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

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. (What is the delta function below?)

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

So, we 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 (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 $x(u)\delta (t-u)$ 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 put into the system the integral of $x(u)\delta (t-u)$ and we would 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 ($v=t-u$, and $dv=-du$) we get $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)h(t-u)du=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi ft}h(t-u)du=-\underset{\mathrm{\infty }}{\overset{-\mathrm{\infty }}{\int }}{e}^{j2\pi f(t-v)}h(v)dv=e^{j2\pi ft}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(v)e^{-j2\pi fv}dv$. 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: $\sum _{n=1}^{\mathrm{\infty }}{\alpha }_{-n}{e}^{\frac{-j2\pi nt}{T}}+{\alpha }_0+\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n{e}^{\frac{j2\pi nt}{T}}$

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 $X(f)$.

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

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

By using the above transforms we can now look at something in the frequency domain or the time domain. We are not limited to just one domain but can use both of them.

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 $cos(2/pi{f}_{0}t)x(t)$?

${ℱ}[cos(2\pi f_{0}t)x(t)]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)cos(2\pi f_{0}t)e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{j2\pi f_{0}t}+e^{-j2\pi f_{0}t}}{2}x(t)e^{-j2\pi ft}dt$

$=\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j2\pi (f-f_0)t}dt+\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{j2\pi (f+f_0)t}dt=\frac{1}{2}X(f-f_0)+\frac{1}{2}X(f+f_0)$

What would happen if we multiplied our time by a constant in $x(t)$? We will substitute $u=at$ and $du=adt$. If $a\ne 0$:

${ℱ}[x(at)]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(at)e^{-j2\pi ft}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)e^{\frac{-j2\pi fu}{a}}\frac{du}{|a|}=\frac{1}{|a|}X(\frac{f}{a})$

Ok, lets take the fourier transform of the fourier series.

${ℱ}[\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n{e}^{\frac{j2\pi nt}{T}}]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n{e}^{\frac{j2\pi nt}{T}}{e}^{-j2\pi ft}dt=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi (f-\frac{n}{T})t}dt=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_n\delta (f-\frac{n}{T})$

Remember: $\delta (f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j2\pi ft}dt==CDPlayer==BelowisadiagramofhowtheinformationonaCDplayerisreadandprocessed.AsyoucanseetheinformationontheCDisprocessedbytheD/Aconverterandthensentthroughalowpassfilterandontothespeaker.Ifyouwererecordingsound,thesoundwouldbecapturedthroughamicrophone.Then,itshouldbesentthroughalowpassfilterandontotheA/DconverterandthenitisreadytobeputontheCD.Recordingsignalsisessentiallythereverseoftheoperationpicturedbelow.[[Image:CDsystem.jpg]{]}^{‴}InTimeDomain{:}^{‴}Le{t}^{\prime }sstartwithasignal<math>h(t)$, 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 $\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\delta (t-nT)$. 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: $h(t)\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\delta (t-nT)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}x(nt)\delta (t-nT)$

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 ${ℱ}[\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\delta (t-nT)]$.

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

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 ${\alpha }_m$.

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

Now we can solve for ${\alpha }_m$.

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

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 $\frac{1}{T}$

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

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