Class Wiki - User contributions [en]

CD Player

2005-11-08T19:27:29Z

Greg: /* Signal in Time and Frequency */

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

2005-11-08T19:27:21Z

Greg: /* Signal in Time and Frequency */

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]]greg

'''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]]

Sampling

2005-11-08T19:24:00Z

Greg: /* Sampling Overview */

==Introduction==
In this chapter we will look at sampling and how it affects signal processing. Sampling is the process of taking a continuous stream of data with infinite points of resolution and getting that data into a form that can be stored in a finite data set.

==Sampling Overview==
Imagine that we have a song that is being played in a studio. We would like to record the song and store it on a computer in form that will allow the song to be played back to sound as much like the original as possible. However, the song being played has an infinite set of points when it is played. We must throw out much of this data and yet keep enough to reconstruct the song. This is where sampling comes in. By taking evenly spaced out samples of the data stream, we can later reconstruct the stream of data. The sampler converts the continuous time signal x(t) into a discrete-time sequence x(n) by taking the values of x(t) at integer multiples of the sampling perioud, T. Typically, discrete signals are formed by periodically sampling a continuous time signal using the form:

<math> \bold x(n) = x_a(n*T_s) </math>

[[Image:Signals_Sampling.JPG]]

There are, however, a few things that we need to keep in mind when trying to sample signals with many frequencies in them. When we sample data, we want to be able to sample enough data to get a correct reconstruction (this is discussed further below), but we don’t want to sample too much data that our sets become unmanageable.

==Sampling Theory==
According to Sampling Theory, you must sample fast enough that you eliminate all possible aliases that can occur from sampling. Aliasing is what happens when you sample a signal that can later be reconstructed with two possible frequencies. This can be seen by the picture shown below. Another example of this can be seen in older western movies when watching the spoked wheels of a stagecoach rotate backward at a slow rate. This happens when each frame of film is taken at a slightly faster rate than that of the rotating wheel. In order to sample fast enough to eliminate all possible aliases, you must sample at a rate greater than twice the maximum frequency found in the signal to be sampled.

[[Image:Sampling_pic1.JPG]]

Going back to our example of audio, we know that humans can hear from about 20 Hz up to about 20,000 Hz. Therefore we know that we want to make sure that we sample fast enough that we will get this entire frequency spectrum. In order to do this, we apply the principle where F is our highest frequency and T is our sampling rate to get a value of T = 40,000 Hz. This is also known as the Nyquist Theorem. This explains why audio that is recorded onto a CD is sampled at a rate of 44,100 Hz (they sample slightly higher than necessary for a 20 KHz max frequency).
Something to note here is the fact that even though we want to sample our signal for frequencies between 20 and 20,000 Hz, most signals have many more frequencies in them due to white noise. In order to make sure that only the desired frequency is sampled you can run your signal through a pre-filter (sometimes called an anti-aliasing filter). This guarantees that the sampled data system receives analog signals having a frequency spectrum no greater than those frequencies allowed by the filter.

To see how the sampled data is reconstructed, refer to my section on how a [[User:santsh|CD Player]] works.

==Sampling Process==

In order to see how the sampling process works, imagine a periodic sequence of impulses,

<math> s_a(t) = \sum_{n=-\infty}^\infty \delta(t-nT_s) </math>

Now multiply your signal by the impulses to form the sampled signal

<math> x_s(t) = x_a(t) * s_a(t) = \sum_{n=-\infty}^\infty x_a(nT_s)*\delta(t-nT_s) </math>

After which the sampled signal is converted into a discrete-time signal by mapping the impulses that are spaced in time by <math> T_s </math> into a sequence x(n) where the sample values are indexed by the integer variable n:

<math> \bold x(n) = x_a(nT_s) </math>

Principle author of this page: [[User:santsh|Shawn Santana]]

Sampling

2005-11-08T19:23:50Z

Greg: /* Sampling Overview */

==Introduction==
In this chapter we will look at sampling and how it affects signal processing. Sampling is the process of taking a continuous stream of data with infinite points of resolution and getting that data into a form that can be stored in a finite data set.

