Formulario de Series de Fourier

July 30, 2024 | Author: Anonymous | Category: N/A
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Formulario de Series de Fourier Ing. Demetrio García Almazán Forma Trigonométrica1

Simetría de Cuarto de Onda Impar

∞ X 1 f (t) = a0 + [an cos(nω0 t) + bn sen(nω0 t)] 2 n=1

f (t) =

b2n−1 =

an =

4 T

T /2

f (t)dt Z

0

bn =

4 T

δ n (t − t0 )g(t)dt =

[bn sen(nω0 t)]

cos(nπ) = (−1)n

0

cos [(2n ± 1)π] = −1 πi sen (2n ± 1) = (−1)n+1 = −(−1)n 2 2π ω0 = T  1  jnω0 t e + e−jnω0 t cos(nω0 t) = 2  1  jnω0 t sen(nω0 t) = e − e−jnω0 t 2j h

{a2n−1 cos [(2n − 1)ω0 t]

n=1

+b2n−1 sen [(2n − 1)ω0 t]} Z

4 = T

Z

T /2

f (t) cos [(2n − 1)ω0 t] dt 0 T /2

e±jt = cos t ± j sen t

f (t) sen [(2n − 1)ω0 t] dt 0

1 1 (an − jbn ) , c0 = a0 2 2 an = 2< [cn ] , a0 = 2c0 , bn = −2= [cn ] cn =

Simetría de Cuarto de Onda Par f (t) =

∞ X

{a2n−1 cos [(2n − 1)ω0 t]}

< = Parte Real = = Parte Imaginaria

n=1

a2n−1

8 = T

Z

(−1)n g n (t0 ), a ≤ t0 ≤ b 0, b < t0 < a

cos(0) = cos(2nπ) = 1

T /2

f (t) sen(nω0 t)dt

4 T



sen(nπ) = sen(0) = sen [(2n ± 1)π] = h πi sen(2nπ) = cos (2n ± 1) =0 2

Simetría de Media Onda ∞ X

f (t)e−jnω0 t dt

−T /2

b

n=1

Z

T /2

Funciones Generales y Equivalencias

0

f (t) =

1 T

Z

a

f (t) cos(nω0 t)dt

∞ X

cn ejnω0 t

La Función Impulso o Delta de Dirac  Z b g(t0 ), a ≤ t0 ≤ b δ(t − t0 )g(t)dt = 0, b < t0 < a a

T /2

Z

4 T

Z

∞ X −∞

Simetría Impar

b2n−1

f (t) sen [(2n − 1)ω0 t] dt 0

f (t) =

∞ X 1 a0 + [an cos(nω0 t)] 2 n=1

a0 =

T /4

Forma Compleja1

Simetría Par

a2n−1 =

Z

8 T

cn =

f (t) =

{b2n−1 sen [(2n − 1)ω0 t]}

n=1

Z 2 T /2 f (t)dt a0 = T −T /2 Z 2 T /2 f (t) cos(nω0 t)dt an = T −T /2 Z 2 T /2 bn = f (t) sen(nω0 t)dt T −T /2

f (t) =

∞ X

T /4

f (t) cos [(2n − 1)ω0 t] dt 0

1 En esta forma se pueden cambiar los límites de integración de 0 a T

1

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