• Author / Uploaded
• Alex Yoel Luna

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En la figura se muestra la carga q(x) sobre las placas de un capacitor en el tiempo x. exprese q(x) como una expresión en serie de Fourier

Hallamos las funciones +0 𝑓(𝑥) = {

𝑥 ;0 < 𝑥 < 𝜋 2𝜋 ; 𝜋 < 𝑥 < 2𝜋

𝑃 = 2𝜋 ; 𝐿 = 𝜋 Calculamos los coeficientes de Fourier 𝑎0 =

2𝜋 1 𝜋 [∫ 𝑥 𝑑𝑥 + ∫ 2𝑥 − 𝑥 𝑑𝑥] 𝜋 0 𝜋 𝜋

2𝜋

1 𝑋2 𝑋2 𝑎0 = {[ ] + [2𝜋𝑥 − ] } 𝜋 2 0 2 𝜋 𝑎0 =

1 𝜋2 4𝜋 2 𝜋2 [ + 0] + [4𝜋 − ] − [2𝜋 2 − ] 𝜋 2 2 2

𝑎0 =

1 𝜋2 4𝜋 2 3𝜋 2 [ +( − )] 𝜋 2 2 2

1 𝜋2 𝜋2 𝑎0 = [ + ] 𝜋 2 2 𝑎0 =

1 2𝜋 2 [ ] 𝜋 2

2𝜋 2 𝑎0 = 2𝜋 𝑎0 = 𝜋

𝑎𝑛 =

2𝜋 1 𝜋 𝑛𝜋𝑥 𝑛𝜋𝑥 [∫ 𝑥 cos ( ) 𝑑𝑥 + ∫ 2𝑥 − 𝑥 cos ( ) 𝑑𝑥] 𝜋 0 𝜋 𝜋 𝜋

2𝜋 1 𝜋 𝑎𝑛 = [∫ 𝑥 cos(𝑛𝑥) 𝑑𝑥 + ∫ 2𝑥 − 𝑥 cos(𝑛𝑥) 𝑑𝑥] 𝜋 0 𝜋

D 1 0

I sin(𝑛𝑥) 𝑛 −cos(𝑛𝑥) 𝑛2

1 𝑥 𝑛𝑥 𝜋 𝑥 cos(𝑛𝑥) 2𝜋 𝑎𝑛 = {[ 𝑠𝑒𝑛(𝑛𝑥) + cos 2 ] + [2𝜋𝑛 − 𝑠𝑒𝑛(𝑛𝑥) − ] } 𝜋 𝑛 𝑛 0 𝑛 𝑛2 𝜋 𝑎𝑛 =

1 𝑥 𝑛𝜋 0 cos(𝑛0) {[ 𝑠𝑒𝑛(𝑛𝜋) + cos 2 ] + [ 𝑠𝑒𝑛(𝑛0) − ]} 𝜋 𝑛 𝑛 𝑛 𝑛2 𝑠𝑒𝑛(2𝜋𝑛) 𝜋 cos(𝑛𝜋) + {[2𝜋(2𝜋) − ] − [2𝜋(𝜋) − 𝑠𝑒𝑛(𝑛𝜋) − ]} 2 𝑛 𝑛 𝑛2 𝜋

(−1)𝑛 1 (−1)𝑛 1 1 𝑎𝑛 = {[ 2 − 2 ] + [4𝜋 2 − 2 − 2𝜋 2 + ]} 𝜋 𝑛 𝑛 0 𝑛 𝑛2 1 (−1)𝑛 1 1 (−1)𝑛 2 𝑎𝑛 = [ 2 − 2 + 2𝜋 − 2 + ] 𝜋 𝑛 𝑛 𝑛 𝑛2 𝑎𝑛 =

1 −2(−1)𝑛 2 [ − 2 + 2𝜋 2 ] 𝜋 𝑛2 𝑛

𝑏𝑛 =

2𝜋 1 𝜋 𝑛𝜋𝑥 𝑛𝜋𝑥 [∫ 𝑥𝑠𝑒𝑛 ( ) 𝑑𝑥 + ∫ 2𝜋 − 𝑥𝑠𝑒𝑛 ( ) 𝑑𝑥] 𝜋 0 𝜋 𝜋 𝜋

2𝜋 1 𝜋 𝑏𝑛 = [∫ 𝑥𝑠𝑒𝑛(𝑛𝑥)𝑑𝑥 + ∫ 2𝜋 − 𝑥𝑠𝑒𝑛(𝑛𝑥)𝑑𝑥] 𝜋 0 𝜋

D x 1

I Sen(nx) cos(𝑛𝑥) 𝑛 −sen(𝑛𝑥) 𝑛2

0

1 −𝑥 𝑠𝑒𝑛(𝑛𝑥) 𝜋 𝑥 𝑠𝑒𝑛(𝑛𝑥) 2𝜋 𝑏𝑛 = [[ 𝑐𝑜𝑠(𝑛𝑥) + ] + [2𝜋𝑥 + cos(𝑛𝑥) − ] ] 𝜋 𝑛 𝑛2 𝑛 𝑛2 0 𝜋 𝑏𝑛 =

𝑏𝑛 =

1 −𝑥 𝑠𝑒𝑛(𝑛𝑥) −0 𝑠𝑒𝑛(𝑛0 {[( 𝑐𝑜𝑠(𝑛𝑥) + ) − ( cos(𝑛0) + )] 2 𝜋 𝑛 𝑛 𝑛 𝑛2 2𝜋 𝑠𝑒𝑛(2𝑛𝜋) + [(2𝜋(2𝜋) + cos(2𝜋𝑛) − ) 𝑛 𝑛2 𝜋 𝑠𝑒𝑛(𝑛𝜋) − (2𝜋(𝜋) + cos(𝑛𝜋) − )]} 𝑛 𝑛2 1 𝜋(−1)𝑛 2𝜋 𝜋(−1)𝑛 {[ ] + [(4𝜋 2 + ) − (2𝜋 2 + )]} 𝜋 𝑛 𝑛 𝑛

1 −𝜋(−1)𝑛 2𝜋 𝜋(−1)𝑛 2 𝑏𝑛 = [ + 2𝜋 + − ] 𝜋 𝑛 𝑛 𝑛 1 −2𝜋(−1)𝑛 2𝜋 2𝜋 𝑏𝑛 = [ + 2𝜋 2 + − ] 𝜋 𝑛 𝑛 𝑛 𝑏𝑛 =

1 −(−1)𝑛 1 [+2𝜋 ( + 𝜋 − )] 𝜋 𝑛 𝑛

−(−1)𝑛 1 𝑏𝑛 = 𝜋 ( +𝜋− ) 𝑛 𝑛

∞

𝑓(𝑡)

1 1 −2(−1)𝑛 −2 𝑛𝜋𝑥 −(−1)𝑛 𝑛𝜋𝑥 𝑛 = 𝜋+∑[ ( − + 2𝜋 ) 𝑐𝑜𝑠 ( ) + 𝜋 ( +𝜋− )] 2 2 2 𝜋 𝑛 𝑛 𝜋 𝑛 𝜋 𝑛=1