==Sampling Overview==
Imagine that we have a song that is being played in a studio. We would like to record the song and store it on a computer in form that will allow the song to be played back to sound as much like the original as possible. However, the song being played has an infinite set of points when it is played. We must throw out much of this data and yet keep enough to reconstruct the song. This is where sampling comes in. By taking evenly spaced out samples of the data stream, we can later reconstruct the stream of data. The sampler converts the continuous time signal x(t) into a discrete-time sequence x(n) by taking the values of x(t) at integer multiples of the sampling perioud, T. Typically, discrete signals are formed by periodically sampling a continuous time signal using the form:

<math> \bold x(n) = x_a(n*T_s) </math>

[[Image:Signals_Sampling.JPG]]greg

There are, however, a few things that we need to keep in mind when trying to sample signals with many frequencies in them. When we sample data, we want to be able to sample enough data to get a correct reconstruction (this is discussed further below), but we don’t want to sample too much data that our sets become unmanageable.

==Sampling Theory==
According to Sampling Theory, you must sample fast enough that you eliminate all possible aliases that can occur from sampling. Aliasing is what happens when you sample a signal that can later be reconstructed with two possible frequencies. This can be seen by the picture shown below. Another example of this can be seen in older western movies when watching the spoked wheels of a stagecoach rotate backward at a slow rate. This happens when each frame of film is taken at a slightly faster rate than that of the rotating wheel. In order to sample fast enough to eliminate all possible aliases, you must sample at a rate greater than twice the maximum frequency found in the signal to be sampled.

[[Image:Sampling_pic1.JPG]]

Going back to our example of audio, we know that humans can hear from about 20 Hz up to about 20,000 Hz. Therefore we know that we want to make sure that we sample fast enough that we will get this entire frequency spectrum. In order to do this, we apply the principle where F is our highest frequency and T is our sampling rate to get a value of T = 40,000 Hz. This is also known as the Nyquist Theorem. This explains why audio that is recorded onto a CD is sampled at a rate of 44,100 Hz (they sample slightly higher than necessary for a 20 KHz max frequency).
Something to note here is the fact that even though we want to sample our signal for frequencies between 20 and 20,000 Hz, most signals have many more frequencies in them due to white noise. In order to make sure that only the desired frequency is sampled you can run your signal through a pre-filter (sometimes called an anti-aliasing filter). This guarantees that the sampled data system receives analog signals having a frequency spectrum no greater than those frequencies allowed by the filter.

To see how the sampled data is reconstructed, refer to my section on how a [[User:santsh|CD Player]] works.

==Sampling Process==

In order to see how the sampling process works, imagine a periodic sequence of impulses,

<math> s_a(t) = \sum_{n=-\infty}^\infty \delta(t-nT_s) </math>

Now multiply your signal by the impulses to form the sampled signal

<math> x_s(t) = x_a(t) * s_a(t) = \sum_{n=-\infty}^\infty x_a(nT_s)*\delta(t-nT_s) </math>

After which the sampled signal is converted into a discrete-time signal by mapping the impulses that are spaced in time by <math> T_s </math> into a sequence x(n) where the sample values are indexed by the integer variable n:

<math> \bold x(n) = x_a(nT_s) </math>

Principle author of this page: [[User:santsh|Shawn Santana]]

Fourier series - by Ray Betz

2005-11-08T19:23:15Z

Greg: /* Linear Systems */

==Fourier Series==
If
# <math> x(t) = x(t + T)</math>
# Dirichlet conditions are satisfied
then we can write
<center>
<math> \bold x(t) = \sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T}</math>
</center>
The above equation is called the complex fourier series. Given <math>x(t)</math>, we may determine <math> \alpha_k </math> by taking the [[inner product]] of <math>\alpha_k</math> with <math>x(t)</math>.
Let us assume a solution for <math>\alpha_k</math> of the form <math>e^ \frac {j 2 \pi n t}{T}</math>. Now we take the inner product of <math>\alpha_k</math> with <math>x(t)</math> over the interval of one period, <math> T </math>.
<math> <\alpha_k|x(t)> = <e^ \frac {j 2 \pi n t}{T}|\sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T}> </math>
<math>= \int_{-\frac{T}{2}}^\frac{T}{2} x(t)e^ \frac {-j 2 \pi n t}{T} dt </math>
<math>= \int_{-\frac{T}{2}}^\frac{T}{2} \sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T}e^ \frac {-j 2 \pi n t}{T} dt </math>
<math>= \sum_{k=-\infty}^\infty \alpha_k \int_{-\frac{T}{2}}^\frac{T}{2} e^ \frac {j 2 \pi (k-n) t}{T} dt </math>

If <math>k=n</math> then,

<math> \int_{-\frac{T}{2}}^\frac{T}{2} e^ \frac {j 2 \pi (k-n) t}{T} dt = \int_{-\frac{T}{2}}^\frac{T}{2} 1 dt = T</math>

If <math>k \ne n </math> then,

<math> \int_{-\frac{T}{2}}^\frac{T}{2} e^ \frac {j 2 \pi (k-n) t}{T} dt = 0 </math>

We can simplify the above two conclusion into one equation. (What is the [[delta function]] below?)

