]<br />
<br />
[[User:Andeda|David Anderson]]<br />
<br />
[[User:Guenan|Anthony Guenterberg]]<br />
<br />
==2005-2006 contributors==<br />
<br />
[[User:GabrielaV|Gabriela Valdivia]]<br />
<br />
[[User:SDiver|Raymond Betz]]<br />
<br />
[[User:chrijen|Jenni Christensen]]<br />
<br />
[[User:wonoje|Jeffrey Wonoprabowo]]<br />
<br />
[[User:wilspa|Paul Wilson]]<br />
<br />
==2006-2007 contributors==<br />
<br />
[[User:Smitry|Ryan J Smith]]<br />
<br />
[[User:Nathan|Nathan Ferch]]<br />
<br />
[[User:Andrew|Andrew Lopez]]<br />
<br />
[[User:Sherna|Nathan Sherman]]<br />
<br />
[[User:Adkich|Chris Adkins]]<br />
<br />
==2007-2008 contributors==<br />
<br />
[[User:baldwin.britton|Baldwin Britton]]<br />
<br />
[[User:Fonggr|Greg Fong]]<br />
<br />
[[User:Harrde|Denver Harris]]<br />
<br />
[[User:Pridma|Mark Priddy]]<br />
<br />
[[User:ChrisRas|Chris Rasmussen]]<br />
<br />
[[User:RothMi|Michael Roth]]<br />
<br />
[[User:Rothsa|Sally Roth]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Signals_and_Systems&diff=4382Signals and Systems2008-02-13T11:03:15Z<p>Wilspa: </p>
<hr />
<div>== Topics ==<br />
[[Fourier series - by Ray Betz|Overview of Signals and Systems]]<br />
<br />
===Individual Subjects===<br />
*[[Linear Time Invarient System|Linear Time Invarient Systems]]<br />
**[[The Game|"The Game"]]<br />
*[[Orthogonal functions|Orthogonal Functions]]<br />
*[[Energy in a signal|Finding the Energy in a Signal]]<br />
**[[Rayleigh's Theorem]]<br />
*[[Fourier series|Fourier Series]]<br />
*[[Fourier transform|Fourier Transforms]]<br />
**[[Discrete Fourier transform]]<br />
*[[Sampling]]<br />
*[[FIR Filter Example]]<br />
*[[Relationship between e, sin and cos]]<br />
<br />
===Course Pages===<br />
[[2005-2006 Assignments]]<br />
<br />
[[2006-2007 Assignments]]<br />
<br />
[http://www.wwc.edu/~frohro/ClassNotes/engr455index.htm Class notes for Signals & Systems]<br />
<br />
<br />
<br />
[[User:Frohro|Instructor: Rob Frohne]]<br />
<br />
==2004-2005 contributors==<br />
<br />
[[User:Barnsa|Sam Barnes]]<br />
<br />
[[User:Santsh|Shawn Santana]]<br />
<br />
[[User:Goeari|Aric Goe]]<br />
<br />
[[User:Caswto|Todd Caswell]]<br />
<br />
[[User:Andeda|David Anderson]]<br />
<br />
[[User:Guenan|Anthony Guenterberg]]<br />
<br />
==2005-2006 contributors==<br />
<br />
[[User:GabrielaV|Gabriela Valdivia]]<br />
<br />
[[User:SDiver|Raymond Betz]]<br />
<br />
[[User:chrijen|Jenni Christensen]]<br />
<br />
[[User:wonoje|Jeffrey Wonoprabowo]]<br />
<br />
[[User:wilspa|Paul Wilson]]<br />
<br />
==2006-2007 contributors==<br />
<br />
[[User:Smitry|Ryan J Smith]]<br />
<br />
[[User:Nathan|Nathan Ferch]]<br />
<br />
[[User:Andrew|Andrew Lopez]]<br />
<br />
[[User:Sherna|Nathan Sherman]]<br />
<br />
[[User:Adkich|Chris Adkins]]<br />
<br />
==2007-2008 contributors==<br />
<br />
[[User:baldwin.britton|Baldwin Britton]]<br />
<br />
[[User:Fonggr|Greg Fong]]<br />
<br />
[[User:Harrde|Denver Harris]]<br />
<br />
[[User:Pridma|Mark Priddy]]<br />
<br />
[[User:ChrisRas|Chris Rasmussen]]<br />
<br />
[[User:RothMi|Michael Roth]]<br />
<br />
[[User:Rothsa|Sally Roth]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Relationship_between_e,_sin_and_cos&diff=4381Relationship between e, sin and cos2008-02-13T11:02:07Z<p>Wilspa: /* Converting from e to sin/cos */</p>
<hr />
<div>==Converting from e to sin/cos==<br />
It is often useful when doing signal processing to understand the relationship between e, sin and cos. Sometimes difficult calculations involving even or odd functions of <math>e</math> can be greatly simplified by using the relationship to simplify things. The relationship is as follows:<br />
<br><br />
<br><br />
<math>e^{j \theta} = cos( \theta ) + j*sin( \theta ). </math><br />
<br />
==Converting from sin/cos to e==<br />
The reverse conversion is also often helpful:<br />
<br />
<math>cos( \theta ) = \frac{e^{j \theta}+e^{-j \theta}}{2}</math><br />
<br />
<math>sin( \theta ) = \frac{e^{j \theta}-e^{-j \theta}}{2j}</math><br />
<br />
We can test to see that this works as follows:<br />
<br />
{|<br />
|-<br />
|<math>{e^{j \theta }}</math><br />
|<math> = cos( \theta ) + j*sin( \theta )</math><br />
|-<br />
|<br />
|<math> = \frac{e^{j \theta}+e^{-j \theta}}{2} + j*\frac{e^{j \theta}-e^{-j \theta}}{2j}</math><br />
|-<br />
|<br />
|<math> = \frac{e^{j \theta}+e^{-j \theta}}{2} + \frac{e^{j \theta}-e^{-j \theta}}{2} </math><br />
|-<br />
|<br />
|<math> = \frac{(e^{j \theta}+e^{-j \theta}) + (e^{j \theta}-e^{-j \theta})}{2} </math><br />
|-<br />
|<br />
|<math> = \frac{2*e^{j \theta}}{2} </math><br />
|-<br />
|<math> e^{j \theta }</math><br />
|<math> = e^{j \theta }</math><br />
|-<br />
|}<br />
<br />
* Back to [[User:Wilspa|my page]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Relationship_between_e,_sin_and_cos&diff=4380Relationship between e, sin and cos2008-02-13T11:01:48Z<p>Wilspa: </p>
<hr />
<div>==Converting from e to sin/cos==<br />
It is often useful when doing signal processing to understand the relationship between e, sin and cos. Sometimes difficult calculations involving even or odd functions of <math>e</math> can be greatly simplified by using the relationship to simplify things. The relationship is as follows:<br />
<br><br />
<math>e^{j \theta} = cos( \theta ) + j*sin( \theta ). </math><br />
<br />
==Converting from sin/cos to e==<br />
The reverse conversion is also often helpful:<br />
<br />
<math>cos( \theta ) = \frac{e^{j \theta}+e^{-j \theta}}{2}</math><br />
<br />
<math>sin( \theta ) = \frac{e^{j \theta}-e^{-j \theta}}{2j}</math><br />
<br />
We can test to see that this works as follows:<br />
<br />
{|<br />
|-<br />
|<math>{e^{j \theta }}</math><br />
|<math> = cos( \theta ) + j*sin( \theta )</math><br />
|-<br />
|<br />
|<math> = \frac{e^{j \theta}+e^{-j \theta}}{2} + j*\frac{e^{j \theta}-e^{-j \theta}}{2j}</math><br />
|-<br />
|<br />
|<math> = \frac{e^{j \theta}+e^{-j \theta}}{2} + \frac{e^{j \theta}-e^{-j \theta}}{2} </math><br />
|-<br />
|<br />
|<math> = \frac{(e^{j \theta}+e^{-j \theta}) + (e^{j \theta}-e^{-j \theta})}{2} </math><br />
|-<br />
|<br />
|<math> = \frac{2*e^{j \theta}}{2} </math><br />
|-<br />
|<math> e^{j \theta }</math><br />
|<math> = e^{j \theta }</math><br />
|-<br />
|}<br />
<br />
* Back to [[User:Wilspa|my page]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Relationship_between_e,_sin_and_cos&diff=4379Relationship between e, sin and cos2008-02-13T11:00:09Z<p>Wilspa: </p>
<hr />
<div>==Converting from e to sin/cos==<br />
It is often useful when doing signal processing to understand the relationship between e, sin and cos. Sometimes difficult calculations involving even or odd functions of <math>e</math> can be greatly simplified by using the relationship to simplify things. The relationship is as follows:<br />
<br><br />
