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Fourier Y Laplace

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Submitted By emiliano19
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Serie de Fourier
El análisis de Fourier fue introducido en 1822 en la “Théorie analyitique de la chaleur” para tratar la solución de problemas de valores en la frontera en la conducción del calor. Más de siglo y medio después las aplicaciones de esta teoría son muy bastas: Sistemas Lineales, Comunicaciones, Física moderna, Electrónica, Óptica y por supuesto, Redes Eléctricas entre muchas otras.
Series de Fourier. 1

Funciones Periódicas Una Función Periódica f(t) cumple la siguiente propiedad para todo valor de t. f(t)=f(t+T) A la constante mínima para la cual se cumple lo anterior se le llama el periodo de la función

Repitiendo la propiedad se puede obtener: Series de Fourier. f(t)=f(t+nT), donde n=0,1,  2, 3,... 2

Funciones Periódicas t t Ejemplo: ¿Cuál es el período de la función cos( 3 )  cos( 4 )? f(t)

Solución.- Si f(t) es periódica se debe cumplir: t t f(t  T)  cos( t T )  cos( t T )  f(t)  cos( 3 )  cos( 4 ) 3 4

Pero como se sabe cos(x+2kp)=cos(x) para cualquier entero k, entonces para que se cumpla la igualdad se requiere que T/3=2k1p, T/4=2k2p Es decir, T = 6k1p = 8k2p Donde k1 y k2 son enteros, El valor mínimo de T se obtiene con k1=4, k2=3, es decir,T=24p Series de Fourier. 3

Funciones Periódicas
Gráfica de la función
3 2 1

t t f(t)  cos( 3 )  cos( 4 )

T

f(t)=cos(t/3)+cos(t/4)

f(t)

0
-1 -2

24p
-3 0 50 100 150 200

t
Series de Fourier. 4

Funciones Periódicas
Podríamos pensar que cualquier suma de funciones seno y coseno produce una función periódica. Esto no es así, por ejemplo, consideremos la función f(t) = cos(w1t)+cos(w2t). Para que sea periódica se requiere encontrar dos enteros m, n tales que w1T= 2pm, w2T=2pn w1 m De donde  w2 n Es decir, la relación w1/ w2 debe ser un número racional.
Series de Fourier. 5

Funciones Periódicas
Ejemplo: la función cos(3t)+cos(p+3)t no es w1 ya 3 periódica,  que no es un número w racional. 2 3  p
2

f(t)=cos(3t)+cos((3+pi)t)

1

f(t)
0 -1 -2 0

5

10

15

t

20

25

30
Series de Fourier. 6

Funciones Periódicas Ejercicio: Encontrar el periodo de las siguientes funciones, si es que son periódicas: 1) f(t) = sen(nt), donde n es un entero. 2) f(t)= sen2(2pt) 3) f(t)= sen(t)+sen(t+p/2) 4) f(t)= sen(w1t)+cos(w2t) 5) f(t)= sen(2 t)
Series de Fourier. 7

Serie Trigonométrica de Fourier Algunas funciones periódicas f(t) de periodo T pueden expresarse por la siguiente serie, llamada Serie Trigonométrica de Fourier f(t) = ½ a0 + a1cos(w0t)+a2cos(2w0t)+... + b1sen(w0t)+b2sen(2w0t)+... Donde w0=2p/T. Es decir,

f ( t )  1 a 0   [a n cos(nw0 t )  b nsen (nw0 t )] 2 n 1
Series de Fourier. 8



Serie Trigonométrica de Fourier Es posible escribir de una manera ligeramente diferente la Serie de Fourier, si observamos que el término ancos(nw0t)+bnsen(nw0t) se puede escribir como   a2 n  b2  n an a 2  b2 n  n cos(nw0 t )  sen (nw0 t )   a 2  b2 n n  bn

Podemos encontrar una manera más compacta para expresar estos coeficientes pensando en un triángulo rectángulo:
Series de Fourier. 9

Serie Trigonométrica de Fourier an b n Cn  a 2  b 2 n n

a2 n a2 n

 

b2 n b2 n

 cos n  senn

n

bn

a n Con lo cual la expresión queda
Cn cos n cos(nw0 t )  senn sen(nw0 t )

 Cn cos(nw0 t  n )
Series de Fourier. 10

Serie Trigonométrica de Fourier Si además definimos C0=a0/2, la serie de Fourier se puede escribir como f ( t )  C0   Cn cos(nw0 t  n ) n 1 

Así, y

Cn  a  b
2 n

2 n

n  tan   a   n
Series de Fourier. 11

1  b n 

Serie Trigonométrica de Fourier Ejercicio: Definir adecuadamente los coeficientes C0, Cn y n, de manera que la serie de Fourier se pueda escribir como f ( t )  C0   Cn sen (nw0 t  n ) n 1 

Series de Fourier. 12

Serie Trigonométrica de Fourier Si además definimos C0=a0/2, la serie de Fourier se puede escribir como f ( t )  C0   Cn cos(nw0 t  n ) n 1 

