This module describes the continuous time Fourier Series (CTFS).
It is based on the following modules:Fourier Series: Eigenfunction Approach at http://cnx.org/content/m10496/latest/ by Justin Romberg,
Derivation of Fourier Coefficients Equation at http://cnx.org/content/m10733/latest/ by Michael Haag,Fourier Series and LTI Systems at http://cnx.org/content/m10752/latest/ by Justin Romberg, and
Fourier Series Wrap-Up at http://cnx.org/content/m10749/latest/ by Michael Haag and Justin Romberg.
Introduction
In this module, we will derive an expansion for
continuous-time, periodic functions, and in doing so, derive the
Continuous Time Fourier Series (CTFS).
Since
complex
exponentials are
eigenfunctions of linear time-invariant (LTI)
systems , calculating the output of an LTI system
$\mathscr{H}$ given
$e^{st}$ as an input amounts to simple multiplication, where
$H(s)\in \mathbb{C}$ is the eigenvalue corresponding to s. As shown in the figure, a simple exponential input would yield the output
$y(t)=H(s)e^{st}$
Using this and the fact that
$\mathscr{H}$ is linear, calculating
$y(t)$ for combinations of complex exponentials is also
straightforward.
The action of
$H$ on an input such
as those in the two equations above is easy to explain.
$\mathscr{H}$ independently
scales each exponential component
$e^{{s}_{n}t}$ by a different complex number
$H({s}_{n})\in \mathbb{C}$ . As such, if we can write a function
$f(t)$ as a combination of complex exponentials it allows us to easily calculate the output of a system.
Fourier series synthesis
Joseph
Fourier demonstrated that an arbitrary
$f(t)$ can be written as a linear combination of harmonic
complex sinusoids
$f(t)=\sum $∞∞cnjω0nt
where
${\omega}_{0}=\frac{2\pi}{T}$ is the fundamental frequency. For almost all
$f(t)$ of practical interest, there exists
${c}_{n}$ to make
[link] true. If
$f(t)$ is finite energy (
$f(t)\in L(0, T)^{2}$ ), then the equality in
[link] holds in the sense of energy convergence; if
$f(t)$ is continuous, then
[link] holds
pointwise. Also, if
$f(t)$ meets some mild conditions (the Dirichlet
conditions), then
[link] holds
pointwise everywhere except at points of discontinuity.
The
${c}_{n}$ - called the Fourier coefficients -
tell us "how much" of the sinusoid
$e^{j{\omega}_{0}nt}$ is in
$f(t)$ .
The formula shows
$f(t)$ as a sum of complex exponentials, each of which is easily processed by an
LTI system (since it is an eigenfunction of
every LTI system). Mathematically,
it tells us that the set ofcomplex exponentials
$\{\forall n, n\in \mathbb{Z}\colon e^{j{\omega}_{0}nt}\}$ form a basis for the space of T-periodic continuous
time functions.
We know from Euler's formula that
$cos\left(\omega t\right)+sin\left(\omega t\right)=\frac{1-j}{2}{e}^{j\omega t}+\frac{1+j}{2}{e}^{-j\omega t}.$
Finding the coefficients of the Fourier series expansion involves some algebraic manipulation of the synthesis formula.
First of all we will multiply both sides of the equation by
$e^{-(j{\omega}_{0}kt)}$ , where
$k\in \mathbb{Z}$ .
$f(t)e^{-(j{\omega}_{0}kt)}=\sum $∞∞cnjω0ntjω0kt
Now integrate both sides over a given period,
$T$ :
Now that we have made this seemingly more complicated, let us
focus on just the integral,
$\int_{0}^{T} e^{j{\omega}_{0}(n-k)t}\,d t$ , on the right-hand side of the above equation.
For this integral we will need to consider two cases:
$n=k$ and
$n\neq k$ . For
$n=k$ we will have:
$\forall n, n=k\colon \int_{0}^{T} e^{j{\omega}_{0}(n-k)t}\,d t=T$
But
$\cos ({\omega}_{0}(n-k)t)$ has an integer number of periods,
$n-k$ , between
$0$ and
$T$ . Imagine a graph of the
cosine; because it has an integer number of periods, there areequal areas above and below the x-axis of the graph. This
statement holds true for
$\sin ({\omega}_{0}(n-k)t)$ as well. What this means is
$\int_{0}^{T} \cos ({\omega}_{0}(n-k)t)\,d t=0$
which also holds for the integral involving the sine function.
Therefore, we conclude the following about our integral ofinterest:
Now let us return our attention to our complicated equation,
[link] , to see if we can finish
finding an equation for our Fourier coefficients. Using thefacts that we have just proven above, we can see that the only
time
[link] will have a nonzero
result is when
$k$ and
$n$ are equal:
$\forall n, n=k\colon \int_{0}^{T} f(t)e^{-(j{\omega}_{0}nt)}\,d t=T{c}_{n}$
Finally, we have our general equation for the Fourier
coefficients:
Because complex exponentials are eigenfunctions of LTI systems, it is often useful to represent signals using a set of complex exponentials as a basis. The continuous time Fourier series synthesis formula expresses a continuous time, periodic function as the sum of continuous time, discrete frequency complex exponentials.
$f(t)=\sum $∞∞cnjω0nt
The continuous time Fourier series analysis formula gives the coefficients of the Fourier series expansion.
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest.
Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.?
How this robot is carried to required site of body cell.?
what will be the carrier material and how can be detected that correct delivery of drug is done
Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?