<math>e^{j \theta} = cos( \theta ) + j*sin( \theta ). </math><br />
<br />
==Converting from sin/cos to e==<br />
The reverse conversion is also often helpful:<br />
<br />
<math>cos( \theta ) = \frac{e^{j \theta}+e^{-j \theta}}{2}</math><br />
<br />
<math>sin( \theta ) = \frac{e^{j \theta}-e^{-j \theta}}{2j}</math><br />
<br />
We can test to see that this works as follows:<br />
<br />
{|<br />
|-<br />
|<math>{e^{j \theta }}</math><br />
|<math> = cos( \theta ) + j*sin( \theta )</math><br />
|-<br />
|<br />
|<math> = \frac{e^{j \theta}+e^{-j \theta}}{2} + j*\frac{e^{j \theta}-e^{-j \theta}}{2j}</math><br />
|-<br />
|<br />
|<math> = \frac{e^{j \theta}+e^{-j \theta}}{2} + \frac{e^{j \theta}-e^{-j \theta}}{2} </math><br />
|-<br />
|<br />
|<math> = \frac{(e^{j \theta}+e^{-j \theta}) + (e^{j \theta}-e^{-j \theta})}{2} </math><br />
|-<br />
|<br />
|<math> = \frac{2*e^{j \theta}}{2} </math><br />
|-<br />
|<math> e^{j \theta }</math><br />
|<math> = e^{j \theta }</math><br />
|-<br />
|}</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Relationship_between_e,_sin_and_cos&diff=4378Relationship between e, sin and cos2008-02-13T10:48:15Z<p>Wilspa: </p>
<hr />
<div>==Converting from e to sin/cos==<br />
It is often useful when doing signal processing to understand the relationship between e, sin and cos. Sometimes difficult calculations involving even or odd functions of <math>e</math> can be greatly simplified by using the relationship to simplify things. The relationship is as follows:<br />
<math>e^{j \theta} = cos( \theta ) + j*sin( \theta ). </math><br />
<br />
==Converting from sin/cos to e==<br />
The reverse conversion is also often helpful:<br />
<br />
<math>cos( \theta ) = \frac{e^{j \theta}+e^{-j \theta}}{2}</math><br />
<br />
<math>sin( \theta ) = \frac{e^{j \theta}-e^{-j \theta}}{2j}</math></div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Relationship_between_e,_sin_and_cos&diff=4377Relationship between e, sin and cos2008-02-13T10:44:45Z<p>Wilspa: New page: ==Converting from e to sin/cos== It is often useful when doing signal processing to understand the relationship between e, sin and cos. Sometimes difficult calculations involving even or o...</p>
<hr />
<div>==Converting from e to sin/cos==<br />
It is often useful when doing signal processing to understand the relationship between e, sin and cos. Sometimes difficult calculations involving even or odd functions of <math>e</math> can be greatly simplified by using the relationship to simplify things.<br />
<br />
==Converting from sin/cos to e==<br />
The reverse conversion is also often helpful:<br />
<br />
<math>cos( \theta ) = \frac{e^{j \theta}+e^{-j \theta}}{2}</math></div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=4376User:Wilspa2008-02-13T10:35:31Z<p>Wilspa: /* My Wiki Contributions */</p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
===My Wiki Contributions===<br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
<br><br />
[[Discrete Fourier Transforms]]<br />
<br><br />
[[Relationship between e, sin and cos]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=4375User:Wilspa2008-02-13T10:34:49Z<p>Wilspa: /* My Wiki Contributions */</p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
===My Wiki Contributions===<br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
<br><br />
[[Discrete Fourier Transforms]]<br />
<br><br />
[[Relationship between <math>e</math>, <math>sin</math> and <math>cos</math>.]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Fourier_transform&diff=4374Fourier transform2008-02-13T10:32:23Z<p>Wilspa: /* A Second Approach to Fourier Transforms */</p>
<hr />
<div>==Fourier Transform==<br />
What is a Fourier Transform? A Fourier Transform is a function that changes a signal or waveform from the time domain into the frequency domain. One simple way to look at it is this: Suppose you are at the beach, watching the waves. You could say that a wave hits the shore at specific times (0 second, 2 seconds, 4 seconds, etc.) that would be describing the waveform in the time domain. If, however, you were to say that the waves hit the beach every two seconds, that would be describing it in the frequency domain. So a Fourier transform would take the data given in the time domain and convert that into the frequency domain. The function that does this is: <math> X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt </math>.<br />
<br />
The reverse is also possible. You can take the information from the frequency domain, and convert it into the time domain using an Inverse Fourier Transform.<br />
<br />
==From the Fourier Transform to the Inverse Fourier Transform==<br />
Lets start with the basic Fourier Transform:<br />
<math><br />
<br />
X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt<br />
<br />
</math><br />
<br><br />
<br><br />
Suppose that we have some function, say <math> \beta (t) </math>, that is nonperiodic and finite in duration.<br><br />
This means that <math> \beta(t)=0 </math> for some <math> T_\alpha < \left | t \right | </math><br />
<br><br><br />
Now let's make a periodic function<br />
<math><br />
\gamma(t)<br />
</math><br />
by repeating<br />
<math><br />
\beta(t)<br />
</math><br />
with a fundamental period<br />
<math><br />
T_\zeta<br />
</math>.<br />
Note that <br />
<math><br />
\lim_{T_\zeta \to \infty}\gamma(t)=\beta(t)<br />
</math><br />
<br><br />
The Fourier Series representation of <math> \gamma(t) </math> is<br />
<br><br />
<math><br />
\gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt}<br />
</math><br />
where<br />
<math> <br />
f={1\over T_\zeta}<br />
</math><br />
<br>and<br />
<math><br />
\alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt<br />
</math><br />
<br><br />
<math> \alpha_k </math> can now be rewritten as<br />
<math><br />
\alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt<br />
</math><br />
<br>From our initial identity then, we can write <math> \alpha_k </math> as<br />
<math><br />
\alpha_k={1\over T_\zeta}\Beta(kf)<br />
</math><br />
<br> and <br />
<math><br />
\gamma(t)<br />
</math><br />
becomes<br />
<math><br />
\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}<br />
</math><br />
<br><br />
Now remember that<br />
<math><br />
\beta(t)=\lim_{T_\zeta \to \infty}\gamma(t)<br />
</math><br />
and<br />
<math><br />
{1\over {T_\zeta}} = f.<br />
</math><br />
<br><br />
Which means that<br />
<math><br />
\beta(t)=\lim_{f \to 0}\gamma(t)=\lim_{f \to 0}\sum_{k=-\infty}^\infty f \Beta(kf) e^{j2\pi fkt}<br />
</math><br />
<br><br />
Which is just to say that<br />
<math><br />
\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df<br />
</math><br />
<br><br />
<br><br />
So we have that<br />
<math><br />
\mathcal{F}[\beta(t)]=\Beta(f)=\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi ft}\, dt<br />
