Class Wiki - User contributions [en]

FIR Filter

2004-12-11T00:01:54Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h \left ( \frac{l T}{2} \right ) \delta \left ( t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is the filter and the term of the right is you data. So how do we do an infinite sum on a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse response of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered on zero. So if you wanted three terms they would be<math> l = -1, 0, 1\,</math> and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2\, </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left (t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left ( \tau - \frac{l T}{2} \right ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) x( k T ) \int_{-\infty}^{\infty} \delta \left (\tau - \frac{l T}{2} \right ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x( k T ) \delta \left ( t- \frac{l T}{2} - k T \right ) </math>

<math>\mbox{Let }n=2k+l\,\mbox{ so }k=\frac{n-l}{2}</math>

<math> \sum_{n= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x\left ( \frac{n-l}{2} T \right ) \delta \left (t- \frac{nT}{2} \right ) </math>
</center>

Where h is your filter coefficients and x is your data. How this actually works is you will be moving down your data point by point, generating a new filtered data point, y, for every unfiltered point, x, you have. The new points will be generated by multiplying the filter coefficients with the unfiltered points before and after where you are in the data stream. For example, if you had three filter coefficients, every new filtered data point would be the fist coefficient times the unfiltered data point right before where you are plus the second coefficient times the unfiltered data point where you are plus the third coefficient times the unfiltered data point after where you are. You will notice that all of this is just a matrix multiply. So if your unfiltered data is x(m), your filter is h(m), your filtered data is y(m), and you have three filter coefficients it would work as follows.

<center>
<math>y(m)=h(m)*x(m)\,</math>

<math>
\begin{matrix} & y(m+2) \\
& \Downarrow \\
y(m)=[h(-1)\ h(0)\ h(1)] & \begin{bmatrix}y(m+1) \\ y(m) \\ y(m-1) \end{bmatrix} \\
& \Downarrow
\end{matrix}
\qquad
\begin{matrix} & y(m+3) \\
& \Downarrow \\
y(m+1)=[h(-1)\ h(0)\ h(1)] & \begin{bmatrix}y(m+2) \\ y(m+1) \\ y(m) \end{bmatrix} \\
& \Downarrow
\end{matrix}
</math></center>

With m counting up as you move down your data and filter more and more points. This is a filter that could be used in a [[CD Player]] to pre-distort and interpolate a signal.

Principle author of this page: [[User:Barnsa|Sam Barnes]]

FIR Filter

2004-12-11T00:00:39Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h \left ( \frac{l T}{2} \right ) \delta \left ( t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is the filter and the term of the right is you data. So how do we do an infinite sum on a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse response of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered on zero. So if you wanted three terms they would be<math> l = -1, 0, 1\,</math> and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2\, </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left (t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left ( \tau - \frac{l T}{2} \right ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) x( k T ) \int_{-\infty}^{\infty} \delta \left (\tau - \frac{l T}{2} \right ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x( k T ) \delta \left ( t- \frac{l T}{2} - k T \right ) </math>

<math>\mbox{Let }n=2k+l\,\mbox{ so }k=\frac{n-l}{2}</math>

<math> \sum_{n= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x\left ( \frac{n-l}{2} T \right ) \delta \left (t- \frac{nT}{2} \right ) </math>
</center>

h is your filter coefficients and x is your data. How this actually works is you will be moving down your data point by point, generating a new filtered data point, y, for every unfiltered point, x, you have. The new points will be generated by multiplying the filter coefficients with the unfiltered points before and after where you are in the data stream. For example, if you had three filter coefficients, every new filtered data point would be the fist coefficient times the unfiltered data point right before where you are plus the second coefficient times the unfiltered data point where you are plus the third coefficient times the unfiltered data point after where you are. You will notice that all of this is just a matrix multiply. So if your unfiltered data is x(m), your filter is h(m), your filtered data is y(m), and you have three filter coefficients it would work as follows.

<center>
<math>y(m)=h(m)*x(m)\,</math>

<math>
\begin{matrix} & y(m+2) \\
& \Downarrow \\
y(m)=[h(-1)\ h(0)\ h(1)] & \begin{bmatrix}y(m+1) \\ y(m) \\ y(m-1) \end{bmatrix} \\
& \Downarrow
\end{matrix}
\qquad
\begin{matrix} & y(m+3) \\
& \Downarrow \\
y(m+1)=[h(-1)\ h(0)\ h(1)] & \begin{bmatrix}y(m+2) \\ y(m+1) \\ y(m) \end{bmatrix} \\
& \Downarrow
\end{matrix}
</math></center>

With m counting up as you move down your data and filter more and more points. This is a filter that could be used in a [[CD Player]] to pre-distort and interpolate a signal.

Principle author of this page: [[User:Barnsa|Sam Barnes]]

FIR Filter

2004-12-10T23:49:04Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h \left ( \frac{l T}{2} \right ) \delta \left ( t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is the filter and the term of the right is you data. So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse response of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered on zero. So if you wanted three terms they would be l = -1, 0, 1 and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2/, </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left (t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left ( \tau - \frac{l T}{2} \right ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) x( k T ) \int_{-\infty}^{\infty} \delta \left (\tau - \frac{l T}{2} \right ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x( k T ) \delta \left ( t- \frac{l T}{2} - k T \right ) </math>

<math>\mbox{Let }n=2k+l\,\mbox{ so }k=\frac{n-l}{2}</math>

<math> \sum_{n= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x\left ( \frac{n-l}{2} T \right ) \delta \left (t- \frac{nT}{2} \right ) </math>
</center>

So the innermost sum from -M to M are the coefficients in you filter. How this actually works is you will be moving down your data point by point, generating a new filtered data point for every unfiltered point you have. The new points will be generated by multiplying the filter coefficients by the unfiltered points before and after where you are in the data stream. So for example if you had three filter coefficients every new filtered data point would be the fist coefficient times the unfiltered data point right before where you are plus the second coefficient times the unfiltered data point where you are plus the third coefficient times the unfiltered data point after where you are. You will notice that all of this is a matrix multiply. So if your unfiltered data is x(m), your filter is h(m), your filtered data is y(m), and you have three filter coefficients it would work as follows.

<center>
<math>y(m)=h(m)*x(m)\,</math>

<math>
\begin{matrix} & y(m+2) \\
& \Downarrow \\
y(m)=[h(-1)\ h(0)\ h(1)] & \begin{bmatrix}y(m+1) \\ y(m) \\ y(m-1) \end{bmatrix} \\
& \Downarrow
\end{matrix}
\qquad
\begin{matrix} & y(m+3) \\
& \Downarrow \\
y(m+1)=[h(-1)\ h(0)\ h(1)] & \begin{bmatrix}y(m+2) \\ y(m+1) \\ y(m) \end{bmatrix} \\
& \Downarrow
\end{matrix}
</math></center>

With m counting up as you move down your data and filter more and more points.

FIR Filter

2004-12-10T23:43:49Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h \left ( \frac{l T}{2} \right ) \delta \left ( t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is the filter and the term of the right is you data. So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse response of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered on zero. So if you wanted three terms they would be l = -1, 0, 1 and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2/, </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left (t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left ( \tau - \frac{l T}{2} \right ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) x( k T ) \int_{-\infty}^{\infty} \delta \left (\tau - \frac{l T}{2} \right ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x( k T ) \delta \left ( t- \frac{l T}{2} - k T \right ) </math>

<math>\mbox{Let }n=2k+l\,\mbox{ so }k=\frac{n-l}{2}</math>

<math> \sum_{n= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x\left ( \frac{n-l}{2} T \right ) \delta \left (t- \frac{nT}{2} \right ) </math>
</center>

So the innermost sum from -M to M are the coefficients in you filter. How this actually works is you will be moving down your data point by point, generating a new filtered data point for every unfiltered point you have. The new points will be generated by multiplying the filter coefficients by the unfiltered points before and after where you are in the data stream. So for example if you had three filter coefficients every new filtered data point would be the fist coefficient times the unfiltered data point right before where you are plus the second coefficient times the unfiltered data point where you are plus the third coefficient times the unfiltered data point after where you are. You will notice that all of this is a matrix multiply. So if your unfiltered data is x(m), your filter is h(m), your filtered data is y(m), and you have three filter coefficients it would work as follows.

