Sensor fusion: Difference between revisions

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The paper referenced uses sensor fusion to estimate attitude, not altitude.
 
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[[Image:Highly oscillatory function.png|right|frame|The Riemann-Lebesgue lemma states that the integral of a function like the above is small. The integral will approach zero as the number of oscillations increases.]]
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In [[mathematics]], the '''Riemann–Lebesgue lemma''', named after [[Bernhard Riemann]] and [[Henri Lebesgue]], is of importance in [[harmonic analysis]] and [[asymptotic analysis]].
 
The lemma says that the [[Fourier transform]] or [[Laplace transform]] of an [[Lp space|''L''<sup>1</sup> function]] vanishes at infinity.
 
== Statement ==
If ''f'' is  [[Lp space|'''L'''<sup>1</sup> integrable]] on '''R'''<sup>''d''</sup>, that is to say, if the Lebesgue integral of |''f''| is finite, then the [[Fourier transform]] of ''f'' satisfies
 
:<math> \hat{f}(z):=\int_{\mathbb{R}^{d}} f(x) e^{-iz \cdot x}\,dx \rightarrow 0\text{ as } |z|\rightarrow \infty.</math>
 
==Other versions==
The Riemann&ndash;Lebesgue lemma holds in a variety of other situations.  
 
* If ''f'' is ''L''<sup>1</sup> integrable and supported on (0,&nbsp;∞), then the Riemann&ndash;Lebesgue lemma also holds for the Laplace transform of ''f''. That is,
::<math>\int_0^\infty f(t) e^{-tz}\,dt \to 0</math>
:as |''z''|&nbsp;&rarr;&nbsp;&infin; within the half-plane Re(''z'')&nbsp;≥&nbsp;0.
 
* A version holds for [[Fourier series]] as well: if ''f'' is an integrable function on an interval, then the [[Fourier coefficient]]s of ''f'' tend to 0 as ''n''&nbsp;→&nbsp;±∞,
::<math>\hat{f}_n \ \to \ 0 .</math>
:This follows by extending ''f'' by zero outside the interval, and then applying the version of the lemma on the entire real line.
 
==Applications==
The Riemann–Lebesgue lemma can be used to prove the validity of asymptotic approximations for integrals. Rigorous treatments of the [[method of steepest descent]] and the [[method of stationary phase]], amongst others, are based on the Riemann–Lebesgue lemma.
 
==Proof==
 
We'll focus on the one dimensional case, the proof in higher dimensions is similar. Suppose first that ''f'' is a [[compact set|compactly]] supported [[smooth function]]. Then integration by parts in each variable yields
 
:<math> \left| \int f(x) e^{-izx}dx\right|=\left|\int \frac{1}{iz} f'(x)e^{-izx}dx\right| \leq \frac{1}{|z|}\int|f'(x)|dx  \rightarrow 0 \mbox{ as } z\rightarrow\pm\infty. </math>
 
If ''f'' is an arbitrary integrable function, it may be approximated in the ''L<sup>1</sup>'' by a compactly supported smooth function ''g''. Pick such a ''g'' so that ''||f-g||<sub>L<sup>1</sup></sub><ε''. Then
 
:<math> \limsup_{z\rightarrow\pm\infty} |\hat{f}(z)| \leq  \limsup_{z\to\pm\infty}  \left|\int (f(x)-g(x))e^{-ixz}dx\right| + \limsup_{z\rightarrow\pm\infty}  \left|\int g(x)e^{-ixz}dx\right| \leq \varepsilon+0=\varepsilon,</math>
and since this holds for any ''ε>0'', the theorem follows.
 
''The case of non-real t''.
Assume first that ''f'' has a compact support on <math>(0,\infty)</math> and that ''f'' is continuously differentiable.
Denote the Fourier/Laplace transforms of ''f'' and <math>f'</math> by ''F'' and ''G'', respectively.
Then <math>F(t)=G(t)/t</math>, hence <math>F(z)\rightarrow 0</math> as <math>|t|\rightarrow\infty</math>.
Because the functions of this form are dense in <math>L^1(0,\infty)</math>, the same holds for every ''f''.
 
==References==
*{{cite book | author =[[Salomon Bochner|Bochner S.]], [[K. S. Chandrasekharan|Chandrasekharan K.]] | title=Fourier Transforms | publisher= Princeton University Press | year=1949}}
* {{mathworld|urlname=Riemann-LebesgueLemma|title=Riemann–Lebesgue Lemma}}
 
{{DEFAULTSORT:Riemann-Lebesgue lemma}}
[[Category:Asymptotic analysis]]
[[Category:Harmonic analysis]]
[[Category:Lemmas]]
[[Category:Theorems in analysis]]
[[Category:Theorems in harmonic analysis]]

Latest revision as of 14:34, 10 October 2014





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