<math> \sum_{k=-\infty}^\infty \alpha_k \int_{-\frac{T}{2}}^\frac{T}{2} e^ \frac {j 2 \pi (k-n) t}{T} dt = \sum_{k=-\infty}^\infty T \delta_{k,n} \alpha_k = T \alpha_n </math>

So, we conclude
<math>\alpha_n = \frac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2} x(t) e^ \frac {-j 2 \pi n t}{T} dt </math>

==Orthogonal Functions==

The function <math> y_n(t) </math> and <math> y_m(t) </math> are orthogonal on <math> (a,b) </math> if and only if <math> <y_n(t)|y_m(t)> = \int_{a}^{b} y_n^*(t)y_m(t) dt = 0 </math>.

The set of functions are orthonormal if and only if <math> <y_n(t)|y_m(t)> = \int_{a}^{b} y_n^*(t)y_m(t) dt = \delta_{m,n} </math>.

==Linear Systems==

Let us say we have a linear time invarient system, where <math> x(t) </math> is the input and <math> y(t) </math> is the output. What outputs do we get as we put different inputs into this system?
[[Image:Linear_System.JPG]]

If we put in an impulse response, <math> \delta(t)</math>, then we get out <math>h(t)</math>. What would happen if we put a time delayed impulse signal, <math> \delta(t-u)</math>, into the system? The output response would be a time delayed <math>h(t)</math>, or <math>h(t-u)</math>, because the system is time invarient. So, no matter when we put in our signal the response would come out the same (just time delayed).

What if we now multiplied our impulse by a coefficient? Since our system is linear, the proportionality property applies. If we put <math> x(u)\delta(t-u)</math> into our system then we should get out <math>x(u)h(t-u)</math>.

By the superposition property(because we have a linear system) we may put into the system the integral of <math> x(u)\delta(t-u)</math> and we would get out <math> \int_{-\infty}^\infty x(u)h(t-u) du</math>. What would we get if we put <math> e^{j 2 \pi f t} </math> into our system? We could find out by plugging <math> e^{j 2 \pi f t} </math> in for <math> x(u) </math> in the integral that we just found the output for above. If we do a change of variables (<math> v = t-u </math>, and <math> dv = -du </math>) we get <math> \int_{-\infty}^\infty x(u)h(t-u) du = \int_{-\infty}^\infty e^{j 2 \pi f t} h(t-u) du = -\int_{\infty}^{-\infty} e^{j 2 \pi f (t-v)} h(v) dv = e^{j 2 \pi f t} \int_{-\infty}^\infty h(v)e^{-j 2 \pi f v} dv</math>. By pulling <math> e^{j 2 \pi f t} </math> out of the integral and calling the remaining integral <math> B_k </math> we get <math> e^{j 2 \pi f t} B_k</math>.

{| style="width:600px; height:100px" border="1"
|-
| '''INPUT'''
| '''OUTPUT'''
| '''REASON'''
|-
| <math> \delta(t)</math>
| <math>h(t)</math>
| Given
|-
| <math> \delta(t-u)</math>
| <math>h(t-u)</math>
| Time Invarient
|-
| <math> x(u)\delta(t-u)</math>
| <math>x(u)h(t-u)</math>
| Proportionality
|-
|<math> \int_{-\infty}^\infty x(u)\delta(t-u) du</math>
|<math> \int_{-\infty}^\infty x(u)h(t-u) du</math>
|Superposition
|-
|<math> \int_{-\infty}^\infty e^{j 2 \pi f t} h(t-u) du</math>
|<math> e^{j 2 \pi f t} \int_{-\infty}^\infty e^{j 2 \pi v t} h(v) dv</math>
|Superposition
|-
|<math> e^{j 2 \pi f t} </math>
|<math> e^{j 2 \pi f t} B_k</math>
|Superposition (from above)
|}

==Fourier Series (indepth)==

I would like to take a closer look at <math> \alpha_k </math> in the Fourier Series. Hopefully this will provide a better understanding of <math> \alpha_k </math>.

