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Este modulo establece los pasos para derivar la ecuación de coeficientes de Fourier de la forma general de las series de Fourier.


Usted debería estar familiarizado con la existencia de la ecuación general de las series de Fourier que es la siguiente:

f t n c n ω 0 n t
De lo que estamos interesados aquíes el como combinar los coeficientes de Fourier, c n , dado a una función f t . En la siguiente explicación lo llevaremos paso por paso por la derivación de la ecuación generar para los coeficientes de Fourier dado a una función.


Para resolver la ecuación para c n , tenemos que hacer una pequeña manipulación algebraica. Primero que todo, tenemos que multiplicar los dos lados de por ω 0 k t , donde k .

f t ω 0 k t n c n ω 0 n t ω 0 k t
Ahora integraremos los dos lados sobre el periodo, T :
t T 0 f t ω 0 k t t T 0 n c n ω 0 n t ω 0 k t
En el lado derecho podemos intercambiar la sumatoria y el integral y sacar la constante fuera del integral.
t T 0 f t ω 0 k t n c n t T 0 ω 0 n k t
Ahora que hemos hecho esto lo que al parecer es mas complicado, nos enfocaremos en tan solo el integral, t T 0 ω 0 n k t , que se encuentra en el lado derecho de la ecuación. Para este integral debemos considerar solo dos casos: n k y n k . Para n k tenemos:
n n k t T 0 ω 0 n k t T
Para n k , tenemos:
n n k t T 0 ω 0 n k t t T 0 ω 0 n k t t T 0 ω 0 n k t
Pero ω 0 n k t tiene periodos con números enteros para, n k , entre 0 y T . Imagine la grafica de un coseno; por que tiene periodos con números enteros, hayáreas de igual valor debajo y arriba del eje de las ordenadas en la grafica. Este hecho es verdadero para ω 0 n k t también. Lo que significa
t T 0 ω 0 n k t 0
También para el integral de una función de seno. Por eso, podemos concluir lo siguiente sobre nuestro integral:
t T 0 ω 0 n k t T n k 0
Regresemos a nuestra complicada ecuación, , para ver si podemos encontrar una ecuación para nuestros coeficientes de Fourier. Usando los hechos que ya hemos probado, podemos ver queúnica vez que la ecuación tiene valores de no cero como resultado es cuando k y n son iguales:
n n k t T 0 f t ω 0 n t T c n
Finalmente nuestra ecuación general para los coeficientes de Fourier es:
c n 1 T t T 0 f t ω 0 n t

Pasos para encontrar los coeficientes de fourier

Para encontrar los coeficientes de Fourier de una f t periódica:

  • Por alguna k , multiplique f t por ω 0 k t , saque elárea por debajo de la curva (dividiendo por T ).
  • Repita el paso (1) para todo k .

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
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I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
is it a question of log
Commplementary angles
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
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Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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how do you translate this in Algebraic Expressions
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Señales y sistemas. OpenStax CNX. Sep 28, 2006 Download for free at http://cnx.org/content/col10373/1.2
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