This module discusses the existence and covergence of the Fourier Series to show that it can be a very good approximation for all signals. The Dirichlet conditions, which are the sufficient conditions to guarantee existence and convergence of the Fourier series, are also discussed.
Before looking at this module, hopefully you have become fully
convinced of the fact that
anyperiodic function,
$f(t)$ , can be represented as a sum of
complex sinusoids . If you
are not, then try looking back at
eigen-stuff in a nutshell or
eigenfunctions of LTI
systems . We have shown that we can represent a signal
as the sum of exponentials through the
Fourier Series equations below:
Joseph
Fourier insisted that these equations were true,
but could not prove it. Lagrange publicly ridiculedFourier, and said that only continuous functions can be
represented by
[link] (indeed he
proved that
[link] holds for
continuous-time functions). However, we know now thatthe real truth lies in between Fourier and Lagrange's
positions.
Understanding the truth
Formulating our question mathematically, let
$$\frac{d {f}_{N}(t)}{d}=\sum_{n=-N}^{N} {c}_{n}e^{i{\omega}_{0}nt}$$ where
${c}_{n}$ equals the Fourier coefficients of
$f(t)$ (see
[link] ).
$\frac{d {f}_{N}(t)}{d}$ is a "partial reconstruction" of
$f(t)$ using the first
$2N+1$ Fourier coefficients.
$\frac{d {f}_{N}(t)}{d}$approximates$f(t)$ , with the approximation getting better and better as
$N$ gets large. Therefore, we can
think of the set
$\{\forall N, N=\{0, 1, \dots \}()\colon \frac{d {f}_{N}(t)}{d}\}()$ as a
sequence of functions , each one
approximating
$f(t)$ better than the one before.
The question is, does this sequence converge to
$f(t)$ ? Does
$\frac{d {f}_{N}(t)}{d}\to f(t)$ as
$N\to $∞ ? We will try to answer this question by thinking
about convergence in two different ways:
Looking at the
energy of the error signal:
$${e}_{N}(t)=f(t)-\frac{d {f}_{N}(t)}{d}$$
Looking at
$\lim_{N\to}N\to $∞fNt at
each point and comparing to
$f(t)$ .
Approach #1
Let
${e}_{N}(t)$ be the difference (
i.e. error) between the signal
$f(t)$ and its partial reconstruction
$\frac{d {f}_{N}(t)}{d}$
${e}_{N}(t)=f(t)-\frac{d {f}_{N}(t)}{d}$
If
$f(t)\in {L}^{2}(\left[0 , T\right]())$ (finite energy), then the energy of
${e}_{N}(t)\to 0$ as
$N\to $∞ is
We can prove this equation using Parseval's relation:
$$\lim_{N\to}N\to $$∞tT0ftfNt2N∞N∞∞ℱnftℱnfNt2N∞nnNcn20 where the last equation before zero is the tail sum of theFourier Series, which approaches zero because
$f(t)\in {L}^{2}(\left[0 , T\right]())$ .Since physical systems respond to energy, the
Fourier Series provides an adequate representation for all
$f(t)\in {L}^{2}(\left[0 , T\right]())$ equaling finite energy over one period.
Approach #2
The fact that
${e}_{N}\to 0$ says nothing about
$f(t)$ and
$\lim_{N\to}N\to $∞fNt being
equal at a given point. Take the
two functions graphed below for example:
Given these two functions,
$f(t)$ and
$g(t)$ , then we can see that for all
$t$ ,
$f(t)\neq g(t)$ , but
$$\int_{0}^{T} \left|f(t)-g(t)\right|^{2}\,d t=0$$ From this we can see the following relationships:
$$\mathrm{energy\; convergence}\neq \mathrm{pointwise\; convergence}$$$$\mathrm{pointwise\; convergence}\implies {\mathrm{convergence\; in\; L}}^{2}(\left[0 , T\right]())$$ However, the reverse of the above statement does not hold true.
It turns out that if
$f(t)$ has a
discontinuity (as can be seen in figure
of
$g(t)$ above) at
${t}_{0}$ , then
$$f({t}_{0})\neq \lim_{N\to}N\to $$∞fNt0 But as long as
$f(t)$ meets some other fairly mild conditions, then
$$f({t}^{\prime})=\lim_{N\to}N\to $$∞fNt′ if
$f(t)$ is
continuous at
$t={t}^{\prime}$ .
These conditions are known as the
Dirichlet Conditions .
Dirichlet conditions
Named after the German mathematician, Peter Dirichlet, the
Dirichlet conditions are the sufficient conditions
to guarantee
existence and
energy convergence of the
Fourier Series.
The weak dirichlet condition for the fourier series
For the Fourier Series to exist, the Fourier coefficients
must be finite. The
Weak Dirichlet Condition guarantees this. It essentially says that the
integral of the absolute value of the signal must befinite.
The coefficients of the Fourier Series are finite if
Weak dirichlet condition for the fourier series
$\int_{0}^{T} \left|f(t)\right|\,d t$∞
This can be shown from the magnitude of the Fourier
Series coefficients:
\left|{c}_{n}()\right|=\left|\frac{1}{T}\int_{0}^{T} f(t)e^{-(i{\omega}_{0}()nt)}\,d t\right|\le \frac{1}{T}\int_{0}^{T} \left|f(t)\right|\left|e^{-(i{\omega}_{0}()nt)}\right|\,d t
Remembering our
complex
exponentials , we know that in the above equation
$\left|e^{-(i{\omega}_{0}()nt)}\right|=1$ , which gives us:
If we have the function:
$$\forall t, 0< t\le T\colon f(t)=\frac{1}{t}$$ then you should note that this function
fails the above condition because:
$$\int_{0}^{T} \left|\frac{1}{t}\right|\,d t$$∞
The strong dirichlet conditions for the fourier series
For the Fourier Series to exist, the
following two conditions must be satisfied (along with the WeakDirichlet Condition):
In one period,
$f(t)$ has only a finite number of minima and maxima.
In one period,
$f(t)$ has only a finite number of discontinuities and
each one is finite.
These are what we refer to as the
Strong
Dirichlet Conditions . In theory we can think of
signals that violate these conditions,
$\sin \lg t$ for instance. However, it is not possible to create a signal
that violates these conditions in a lab. Therefore, anyreal-world signal will have a Fourier representation.
Let us assume we have the following function and equality:
$\frac{d f(t)}{d}=\lim_{N\to}N\to $∞fNt
If
$f(t)$ meets all three conditions of the Strong Dirichlet
Conditions, then
$$f(\tau )=\frac{d f(\tau )}{d}$$ at every
$\tau $ at which
$f(t)$ is continuous. And where
$f(t)$ is discontinuous,
$\frac{d f(t)}{d}$ is the
average of the values
on the right and left.
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Rafiq
Rafiq
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Mahi
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Rafiq
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brayan
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Damian
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Tarell
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Damian
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Tarell
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