</math><br />
<br><br />
Further<br />
<math><br />
\mathcal{F}^{-1}[\Beta(f)]=\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df<br />
</math><br />
<br />
==Some Useful Fourier Transform Pairs==<br />
<math><br />
\mathcal{F}[\alpha(t)]=\frac{1}{\mid \alpha \mid}f(\frac{\omega}{\alpha})<br />
</math><br />
<br><br />
{|<br />
|-<br />
|<math>\mathcal{F}[c_1\alpha(t)+c_2\beta(t)]</math><br />
|<math>=\int_{-\infty}^{\infty} (c_1\alpha(t)+c_2\beta(t)) e^{-j2\pi ft}\, dt</math><br />
|-<br />
|<br />
|<math>=\int_{-\infty}^{\infty}c_1\alpha(t)e^{-j2\pi ft}\, dt+\int_{-\infty}^{\infty}c_2\beta(t)e^{-j2\pi ft}\, dt</math><br />
|-<br />
|<br />
|<math>=c_1\int_{-\infty}^{\infty}\alpha(t)e^{-j2\pi ft}\, dt+c_2\int_{-\infty}^{\infty}\beta(t)e^{-j2\pi ft}\, dt=c_1\Alpha(f)+c_2\Beta(f)</math><br />
|-<br />
|}<br />
<br><br />
<math><br />
\mathcal{F}[\alpha(t-\gamma)]=e^{-j2\pi f\gamma}\Alpha(f)<br />
</math><br />
<br><br />
<math><br />
\mathcal{F}[\alpha(t)*\beta(t)]=\Alpha(f)\Beta(f)<br />
</math><br />
<br><br />
<math><br />
\mathcal{F}[\alpha(t)\beta(t)]=\Alpha(f)*\Beta(f)<br />
</math><br />
<br><br />
Some other usefull pairs can be found here: [[Fourier Transforms]]<br />
<br />
==Another look at Fourier Transforms==<br />
*[[Fourier Transforms]]<br />
<br />
Return to [[Signals and systems|Signals and Systems]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Fourier_transform&diff=4373Fourier transform2008-02-13T10:30:43Z<p>Wilspa: /* Fourier Transform */</p>
<hr />
<div>==Fourier Transform==<br />
What is a Fourier Transform? A Fourier Transform is a function that changes a signal or waveform from the time domain into the frequency domain. One simple way to look at it is this: Suppose you are at the beach, watching the waves. You could say that a wave hits the shore at specific times (0 second, 2 seconds, 4 seconds, etc.) that would be describing the waveform in the time domain. If, however, you were to say that the waves hit the beach every two seconds, that would be describing it in the frequency domain. So a Fourier transform would take the data given in the time domain and convert that into the frequency domain. The function that does this is: <math> X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt </math>.<br />
<br />
The reverse is also possible. You can take the information from the frequency domain, and convert it into the time domain using an Inverse Fourier Transform.<br />
<br />
==From the Fourier Transform to the Inverse Fourier Transform==<br />
Lets start with the basic Fourier Transform:<br />
<math><br />
<br />
X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt<br />
<br />
</math><br />
<br><br />
<br><br />
Suppose that we have some function, say <math> \beta (t) </math>, that is nonperiodic and finite in duration.<br><br />
This means that <math> \beta(t)=0 </math> for some <math> T_\alpha < \left | t \right | </math><br />
<br><br><br />
Now let's make a periodic function<br />
<math><br />
\gamma(t)<br />
</math><br />
by repeating<br />
<math><br />
\beta(t)<br />
</math><br />
with a fundamental period<br />
<math><br />
T_\zeta<br />
</math>.<br />
Note that <br />
<math><br />
\lim_{T_\zeta \to \infty}\gamma(t)=\beta(t)<br />
</math><br />
<br><br />
The Fourier Series representation of <math> \gamma(t) </math> is<br />
<br><br />
<math><br />
\gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt}<br />
</math><br />
where<br />
<math> <br />
f={1\over T_\zeta}<br />
</math><br />
<br>and<br />
<math><br />
\alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt<br />
</math><br />
<br><br />
<math> \alpha_k </math> can now be rewritten as<br />
<math><br />
\alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt<br />
</math><br />
<br>From our initial identity then, we can write <math> \alpha_k </math> as<br />
<math><br />
\alpha_k={1\over T_\zeta}\Beta(kf)<br />
</math><br />
<br> and <br />
<math><br />
\gamma(t)<br />
</math><br />
becomes<br />
<math><br />
\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}<br />
</math><br />
<br><br />
Now remember that<br />
<math><br />
\beta(t)=\lim_{T_\zeta \to \infty}\gamma(t)<br />
</math><br />
and<br />
<math><br />
{1\over {T_\zeta}} = f.<br />
</math><br />
<br><br />
Which means that<br />
<math><br />
\beta(t)=\lim_{f \to 0}\gamma(t)=\lim_{f \to 0}\sum_{k=-\infty}^\infty f \Beta(kf) e^{j2\pi fkt}<br />
</math><br />
<br><br />
Which is just to say that<br />
<math><br />
\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df<br />
</math><br />
<br><br />
<br><br />
So we have that<br />
<math><br />
\mathcal{F}[\beta(t)]=\Beta(f)=\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi ft}\, dt<br />
</math><br />
<br><br />
Further<br />
<math><br />
\mathcal{F}^{-1}[\Beta(f)]=\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df<br />
</math><br />
<br />
==Some Useful Fourier Transform Pairs==<br />
<math><br />
\mathcal{F}[\alpha(t)]=\frac{1}{\mid \alpha \mid}f(\frac{\omega}{\alpha})<br />
</math><br />
<br><br />
{|<br />
|-<br />
|<math>\mathcal{F}[c_1\alpha(t)+c_2\beta(t)]</math><br />
|<math>=\int_{-\infty}^{\infty} (c_1\alpha(t)+c_2\beta(t)) e^{-j2\pi ft}\, dt</math><br />
|-<br />
|<br />
|<math>=\int_{-\infty}^{\infty}c_1\alpha(t)e^{-j2\pi ft}\, dt+\int_{-\infty}^{\infty}c_2\beta(t)e^{-j2\pi ft}\, dt</math><br />
|-<br />
|<br />
|<math>=c_1\int_{-\infty}^{\infty}\alpha(t)e^{-j2\pi ft}\, dt+c_2\int_{-\infty}^{\infty}\beta(t)e^{-j2\pi ft}\, dt=c_1\Alpha(f)+c_2\Beta(f)</math><br />
|-<br />
|}<br />
<br><br />
<math><br />
\mathcal{F}[\alpha(t-\gamma)]=e^{-j2\pi f\gamma}\Alpha(f)<br />
</math><br />
<br><br />
<math><br />
\mathcal{F}[\alpha(t)*\beta(t)]=\Alpha(f)\Beta(f)<br />
</math><br />
<br><br />
<math><br />
\mathcal{F}[\alpha(t)\beta(t)]=\Alpha(f)*\Beta(f)<br />
</math><br />
<br><br />
Some other usefull pairs can be found here: [[Fourier Transforms]]<br />
<br />
==A Second Approach to Fourier Transforms==<br />
*[[Fourier Transforms]]<br />
<br />
Return to *[[Signals and systems|Signals and Systems]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Fourier_transform&diff=4372Fourier transform2008-02-13T10:30:06Z<p>Wilspa: </p>
<hr />
<div>==Fourier Transform==<br />
What is a Fourier Transform? A Fourier Transform is a function that changes a signal or waveform from the time domain into the frequency domain. One simple way to look at it is this: Suppose you are at the beach, watching the waves. You could say that a wave hits the shore at specific times (0 second, 2 seconds, 4 seconds, etc.) that would be describing the waveform in the time domain. If, however, you were to say that the waves hit the beach every two seconds, that would be describing it in the frequency domain. So a Fourier transform would take the data given in the time domain and convert that into the frequency domain. The function that does this is: <math> X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt<br />