We will seperate x(t) into three parts; where <math> \alpha_k </math> is negative, zero, and positive.
<math> \bold x(t) = \sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T} = \sum_{k=-\infty}^{-1} \alpha_k e^ \frac {j 2 \pi k t}{T} + \alpha_0 + \sum_{k=1}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T}</math>

Now, by substituting <math> n = -k </math> into the summation where <math> k </math> is negative and substituting <math> n = k </math> into the summation where <math> k </math> is positive we get:
<math> \sum_{n=1}^{\infty} \alpha_{-n} e^ \frac {-j 2 \pi n t}{T} + \alpha_0 + \sum_{n=1}^\infty \alpha_n e^ \frac {j 2 \pi n t}{T} </math>

Recall that <math>\alpha_n = \frac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2} x(u) e^ \frac {-j 2 \pi n t}{T} dt </math>

If <math> x(t) </math> is real, then <math> \alpha_n^* = \alpha_{-n} </math>. Let us assume that <math> x(t) </math> is real.

<math> x(t) = \alpha_0 +\sum_{n=1}^\infty (\alpha_n e^ \frac {j 2 \pi n t}{T} + \alpha_n^* e^ \frac {-j 2 \pi n t}{T}) </math>

Recall that <math> y + y^* = 2Re(y) </math> [[Here is further clarification on this property]]

So, we may write:

<math> x(t) = \alpha_0 +\sum_{n=1}^\infty 2Re(\alpha_n e^ \frac {j 2 \pi n t}{T}) </math>

==Fourier Transform==

Fourier transforms emerge because we want to be able to make Fourier expressions of non-periodic functions. We can take the limit of those non-periodic functions to get a fourier expression for the function.

Remember that:
<math>x(t)=x(t+T)= \sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T} = \sum_{k=-\infty}^\infty 1/T \int_{-\frac{T}{2}}^\frac{T}{2} x(u)e^ \frac {-j 2 \pi k u }{T} du e^ \frac {j 2 \pi k t}{T} </math>

So,
<math> \lim_{x \to \infty}x(t)= \int_{-\infty}^\infty (\int_{-\infty}^\infty x(u) e^{-j 2 \pi f u} du) e^{j 2 \pi f t} df</math>

From the above limit we define <math> x(t)</math> and <math> X(f) </math>.

<math> x(t) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^ {j 2 \pi f t} df</math>

<math> X(f) = \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^ {j 2 \pi f t} dt</math>

We can take the derivitive of <math> x(t) </math> and then put in terms of the reverse fourier transform.

<math> \frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi f X(f) e^ {j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]

</math>

What happens if we just shift the time of <math> x(t) </math>?

<math> x(t-t_0) = \int_{-\infty}^\infty X(f) e^{j 2 \pi f(t-t_0)} df = \int_{-\infty}^\infty e^{-j 2 \pi f t_0} X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[e^{-j 2 \pi f t_0} X(f)] </math>

In the same way, if we shift the frequency we get:

<math> X(f-f_0) = \int_{-\infty}^\infty x(t) e^{j 2 \pi (f-f_0)t} dt = \int_{-\infty}^\infty e^{-j 2 \pi t f_0} x(t) e^{j 2 \pi f t} df = \mathcal{F} [e^{-j 2 \pi t f_0} x(t)] </math>

What would be the Fourier transform of <math> cos(2 /pi f_0 t) x(t) </math>?

<
==CD Player==

Below is a diagram of how the information on a CD player is read and processed. As you can see the information on the CD is processed by the D/A converter and then sent through a low pass filter and on to the speaker. If you were recording sound, the sound would be captured through a microphone. Then, it should be sent through a low pass filter and onto the A/D converter and then it is ready to be put on the CD. Recording signals is essentially the reverse of the operation pictured below.