<center>
<math>y(m)=h(m)*x(m)\,</math>

<math>
\begin{matrix} & & & y(m+2) \\
& & & \Downarrow \\
y(m) & = & [h(-1)\ h(0)\ h(1)] & \begin{bmatrix}y(m+1) \\ y(m) \\ y(m-1) \end{bmatrix} \\
& & & \Downarrow
\end{matrix}
</math>

<math>
\begin{matrix} & & & y(m+3) \\
& & & \Downarrow \\
y(m) & = & [h(-1)\ h(0)\ h(1)] & \begin{bmatrix}y(m+2) \\ y(m+1) \\ y(m) \end{bmatrix} \\
& & & \Downarrow
\end{matrix}
</math>

With m counting up as you move down your data and filter more and more points.

FIR Filter

2004-12-10T23:29:54Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h \left ( \frac{l T}{2} \right ) \delta \left ( t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is the filter and the term of the right is you data. So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse response of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered on zero. So if you wanted three terms they would be l = -1, 0, 1 and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2/, </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left (t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left ( \tau - \frac{l T}{2} \right ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) x( k T ) \int_{-\infty}^{\infty} \delta \left (\tau - \frac{l T}{2} \right ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x( k T ) \delta \left ( t- \frac{l T}{2} - k T \right ) </math>

<math>\mbox{Let }n=2k+l\,\mbox{ so }k=\frac{n-l}{2}</math>

<math> \sum_{n= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x\left ( \frac{n-l}{2} T \right ) \delta \left (t- \frac{nT}{2} \right ) </math>
</center>

So the innermost sum from -M to M are the coefficients in you filter. How this actually works is you will be moving down your data point by point, generating a new filtered data point for every unfiltered point you have. The new points will be generated by multiplying the filter coefficients by the unfiltered points before and after where you are in the data stream. So for example if you had three filter coefficients every new filtered data point would be the fist coefficient times the unfiltered data point right before where you are plus the second coefficient times the unfiltered data point where you are plus the third coefficient times the unfiltered data point after where you are. You will notice that all of this is a matrix multiply. So if your unfiltered data is x(m), your filter is h(m), your filtered data is y(m), and you have three filter coefficients it would work as follows.

<center>
<math>y(m)=h(m)*x(m)</math>

<math>y(m)=[h(-1)\ h(0)\ h(1)] \begin{bmatrix} \Downarrow \\ y(m+1) \\ y(m) \\ y(m-1) \\ \Downarrow \end{bmatrix}</math>

<math>y(m+1)=[h(-1)\ h(0)\ h(1)] \begin{bmatrix} y(m+2) \\ y(m+1) \\ y(m) \end{bmatrix}
</math></center>

With m counting up as you move down your data and filter more and more points.

FIR Filter

2004-12-10T23:26:16Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h \left ( \frac{l T}{2} \right ) \delta \left ( t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is the filter and the term of the right is you data. So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse response of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered on zero. So if you wanted three terms they would be l = -1, 0, 1 and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2/, </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left (t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left ( \tau - \frac{l T}{2} \right ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) x( k T ) \int_{-\infty}^{\infty} \delta \left (\tau - \frac{l T}{2} \right ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x( k T ) \delta \left ( t- \frac{l T}{2} - k T \right ) </math>

<math>\mbox{Let }n=2k+l\,\mbox{ so }k=\frac{n-l}{2}</math>

<math> \sum_{n= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x\left ( \frac{n-l}{2} T \right ) \delta \left (t- \frac{nT}{2} \right ) </math>
</center>

So the innermost sum from -M to M are the coefficients in you filter. How this actually works is you will be moving down your data point by point, generating a new filtered data point for every unfiltered point you have. The new points will be generated by multiplying the filter coefficients by the unfiltered points before and after where you are in the data stream. So for example if you had three filter coefficients every new filtered data point would be the fist coefficient times the unfiltered data point right before where you are plus the second coefficient times the unfiltered data point where you are plus the third coefficient times the unfiltered data point after where you are. You will notice that all of this is a matrix multiply. So if your unfiltered data is x(m), your filter is h(m), your filtered data is y(m), and you have three filter coefficients it would work as follows.

<center><math>
y(m)=h(m)*x(m)

y(m)=[h(-1) h(0) h(1)] \begin{bmatrix} y(m+1) \\ y(m) \\ y(m-1) \end{bmatrix}

y(m+1)=[h(-1) h(0) h(1)] \begin{bmatrix} y(m+2) \\ y(m+1) \\ y(m) \end{bmatrix}
</math></center>

With m counting up as you move down your data and filter more and more points.

FIR Filter

2004-12-10T23:23:27Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h \left ( \frac{l T}{2} \right ) \delta \left ( t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is ther filter and the term of the right is you data. So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse responce of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered around zero. So if you wanted three terms they would be l = -1, 0, 1 and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2/, </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left (t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left ( \tau - \frac{l T}{2} \right ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) x( k T ) \int_{-\infty}^{\infty} \delta \left (\tau - \frac{l T}{2} \right ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x( k T ) \delta \left ( t- \frac{l T}{2} - k T \right ) </math>

<math>\mbox{Let }n=2k+l\,\mbox{ so }k=\frac{n-l}{2}</math>

<math> \sum_{n= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x\left ( \frac{n-l}{2} T \right ) \delta \left (t- \frac{nT}{2} \right ) </math>

So the inner most summat from -M to M are the coefficients in you filter. How this actually works is you will be moving down your data point by point, generating a new filtered data point for every unfiltered point you have. The new points will be generated by multiplying the filter coefficients by the unfiltered points before and after where you are in the data stream. So for example if you had three filter coefficients every new filtered data point would be the fist coefficient times the unfiltered data point right before where you are plus the second coefficient times the unfiltered data point where you are plus the third coefficient times the unfiltered data point after where you are. You will notice that all this is is a matrix multiply. So if your unfiltered data is x(m), your filter is h(m), your filtered data is y(m), and you have three filter coefficients it would work as follows.

<center><math>
y(m)=h(m)*x(m)
y(m)=[h(-1) h(0) h(1)] \begin{bmatrix} y(m+1) \\ y(m) \\ y(m-1) \end{bmatrix}
y(m+1)=[h(-1) h(0) h(1)] \begin{bmatrix} y(m+2) \\ y(m+1) \\ y(m) \end{bmatrix}
</math>
</center>

With m counting up as you move down your data and filter more and more points.