Así, y

Cn  a  b
2 n

2 n

n  tan   a   n
Series de Fourier. 13

1  b n 

Componentes y armónicas
Así, una función periódica f(t) se puede escribir como la suma de componentes sinusoidales de diferentes frecuencias wn=nw0. A la componente sinusoidal de frecuencia nw0: Cncos(nw0t+n) se le llama la enésima armónica de f(t). A la primera armónica (n=1) se le llama la componente fundamental y su periodo es el mismo que el de f(t)
A la frecuencia w0=2pf0=2p/T se le llama frecuencia angular fundamental.
Series de Fourier. 14

Componentes y armónicas A la componente de frecuencia cero C0, se le llama componente de corriente directa (cd) y corresponde al valor promedio de f(t) en cada periodo. Los coeficientes Cn y los ángulos n son respectiva-mente las amplitudes y los ángulos de fase de las armónicas.

Series de Fourier. 15

Componentes y armónicas t t Ejemplo: La función f(t)  cos( 3 )  cos( 4 ) Como ya se mostró tiene un periodo T=24p, por lo tanto su frecuencia fundamental es w0=1/12 rad/seg. 3 Componente fundamental es def(t)=cos(t/3)+cos(t/4) la forma: 2 0*cos(t/12). 1 Tercer armónico: 0 cos(3t/12)=cos(t/4) -1 Cuarto armónico: -2 24p Cos(4t/12)=cos(t/3) -3 f(t) 0 50 100

t

150

200

Series de Fourier. 16

Componentes y armónicas
Ejemplo: Como puede verse, la función anterior tiene tantas partes positivas como negativas, por lo tanto su componente de cd es cero, en cambio t t f(t)  1  cos( 3 )  cos( 4 )
3
2 1

Tiene tantas partes arriba como abajo de 1 por lo tanto, su componente de cd es 1.

f(t)

0

-1

f(t)=1+cos(t/3)+cos(t/4)
-2 -3 0

24p
50 100

t

150

200

Series de Fourier. 17

Componentes y armónicas Ejercicio: ¿Cuál es la componente fundamental, las armónicas distintas de cero y la componente de directa de a) f(t) = sen2t b) f(t) = cos2t ? Justifícalo además mostrando la gráfica de las funciones y marcando en ellas el periodo fundamental y la componente de cd.

Series de Fourier. 18

Ortogonalidad de senos y cosenos Se dice que un conjunto de funciones fk(t) son ortogonales en el intervalo a

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...sample l.43 π 2 y ( t ) =  3 cos  200t + --     6  2 π = 9 cos  200t + --   6 9 π = -- cos  400t + --  1  2 3 9 (a) DC component = -2 9 π (b) Sinusoidal component = -- cos  400t + --   2 3 9 Amplitude = -2 1 200 Fundamental frequency = -------- Hz π 1.44 The RMS value of sinusoidal x(t) is A ⁄ 2 . Hence, the average power of x(t) in a 1-ohm 2 resistor is ( A ⁄ 2 ) = A2/2. 1.45 Let N denote the fundamental period of x[N]. which is defined by 2π N = ----Ω The average power of x[n] is therefore N -1 1 2 P = --- ∑ x [ n ] N 1 = --N n=0 N -1 ∑A n=0 2 N -1 A = ----N 1.46 2 2 2πn cos  --------- + φ  N  ∑ cos n=0 2  2πn + φ -------- N  The energy of the raised cosine pulse is E = π⁄ω 1 ∫–π ⁄ ω -- ( cos ( ωt ) + 1 ) 4 2 dt 1 π⁄ω 2 = -- ∫ ( cos ( ωt ) + 2 cos ( ωt ) + 1 ) dt 2 0 1 π ⁄ ω 1 1 -- cos ( 2ωt ) + -- + 2 cos ( ωt ) + 1 dt = -- ∫  2 0 2 2 1 3 π - - = --  --  --- = 3π ⁄ 4ω 2  2  ω 1.47 The signal x(t) is even; its total energy is therefore 5 2 E = 2 ∫ x ( t ) dt 0 2 4 5 = 2 ∫ ( 1 ) dt + 2 ∫ ( 5 – t ) dt 2 0 2 4 5 1 4 3 = 2 [ t ] t=0 + 2 – -- ( 5 – t ) 3 t=4 2 26 = 8 + -- = ----3 3 1.48 (a) The differentiator output is  1  y ( t ) =  –1   0 for – 5 < t < – 4 for 4 < t < 5 otherwise (b) The energy of y(t) is E = –4 ∫–5 5 ( 1 ) dt + ∫ ( – 1 ) dt 2 2 4 = 1+1 = 2 1.49 The output of the integrator is t y ( t ) = A ∫ τ dτ = At for 0 ≤ t ≤ T 0 Hence the energy of y(t) is E = 1.50 T...

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