[[Image:CDsystem.jpg]]

'''In Time Domain:'''

Let's start with a signal <math> h(t) </math>, as shown below. In this signal there is an infinite amount of information. Obviously, we can't hold it all in a computer, but we could take samples every <math> T </math>. Lets do that by multiplying <math> h(t) </math> by <math> \sum_{n=-\infty}^\infty \delta (t-nT) </math>. Since the magnetude of our delta function is one, we get a series of delta functions that record the value of <math> h(t) </math> at intervals of <math> T </math>. This gives us a result that looks like: <math> h(t)\sum_{n=-\infty}^\infty \delta (t-nT) = \sum_{n=-\infty}^\infty x(nt) \delta (t-nT)</math>

'''In Frequency Domain:'''

In the frequency domain we start with <math> H(f) </math>. Now we are in frequency, so we must convolve instead of multiply like we did in the time domain. We would have to convolve <math> H(f) </math> with <math> \mathcal{F}[ \sum_{n=-\infty}^\infty \delta (t-nT) ]</math>.

Aside:<math> \mathcal{F}[ \sum_{n=-\infty}^\infty \delta (t-nT) ] = \int_{-\infty}^\infty \sum_{n=-\infty}^\infty \delta (t-nT) e^{j 2 \pi f t} dt = \sum_{n=-\infty}^\infty \int_{-\infty}^\infty \delta (t-nT) e^{j 2 \pi f t} dt = \sum_{n=-\infty}^\infty e^{j 2 \pi f n T}</math>

This result looks it could be a fourier series. We would like to get our result in terms of delta functions. As shown below, the periodic delta functions could be represented as a fourier series with coefficients <math> \alpha_m </math>.

<math> \sum_{n=-\infty}^\infty \delta (t-nT) = \sum_{m=-\infty}^\infty \alpha_m e^ {j 2 \pi m t} </math>

Now we can solve for <math> \alpha_m </math>.

<math> \alpha_m = \frac {1}{T} \int_{\frac{-T}{2}}^{\frac{T}{2}} \sum_{n=-\infty}^\infty \delta (t-nT) \frac {j 2 \pi m t}{T} dt = \frac {1}{T} \int_{\frac{-T}{2}}^{\frac{-T}{2}} \delta (t) \frac {j 2 \pi m t}{T} dt = \frac {1}{T} </math>

Since the only delta function within the integration limits is the delta function at <math> t=0 </math>, we can take out the summation and just leave one delta function. Then, evaluating the integral at <math> t=0 </math> we get <math> \frac{1}{T} </math>

<math> \sum_{n=-\infty}^\infty \delta (t-nT) = \sum_{n=-\infty}^\infty \frac {1}{T} e^ \frac {j 2 \pi k t}{T} </math>
<math> \mathcal{F} \sum_{n=-\infty}^\infty \delta (t-nT) = \mathcal{F} \sum_{n=-\infty}^\infty \frac {1}{T} e^ \frac {j 2 \pi k t}{T} = \sum_{n=-\infty}^\infty \frac {1}{T} \int_{-\infty}^\infty e^ \frac {j 2 \pi k t}{T} e^ {-j 2 \pi f t} dt= \frac {1}{T} \sum_{n=-\infty}^\infty \int_{-\infty}^\infty e^ {-j 2 \pi (f-\frac{m}{T} t} dt</math>

[[Image:barnsasample.jpg|Picture uploaded by Sam Barnes]]

[[Image:barnsaDA.jpg|Picture uploaded by Sam Barnes]]

Fourier series - by Ray Betz

2005-11-08T19:23:05Z

Greg: /* Linear Systems */

==Fourier Series==
If
# <math> x(t) = x(t + T)</math>
# Dirichlet conditions are satisfied
then we can write
<center>
<math> \bold x(t) = \sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T}</math>
</center>
The above equation is called the complex fourier series. Given <math>x(t)</math>, we may determine <math> \alpha_k </math> by taking the [[inner product]] of <math>\alpha_k</math> with <math>x(t)</math>.
Let us assume a solution for <math>\alpha_k</math> of the form <math>e^ \frac {j 2 \pi n t}{T}</math>. Now we take the inner product of <math>\alpha_k</math> with <math>x(t)</math> over the interval of one period, <math> T </math>.
<math> <\alpha_k|x(t)> = <e^ \frac {j 2 \pi n t}{T}|\sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T}> </math>
<math>= \int_{-\frac{T}{2}}^\frac{T}{2} x(t)e^ \frac {-j 2 \pi n t}{T} dt </math>
<math>= \int_{-\frac{T}{2}}^\frac{T}{2} \sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T}e^ \frac {-j 2 \pi n t}{T} dt </math>
<math>= \sum_{k=-\infty}^\infty \alpha_k \int_{-\frac{T}{2}}^\frac{T}{2} e^ \frac {j 2 \pi (k-n) t}{T} dt </math>

If <math>k=n</math> then,

<math> \int_{-\frac{T}{2}}^\frac{T}{2} e^ \frac {j 2 \pi (k-n) t}{T} dt = \int_{-\frac{T}{2}}^\frac{T}{2} 1 dt = T</math>

If <math>k \ne n </math> then,

<math> \int_{-\frac{T}{2}}^\frac{T}{2} e^ \frac {j 2 \pi (k-n) t}{T} dt = 0 </math>

We can simplify the above two conclusion into one equation. (What is the [[delta function]] below?)