FIR Filter

2004-12-10T23:08:48Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h \left ( \frac{l T}{2} \right ) \delta \left ( t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is ther filter and the term of the right is you data. So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse responce of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered around zero. So if you wanted three terms they would be l = -1, 0, 1 and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2/, </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left (t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left ( \tau - \frac{l T}{2} \right ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) x( k T ) \int_{-\infty}^{\infty} \delta \left (\tau - \frac{l T}{2} \right ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x( k T ) \delta \left ( t- \frac{l T}{2} - k T \right ) </math>

<math>\mbox{Let }n=2k+l\,\mbox{ so }k=\frac{n-l}{2}</math>

<math> \sum_{n= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x\left ( \frac{n-l}{2} T \right ) \delta \left (t- \frac{nT}{2} \right ) </math>

</center>

FIR Filter

2004-12-10T23:05:09Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h \left ( \frac{l T}{2} \right ) \delta \left ( t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is ther filter and the term of the right is you data. So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse responce of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered around zero. So if you wanted three terms they would be <math>l = -1, 0, 1 </math> and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2 </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left (t- \frac{l T}{2} \right ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) \delta \left ( \tau - \frac{l T}{2} \right ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h \left ( \frac{l T}{2} \right ) x( k T ) \int_{-\infty}^{\infty} \delta \left (\tau - \frac{l T}{2} \right ) \delta (t- \tau - k T ) \, d \tau</math>

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x( k T ) \delta \left ( t- \frac{l T}{2} - k T \right ) </math>

Let <math> n=2k+l\, </math> so <math> k=\frac{n-l}{2}</math>

<math> \sum_{n= -\infty}^ \infty \sum_{l= -M}^ M h\left ( \frac{l T}{2} \right ) x\left ( \frac{n-l}{2} T \right ) \delta \left (t- \frac{nT}{2} \right ) </math>

</center>

FIR Filter

2004-12-10T21:05:41Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is ther filter and the term of the right is you data. So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse responce of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered around zero. So if you wanted three terms they would be <math>l = -1, 0, 1 </math> and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2 </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>.

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h( \frac{l T}{2} ) \delta (\tau - \frac{l T}{2} ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>.

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h( \frac{l T}{2} ) x( k T ) \int_{-\infty}^{\infty} \delta (\tau - \frac{l T}{2} ) \delta (t- \tau - k T ) \, d \tau</math>.

</center>

FIR Filter

2004-12-10T21:05:02Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>
</center>

Where the term on the left is ther filter and the term of the right is you data. So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.
The above sum for the filter is the impulse responce of the filter. This is an infinite sum, and for your filter to work exactly as described you would need all of the terms. Since that is not possible you want to keep as many of the most important terms as you can. The most important terms are the ones centered around zero. So if you wanted three terms they would be <math>l = -1, 0, 1 </math> and if you wanted 5 terms <math> l = -2, -1, 0, 1, 2 </math> etc. Once you have your filter limited to a certain number of terms you can proceed as follows with the convolution.

<center>
<math> \sum_{l= -M}^ M h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>.

<math> \int_{-\infty}^{\infty} \sum_{l= -M}^ M h( \frac{l T}{2} ) \delta (\tau - \frac{l T}{2} ) \sum_{k= -\infty}^ \infty x( k T ) \delta (t- \tau - k T ) \, d \tau</math>.

<math> \sum_{k= -\infty}^ \infty \sum_{l= -M}^ M h( \frac{l T}{2} ) x( k T ) \int_{-\infty}^{\infty} \delta (\tau - \frac{l T}{2} ) \delta (t- \tau - k T ) \, d \tau</math>.

</center>

FIR Filter

2004-12-10T18:59:02Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this this is done by the following convolution:

<center>
<math> \sum_{l= -\infty}^ \infty h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>.
</center>

So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.

FIR Filter

2004-12-10T18:58:29Z

Barnsa:

A FIR filter is a type of digital interpolating filter. We will be looking at its use to both distort and interpolate a digital signal as in a [[CD Player]]. Mathematically this this is done by convolving the following functions:

<center>
<math> \sum_{l= -\infty}^ \infty h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) * \sum_{k= -\infty}^ \infty x( k T ) \delta (t- k T )</math>.
</center>

So how do we do an infinite sum in a computer? Well we can't because we can't fit an infinite amount of data in a computer, so we have to trim it down.

CD Player

2004-12-10T18:50:52Z

Barnsa:

A CD is a digitally recorded sound wave. The CD has a whole bunch of numbers on it. These numbers represent the magnitude of the sound wave at certain points in time. If your sound was a sine wave for example and you looked at it graphically you would see a smooth wave going up and down. A CD of this wave would just be the magnitude, or height, of that wave every so often.
The basic idea in turning these numbers into a sound wave you can here is to run them through a digital to analog converter. The D-to-A will read in a value then output and hold a corresponding voltage until it gets the next value, then it will output and hold that voltage until the next one, etc. The end result will be a stair step representation of the original wave.

You can then run this through a low pass filter to smooth it out a little and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

To get a better idea of why the signal becomes distorted and how to correct it, it helps to look at the signal in the time and the frequency domain. To represent the sampled signal in the time domain we will show a string of impulse functions whose area represents the magnitude of the signal at that point. While there is actually no impulse function physically in the system it is a useful mathematical construct to look at what is happening in the frequency domain.

[[Image:barnsasample.jpg|Sampling a signal]]

'''Figure 1'''

[[Image:barnsaDA.jpg|Digital to analog conversion]]

'''Figure 2'''

This shows the progression of your signal both in the time and frequency domains, as it is sampled (figure 1), and then ran through the D-to-A converter (figure 2). You will notice that the frequency representation of the stair step wave does not look like the original signal. It is too tall in the middle and drops off too fast of the edges; it also has too many high frequency (both positive and negative) components that aren't wanted. These high frequencies components need to be filtered off (which we stated above, the equivalent of smoothing out the stair step function in time) but you would still have to problem of the signal being too tall in the middle and dropping off too fast. In addition in order to filter out the high frequency parts and leave your signal unchanged you need a low pass filter with a very very sharp edge. This is a high precision and very spendy component that we would like to eliminate the need for.
We have two basic problems, the distorted signal, and the need to filter out the high frequency components. To fix this we need to run the signal through some sort of filter that will both pre-distort the signal (increase its wanted high frequency components) and flatten out the unwanted high frequency components that are very close to the signal. This leads us into oversampling.

==Over Sampling==
The basic idea behind oversampling is to run the signal through an interpolation filter that will effectively double (or triple, or quadruple, etc) the number of points in the time signal. Suppose that there is a signal:

0 @ t=0

4 @ t=T

8 @ t=2T

The filter would guess at what the signal would be at 1/2T and 3/2T so the signal would become something like:

0 @ t=0

2 @ t=1/2T

4 @ t=T

6 @ t=3/2T

8 @ t=T

This is a very simple case, in actuality the filter would take a preset linear combination of the data points before and after to determine what the inserted value should be. Mathematically this is a convolution of the signal from the CD with a function of the form:

<center>
<math> f(t) = \sum_{l= -\infty}^ \infty h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) </math>.
</center>

To see why this solves the problems stated above we need to look at the frequency domain. In the frequency domain we needed to filter out the unwanted higher frequencies to get our original signal back. To make this easier it would be nice to eliminate the frequencies that are really close to the signal you are trying to keep, and at the same time we wanted to pre-emphasize the high frequencies of our signal. This would mean multiplying them in frequency by a domain by a filtering function as shown below.

[[Image:Barnsapredistort.jpg|Oversampling]]

Multiplication in the frequency domain corresponds to a convolution in the time domain, and the function that we used above in the frequency domain corresponds to a string of impulses in time. Convolving by these impulses in time is the over sampling described above.
This makes since because we have effectively doubled the frequency in the time domain which will space things out in the frequency domain and that is exactly what we see. The next logical question is, well that is nice but how do I actually do this on a computer. One way is with a Finite Impulse Response Filter of [[FIR Filter]].