<br />
The reverse is also possible. You can take the information from the frequency domain, and convert it into the time domain using an Inverse Fourier Transform.<br />
<br />
==From the Fourier Transform to the Inverse Fourier Transform==<br />
Lets start with the basic Fourier Transform:<br />
<math><br />
<br />
X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt<br />
<br />
</math><br />
<br><br />
<br><br />
Suppose that we have some function, say <math> \beta (t) </math>, that is nonperiodic and finite in duration.<br><br />
This means that <math> \beta(t)=0 </math> for some <math> T_\alpha < \left | t \right | </math><br />
<br><br><br />
Now let's make a periodic function<br />
<math><br />
\gamma(t)<br />
</math><br />
by repeating<br />
<math><br />
\beta(t)<br />
</math><br />
with a fundamental period<br />
<math><br />
T_\zeta<br />
</math>.<br />
Note that <br />
<math><br />
\lim_{T_\zeta \to \infty}\gamma(t)=\beta(t)<br />
</math><br />
<br><br />
The Fourier Series representation of <math> \gamma(t) </math> is<br />
<br><br />
<math><br />
\gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt}<br />
</math><br />
where<br />
<math> <br />
f={1\over T_\zeta}<br />
</math><br />
<br>and<br />
<math><br />
\alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt<br />
</math><br />
<br><br />
<math> \alpha_k </math> can now be rewritten as<br />
<math><br />
\alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt<br />
</math><br />
<br>From our initial identity then, we can write <math> \alpha_k </math> as<br />
<math><br />
\alpha_k={1\over T_\zeta}\Beta(kf)<br />
</math><br />
<br> and <br />
<math><br />
\gamma(t)<br />
</math><br />
becomes<br />
<math><br />
\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}<br />
</math><br />
<br><br />
Now remember that<br />
<math><br />
\beta(t)=\lim_{T_\zeta \to \infty}\gamma(t)<br />
</math><br />
and<br />
<math><br />
{1\over {T_\zeta}} = f.<br />
</math><br />
<br><br />
Which means that<br />
<math><br />
\beta(t)=\lim_{f \to 0}\gamma(t)=\lim_{f \to 0}\sum_{k=-\infty}^\infty f \Beta(kf) e^{j2\pi fkt}<br />
</math><br />
<br><br />
Which is just to say that<br />
<math><br />
\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df<br />
</math><br />
<br><br />
<br><br />
So we have that<br />
<math><br />
\mathcal{F}[\beta(t)]=\Beta(f)=\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi ft}\, dt<br />
</math><br />
<br><br />
Further<br />
<math><br />
\mathcal{F}^{-1}[\Beta(f)]=\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df<br />
</math><br />
==Some Useful Fourier Transform Pairs==<br />
<math><br />
\mathcal{F}[\alpha(t)]=\frac{1}{\mid \alpha \mid}f(\frac{\omega}{\alpha})<br />
</math><br />
<br><br />
{|<br />
|-<br />
|<math>\mathcal{F}[c_1\alpha(t)+c_2\beta(t)]</math><br />
|<math>=\int_{-\infty}^{\infty} (c_1\alpha(t)+c_2\beta(t)) e^{-j2\pi ft}\, dt</math><br />
|-<br />
|<br />
|<math>=\int_{-\infty}^{\infty}c_1\alpha(t)e^{-j2\pi ft}\, dt+\int_{-\infty}^{\infty}c_2\beta(t)e^{-j2\pi ft}\, dt</math><br />
|-<br />
|<br />
|<math>=c_1\int_{-\infty}^{\infty}\alpha(t)e^{-j2\pi ft}\, dt+c_2\int_{-\infty}^{\infty}\beta(t)e^{-j2\pi ft}\, dt=c_1\Alpha(f)+c_2\Beta(f)</math><br />
|-<br />
|}<br />
<br><br />
<math><br />
\mathcal{F}[\alpha(t-\gamma)]=e^{-j2\pi f\gamma}\Alpha(f)<br />
</math><br />
<br><br />
<math><br />
\mathcal{F}[\alpha(t)*\beta(t)]=\Alpha(f)\Beta(f)<br />
</math><br />
<br><br />
<math><br />
\mathcal{F}[\alpha(t)\beta(t)]=\Alpha(f)*\Beta(f)<br />
</math><br />
<br><br />
Some other usefull pairs can be found here: [[Fourier Transforms]]<br />
<br />
==A Second Approach to Fourier Transforms==<br />
*[[Fourier Transforms]]<br />
<br />
Return to *[[Signals and systems|Signals and Systems]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Fourier_series&diff=4371Fourier series2008-02-13T10:21:23Z<p>Wilspa: /* The Fourier Series */</p>
<hr />
<div>*[[Signals and systems|Signals and Systems]]<br />
==Introduction==<br />
Born in Auxerre, France in 1768, Jean Baptiste Joseph Fourier was orphaned at the age of eight. Later he was instructed by Benedictine monks who taught at and ran a military college. <br />
<br />
Years later (1822) Fourier's genius became evident when he discovered that he could represent a periodic function as a sum of sinusoids. It may be interesting to note that what has come to be known as the Fourier series was invented while Fourier was studying heat flow.<br />
<br />
Even though Fourier had discovered a powerful tool, his peers were slow in accepting it. This may be because he provided to rigorous proof to show that his series was an accurate representation of a periodic function. Later, P.G.L. Dirichlet was able to present an acceptable proof.<br />
<br />
<br />
<small>Information used in the introduction has been adapted from <u>Linear Circuit Analysis</u> by DeCarlo & Lin and <u>Fundamentals of Electric Circuits</u> by Alexander and Sadiku.</small><br />
<br />
==Periodic Functions==<br />
A continuous time signal <math>x(t)</math> is said to be periodic if there is a positive nonzero value of T such that <br />
<center><br />
<math> s(t + T) = s(t)</math> for all <math>t</math><br />
</center><br />
==Dirichlet Conditions==<br />
The conditions for a periodic function <math>f</math> with period 2L to have a convergent Fourier series are as follows:<br />
<br />
===''Theorem:''===<br />
<br />
Let <math>f</math> be a piecewise regular real-valued function defined on some interval [-L,L], such that <math>f</math> has<br />
''only a finite number of discontinuities and extrema'' in [-L,L]. Then the Fourier series of this function converges to <math>f</math> when <math>f</math> is continuous and to the arithmetic mean of the left-handed and right-handed limit of <math>f</math> at a point where it is discontinuous.<br />
<br />
==The Fourier Series==<br />
A Fourier series is an expansion of a periodic function <math>f</math> in terms of an infinite sum of sines and cosines. Fourier series make use of the orthogonality relationships of the sine and cosine functions. One way to represent a Fourier series is<br />
<center><br />
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>. <br />
</center><br />
<br />
==See Also==<br />
*[[Orthogonal Functions]]<br />
*[[Fourier Transforms]]<br />
<br />
==Contributors==<br />
Principle author of this page: [[User:Goeari|Aric Goe]]<br />
<br />
Introduction added on 10/06/05 by [[User:wonoje|Jeff W]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Fourier_series&diff=4370Fourier series2008-02-13T10:20:08Z<p>Wilspa: /* Dirichlet Conditions */</p>
<hr />
<div>*[[Signals and systems|Signals and Systems]]<br />
==Introduction==<br />