<math> \sum_{k=-\infty}^\infty \alpha_k \int_{-\frac{T}{2}}^\frac{T}{2} e^ \frac {j 2 \pi (k-n) t}{T} dt = \sum_{k=-\infty}^\infty T \delta_{k,n} \alpha_k = T \alpha_n </math>

So, we conclude
<math>\alpha_n = \frac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2} x(t) e^ \frac {-j 2 \pi n t}{T} dt </math>

==Orthogonal Functions==

The function <math> y_n(t) </math> and <math> y_m(t) </math> are orthogonal on <math> (a,b) </math> if and only if <math> <y_n(t)|y_m(t)> = \int_{a}^{b} y_n^*(t)y_m(t) dt = 0 </math>.

The set of functions are orthonormal if and only if <math> <y_n(t)|y_m(t)> = \int_{a}^{b} y_n^*(t)y_m(t) dt = \delta_{m,n} </math>.

==Linear Systems==

Let us say we have a linear time invarient system, where <math> x(t) </math> is the input and <math> y(t) </math> is the output. What outputs do we get as we put different inputs into this system?
[[Image:Linear_System.JPG]]sdflkjjlkfsdljkfsdljk

If we put in an impulse response, <math> \delta(t)</math>, then we get out <math>h(t)</math>. What would happen if we put a time delayed impulse signal, <math> \delta(t-u)</math>, into the system? The output response would be a time delayed <math>h(t)</math>, or <math>h(t-u)</math>, because the system is time invarient. So, no matter when we put in our signal the response would come out the same (just time delayed).

What if we now multiplied our impulse by a coefficient? Since our system is linear, the proportionality property applies. If we put <math> x(u)\delta(t-u)</math> into our system then we should get out <math>x(u)h(t-u)</math>.

By the superposition property(because we have a linear system) we may put into the system the integral of <math> x(u)\delta(t-u)</math> and we would get out <math> \int_{-\infty}^\infty x(u)h(t-u) du</math>. What would we get if we put <math> e^{j 2 \pi f t} </math> into our system? We could find out by plugging <math> e^{j 2 \pi f t} </math> in for <math> x(u) </math> in the integral that we just found the output for above. If we do a change of variables (<math> v = t-u </math>, and <math> dv = -du </math>) we get <math> \int_{-\infty}^\infty x(u)h(t-u) du = \int_{-\infty}^\infty e^{j 2 \pi f t} h(t-u) du = -\int_{\infty}^{-\infty} e^{j 2 \pi f (t-v)} h(v) dv = e^{j 2 \pi f t} \int_{-\infty}^\infty h(v)e^{-j 2 \pi f v} dv</math>. By pulling <math> e^{j 2 \pi f t} </math> out of the integral and calling the remaining integral <math> B_k </math> we get <math> e^{j 2 \pi f t} B_k</math>.

{| style="width:600px; height:100px" border="1"
|-
| '''INPUT'''
| '''OUTPUT'''
| '''REASON'''
|-
| <math> \delta(t)</math>
| <math>h(t)</math>
| Given
|-
| <math> \delta(t-u)</math>
| <math>h(t-u)</math>
| Time Invarient
|-
| <math> x(u)\delta(t-u)</math>
| <math>x(u)h(t-u)</math>
| Proportionality
|-
|<math> \int_{-\infty}^\infty x(u)\delta(t-u) du</math>
|<math> \int_{-\infty}^\infty x(u)h(t-u) du</math>
|Superposition
|-
|<math> \int_{-\infty}^\infty e^{j 2 \pi f t} h(t-u) du</math>
|<math> e^{j 2 \pi f t} \int_{-\infty}^\infty e^{j 2 \pi v t} h(v) dv</math>
|Superposition
|-
|<math> e^{j 2 \pi f t} </math>
|<math> e^{j 2 \pi f t} B_k</math>
|Superposition (from above)
|}

==Fourier Series (indepth)==

I would like to take a closer look at <math> \alpha_k </math> in the Fourier Series. Hopefully this will provide a better understanding of <math> \alpha_k </math>.