Principle author of this page: [[User:Barnsa|Sam Barnes]]

User:Barnsa

2004-12-10T18:50:22Z

Barnsa:

[[Image:Sam.JPG|thumb|Sam's photo]]
This is Sam's Page. Yeeehaaaa

Homework #10 an explination on how a [[CD Player]] works

Last homework an explination of an [[FIR Filter]]

CD Player

2004-12-10T18:48:57Z

Barnsa:

A CD is a digitally recorded sound wave. The CD has a whole bunch of numbers on it. These numbers represent the magnitude of the sound wave at certain points in time. If your sound was a sine wave for example and you looked at it graphically you would see a smooth wave going up and down. A CD of this wave would just be the magnitude, or height, of that wave every so often.
The basic idea in turning these numbers into a sound wave you can here is to run them through a digital to analog converter. The D-to-A will read in a value then output and hold a corresponding voltage until it gets the next value, then it will output and hold that voltage until the next one, etc. The end result will be a stair step representation of the original wave.

You can then run this through a low pass filter to smooth it out a little and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

To get a better idea of why the signal becomes distorted and how to correct it, it helps to look at the signal in the time and the frequency domain. To represent the sampled signal in the time domain we will show a string of impulse functions whose area represents the magnitude of the signal at that point. While there is actually no impulse function physically in the system it is a useful mathematical construct to look at what is happening in the frequency domain.

[[Image:barnsasample.jpg|Sampling a signal]]

'''Figure 1'''

[[Image:barnsaDA.jpg|Digital to analog conversion]]

'''Figure 2'''

This shows the progression of your signal both in the time and frequency domains, as it is sampled (figure 1), and then ran through the D-to-A converter (figure 2). You will notice that the frequency representation of the stair step wave does not look like the original signal. It is too tall in the middle and drops off too fast of the edges; it also has too many high frequency (both positive and negative) components that aren't wanted. These high frequencies components need to be filtered off (which we stated above, the equivalent of smoothing out the stair step function in time) but you would still have to problem of the signal being too tall in the middle and dropping off too fast. In addition in order to filter out the high frequency parts and leave your signal unchanged you need a low pass filter with a very very sharp edge. This is a high precision and very spendy component that we would like to eliminate the need for.
We have two basic problems, the distorted signal, and the need to filter out the high frequency components. To fix this we need to run the signal through some sort of filter that will both pre-distort the signal (increase its wanted high frequency components) and flatten out the unwanted high frequency components that are very close to the signal. This leads us into oversampling.

==Over Sampling==
The basic idea behind oversampling is to run the signal through an interpolation filter that will effectively double (or triple, or quadruple, etc) the number of points in the time signal. Suppose that there is a signal:

0 @ t=0

4 @ t=T

8 @ t=2T

The filter would guess at what the signal would be at 1/2T and 3/2T so the signal would become something like:

0 @ t=0

2 @ t=1/2T

4 @ t=T

6 @ t=3/2T

8 @ t=T

This is a very simple case, in actuality the filter would take a preset linear combination of the data points before and after to determine what the inserted value should be. Mathematically this is a convolution of the signal from the CD with a function of the form:

<center>
<math> f(t) = \sum_{l= -\infty}^ \infty h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) </math>.
</center>

To see why this solves the problems stated above we need to look at the frequency domain. In the frequency domain we needed to filter out the unwanted higher frequencies to get our original signal back. To make this easier it would be nice to eliminate the frequencies that are really close to the signal you are trying to keep, and at the same time we wanted to pre-emphasize the high frequencies of our signal. This would mean multiplying them in frequency by a domain by a filtering function as shown below.

[[Image:Barnsapredistort.jpg|Oversampling]]

Multiplication in the frequency domain corresponds to a convolution in the time domain, and the function that we used above in the frequency domain corresponds to a string of impulses in time. Convolving by these impulses in time is the over sampling described above.
This makes since because we have effectively doubled the frequency in the time domain which will space things out in the frequency domain and that is exactly what we see. The next logical question is, well that is nice but how do I actually do this on a computer. One way is with a Finite Impulse Response Filter of [FIR Filter].

Principle author of this page: [[User:Barnsa|Sam Barnes]]

CD Player

2004-12-10T18:44:33Z

Barnsa:

A CD is a digitally recorded sound wave. The CD has a whole bunch of numbers on it. These numbers represent the magnitude of the sound wave at certain points in time. If your sound was a sine wave for example and you looked at it graphically you would see a smooth wave going up and down. A CD of this wave would just be the magnitude, or height, of that wave every so often.
The basic idea in turning these numbers into a sound wave you can here is to run them through a digital to analog converter. The D-to-A will read in a value then output and hold a corresponding voltage until it gets the next value, then it will output and hold that voltage until the next one, etc. The end result will be a stair step representation of the original wave.

You can then run this through a low pass filter to smooth it out a little and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

To get a better idea of why the signal becomes distorted and how to correct it, it helps to look at the signal in the time and the frequency domain. To represent the sampled signal in the time domain we will show a string of impulse functions whose area represents the magnitude of the signal at that point. While there is actually no impulse function physically in the system it is a useful mathematical construct to look at what is happening in the frequency domain.

[[Image:barnsasample.jpg|Sampling a signal]]

'''Figure 1'''

[[Image:barnsaDA.jpg|Digital to analog conversion]]

'''Figure 2'''

This shows the progression of your signal both in the time and frequency domains, as it is sampled (figure 1), and then ran through the D-to-A converter (figure 2). You will notice that the frequency representation of the stair step wave does not look like the original signal. It is too tall in the middle and drops off too fast of the edges; it also has too many high frequency (both positive and negative) components that aren't wanted. These high frequencies components need to be filtered off (which we stated above, the equivalent of smoothing out the stair step function in time) but you would still have to problem of the signal being too tall in the middle and dropping off too fast. In addition in order to filter out the high frequency parts and leave your signal unchanged you need a low pass filter with a very very sharp edge. This is a high precision and very spendy component that we would like to eliminate the need for.
We have two basic problems, the distorted signal, and the need to filter out the high frequency components. To fix this we need to run the signal through some sort of filter that will both pre-distort the signal (increase its wanted high frequency components) and flatten out the unwanted high frequency components that are very close to the signal. This leads us into oversampling.

==Over Sampling==
The basic idea behind oversampling is to run the signal through an interpolation filter that will effectively double (or triple, or quadruple, etc) the number of points in the time signal. Suppose that there is a signal:

0 @ t=0

4 @ t=T

8 @ t=2T

The filter would guess at what the signal would be at 1/2T and 3/2T so the signal would become something like:

0 @ t=0

2 @ t=1/2T

4 @ t=T

6 @ t=3/2T

8 @ t=T

This is a very simple case, in actuality the filter would take a preset linear combination of the data points before and after to determine what the inserted value should be. Mathematically this is a convolution of the signal from the CD with a function of the form:

<center>
<math> f(t) = \sum_{l= -\infty}^ \infty h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) </math>.
</center>

To see why this solves the problems stated above we need to look at the frequency domain. In the frequency domain we needed to filter out the unwanted higher frequencies to get our original signal back. To make this easier it would be nice to eliminate the frequencies that are really close to the signal you are trying to keep, and at the same time we wanted to pre-emphasize the high frequencies of our signal. This would mean multiplying them in frequency by a domain by a filtering function as shown below.

[[Image:Barnsapredistort.jpg|Oversampling]]

Multiplication in the frequency domain corresponds to a convolution in the time domain, and the function that we used above in the frequency domain corresponds to a string of impulses in time. Convolving by these impulses in time is the over sampling described above.
This makes since because we have effectively doubled the frequency in the time domain which will space things out in the frequency domain and that is exactly what we see.