Born in Auxerre, France in 1768, Jean Baptiste Joseph Fourier was orphaned at the age of eight. Later he was instructed by Benedictine monks who taught at and ran a military college. <br />
<br />
Years later (1822) Fourier's genius became evident when he discovered that he could represent a periodic function as a sum of sinusoids. It may be interesting to note that what has come to be known as the Fourier series was invented while Fourier was studying heat flow.<br />
<br />
Even though Fourier had discovered a powerful tool, his peers were slow in accepting it. This may be because he provided to rigorous proof to show that his series was an accurate representation of a periodic function. Later, P.G.L. Dirichlet was able to present an acceptable proof.<br />
<br />
<br />
<small>Information used in the introduction has been adapted from <u>Linear Circuit Analysis</u> by DeCarlo & Lin and <u>Fundamentals of Electric Circuits</u> by Alexander and Sadiku.</small><br />
<br />
==Periodic Functions==<br />
A continuous time signal <math>x(t)</math> is said to be periodic if there is a positive nonzero value of T such that <br />
<center><br />
<math> s(t + T) = s(t)</math> for all <math>t</math><br />
</center><br />
==Dirichlet Conditions==<br />
The conditions for a periodic function <math>f</math> with period 2L to have a convergent Fourier series are as follows:<br />
<br />
===''Theorem:''===<br />
<br />
Let <math>f</math> be a piecewise regular real-valued function defined on some interval [-L,L], such that <math>f</math> has<br />
''only a finite number of discontinuities and extrema'' in [-L,L]. Then the Fourier series of this function converges to <math>f</math> when <math>f</math> is continuous and to the arithmetic mean of the left-handed and right-handed limit of <math>f</math> at a point where it is discontinuous.<br />
<br />
==The Fourier Series==<br />
A Fourier series is an expansion of a periodic function <math>f</math> in terms of an infinite sum of sines and cosines. Fourier series make use of the orthogonality relationships of the sine and cosine functions.<br />
<center><br />
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>. <br />
</center><br />
<br />
<br />
==See Also==<br />
*[[Orthogonal Functions]]<br />
*[[Fourier Transforms]]<br />
<br />
==Contributors==<br />
Principle author of this page: [[User:Goeari|Aric Goe]]<br />
<br />
Introduction added on 10/06/05 by [[User:wonoje|Jeff W]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=3823User:Wilspa2005-12-13T20:36:42Z<p>Wilspa: </p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
===My Wiki Contributions===<br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
<br><br />
[[Discrete Fourier Transforms]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=My_FIR_Filter_page&diff=2688My FIR Filter page2005-12-13T20:35:47Z<p>Wilspa: </p>
<hr />
<div>[[User:Wilspa|Back to my page]]<br />
==FIR Filters==<br />
What is an FIR Filter? FIR stands for Finite Impulse Response, which means that the response to an individual impulse is going to stop at some point in time, where and Infinite Impulse Response is going to have a response that, at least in theory, never stops. The FIR Filter is almost always more useful than the IIR filter, so FIR filters are more widely used.<br />
===What is it used for?===<br />
An FIR Filter is used in Digital Signal Processing. A digital signal is really a series of impulse functions that were sampled from the orignal sound/radio/whatever wave. From these impulse functions the computer can re-create the original waveform. This all works great, but what if you want to change something about the wave, like it's frequency response, or maybe the phase delay? This is where FIR filters come into play.<br />
===How does it work?===<br />
An FIR Filter is basically a convolution of two series of impulse functions. It takes the impulse functions stored digitally and convolves them with another series of impulse functions which represent a certain frequency response. This convolution in time multiplies the original frequency response with the frequency response of the FIR Filter to give the desired output from the filter. Ideally you would use an infinite convolution like this:<br />
<math>\sum_{n=-\infty}^\infty \sum_{m=-M}^M x(nT) h\left (\frac{mT}{K}\right ) \delta\left (t-nT-\frac{mT}{K}\right )</math> where K is a variable for the amount of interpolation desired in the filter.<br />
This isn't possible in real life. We can't store an infinite amount of data points, so we compromise and store as many as we can reasonably store while still allowing for time constraints. What we end up with looks more like this:<br />
<math>\sum_{n=-(N-1)}^0 \sum_{m=-M}^M x(nT) h\left (\frac{mT}{K}\right ) \delta\left (t-nT-\frac{mT}{K}\right )</math> The reason the outer sum goes from -(N-1) to 0 is because you are storing the previous N-1 impulses and using those data points in your convolution.<br />
<br />
==Adaptive Filters==<br />
<br />
An adaptive filter does what its name implies. It adapts. Basically, an adaptive filter will try to make it's characteristics look exactly like the characteristics of an unknown LTI system by trying to make the output of the adaptive filter look like the output of the unknow LTI system.<br />
<br />
===How does it work?===<br />
<br />
An adaptive filter works by taking the output of both the unknown system, and the adaptive FIR, finding the difference, and then using that difference to calculate new coefficients for the adaptive filter. By repeating this process over and over again, eventually the differences between the two will be minimized. Usually, we want this done quickly, so most of the time, it is best to decrease the difference between the systems as fast as possible. The best way to do this is by taking the gradient of the system, then plug in the error, and use that to calculate the next set of coefficients, stepping closer and closer to the point of no error (the smaller the step size, the slower the process, but the more accurate the results). The block diagram looks something like this.<br />
--Still having trouble getting pictures in here--<br />
Now, this isn't something that you could do by hand very easily, to do an adaptive filter, you really need to use a computer and have the computer do all the calculations for you, because it is just too much number crunching for one person to do. So this wasn't really available until computers became available. So, now that it is available, what is it used for?<br />
<br />
===What is it used for?===<br />
<br />