We will seperate x(t) into three parts; where <math> \alpha_k </math> is negative, zero, and positive.
<math> \bold x(t) = \sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T} = \sum_{k=-\infty}^{-1} \alpha_k e^ \frac {j 2 \pi k t}{T} + \alpha_0 + \sum_{k=1}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T}</math>

Now, by substituting <math> n = -k </math> into the summation where <math> k </math> is negative and substituting <math> n = k </math> into the summation where <math> k </math> is positive we get:
<math> \sum_{n=1}^{\infty} \alpha_{-n} e^ \frac {-j 2 \pi n t}{T} + \alpha_0 + \sum_{n=1}^\infty \alpha_n e^ \frac {j 2 \pi n t}{T} </math>

Recall that <math>\alpha_n = \frac{1}{T}\int_{-\frac{T}{2}}^\frac{T}{2} x(u) e^ \frac {-j 2 \pi n t}{T} dt </math>

If <math> x(t) </math> is real, then <math> \alpha_n^* = \alpha_{-n} </math>. Let us assume that <math> x(t) </math> is real.

<math> x(t) = \alpha_0 +\sum_{n=1}^\infty (\alpha_n e^ \frac {j 2 \pi n t}{T} + \alpha_n^* e^ \frac {-j 2 \pi n t}{T}) </math>

Recall that <math> y + y^* = 2Re(y) </math> [[Here is further clarification on this property]]

So, we may write:

<math> x(t) = \alpha_0 +\sum_{n=1}^\infty 2Re(\alpha_n e^ \frac {j 2 \pi n t}{T}) </math>

==Fourier Transform==

Fourier transforms emerge because we want to be able to make Fourier expressions of non-periodic functions. We can take the limit of those non-periodic functions to get a fourier expression for the function.

Remember that:
<math>x(t)=x(t+T)= \sum_{k=-\infty}^\infty \alpha_k e^ \frac {j 2 \pi k t}{T} = \sum_{k=-\infty}^\infty 1/T \int_{-\frac{T}{2}}^\frac{T}{2} x(u)e^ \frac {-j 2 \pi k u }{T} du e^ \frac {j 2 \pi k t}{T} </math>

So,
<math> \lim_{x \to \infty}x(t)= \int_{-\infty}^\infty (\int_{-\infty}^\infty x(u) e^{-j 2 \pi f u} du) e^{j 2 \pi f t} df</math>

From the above limit we define <math> x(t)</math> and <math> X(f) </math>.

<math> x(t) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^ {j 2 \pi f t} df</math>

<math> X(f) = \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^ {j 2 \pi f t} dt</math>

We can take the derivitive of <math> x(t) </math> and then put in terms of the reverse fourier transform.

<math> \frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi f X(f) e^ {j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]

</math>

What happens if we just shift the time of <math> x(t) </math>?

<math> x(t-t_0) = \int_{-\infty}^\infty X(f) e^{j 2 \pi f(t-t_0)} df = \int_{-\infty}^\infty e^{-j 2 \pi f t_0} X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[e^{-j 2 \pi f t_0} X(f)] </math>

In the same way, if we shift the frequency we get:

<math> X(f-f_0) = \int_{-\infty}^\infty x(t) e^{j 2 \pi (f-f_0)t} dt = \int_{-\infty}^\infty e^{-j 2 \pi t f_0} x(t) e^{j 2 \pi f t} df = \mathcal{F} [e^{-j 2 \pi t f_0} x(t)] </math>

What would be the Fourier transform of <math> cos(2 /pi f_0 t) x(t) </math>?

<
==CD Player==

Below is a diagram of how the information on a CD player is read and processed. As you can see the information on the CD is processed by the D/A converter and then sent through a low pass filter and on to the speaker. If you were recording sound, the sound would be captured through a microphone. Then, it should be sent through a low pass filter and onto the A/D converter and then it is ready to be put on the CD. Recording signals is essentially the reverse of the operation pictured below.