Principle author of this page: [[User:Barnsa|Sam Barnes]]

CD Player

2004-12-10T18:43:54Z

Barnsa:

A CD is a digitally recorded sound wave. The CD has a whole bunch of numbers on it. These numbers represent the magnitude of the sound wave at certain points in time. If your sound was a sine wave for example and you looked at it graphically you would see a smooth wave going up and down. A CD of this wave would just be the magnitude, or height, of that wave every so often.
The basic idea in turning these numbers into a sound wave you can here is to run them through a digital to analog converter. The D-to-A will read in a value then output and hold a corresponding voltage until it gets the next value, then it will output and hold that voltage until the next one, etc. The end result will be a stair step representation of the original wave.

You can then run this through a low pass filter to smooth it out a little and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

To get a better idea of why the signal becomes distorted and how to correct it, it helps to look at the signal in the time and the frequency domain. To represent the sampled signal in the time domain we will show a string of impulse functions whose area represents the magnitude of the signal at that point. While there is actually no impulse function physically in the system it is a useful mathematical construct to look at what is happening in the frequency domain.

[[Image:barnsasample.jpg|Sampling a signal]]

'''Figure 1'''

[[Image:barnsaDA.jpg|Digital to analog conversion]]

'''Figure 2'''

This shows the progression of your signal both in the time and frequency domains, as it is sampled (figure 1), and then ran through the D-to-A converter (figure 2). You will notice that the frequency representation of the stair step wave does not look like the original signal. It is too tall in the middle and drops off too fast of the edges; it also has too many high frequency (both positive and negative) components that aren't wanted. These high frequencies components need to be filtered off (which we stated above, the equivalent of smoothing out the stair step function in time) but you would still have to problem of the signal being too tall in the middle and dropping off too fast. In addition in order to filter out the high frequency parts and leave your signal unchanged you need a low pass filter with a very very sharp edge. This is a high precision and very spendy component that we would like to eliminate the need for.
We have two basic problems, the distorted signal, and the need to filter out the high frequency components. To fix this we need to run the signal through some sort of filter that will both pre-distort the signal (increase its wanted high frequency components) and flatten out the unwanted high frequency components that are very close to the signal. This leads us into oversampling.

==Over Sampling==
The basic idea behind oversampling is to run the signal through an interpolation filter that will effectively double (or triple, or quadruple, etc) the number of points in the time signal. Suppose that there is a signal:

0 @ t=0

4 @ t=T

8 @ t=2T

The filter would guess at what the signal would be at 1/2T and 3/2T so the signal would become something like:

0 @ t=0

2 @ t=1/2T

4 @ t=T

6 @ t=3/2T

8 @ t=T

This is a very simple case, in actuality the filter would take a preset linear combination of the data points before and after to determine what the inserted value should be. Mathematically this is a convolution of the signal from the CD with a function of the form:

<center>
<math> f(t) = \sum_{l= -\infty}^ \infty h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) </math>.
</center>

To see why this solves the problems stated above we need to look at the frequency domain. In the frequency domain we needed to filter out the unwanted higher frequencies to get our original signal back. To make this easier it would be nice to eliminate the frequencies that are really close to the signal you are trying to keep, and at the same time we wanted to pre-emphasize the high frequencies of our signal. This would mean multiplying them in frequency by a domain by a filtering function as shown below.

[[Image:Barnsapredistort.jpg|Oversampling]]

Multiplication in the frequency domain corresponds to a convolution in the time domain, and the function that we used above in the frequency domain corresponds to a string of impulses in time. Convolving by these impulses in time is the over sampling described above.
This makes since because we have effectively doubled the frequency in the time domain which will space things out in the frequency domain and that is exactly what we see.

Principle author of this page: [[User:Barnsa|Sam Barnes]]

The following is only included for my sake, so I can look at the syntax.
==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

CD Player

2004-12-10T18:41:26Z

Barnsa:

A CD is a digitally recorded sound wave. The CD has a whole bunch of numbers on it. These numbers represent the magnitude of the sound wave at certain points in time. If your sound was a sine wave for example and you looked at it graphically you would see a smooth wave going up and down. A CD of this wave would just be the magnitude, or height, of that wave every so often.
The basic idea in turning these numbers into a sound wave you can here is to run them through a digital to analog converter. The D-to-A will read in a value then output and hold a corresponding voltage until it gets the next value, then it will output and hold that voltage until the next one, etc. The end result will be a stair step representation of the original wave.

You can then run this through a low pass filter to smooth it out a little and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

To get a better idea of why the signal becomes distorted and how to correct it, it helps to look at the signal in the time and the frequency domain. To represent the sampled signal in the time domain we will show a string of impulse functions whose area represents the magnitude of the signal at that point. While there is actually no impulse function physically in the system it is a useful mathematical construct to look at what is happening in the frequency domain.

[[Image:barnsasample.jpg|Sampling a signal]]

'''Figure 1'''

[[Image:barnsaDA.jpg|Digital to analog conversion]]

'''Figure 2'''

This shows the progression of your signal both in the time and frequency domains, as it is sampled (figure 1), and then ran through the D-to-A converter (figure 2). You will notice that the frequency representation of the stair step wave does not look like the original signal. It is too tall in the middle and drops off too fast of the edges; it also has too many high frequency (both positive and negative) components that aren't wanted. These high frequencies components need to be filtered off (which we stated above, the equivalent of smoothing out the stair step function in time) but you would still have to problem of the signal being too tall in the middle and dropping off too fast. In addition in order to filter out the high frequency parts and leave your signal unchanged you need a low pass filter with a very very sharp edge. This is a high precision and very spendy component that we would like to eliminate the need for.
We have two basic problems, the distorted signal, and the need to filter out the high frequency components. To fix this we need to run the signal through some sort of filter that will both pre-distort the signal (increase its wanted high frequency components) and flatten out the unwanted high frequency components that are very close to the signal. This leads us into oversampling.

==Over Sampling==
The basic idea behind oversampling is to run the signal through an interpolation filter that will effectively double (or triple, or quadruple, etc) the number of points in the time signal. Suppose that there is a signal:

0 @ t=0

4 @ t=T

8 @ t=2T

The filter would guess at what the signal would be at 1/2T and 3/2T so the signal would become something like:

0 @ t=0

2 @ t=1/2T

4 @ t=T

6 @ t=3/2T

8 @ t=T

This is a very simple case, in actuality the filter would take a preset linear combination of the data points before and after to determine what the inserted value should be. Mathematically this is a convolution of the signal from the CD with a function of the form:

<center>
<math> f(t) = \sum_{l= -\infty}^ \infty h( \frac{l T}{2} ) \delta (t- \frac{l T}{2} ) </math>.
</center>

To see why this solves the problems stated above we need to look at the frequency domain. In the frequency domain we needed to filter out the unwanted higher frequencies to get our original signal back. To make this easier it would be nice to eliminate the frequencies that are really close to the signal you are trying to keep, and at the same time we wanted to pre-emphasize the high frequencies of our signal. This would mean multiplying them in frequency by a domain by a filtering function as shown below.

Insert Picture

Multiplication in the frequency domain corresponds to a convolution in the time domain, and the function that we used above in the frequency domain corresponds to a string of impulses in time. Convolving by these impulses in time is the over sampling described above.
This makes since because we have effectively doubled the frequency in the time domain which will space things out in the frequency domain and that is exactly what we see.

