View source for Fourier series - by Ray Betz

==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>.
<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.

<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 may 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.  

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 take the integral of <math> x(u)\delta(t-u)</math> and we 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, and dv = -du </math>) we get <math> \int_{-\infty}^\infty x(u)h(t-u) du</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_{k=1}^{\infty} \alpha_{-n} e^ \frac {-j 2 \pi n t}{T} + \alpha_0 + \sum_{k=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]]