Adaptive filters are used in many areas, but perhaps the biggest area is noise cancelling. The way this is done is quite brilliant. If you say that your unknown LTI system is silence, then the FIR will try to make its output to be zero. This is very useful if you have a signal that is made of two parts, a very quiet portion that you want, like someone's voice, and a portion that is loud that you don't want, like water running. If you put the signal into an adaptive filter with the unknown system being silence, it will try to make the incoming signal the same. The consant sound of the water will become filtered out, but the ever-changing quality of the voice will not be filtered out because the filter won't be able to keep up with the change.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=My_FIR_Filter_page&diff=1493My FIR Filter page2005-12-06T21:44:39Z<p>Wilspa: </p>
<hr />
<div>[[User:Wilspa|Back to my page]]<br />
==FIR Filters==<br />
What is an FIR Filter? FIR stands for Finite Impulse Response, which means that the response to an individual impulse is going to stop at some point in time, where and Infinite Impulse Response is going to have a response that, at least in theory, never stops. The FIR Filter is almost always more useful than the IIR filter, so FIR filters are more widely used.<br />
===What is it used for?===<br />
An FIR Filter is used in Digital Signal Processing. A digital signal is really a series of impulse functions that were sampled from the orignal sound/radio/whatever wave. From these impulse functions the computer can re-create the original waveform. This all works great, but what if you want to change something about the wave, like it's frequency response, or maybe the phase delay? This is where FIR filters come into play.<br />
===How does it work?===<br />
An FIR Filter is basically a convolution of two series of impulse functions. It takes the impulse functions stored digitally and convolves them with another series of impulse functions which represent a certain frequency response. This convolution in time multiplies the original frequency response with the frequency response of the FIR Filter to give the desired output from the filter. Ideally you would use an infinite convolution like this:<br />
<math>\sum_{n=-\infty}^\infty \sum_{m=-M}^M x(nT) h\left (\frac{mT}{K}\right ) \delta\left (t-nT-\frac{mT}{K}\right )</math> where K is a variable for the amount of interpolation desired in the filter.<br />
This isn't possible in real life. We can't store an infinite amount of data points, so we compromise and store as many as we can reasonably store while still allowing for time constraints. What we end up with looks more like this:<br />
<math>\sum_{n=-(N-1)}^0 \sum_{m=-M}^M x(nT) h\left (\frac{mT}{K}\right ) \delta\left (t-nT-\frac{mT}{K}\right )</math> The reason the outer sum goes from -(N-1) to 0 is because you are storing the previous N-1 impulses and using those data points in your convolution.<br />
<br />
==Adaptive Filters==</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Discrete_Fourier_Transforms&diff=2683Discrete Fourier Transforms2005-12-06T21:41:01Z<p>Wilspa: /* What is it used for? */</p>
<hr />
<div>==Paul's DFT Page==<br />
One of the major tools used in signal processing is the DFT, which stands for Discrete Fourier Transform. The reason we need to to a DFT instead of a Fourier Transform is that our computers are limited in their abilites. They use sampling, and they have limited memory, so we have to adapt to the computers.<br />
===What is a DFT?===<br />
A DFT is like doing a Fourier Transform, but instead of doing it with an integral, we do it with discrete values and a sum. A Fourier Transform looks like this:<br><br />
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt<br />
</math><br><br />
Which uses an integral, while the DFT which looks like this:<br><br />
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}<br />
</math><br><br />
Which is using a sum and a noncontinous series of delta functions x(n) instead of the continuous function x(t).<br />
===What is it used for?===<br />
The DFT is used in a wide variety of areas. Not only is it useful for one dimensional applications suc as signal processing, it is also used for 2D applications such as CAT scans. The DFT allows data to be transformed from one domain into another domain so that minipulation of the data is easier, much the same as a Fourier Transform, only in a Discrete setting. Then, when the minipulations are finished, the IDFT (Inverse DFT) can be used to convert back to the original domain if that is required.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=Discrete_Fourier_Transforms&diff=1398Discrete Fourier Transforms2005-12-06T18:31:30Z<p>Wilspa: </p>
<hr />
<div>==Paul's DFT Page==<br />
One of the major tools used in signal processing is the DFT, which stands for Discrete Fourier Transform. The reason we need to to a DFT instead of a Fourier Transform is that our computers are limited in their abilites. They use sampling, and they have limited memory, so we have to adapt to the computers.<br />
===What is a DFT?===<br />
A DFT is like doing a Fourier Transform, but instead of doing it with an integral, we do it with discrete values and a sum. A Fourier Transform looks like this:<br><br />
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt<br />
</math><br><br />
Which uses an integral, while the DFT which looks like this:<br><br />
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}<br />
</math><br><br />
Which is using a sum and a noncontinous series of delta functions x(n) instead of the continuous function x(t).<br />
===What is it used for?===</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=1494User:Wilspa2005-12-06T18:15:54Z<p>Wilspa: </p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
===Myself===<br />
I don't know what to say really.<br />
===My Wiki Contributions===<br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
<br><br />
[[Discrete Fourier Transforms]]<br />
----<br />
<br />
http://www.923thefort.com/images/shannon/Red%20Wings.jpg<br />
<br><br />
This is the Detroit Red Wings logo, they are my favorite hockey team, so I had to add it.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=My_FIR_Filter_page&diff=1400My FIR Filter page2005-12-06T18:13:30Z<p>Wilspa: </p>
<hr />
<div>[[User:Wilspa|Back to my page]]<br />
==FIR Filters==<br />
What is an FIR Filter? FIR stands for Finite Impulse Response, which means that the response to an individual impulse is going to stop at some point in time, where and Infinite Impulse Response is going to have a response that, at least in theory, never stops. The FIR Filter is almost always more useful than the IIR filter, so FIR filters are more widely used.<br />
===What is it used for?===<br />
An FIR Filter is used in Digital Signal Processing. A digital signal is really a series of impulse functions that were sampled from the orignal sound/radio/whatever wave. From these impulse functions the computer can re-create the original waveform. This all works great, but what if you want to change something about the wave, like it's frequency response, or maybe the phase delay? This is where FIR filters come into play.<br />