Principle author of this page: [[User:Barnsa|Sam Barnes]]

The following is only included for my sake, so I can look at the syntax.
==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

CD Player

2004-11-17T18:59:35Z

Barnsa:

A CD is a digitally recorded sound wave. The CD has a whole bunch of numbers on it. These numbers represent the magnitude of the sound wave at certain points in time. If your sound was a sine wave for example and you looked at it graphically you would see a smooth wave going up and down. A CD of this wave would just be the magnitude, or height, of that wave every so often.
The basic idea in turning these numbers into a sound wave you can here is to run them through a digital to analog converter. The D-to-A will read in a value then output and hold a corresponding voltage until it gets the next value, then it will output and hold that voltage until the next one, etc. The end result will be a stair step representation of the original wave similar to the one shown below.

Insert Image

You can then run this through a low pass filter to smooth it out a little and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

To get a better idea of why the signal becomes distorted and how to correct it, it helps to look at the signal in the time and the frequency domain. To represent the sampled signal in the time domain we will show a string of impulse functions whose area represents the magnitude of the signal at that point. While there is actually no impulse function physically in the system it is a useful mathematical construct to look at what is happening in the frequency domain.

[[Image:barnsasample.jpg|Sampling a signal]]
[[Image:barnsaDA.jpg|Digital to analog conversion]]

This shows the progression of your signal both in the time and frequency domains, as it is sampled (figure 1), and then ran through the D-to-A converter (figure 2). You will notice that the frequency representation of the stair step wave does not look like the original signal. It is too tall in the middle and drops off too fast of the edges; it also has too many high frequency (both positive and negative) components that aren't wanted. These high frequencies components need to be filtered off (which we stated above, the equivalent of smoothing out the stair step function in time) but you would still have to problem of the signal being too tall in the middle and dropping off too fast. In addition in order to filter out the high frequency parts and leave your signal unchanged you need a low pass filter with a very very sharp edge. This is a high precision and very spendy component that we would like to eliminate the need for.
We have two basic problems, the distorted signal, and the need to filter out the high frequency components. To fix this we need to run the signal through some sort of filter that will both pre-distort the signal (increase its wanted high frequency components) and flatten out the unwanted high frequency components that are very close to the signal. This leads us into oversampling.

==Over Sampling==
The basic idea behind oversampling is to run the signal through an interpolation filter that will effectively double (or triple, or quadruple, etc) the number of points (frequency) in the time signal. Suppose that there is a signal:
0 @ t=0
4 @ t=T
8 @ t=2T
The filter would guess at what the signal would be at 1/2T and 3/2T so the signal would become something like:
0 @ t=0
2 @ t=1/2T
4 @ t=T
6 @ t=3/2T
8 @ t=T
This is a very simple case, in actuality the filter would take a preset linear combination of the data points before and after to determine what the inserted value should be. Mathematically this is a convolution of the signal from the CD with a function of the form:

Insert Math

To see why this solves the problems stated above we need to look at the frequency domain. In the frequency domain we needed to filter out the unwanted higher frequencies to get our original signal back. To make this easier it would be nice to eliminate the frequencies that are really close to the signal you are trying to keep, and at the same time we wanted to pre-emphasize the high frequencies of our signal. This would mean multiplying the in frequency by a domain by a filtering function as shown below.

Insert Picture

Multiplication in the frequency domain corresponds to a convolution in the time domain, and the function that we used above in the frequency domain corresponds to a string of impulses in time. Convolving by these impulses in time is the over sampling described above.
This makes since because we have effectively doubled the frequency in the time domain which will space things out in the frequency domain and that is exactly what we see.

Principle author of this page: [[User:Barnsa|Sam Barnes]]

The following is only included for my sake, so I can look at the syntax.
==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

CD Player

2004-11-17T18:58:23Z

Barnsa:

A CD is a digitally recorded sound wave. The CD has a whole bunch of numbers on it. These numbers represent the magnitude of the sound wave at certain points in time. If your sound was a sine wave for example and you looked at it graphically you would see a smooth wave going up and down. A CD of this wave would just be the magnitude, or height, of that wave every so often.
The basic idea in turning these numbers into a sound wave you can here is to run them through a digital to analog converter. The D-to-A will read in a value then output and hold a corresponding voltage until it gets the next value, then it will output and hold that voltage until the next one, etc. The end result will be a stair step representation of the original wave similar to the one shown below.

Insert Image

You can then run this through a low pass filter to smooth it out a little and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

To get a better idea of why the signal becomes distorted and how to correct it, it helps to look at the signal in the time and the frequency domain. To represent the sampled signal in the time domain we will show a string of impulse functions whose area represents the magnitude of the signal at that point. While there is actually no impulse function physically in the system it is a useful mathematical construct to look at what is happening in the frequency domain.

[[Image:barnsasample.jpg|thumb|Sampling a signal]]
[[Image:barnsada.jpg|thumb|Digital to analog conversion]]

This shows the progression of your signal both in the time and frequency domains, as it is sampled (figure 1), and then ran through the D-to-A converter (figure 2). You will notice that the frequency representation of the stair step wave does not look like the original signal. It is too tall in the middle and drops off too fast of the edges; it also has too many high frequency (both positive and negative) components that aren't wanted. These high frequencies components need to be filtered off (which we stated above, the equivalent of smoothing out the stair step function in time) but you would still have to problem of the signal being too tall in the middle and dropping off too fast. In addition in order to filter out the high frequency parts and leave your signal unchanged you need a low pass filter with a very very sharp edge. This is a high precision and very spendy component that we would like to eliminate the need for.
We have two basic problems, the distorted signal, and the need to filter out the high frequency components. To fix this we need to run the signal through some sort of filter that will both pre-distort the signal (increase its wanted high frequency components) and flatten out the unwanted high frequency components that are very close to the signal. This leads us into oversampling.

==Over Sampling==
The basic idea behind oversampling is to run the signal through an interpolation filter that will effectively double (or triple, or quadruple, etc) the number of points (frequency) in the time signal. Suppose that there is a signal:
0 @ t=0
4 @ t=T
8 @ t=2T
The filter would guess at what the signal would be at 1/2T and 3/2T so the signal would become something like:
0 @ t=0
2 @ t=1/2T
4 @ t=T
6 @ t=3/2T
8 @ t=T
This is a very simple case, in actuality the filter would take a preset linear combination of the data points before and after to determine what the inserted value should be. Mathematically this is a convolution of the signal from the CD with a function of the form:

Insert Math

To see why this solves the problems stated above we need to look at the frequency domain. In the frequency domain we needed to filter out the unwanted higher frequencies to get our original signal back. To make this easier it would be nice to eliminate the frequencies that are really close to the signal you are trying to keep, and at the same time we wanted to pre-emphasize the high frequencies of our signal. This would mean multiplying the in frequency by a domain by a filtering function as shown below.

Insert Picture

Multiplication in the frequency domain corresponds to a convolution in the time domain, and the function that we used above in the frequency domain corresponds to a string of impulses in time. Convolving by these impulses in time is the over sampling described above.
This makes since because we have effectively doubled the frequency in the time domain which will space things out in the frequency domain and that is exactly what we see.

Principle author of this page: [[User:Barnsa|Sam Barnes]]

The following is only included for my sake, so I can look at the syntax.
==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

CD Player

2004-11-17T18:57:38Z

Barnsa:

A CD is a digitally recorded sound wave. The CD has a whole bunch of numbers on it. These numbers represent the magnitude of the sound wave at certain points in time. If your sound was a sine wave for example and you looked at it graphically you would see a smooth wave going up and down. A CD of this wave would just be the magnitude, or height, of that wave every so often.
The basic idea in turning these numbers into a sound wave you can here is to run them through a digital to analog converter. The D-to-A will read in a value then output and hold a corresponding voltage until it gets the next value, then it will output and hold that voltage until the next one, etc. The end result will be a stair step representation of the original wave similar to the one shown below.

