Fourier series - by Ray Betz

Fourier Series

If

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

then we can write

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

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 $x(t)$. $<{\alpha }_k|x(t)>=<e^{\frac{j2\pi nt}{T}}|\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_k{e}^{\frac{j2\pi kt}{T}}>$ $=\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t)e^{\frac{-j2\pi nt}{T}}dt$ $=\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_k{e}^{\frac{j2\pi kt}{T}}{e}^{\frac{-j2\pi nt}{T}}dt$ $=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_k\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}{e}^{\frac{j2\pi (k-n)t}{T}}dt$

If $k=n$ then,

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

If $k\ne n$ then,

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

We can simplify the above two conclusion into one equation.

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

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

Orthogonal Functions

The function $y_n(t)$ and $y_m(t)$ are orthogonal on $(a,b)$ if and only if $<y_n(t)|y_m(t)>=\underset{a}{\overset{b}{\int }}{y}_n^{*}(t)y_m(t)dt=0$.

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

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 $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)h(t-u)du$. 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 $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)h(t-u)du$. 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
$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)\delta (t-u)du$	$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)h(t-u)du$	Superposition
$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi ft}h(t-u)du$	$e^{j2\pi ft}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi vt}h(v)dv$	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. $\mathbf{x}(t)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_k{e}^{\frac{j2\pi kt}{T}}=\sum _{k=-\mathrm{\infty }}^{-1}{\alpha }_k{e}^{\frac{j2\pi kt}{T}}+{\alpha }_0+\sum _{k=1}^{\mathrm{\infty }}{\alpha }_k{e}^{\frac{j2\pi kt}{T}}$

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: $\sum _{k=1}^{\mathrm{\infty }}{\alpha }_{-n}{e}^{\frac{-j2\pi nt}{T}}+{\alpha }_0+\sum _{k=1}^{\mathrm{\infty }}{\alpha }_n{e}^{\frac{j2\pi nt}{T}}$

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

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

$x(t)={\alpha }_0+\sum _{n=1}^{\mathrm{\infty }}({\alpha }_n{e}^{\frac{j2\pi nt}{T}}+{\alpha }_n^{*}{e}^{\frac{-j2\pi nt}{T}})$

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

So, we may write:

$x(t)={\alpha }_0+\sum _{n=1}^{\mathrm{\infty }}2Re({\alpha }_n{e}^{\frac{j2\pi nt}{T}})$

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

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

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

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

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

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

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

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

$x(t-t_0)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f)e^{j2\pi f(t-t_0)}df=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi f{t}_0}X(f)e^{j2\pi ft}df={{ℱ}}^{-1}[e^{-j2\pi f{t}_0}X(f)]$

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

$X(f-f_0)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{j2\pi (f-f_0)t}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi t{f}_0}x(t)e^{j2\pi ft}df={ℱ}[e^{-j2\pi t{f}_0}x(t)]$

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

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CD Player

Below is a diagram of how the information on a CD player is read and processed. As you can see

File:CDsystem.jpg