===How does it work?===<br />
An FIR Filter is basically a convolution of two series of impulse functions. It takes the impulse functions stored digitally and convolves them with another series of impulse functions which represent a certain frequency response. This convolution in time multiplies the original frequency response with the frequency response of the FIR Filter to give the desired output from the filter. Ideally you would use an infinite convolution like this:<br />
<math>\sum_{n=-\infty}^\infty \sum_{m=-M}^M x(nT) h\left (\frac{mT}{K}\right ) \delta\left (t-nT-\frac{mT}{K}\right )</math> where K is a variable for the amount of interpolation desired in the filter.<br />
This isn't possible in real life. We can't store an infinite amount of data points, so we compromise and store as many as we can reasonably store while still allowing for time constraints. What we end up with looks more like this:<br />
<math>\sum_{n=-(N-1)}^0 \sum_{m=-M}^M x(nT) h\left (\frac{mT}{K}\right ) \delta\left (t-nT-\frac{mT}{K}\right )</math> The reason the outer sum goes from -(N-1) to 0 is because you are storing the previous N-1 impulses and using those data points in your convolution.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=1394User:Wilspa2005-12-06T18:04:06Z<p>Wilspa: </p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
===Myself===<br />
I don't know what to say really.<br />
===My Wiki Contributions===<br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
<br><br />
[[The DFT]]<br />
----<br />
<br />
http://www.923thefort.com/images/shannon/Red%20Wings.jpg<br />
<br><br />
This is the Detroit Red Wings logo, they are my favorite hockey team, so I had to add it.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=My_FIR_Filter_page&diff=1393My FIR Filter page2005-11-14T01:18:55Z<p>Wilspa: </p>
<hr />
<div>[[User:Wilspa|Back to my page]]<br />
==FIR Filters==<br />
What is an FIR Filter? FIR stands for Finite Impulse Response, which means that the response to an individual impulse is going to stop at some point in time, where and Infinite Impulse Response is going to have a response that, at least in theory, never stops. The FIR Filter is almost always more useful than the IIR filter, so FIR filters are more widely used.<br />
===What is it used for?===<br />
An FIR Filter is used in Digital Signal Processing. A digital signal is really a series of impulse functions that were sampled from the orignal sound/radio/whatever wave. From these impulse functions the computer can re-create the original waveform. This all works great, but what if you want to change something about the wave, like it's frequency response, or maybe the phase delay? This is where FIR filters come into play.<br />
===How does it work?===<br />
An FIR Filter is basically a convolution of two series of impulse functions. It takes the impulse functions stored digitally and convolves them with another series of impulse functions which represent a certain frequency response. This convolution in time multiplies the original frequency response with the frequency response of the FIR Filter to give the desired output from the filter. Ideally you would use an infinite convolution like this:<br />
<math>\sum_{n=-\infty}^\infty \sum_{m=-M}^M x(nT) h\left (\frac{mT}{K}\right ) \delta\left (t-nT-\frac{mT}{K}\right )</math> where K is a variable for the amount of interpolation desired in the filter.<br />
Unfortunately, we can't do an infinite convolution like this so we end up doing a discrete convolution where we let n in the outer sum be as big or as small as we like, as long as we get the response that we think is good enough (the bigger N is, the better the response.) It looks like this: <math>\sum_{n=-N}^N \sum_{m=-M}^M x(nT) h\left (\frac{mT}{K}\right ) \delta\left (t-nT-\frac{mT}{K}\right )</math></div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=How_a_CD_player_works&diff=1303How a CD player works2005-11-14T00:45:49Z<p>Wilspa: </p>
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<div>[[User:Wilspa|Back to my page]]<br />
==How a CD player works==<br />
So, what is actually happening when you put a CD into your CD player? The disk spins and music comes out, but how exactly does that work? The answer is that it works by combining the properties of sampling and Fourier Transforms to change the data on the CD into the music coming out of the CD player.<br />
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===Nyquist Theorem===<br />
The problem with a computer is that when it records the music, it actually is only sampling the wave over and over again, and then it stores those samples instead of the actual wave itself. The samples are usually expressed as a series of delta functions multiplied by their respective coefficients: <br><math>\sum_{n=-\infty}^\infty x(nT) \delta(t-nT)</math> <br>However, from these samples it is possible to recreate the original wave if the sampling rate is greater than or equal to twice the frequency of the maximum frequency of the wave being sampled. So,<br><br />
<math>T \ge 2f_{max}</math> <br>This is called the Nyquist Theorem, and it allows us to select the sampling rate we need in order to recreate the music.<br />
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===Sampling===<br />
Because a computer must use sampling in order to store music, it is necessary to have ways of turning the sampled data back into music. When we sample in time, it changes the characteristics of the frequency domain. This changes the music, so we need to figure out how to undo those changes and bring back the original music so that it will sound good to listen to. The change the occurs is called replication and it means that when we sample in time, the original frequency response is repeated every 1/T Hz, which means that you are adding in a whole bunch of high frequency responses that shouldn't be there. How do we fix this? Well, one method is to send the output (a series of impulse functions corresponding to the sampled values) through a perfect brickwall (basically brickwall means perfect) low pass filter that will take out all of the high frequency components and just leave the original sound wave. This would be nice, but it doesn't exactly work that way, since it isn't humanly possible to create a perfect brickwall filter. So, instead we use other methods to create the output we are looking for, starting with the Digital to Analog converter.<br />
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===D/A converter===<br />