[[Image:CDsystem.jpg]]

'''In Time Domain:'''

Let's start with a signal <math> h(t) </math>, as shown below. In this signal there is an infinite amount of information. Obviously, we can't hold it all in a computer, but we could take samples every <math> T </math>. Lets do that by multiplying <math> h(t) </math> by <math> \sum_{n=-\infty}^\infty \delta (t-nT) </math>. Since the magnetude of our delta function is one, we get a series of delta functions that record the value of <math> h(t) </math> at intervals of <math> T </math>. This gives us a result that looks like: <math> h(t)\sum_{n=-\infty}^\infty \delta (t-nT) = \sum_{n=-\infty}^\infty x(nt) \delta (t-nT)</math>

'''In Frequency Domain:'''

In the frequency domain we start with <math> H(f) </math>. Now we are in frequency, so we must convolve instead of multiply like we did in the time domain. We would have to convolve <math> H(f) </math> with <math> \mathcal{F}[ \sum_{n=-\infty}^\infty \delta (t-nT) ]</math>.

Aside:<math> \mathcal{F}[ \sum_{n=-\infty}^\infty \delta (t-nT) ] = \int_{-\infty}^\infty \sum_{n=-\infty}^\infty \delta (t-nT) e^{j 2 \pi f t} dt = \sum_{n=-\infty}^\infty \int_{-\infty}^\infty \delta (t-nT) e^{j 2 \pi f t} dt = \sum_{n=-\infty}^\infty e^{j 2 \pi f n T}</math>

This result looks it could be a fourier series. We would like to get our result in terms of delta functions. As shown below, the periodic delta functions could be represented as a fourier series with coefficients <math> \alpha_m </math>.

<math> \sum_{n=-\infty}^\infty \delta (t-nT) = \sum_{m=-\infty}^\infty \alpha_m e^ {j 2 \pi m t} </math>

Now we can solve for <math> \alpha_m </math>.

<math> \alpha_m = \frac {1}{T} \int_{\frac{-T}{2}}^{\frac{T}{2}} \sum_{n=-\infty}^\infty \delta (t-nT) \frac {j 2 \pi m t}{T} dt = \frac {1}{T} \int_{\frac{-T}{2}}^{\frac{-T}{2}} \delta (t) \frac {j 2 \pi m t}{T} dt = \frac {1}{T} </math>

Since the only delta function within the integration limits is the delta function at <math> t=0 </math>, we can take out the summation and just leave one delta function. Then, evaluating the integral at <math> t=0 </math> we get <math> \frac{1}{T} </math>

<math> \sum_{n=-\infty}^\infty \delta (t-nT) = \sum_{n=-\infty}^\infty \frac {1}{T} e^ \frac {j 2 \pi k t}{T} </math>
<math> \mathcal{F} \sum_{n=-\infty}^\infty \delta (t-nT) = \mathcal{F} \sum_{n=-\infty}^\infty \frac {1}{T} e^ \frac {j 2 \pi k t}{T} = \sum_{n=-\infty}^\infty \frac {1}{T} \int_{-\infty}^\infty e^ \frac {j 2 \pi k t}{T} e^ {-j 2 \pi f t} dt= \frac {1}{T} \sum_{n=-\infty}^\infty \int_{-\infty}^\infty e^ {-j 2 \pi (f-\frac{m}{T} t} dt</math>

[[Image:barnsasample.jpg|Picture uploaded by Sam Barnes]]

[[Image:barnsaDA.jpg|Picture uploaded by Sam Barnes]]

User:Wonoje

2005-11-08T19:21:25Z

Greg:

==Jeffrey Wonoprabowo==
Making a page about myself is the first assignment for the class; although it isn't exactly the easiest thing to do as I never know what to write for these things.

This is my fourth and final year at Walla Walla College. When I first came my concentration was in computer engineering. However, towards the end of my second year, I switched to Bioengineering and am currently applying to medical schools.

At this point I am very sad because I cannot tell you how a CD player works... All I can do is write about myself. So below you'll find a modified picture of the one I put up on my hall in Meske; and the only reason I put one up is cause the RAs were asked to do so... I did not just put up a poster of myself up for fun.

[[Image:JeffSS.jpg]]

User:Wonoje

2005-11-08T19:21:10Z

Greg:

User:Greg

2005-11-08T19:09:08Z

Greg:

[[Image:Greg.jpg]]

File:Greg.jpg

2005-11-08T18:58:30Z

Greg:

File:Greg.jpg

2005-11-08T18:56:01Z

Greg: My picture

My picture