Insert Image

You can then run this through a low pass filter to smooth it out a little and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

To get a better idea of why the signal becomes distorted and how to correct it, it helps to look at the signal in the time and the frequency domain. To represent the sampled signal in the time domain we will show a string of impulse functions whose area represents the magnitude of the signal at that point. While there is actually no impulse function physically in the system it is a useful mathematical construct to look at what is happening in the frequency domain.

[[Image:barnsasample.jpg|Sampling a signal]]
[[Image:barnsada.jpg|Digital to analog conversion]]

This shows the progression of your signal both in the time and frequency domains, as it is sampled (figure 1), and then ran through the D-to-A converter (figure 2). You will notice that the frequency representation of the stair step wave does not look like the original signal. It is too tall in the middle and drops off too fast of the edges; it also has too many high frequency (both positive and negative) components that aren't wanted. These high frequencies components need to be filtered off (which we stated above, the equivalent of smoothing out the stair step function in time) but you would still have to problem of the signal being too tall in the middle and dropping off too fast. In addition in order to filter out the high frequency parts and leave your signal unchanged you need a low pass filter with a very very sharp edge. This is a high precision and very spendy component that we would like to eliminate the need for.
We have two basic problems, the distorted signal, and the need to filter out the high frequency components. To fix this we need to run the signal through some sort of filter that will both pre-distort the signal (increase its wanted high frequency components) and flatten out the unwanted high frequency components that are very close to the signal. This leads us into oversampling.

==Over Sampling==
The basic idea behind oversampling is to run the signal through an interpolation filter that will effectively double (or triple, or quadruple, etc) the number of points (frequency) in the time signal. Suppose that there is a signal:
0 @ t=0
4 @ t=T
8 @ t=2T
The filter would guess at what the signal would be at 1/2T and 3/2T so the signal would become something like:
0 @ t=0
2 @ t=1/2T
4 @ t=T
6 @ t=3/2T
8 @ t=T
This is a very simple case, in actuality the filter would take a preset linear combination of the data points before and after to determine what the inserted value should be. Mathematically this is a convolution of the signal from the CD with a function of the form:

Insert Math

To see why this solves the problems stated above we need to look at the frequency domain. In the frequency domain we needed to filter out the unwanted higher frequencies to get our original signal back. To make this easier it would be nice to eliminate the frequencies that are really close to the signal you are trying to keep, and at the same time we wanted to pre-emphasize the high frequencies of our signal. This would mean multiplying the in frequency by a domain by a filtering function as shown below.

Insert Picture

Multiplication in the frequency domain corresponds to a convolution in the time domain, and the function that we used above in the frequency domain corresponds to a string of impulses in time. Convolving by these impulses in time is the over sampling described above.
This makes since because we have effectively doubled the frequency in the time domain which will space things out in the frequency domain and that is exactly what we see.

Principle author of this page: [[User:Barnsa|Sam Barnes]]

The following is only included for my sake, so I can look at the syntax.
==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

File:Barnsasample.jpg

2004-11-17T18:53:07Z

Barnsa:

File:Barnsapredistort.jpg

2004-11-17T18:52:53Z

Barnsa:

File:BarnsaDA.jpg

2004-11-17T18:52:04Z

Barnsa:

CD Player

2004-11-17T18:49:30Z

Barnsa:

A CD is a digitally recorded sound wave. The CD has a whole bunch of numbers on it. These numbers represent the magnitude of the sound wave at certain points in time. If your sound was a sine wave for example and you looked at it graphically you would see a smooth wave going up and down. A CD of this wave would just be the magnitude, or height, of that wave every so often.
The basic idea in turning these numbers into a sound wave you can here is to run them through a digital to analog converter. The D-to-A will read in a value then output and hold a corresponding voltage until it gets the next value, then it will output and hold that voltage until the next one, etc. The end result will be a stair step representation of the original wave similar to the one shown below.

Insert Image

You can then run this through a low pass filter to smooth it out a little and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

To get a better idea of why the signal becomes distorted and how to correct it, it helps to look at the signal in the time and the frequency domain. To represent the sampled signal in the time domain we will show a string of impulse functions whose area represents the magnitude of the signal at that point. While there is actually no impulse function physically in the system it is a useful mathematical construct to look at what is happening in the frequency domain.

Insert Image

This shows the progression of your signal both in the time and frequency domains, as it is sampled (figure 1), and then ran through the D-to-A converter (figure 2). You will notice that the frequency representation of the stair step wave does not look like the original signal. It is too tall in the middle and drops off too fast of the edges; it also has too many high frequency (both positive and negative) components that aren't wanted. These high frequencies components need to be filtered off (which we stated above, the equivalent of smoothing out the stair step function in time) but you would still have to problem of the signal being too tall in the middle and dropping off too fast. In addition in order to filter out the high frequency parts and leave your signal unchanged you need a low pass filter with a very very sharp edge. This is a high precision and very spendy component that we would like to eliminate the need for.
We have two basic problems, the distorted signal, and the need to filter out the high frequency components. To fix this we need to run the signal through some sort of filter that will both pre-distort the signal (increase its wanted high frequency components) and flatten out the unwanted high frequency components that are very close to the signal. This leads us into oversampling.

==Over Sampling==
The basic idea behind oversampling is to run the signal through an interpolation filter that will effectively double (or triple, or quadruple, etc) the number of points (frequency) in the time signal. Suppose that there is a signal:
0 @ t=0
4 @ t=T
8 @ t=2T
The filter would guess at what the signal would be at 1/2T and 3/2T so the signal would become something like:
0 @ t=0
2 @ t=1/2T
4 @ t=T
6 @ t=3/2T
8 @ t=T
This is a very simple case, in actuality the filter would take a preset linear combination of the data points before and after to determine what the inserted value should be. Mathematically this is a convolution of the signal from the CD with a function of the form:

Insert Math

To see why this solves the problems stated above we need to look at the frequency domain. In the frequency domain we needed to filter out the unwanted higher frequencies to get our original signal back. To make this easier it would be nice to eliminate the frequencies that are really close to the signal you are trying to keep, and at the same time we wanted to pre-emphasize the high frequencies of our signal. This would mean multiplying the in frequency by a domain by a filtering function as shown below.

Insert Picture

Multiplication in the frequency domain corresponds to a convolution in the time domain, and the function that we used above in the frequency domain corresponds to a string of impulses in time. Convolving by these impulses in time is the over sampling described above.
This makes since because we have effectively doubled the frequency in the time domain which will space things out in the frequency domain and that is exactly what we see.

The following is only included for my sake, so I can look at the syntax.
==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

Principle author of this page: [[User:Goeari|Aric Goe]]

CD Player

2004-11-16T08:19:57Z

Barnsa: Mathematics of a CD player

A CD is a digitally recorded sound wave. Since the CD is digital it must have definite numerical values for the magnitude of that wave at any given time, and since there is only a finite amount of storage on a CD those definite values only change every so often. So if you were to look at a what is stored on the CD you would see a stair stepping function similar to the one below.

