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• Alex Orozco

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889

11/1/2018

10 Ejercicios Series de Fourier 1. Hallar la serie de fourier de: f(t)

𝜋2

𝑇 = 2𝜋 2

f(t) = π − t

2

Señal par a0 = an =? bn = 0 ∞

a0 f(t) = + ∑ an cos nt 2 1

a0 =

T 2

2 ∫ f(t)dt T −T 2

a0 =

2 π 2 ∫ (π − t 2 )dt π 0 π

2 t3 a0 = [π2 t − ] π 3 0 a0 =

2 3 π3 [π − ] π 3

a0 =

2 3π3 − π3 [ ] π 3

a0 =

4π2 3 T

2 2 an = ∫ f(t) cos(nt) dt T −T 2

an =

2 π 2 ∫ (π − t 2 ) cos(nt) dt π 0

an =

π 2 π 2 [∫ π cos(nt) dt − ∫ t 2 cos(nt) dt] π 0 0

u = t2

dv = cos nt

du = 2tdt

v=

t2

sin nt n

sin nt 2 − ∫ t sin nt dt n n

u=t

dv = sin nt

du = dt

v=−

t2

cos nt n

sin nt 2 cos nt 1 − [−t + ∫ cos nt dt] n n n n

an =

2 2 sin nt sin nt 2tcos nt 2 sin nt π − t2 − + [π ] π n n n2 n3 0

an =

2 −2π(−1)n ( ) π n2

an = − an =

4(−1)n n2

4(−1)n+1 n2 ∞

4π2 4(−1)n+1 f(t) = +∑ cos nt 6 n2 1

2.- Hallar la serie de exponencial de fourier de f(x) = {

1, 4 < x ≤ 5 2, 5 < x ≤ 6

w=

2π T

w=π ∞

f(t) = C0 + ∑ Cn ejnωt + C−n e−jnωt 1 T

1 2 C0 = ∫ f(x)dx T −T 2

1 6 C0 = ∫ f(x)dx 2 4 6 1 5 C0 = [∫ 1dx + ∫ 2dx] 2 4 5

1 C0 = [x|54 + 2x|65 ] 2 1 C0 = [5 − 4 + 12 − 10] 2 C0 =

3 2 T

1 2 Cn = ∫ f(x)e−jnwx dx T −T 2

1 6 Cn = ∫ f(x)e−jnπ(x−5) dx 2 4 6 1 5 Cn = [∫ e−jnπ(x−5) dx + ∫ 2e−jnπ(x−5) dx] 2 4 5 5

6

1 e−jnπ(x−5) 2e−jnπ(x−5) Cn = [ ] +[ ] 2 −jnπ 4 −jnπ 5 1 1 ejnπ 2e−jnπ 2 Cn = [− + − + ] 2 jnπ jnπ jnπ jnπ

1 1 1 2 Cn = [ − + ] 2 jnπ jnπ jnπ Cn =

1 jπn

Cn =

1 jπn

1 6 C−n = ∫ f(x)ejnπ(x−5) dx 2 4 6 1 5 C−n = [∫ ejnπ(x−5) dx + ∫ 2ejnπ(x−5) dx] 2 4 5 5

6

C−n

1 ejnπ(x−5) 2ejnπ(x−5) = [ ] +[ ] 2 jnπ jnπ 4 5

C−n

1 1 e−jnπ 2ejnπ 2 = [ − + − ] 2 jnπ jnπ jnπ jnπ

1 1 1 2 C−n = [− + − ] 2 jnπ jnπ jnπ C−n = − C−n =

1 jπn

j πn ∞

3 1 jnωt j −jnωt f(t) = + ∑ e + e 2 jπn πn 1

3.- Hallar la serie de fourier de:

𝜋

−𝜋

𝑇 = 2𝜋

f(t) = t Señal impar a0 = an = 0 bn =?

∞

f(t) = ∑ bn sin nt 1 T

2 2 bn = ∫ f(t) sin nt dt T −T 2

bn =

1 π ∫ t sin nt dt π −π

u=t

dv = sin nt

du = dt

v=−

bn =

cos nt n

1 cos nt 1 + ∫ cos nt dt] [−t π n n

1 cos nt sin nt π bn = [−t + 2 ] π n n −π bn =

1 π(−1)n π(−1)n + ( ) π n n

2(−1)n bn = n ∞

f(t) = ∑ 1

2(−1)n sin nt n

4.- Hallar la serie de fourier de f(t) = |Asin w0 t|:

𝐴

−𝜋

𝜋 𝑇 = 2𝜋

f(t) = |Asin w0 t| Señal par a0 = an =? bn = 0

∞

a0 f(t) = + ∑ an cos nt 2 1

T

2 2 a0 = ∫ f(t)dt T −T 2

a0 =

2 π ∫ Asin w0 t dt π 0

a0 =

2A [cos w0 t]π0 π

a0 =

2A [2] π

a0 =

4A π T

2 2 an = ∫ f(t) cos(nt) dt T −T 2

an =

2A π ∫ sin w0 t cos(nt) dt π 0 w0 = 1

an =

2A π sin(nt + t) + sin(nt − t) [∫ dt] π 0 2

an =

A cos(n + 1)t cos(n − 1)t π − [− ] π n+1 n−1 0

an =

A (−1)n (−1)n 1 1 [− − + + ] π n+1 n−1 n+1 n−1

an =

A (−1)n+1 − 1 1 − (−1)n+1 [ + ] π n−1 n+1

2(−1)n+1 − 2 an = [ ] π(n2 − 1) a1 =

2A π ∫ sin t cos(t) dt π 0

u = sin t a1 =

du = cos t dt

2A π du ∫ u cos t π 0 cos t

2A π a1 = ∫ udu π 0 a1 =

2A u2 π 2

a1 =

A [sin2 t]π0 π

a1 = 0 ∞

2A 2(−1)n+1 − 2 f(t) = + ∑[ ] cos nt π π(n2 − 1) 2

5.- Hallar la serie de fourier de f(x) = sen(3x) ,

−π