A Digital to Analog converter doesn't send out impulse functions, instead it sends out a function that will step along with the height of the steps being the height of the impulse functions, basically convolving the impulse functions with a pulse of period T and amplitude 1. This changes things a bit. Because the impulse functions are being convolved with the pulse, their fourier transfers are being multiplied. The fourier transform of the pulse, when multiplied by the fourier transform of the impulse functions, changes the output in two ways. It smushes the response that we want, the low frequency components of the wave, but it also cuts out a big portion of the high frequency components we don't want since the fourier transform of the pulse is zero in the middle of the higher frequency responses, thereby making them much much smaller and less significant. This is very good because we can correct for the smushing of the response we want by changing our low pass filter, and we don't have to worry as much about the stuff we don't want, since it is mostly removed by the pulse function. The shape of the low pass filter is determined by the shape of the fourier transform of the pulse function. Then, after the signal passes through the low pass filter, it comes out of the speaker as music!<br />
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==Two times oversampling==<br />
If you aren't satisfied by the pretty good method described above, then take a look at 2x oversampling. The biggest advantage given by 2x oversampling is that it completely knocks out the lowest replica of the frequency response that is created by sampling. This makes it much much easier to make a low pass filter that gives you an output that is the same, or at least very close to the same, as the original.<br />
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===How it works===<br />
It starts the same way, by sampling the incoming signal. But then, in order to get a nicer wave to work with, we convolve the sampled values with another series of impulse functions. This series of impulse functions has a fourier transform that does two things. It is zero in the area of the first replica on either side, and it is shaped so that the top is slightly dished to compensate for the affects of the pulse function later on. The formula for these delta functions is:<br><br />
<math>h(t)=\sum_{m=-M}^M h\left (\frac{mT}{2}\right )\delta\left (t- \frac{mT}{2} \right )</math><br />
*Note, the period between these impulse functions is T/2, which is where the 2x comes from in 2x oversampling.<br />
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Also, the frequency response of this is:<br />
<br><br />
<math>H(f)=\mathcal{F}[\sum_{m=-\infty}^\infty h\left (\frac{mT}{2}\right )\delta\left (t-\frac{mT}{2}\right )]</math><br />
<br><br />
<math>=\sum_{m=-\infty}^\infty h\left (\frac{mT}{2}\right )e^{-j2\pi \frac{mT}{2}}</math><br />
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You can choose to make <math>h\left (\frac{mT}{2}\right )</math> be the shape you want so that the later affects of D/A conversion are compensated for. After this is done, it is exactly the same as before, you send the impulse functions to the D/A converter, and then from there to the low pass filter, and then out to the speakers to hear the music.<br />
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===Other info===<br />
Oversampling does not have to be 2x oversampling, there is also 1x oversampling, 3x oversampling, etc. The only difference between those and this is the period between the impulse functions that you convolve your data with, for 2x oversampling it is <math>\frac{T}{2}</math> and for Nx oversampling it is <math>\frac{T}{N}</math>.<br />
<br>[[User:Wilspa|Back to my page]]</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=1392User:Wilspa2005-11-14T00:40:07Z<p>Wilspa: </p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
<br />
===My Wiki Experience...so far===<br />
So far, this seems to be way easy, like Aric said last year, it is far too easy to make stuff work, at least so far. However, I haven't gotten very far in to it yet.<br />
<br />
===Some info about myself===<br />
I am nerd. I love being a nerd, I embrace it. I am also kind of a Jock. I play sports, I hang have lots of friends who are on sports teams, and I love to watch sports. So, maybe I am more of a Jerd, or a Nock, or a Neck, who knows? So, my favorite sport is Hockey, and in order to test my newfound Wiki abilities, I am going to try to add a link here to [http://www.nhl.com NHL.com]. The most exciting part of this is that the NHL is BACK. Yes, and I have cable this year so I can watch. I am excited. In other news, I am experimenting on my computer with Linux. Right now I have SuSe 9.? on it, but I am considering trying Ubuntu (sp?) soon. I just have to find a time to install it. So anyway, thats pretty much it, I hope that you find this far more entertaining than I did, otherwise, you may end up falling asleep.<br />
===My Wiki Contributions===<br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[My FIR Filter page]]<br />
----<br />
<br />
http://www.923thefort.com/images/shannon/Red%20Wings.jpg<br />
<br><br />
This is the Detroit Red Wings logo, they are my favorite hockey team, so I had to add it.</div>Wilspahttps://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wilspa&diff=1294User:Wilspa2005-11-14T00:39:27Z<p>Wilspa: </p>
<hr />
<div><center><br />
==Paul Wilson==<br />
CpE in an EE class!<br><br />
</center><br />
<br><br />
<br />
===My Wiki Experience...so far===<br />
So far, this seems to be way easy, like Aric said last year, it is far too easy to make stuff work, at least so far. However, I haven't gotten very far in to it yet.<br />
<br />
===Some info about myself===<br />
I am nerd. I love being a nerd, I embrace it. I am also kind of a Jock. I play sports, I hang have lots of friends who are on sports teams, and I love to watch sports. So, maybe I am more of a Jerd, or a Nock, or a Neck, who knows? So, my favorite sport is Hockey, and in order to test my newfound Wiki abilities, I am going to try to add a link here to [http://www.nhl.com NHL.com]. The most exciting part of this is that the NHL is BACK. Yes, and I have cable this year so I can watch. I am excited. In other news, I am experimenting on my computer with Linux. Right now I have SuSe 9.? on it, but I am considering trying Ubuntu (sp?) soon. I just have to find a time to install it. So anyway, thats pretty much it, I hope that you find this far more entertaining than I did, otherwise, you may end up falling asleep.<br />
===My Wiki Contributions===<br />
[[Fourier Transforms]]<br />
<br><br />
[[How a CD player works]]<br />
<br><br />
[[FIR Filters]]<br />
----<br />
<br />
http://www.923thefort.com/images/shannon/Red%20Wings.jpg<br />
<br><br />
This is the Detroit Red Wings logo, they are my favorite hockey team, so I had to add it.</div>Wilspa