Insert Image

The basic idea of turning this stair step into the smooth sound wave that it used to be is to run it through a digital to analog converter then through a low pass filter and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

Insert Image

This shows the progression of your signal both in the time and frequency domains. You will notice that the frequency representation of your stair step wave, which is stored on the CD, does not look like the original signal. It is too tall in the middle and drops off too fast of the edges, it also has too many high frequency (both positive and negative) components that aren't wanted. The basic idea to fix this is to pre-distort the signal by multiplying by a function that would increase the edges of the signal and shrink the middle so that when it gets distorted it ends up being what you want. But there is also the problem of filtering out the unwanted high frequencies. To do this we would need a low pass filter with a very very sharp edge, this is an expensive piece of electronics and we would like to eliminate it if we could, and as I'm sure you guessed, we can.
==Over Sampling==

The following is only included for my sake, so I can look at the syntax.
==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

Principle author of this page: [[User:Goeari|Aric Goe]]

CD Player

2004-11-16T08:16:55Z

Barnsa:

A CD is a digitally recorded sound wave. Since the CD is digital it must have definite numerical values for the magnitude of that wave at any given time, and since there is only a finite amount of storage on a CD those definite values only change every so often. So if you were to look at a what is stored on the CD you would see a stair stepping function similar to the one below.

Insert Image

The basic idea of turning this stair step into the smooth sound wave that it used to be is to run it through a digital to analog converter then through a low pass filter and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

Insert Image

This shows the progression of your signal both in the time and frequency domains. You will notice that the frequency representation of your stair step wave, which is stored on the CD, does not look like the original signal. It is too tall in the middle and drops off too fast of the edges. It also how many high frequency (both positive and negative) components that aren't wanted. The basic idea to fix this is to predistort the signal by multplying by a function that would increase the edges of the signal and shrink the middle so that when it gets distorted it ends up being what you want. But there is also the problem of filtering out the unwanted high frequencies. To do this we would need a low pass filter with a very very sharp edge, this is an expensive piece of electronics and we would like to eliminate it if we could, and as I'm sure you guessed, we can.
==Oversampling==

The following is only included for my sake, so I can look at the syntax.
==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

Principle author of this page: [[User:Goeari|Aric Goe]]

CD Player

2004-11-16T08:16:00Z

Barnsa:

A CD is a digitally recorded sound wave. Since the CD is digital it must have definite numerical values for the magnitude of that wave at any given time, and since there is only a finite amount of storage on a CD those definite values only change every so often. So if you were to look at a what is stored on the CD you would see a stair stepping function similar to the one below.

Insert Image

The basic idea of turning this stair step into the smooth sound wave that it used to be is to run it through a digital to analog converter then through a low pass filter and then out to a speaker. While this would work your signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few different ways to compensate for this.

=====Signal in Time and Frequency=====

This shows the progression of your signal both in the time and frequency domains. You will notice that the frequency representation of your stair step wave, which is stored on the CD, does not look like the original signal. It is too tall in the middle and drops off too fast of the edges. It also how many high frequency (both positive and negative) components that aren't wanted. The basic idea to fix this is to predistort the signal by multplying by a function that would increase the edges of the signal and shrink the middle so that when it gets distorted it ends up being what you want. But there is also the problem of filtering out the unwanted high frequencies. To do this we would need a low pass filter with a very very sharp edge, this is an expensive piece of electronics and we would like to eliminate it if we could, and as I'm sure you guessed, we can.
==Oversampling==

The following is only included for my sake, so I can look at the syntax.
==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

Principle author of this page: [[User:Goeari|Aric Goe]]

CD Player

2004-11-16T08:02:20Z

Barnsa:

A CD is a digitally recorded sound wave. Since the CD is digital it must have deffinate numerical values for the magnitude of that wave at any given time, and since there is only a finite amout of storage on a CD those deffinite values only change every so often. So if you were to look at a what is stored on the CD you would see a stair stepping function similar to the one below.

Insert Image

The basic idea of turning this stair step into the smooth sound wave that it used to be is to run it through a digital to analog converter then through a low pass filter WHY?? and then out to a speaker. While this would work you signal would be substantially distorted due to its conversion to and then from a digital signal. We will look at a few diferent ways to compensate for this.

The following is only included for my sake, so I can look at the syntax.
==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

Principle author of this page: [[User:Goeari|Aric Goe]]

User:Barnsa

2004-11-16T07:54:16Z

Barnsa:

[[Image:Sam.JPG|thumb|Sam's photo]]
This is Sam's Page. Yeeehaaaa

[[CD Player]]

CD Player Exlanation

2004-11-16T07:53:42Z

Barnsa:

CD Player Exlanation

2004-11-16T07:47:27Z

Barnsa:

This will contain my explataion on how a CD player works. Right now I am just trying to figure out the syntax of this program.

==The Fourier Series==
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.
<center>
<math> f(t) = \sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T} </math>.
</center>

see also:[[Orthogonal Functions]]

Principle author of this page: [[User:Goeari|Aric Goe]]

CD Player Exlanation

2004-11-16T07:45:05Z

Barnsa:

This will contain my explataion on how a CD player works. Right now I am just trying to figure out the syntax of this program.

User:Barnsa

2004-11-16T07:44:23Z

Barnsa:

[[Image:Sam.JPG|thumb|Sam's photo]]
This is Sam's Page. Yeeehaaaa

[[CD Player Exlanation]]

User:Barnsa

2004-11-16T07:42:32Z

Barnsa:

[[Image:sam.jpg|thumb|Sam's photo]]
This is Sam's Page. Yeeehaaaa
[[CD Player Exlanation]]

Signals and Systems

2004-11-16T07:40:57Z

Barnsa:

[http://www.wwc.edu/~frohro/ClassNotes/engr455index.htm Class notes for Signals & Systems]

== Topics ==
*[[Orthogonal functions]]
*[[Fourier series]]
*[[Fourier transform]]
*[[Sampling]]
*[[Discrete Fourier transform]]

I couldn't figure out how to get to others Users pages easily so I decided to start posting them here, please add yours:

[[User:Frohro|Rob Frohne]]

[[User:Barnsa|Sam Barnes]]

[[User:Santsh|Shawn Santana]]

Signals and Systems

2004-11-16T07:40:39Z

Barnsa:

[http://www.wwc.edu/~frohro/ClassNotes/engr455index.htm Class notes for Signals & Systems]

== Topics ==
*[[Orthogonal functions]]
*[[Fourier series]]
*[[Fourier transform]]
*[[Sampling]]
*[[Discrete Fourier transform]]

I couldn't figure out how to get to others Users pages easily so I decided to start posting them here, please add yours:

[[User:Frohro|Rob Frohne]]

[[User:Barnsa|Sam Barnes]]

[[User:Santsh|Shawn Santana]]

Signals and Systems

2004-11-16T07:39:40Z

Barnsa:

[http://www.wwc.edu/~frohro/ClassNotes/engr455index.htm Class notes for Signals & Systems]

== Topics ==
*[[Orthogonal functions]]
*[[Fourier series]]
*[[Fourier transform]]
*[[Sampling]]
*[[Discrete Fourier transform]]

I couldn't figure out how to get to others Users pages easily so I decided to start posting them here, please add yours:
[[User:Frohro|Rob Frohne]]
[[User:Barnsa|Sam Barnes]]
[[User:Santsh|Shawn Santana]]

Signals and Systems

2004-11-16T07:30:46Z

Barnsa:

[http://www.wwc.edu/~frohro/ClassNotes/engr455index.htm Class notes for Signals & Systems]

== Topics ==
*[[Orthogonal functions]]
*[[Fourier series]]
*[[Fourier transform]]
*[[Sampling]]
*[[Discrete Fourier transform]]

*[[Sam's CD-Player Explanation]]

User:Barnsa

2004-09-29T05:41:15Z

Barnsa:

[[Image:sam.JPG]]
This is Sam's Page. Yeeehaaaa

File:Sam.JPG

2004-09-29T05:40:52Z

Barnsa:

User:Barnsa

2004-09-29T05:40:18Z

Barnsa:

[[Image:me.JPG]]
This is Sam's Page. Yeeehaaaa

File:P1010008.JPG

2004-09-29T05:37:34Z

Barnsa: