Fourier series - by Ray Betz

Fourier Series

If

1. $x(t)=x(t+T)$
2. Dirichlet conditions are satisfied

then we can write

${\mathbf {x}}(t)=\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}$

The above equation is called the complex fourier series. Given $x(t)$, we may determine $\alpha _{k}$ by taking the inner product of $\alpha _{k}$ with $x(t)$. Let us assume a solution for $\alpha _{k}$ of the form $e^{\frac {j2\pi nt}{T}}$. Now we take the inner product of $\alpha _{k}$ with $x(t)$ over the interval of one period, $T$. $<\alpha _{k}|x(t)>=<e^{\frac {j2\pi nt}{T}}|\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}>$ $=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t)e^{\frac {-j2\pi nt}{T}}dt$ $=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}e^{\frac {-j2\pi nt}{T}}dt$ $=\sum _{k=-\infty }^{\infty }\alpha _{k}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt$

If $k=n$ then,

$\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}1dt=T$

If $k\neq n$ then,

$\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt=0$

We can simplify the above two conclusion into one equation. [[What is $\delta _{k,n}$?]]

$\sum _{k=-\infty }^{\infty }\alpha _{k}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{\frac {j2\pi (k-n)t}{T}}dt=\sum _{k=-\infty }^{\infty }T\delta _{k,n}\alpha _{k}=T\alpha _{n}$

So, we conclude $\alpha _{n}={\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t)e^{\frac {-j2\pi nt}{T}}dt$

Orthogonal Functions

The function $y_{n}(t)$ and $y_{m}(t)$ are orthogonal on $(a,b)$ if and only if $<y_{n}(t)|y_{m}(t)>=\int _{a}^{b}y_{n}^{*}(t)y_{m}(t)dt=0$.

The set of functions are orthonormal if and only if $<y_{n}(t)|y_{m}(t)>=\int _{a}^{b}y_{n}^{*}(t)y_{m}(t)dt=\delta _{m,n}$.

Linear Systems

Let us say we have a linear time invarient system, where $x(t)$ is the input and $y(t)$ is the output. What outputs do we get as we put different inputs into this system? File:Linear System.JPG

If we put in an impulse response, $\delta (t)$, then we get out $h(t)$. What would happen if we put a time delayed impulse signal ($\delta (t-u)$) into the system. The output response would be a time delayed $h(t)$, or $h(t-u)$, 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 $x(u)\delta (t-u)$ into our system then we should get out $x(u)h(t-u)$.

By the superposition property(because we have a linear system) we may take the integral of $x(u)\delta (t-u)$ and we get out $\int _{-\infty }^{\infty }x(u)h(t-u)du$. What would we get if we put $e^{j2\pi ft}$ into our system. We could find out by plugging $e^{j2\pi ft}$ in for $x(u)$ in the integral that we just found the output for above. If we do a change of variables ($v=t-u,anddv=-du$) we get $\int _{-\infty }^{\infty }x(u)h(t-u)du$. By pulling $e^{j2\pi ft}$ out of the integral and calling the remaining integral $B_{k}$ we get $e^{j2\pi ft}B_{k}$.

INPUT	OUTPUT	REASON
$\delta (t)$	$h(t)$	Given
$\delta (t-u)$	$h(t-u)$	Time Invarient
$x(u)\delta (t-u)$	$x(u)h(t-u)$	Proportionality
$\int _{-\infty }^{\infty }x(u)\delta (t-u)du$	$\int _{-\infty }^{\infty }x(u)h(t-u)du$	Superposition
$\int _{-\infty }^{\infty }e^{j2\pi ft}h(t-u)du$	$e^{j2\pi ft}\int _{-\infty }^{\infty }e^{j2\pi vt}h(v)dv$	Superposition
$e^{j2\pi ft}$	$e^{j2\pi ft}B_{k}$	Superposition (from above)

Fourier Series (indepth)

I would like to take a closer look at $\alpha _{k}$ in the Fourier Series. Hopefully this will provide a better understanding of $\alpha _{k}$.

We will seperate x(t) into three parts; where $\alpha _{k}$ is negative, zero, and positive. ${\mathbf {x}}(t)=\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}=\sum _{k=-\infty }^{-1}\alpha _{k}e^{\frac {j2\pi kt}{T}}+\alpha _{0}+\sum _{k=1}^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}$

Now, by substituting $n=-k$ into the summation where $k$ is negative and substituting $n=k$ into the summation where $k$ is positive we get: $\sum _{k=1}^{\infty }\alpha _{-n}e^{\frac {-j2\pi nt}{T}}+\alpha _{0}+\sum _{k=1}^{\infty }\alpha _{n}e^{\frac {j2\pi nt}{T}}$

Recall that $\alpha _{n}={\frac {1}{T}}\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(u)e^{\frac {-j2\pi nt}{T}}dt$

If $x(t)$ is real, then $\alpha _{n}^{*}=\alpha _{-n}$. Let us assume that $x(t)$ is real.

$x(t)=\alpha _{0}+\sum _{n=1}^{\infty }(\alpha _{n}e^{\frac {j2\pi nt}{T}}+\alpha _{n}^{*}e^{\frac {-j2\pi nt}{T}})$

Recall that $y+y^{*}=2Re(y)$ Here is further clarification on this property

So, we may write:

$x(t)=\alpha _{0}+\sum _{n=1}^{\infty }2Re(\alpha _{n}e^{\frac {j2\pi nt}{T}})$

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: $x(t)=x(t+T)=\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}=\sum _{k=-\infty }^{\infty }1/T\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(u)e^{\frac {-j2\pi ku}{T}}due^{\frac {j2\pi kt}{T}}$

So, $\lim _{x\to \infty }x(t)=\int _{-\infty }^{\infty }(\int _{-\infty }^{\infty }x(u)e^{-j2\pi fu}du)e^{j2\pi ft}df$

From the above limit we define $x(t)$ and $X(f)$.

$x(t)={\mathcal {F}}^{-1}[X(f)]=\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df$

$X(f)={\mathcal {F}}[x(t)]=\int _{-\infty }^{\infty }x(t)e^{j2\pi ft}dt$

We can take the derivitive of $x(t)$ and then put in terms of the reverse fourier transform.

${\frac {dx}{dt}}=\int _{-\infty }^{\infty }j2\pi fX(f)e^{j2\pi ft}df={\mathcal {F}}^{-1}[j2\pi fX(f)]$

What happens if we just shift the time of $x(t)$?

$x(t-t_{0})=\int _{-\infty }^{\infty }X(f)e^{j2\pi f(t-t_{0})}df=\int _{-\infty }^{\infty }e^{-j2\pi ft_{0}}X(f)e^{j2\pi ft}df={\mathcal {F}}^{-1}[e^{-j2\pi ft_{0}}X(f)]$

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

$X(f-f_{0})=\int _{-\infty }^{\infty }x(t)e^{j2\pi (f-f_{0})t}dt=\int _{-\infty }^{\infty }e^{-j2\pi tf_{0}}x(t)e^{j2\pi ft}df={\mathcal {F}}[e^{-j2\pi tf_{0}}x(t)]$

What would be the Fourier transform of $cos(2/pif_{0}t)x(t)$?

<

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.

File:CDsystem.jpg

In Time Domain:

Let's start with a signal $h(t)$, 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 $T$. Lets do that by multiplying $h(t)$ by $\sum _{n=-\infty }^{\infty }\delta (t-nT)$. Since the magnetude of our delta function is one, we get a series of delta functions that record the value of $h(t)$ at intervals of $T$. This gives us a result that looks like: $h(t)\sum _{n=-\infty }^{\infty }\delta (t-nT)=\sum _{n=-\infty }^{\infty }x(nt)\delta (t-nT)$

In Frequency Domain:

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

Aside:${\mathcal {F}}[\sum _{n=-\infty }^{\infty }\delta (t-nT)]=\int _{-\infty }^{\infty }\sum _{n=-\infty }^{\infty }\delta (t-nT)e^{j2\pi ft}dt=\sum _{n=-\infty }^{\infty }\int _{-\infty }^{\infty }\delta (t-nT)e^{j2\pi ft}dt=\sum _{n=-\infty }^{\infty }e^{j2\pi fnT}$

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 $\alpha _{m}$.

$\sum _{n=-\infty }^{\infty }\delta (t-nT)=\sum _{m=-\infty }^{\infty }\alpha _{m}e^{j2\pi mt}$

Now we can solve for $\alpha _{m}$.

$\alpha _{m}={\frac {1}{T}}\int _{\frac {-T}{2}}^{\frac {T}{2}}\sum _{n=-\infty }^{\infty }\delta (t-nT){\frac {j2\pi mt}{T}}dt={\frac {1}{T}}\int _{\frac {-T}{2}}^{\frac {-T}{2}}\delta (t){\frac {j2\pi mt}{T}}dt={\frac {1}{T}}$

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

$\sum _{n=-\infty }^{\infty }\delta (t-nT)=\sum _{n=-\infty }^{\infty }{\frac {1}{T}}e^{\frac {j2\pi kt}{T}}$ ${\mathcal {F}}\sum _{n=-\infty }^{\infty }\delta (t-nT)={\mathcal {F}}\sum _{n=-\infty }^{\infty }{\frac {1}{T}}e^{\frac {j2\pi kt}{T}}=\sum _{n=-\infty }^{\infty }{\frac {1}{T}}\int _{-\infty }^{\infty }e^{\frac {j2\pi kt}{T}}e^{-j2\pi ft}dt={\frac {1}{T}}\sum _{n=-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-j2\pi (f-{\frac {m}{T}}t}dt$

File:Barnsasample.jpg

File:BarnsaDA.jpg