SuperNotes by yuri.rodrix

Notas de Yuri.Rodrix

Página tipo blog en el que voy a publicar mis notas de aprendizaje, en especial de temas como matemáticas, física y quizá algo de programación

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Transformada y series de Fourier

Serie de Fourier de f(x)\Large f(x)

Toda funcion periódica puede ser expresada como serie de Fourier

f(x)=∑n=0∞(Ancos⁡(2nπxT)+Bnsin⁡(2nπxT))\Large f(x)=\sum_{n=0}^{\infty } \left( A_n \cos( \frac{2n\pi x}{T} )+B_n\sin( \frac{2n\pi x}{T}) \right)

Simil con el vector

Base ortonormal canónica
{en}n=1∞\Large \{e_n\}^\infty_{n=1}
Base ortonormal Serie de Fourier
{cos⁡(nx),sin⁡(nx)}n=1∞\Large \{\cos(nx), \sin(nx) \}^\infty_{n=1}
Vector genérico en dicha base de ejemplo:
v→=∑n=1∞vnen→\overrightarrow{v} = \sum_{n=1}^{\infty} \textcolor{#3DBBE9}{v_n} \textcolor{#ffb764}{\overrightarrow{e_n}}
f(x)=∑n=0∞(Ancos⁡(nx)+Bnsin⁡(nx))f(x) = \sum_{n=0}^{\infty } \left( \textcolor{#3DBBE9}{A_n} \textcolor{#ffb764}{\cos( n x )} + \textcolor{#3DBBE9}{B_n} \textcolor{#ffb764}{\sin(nx)} \right)
Hallar cierto coeficiente:
v→⋅em→=∑n=1∞vnen→⋅em→\overrightarrow{v} \cdot \overrightarrow{e_{\textcolor{#FF6040}m}} = \sum_{n=1}^{\infty} \textcolor{#3DBBE9}{v_n} \textcolor{#ffb764}{\overrightarrow{e_n}} \cdot \overrightarrow{e_{\textcolor{#FF6040}m}}
en→⋅em→=δnm\overrightarrow{e_n} \cdot \overrightarrow{e_m} = \delta_{nm}
v→⋅em→=∑n=1∞vnδnm\overrightarrow{v} \cdot \overrightarrow{e_{\textcolor{#FF6040}m}} = \sum_{n=1}^{\infty} \textcolor{#3DBBE9}{v_n} \delta_{\textcolor{#ffb764}n\textcolor{#FF6040}m} v→⋅em→=vm\overrightarrow{v} \cdot \overrightarrow{e_{\textcolor{#FF6040}m}} = \textcolor{#FF6040}{v_m}
f(x)⋅cos⁡(mx)=∑n=1∞(Anc(nx)+Bns(nx))⋅cos⁡(mx)f(x) \cdot \cos(\textcolor{#FF6040}mx) = \sum_{n=1}^{\infty } \left( \textcolor{#3DBBE9}{A_n} \textcolor{#ffb764}{c( n x ) } + \textcolor{#3DBBE9}{B_n} \textcolor{#ffb764}{s(nx)} \right)\cdot \cos(\textcolor{#FF6040}mx)
cos⁡(nx)⋅cos⁡(mx)=(2π/2)δnm=πδnm\cos(nx) \cdot \cos(mx) = (2\pi /2) \delta_{nm} = \pi \delta_{nm}cos⁡(nx)⋅sin⁡(mx)=0\cos(nx) \cdot \sin(mx) = 0sin⁡(nx)⋅sin⁡(mx)=(2π/2)δnm=πδnm\sin(nx) \cdot \sin(mx) = (2\pi /2) \delta_{nm}= \pi \delta_{nm}
f(x)⋅cos⁡(mx)=∑n=1∞(Anπδnm+Bn0)f(x) \cdot \cos(\textcolor{#FF6040}mx) = \sum_{n=1}^{\infty } \left( \textcolor{#3DBBE9}{A_n} \pi \delta_{\textcolor{#ffb764}n\textcolor{#FF6040}m} + \textcolor{#3DBBE9}{B_n}0 \right)f(x)⋅cos⁡(mx)=∑n=1∞Anπδnmf(x) \cdot \cos(\textcolor{#FF6040}mx) = \sum_{n=1}^{\infty } \textcolor{#3DBBE9}{A_n}\pi \delta_{\textcolor{#ffb764}n\textcolor{#FF6040}m} f(x)⋅cos⁡(mx)=Amπf(x) \cdot \cos(\textcolor{#FF6040}mx) = \textcolor{#FF6040}{A_m} \pi

De forma general para hallar los coeficientes de Fourier
f(x)=∑n=0∞(Ancos⁡(nx)+Bnsin⁡(nx))f(x) = \sum_{n=0}^{\infty } \left( \textcolor{#3DBBE9}{A_n} \textcolor{#ffb764}{\cos( n x )} + \textcolor{#3DBBE9}{B_n} \textcolor{#ffb764}{\sin(nx)} \right)
T=2πω  , x∈(0,T)T=\frac{2\pi}{\omega} \ \ , \ x \in (0,T)
f(x)=A0cos⁡(2(0)πxT)+B0sin⁡(2(0)πxT)+∑n=1∞(Ancos⁡(2nπxT)+Bnsin⁡(2nπxT))\small f(x) = A_0 \cos(\frac{2 (0)\pi x}{T}) + B_0\sin(\frac{2 (0)\pi x}{T})+ \sum_{n=1}^{\infty } \left( \textcolor{#3DBBE9}{A_n} \textcolor{#ffb764}{\cos( \frac{2n \pi x}{T})} + \textcolor{#3DBBE9}{B_n} \textcolor{#ffb764}{\sin(\frac{2n \pi x}{T})} \right)f(x)=A0(1)+B0(0)+∑n=1∞(Ancos⁡(2nπxT)+Bnsin⁡(2nπxT))\small f(x) = A_0 (1) + B_0(0)+ \sum_{n=1}^{\infty } \left( \textcolor{#3DBBE9}{A_n} \textcolor{#ffb764}{\cos( \frac{2n \pi x}{T})} + \textcolor{#3DBBE9}{B_n} \textcolor{#ffb764}{\sin(\frac{2n \pi x}{T})} \right)f(x)=A0+∑n=1∞(Ancos⁡(2nπxT)+Bnsin⁡(2nπxT))\large f(x) = A_0 + \sum_{n=1}^{\infty } \left( \textcolor{#3DBBE9}{A_n} \textcolor{#ffb764}{\cos( \frac{2n \pi x}{T})} + \textcolor{#3DBBE9}{B_n} \textcolor{#ffb764}{\sin(\frac{2n \pi x}{T})} \right)
Hallando AnA_n
f(x)⋅cos⁡(2mπxT)=A0cos⁡(2mπxT)⋅1+∑n=1∞(AnT2δnm+Bn0)f(x) \cdot \cos(\frac{2m \pi x}{T}) = A_0 \cos(\frac{2m \pi x}{T}) \cdot 1 + \sum_{n=1}^{\infty } \left( \textcolor{#3DBBE9}{A_n} \frac{T}2 \delta_{nm} + \textcolor{#3DBBE9}{B_n} 0 \right)f(x)⋅cos⁡(2mπxT)=AmT2f(x) \cdot \cos(\frac{2m \pi x}{T}) = A_m \frac{T}2
2T∫0Tcos⁡(2nπxT)f(x)dx=An\frac{2}{T} \int^{T}_0 \cos(\frac{2n \pi x}{T}) f(x) dx= A_n

Hallando BnB_n
f(x)⋅sin⁡(2mπxT)=A0sin⁡(2mπxT)⋅1+∑n=1∞(An0+BnT2δnm)f(x) \cdot \sin(\frac{2m \pi x}{T}) = A_0 \sin(\frac{2m \pi x}{T}) \cdot 1 + \sum_{n=1}^{\infty } \left( \textcolor{#3DBBE9}{A_n} 0 + \textcolor{#3DBBE9}{B_n} \frac{T}{2} \delta_{nm} \right)f(x)⋅sin⁡(2mπxT)=BmT2f(x) \cdot \sin(\frac{2m \pi x}{T}) = B_m \frac{T}{2}∫0Tsin⁡(2mπxT)f(x)dx=BmT2 \int^{T}_0 \sin(\frac{2m \pi x}{T}) f(x) dx= B_m \frac{T}{2}
2T∫0Tsin⁡(2nπxT)f(x)dx=Bn\frac{2}{T} \int^{T}_0 \sin(\frac{2n \pi x}{T}) f(x) dx= B_n

Hallando A0A_0
f(x)⋅1=A0⋅1+∑n=1∞(An0+Bn0)f(x) \cdot 1 = A_0 \cdot 1 + \sum_{n=1}^{\infty } \left( \textcolor{#3DBBE9}{A_n} 0 + \textcolor{#3DBBE9}{B_n} 0 \right)f(x)⋅1=A01⋅1f(x) \cdot 1 = A_0 1 \cdot 1∫0T1f(x)dx=A0∫0T1dx=A0(T−0)dx=AT \int^{T}_0 1 f(x) dx= A_0 \int^{T}_0 1 dx = A_0 (T-0) dx = A T
1T∫0T1f(x)dx=A0\frac{1}{T} \int^{T}_0 1 f(x) dx= A_0

"Complejizando" la serie de Fourier

f(x)=∑n=0∞(Ancos⁡(nx)+Bnsin⁡(nx))f(x) = \sum_{n=0}^{\infty } \left( A_n \cos( n x )+B_n\sin( n x) \right)
cos⁡(nx)=einx+e−inx2\cos(nx) = \frac{e^{i n x} + e^{-i n x}}{2}sin⁡(nx)=einx−e−inx2i\sin(nx) = \frac{e^{i n x} - e^{-i n x}}{2i}
f(x)=∑n=0∞(Aneinx+e−inx2+Bneinx−e−inx2i)f(x) = \sum_{n=0}^{\infty } \left( A_n \frac{e^{i n x} + e^{-i n x}}{2} + B_n \frac{e^{i n x} - e^{-i n x}}{2i} \right)f(x)=∑n=0∞(An2einx+An2e−inx+Bn2ieinx−Bn2ie−inx)f(x) = \sum_{n=0}^{\infty } \left( \frac{A_n}{2} e^{i n x} + \frac{A_n}{2} e^{-i n x} + \frac{B_n}{2i} e^{i n x} - \frac{B_n}{2i} e^{-i n x} \right)f(x)=∑n=0∞((An2+Bn2i)einx+(An2−Bn2i)e−inx)f(x) = \sum_{n=0}^{\infty } \left( \left( \frac{A_n}{2} + \frac{B_n}{2i} \right) e^{i n x} + \left( \frac{A_n}{2} - \frac{B_n}{2i} \right) e^{-i n x} \right)

Recordando el conjugado

f(x)=∑n=0∞((An2+Bn2i)einx+(An2+Bn2i)einx‾)f(x) = \sum_{n=0}^{\infty } \left( \textcolor{#80BBFF}{\left( \frac{A_n}{2} + \frac{B_n}{2i} \right) e^{i n x} } + \overline{ \textcolor{#80BBFF}{ \left( \frac{A_n}{2} + \frac{B_n}{2i} \right) e^{i n x} } } \right)

Aquí entiende al n como el ángulo, de tal forma que:

arg⁡(z)=n⇔arg⁡(zˉ)=−n\arg( z )= n ⇔ \arg( \bar{z} )= -n

f(x)=∑n=0∞(An2+Bn2i)einx+∑n=0∞(An2+Bn2i)einx‾f(x) = \sum_{n=0}^{\infty } \textcolor{#80BBFF}{\left( \frac{A_n}{2} + \frac{B_n}{2i} \right) e^{i n x} } + \sum_{n=0}^{\infty } \overline{ \textcolor{#80BBFF}{ \left( \frac{A_n}{2} + \frac{B_n}{2i} \right) e^{i n x} } } f(x)=∑n=0∞(An2+Bn2i)einx+∑n=−1−∞(An2+Bn2i)einxf(x) = \sum_{n=0}^{\infty } \textcolor{#80BBFF}{\left( \frac{A_n}{2} + \frac{B_n}{2i} \right) e^{i n x} } + \sum_{n=-1}^{-\infty } \textcolor{#80BBFF}{ \left( \frac{A_n}{2} + \frac{B_n}{2i} \right) e^{i n x} } f(x)=∑n=−∞∞(An2+Bn2i)einxf(x) = \sum_{n=-\infty}^{\infty } \textcolor{#80BBFF}{\left( \frac{A_n}{2} + \frac{B_n}{2i} \right) e^{i n x} }
cn=(An2+Bn2i)einxc_n = \left( \frac{A_n}{2} + \frac{B_n}{2i} \right) e^{i n x}
y asıˊ\text{y así}
c−n=cn‾=(An2−Bn2i)e−inx c_{-n} = \overline{c_n}= \left( \frac{A_n}{2} - \frac{B_n}{2i} \right) e^{-i n x}

f(x)=∑n=−∞∞Cneinxf(x) = \sum_{n=-\infty}^{\infty} C_n e^{i n x}
Cn=1T∫0Tf(x)e−inxdxC_n = \frac{1}{T} \int_0^T f(x) e^{-i n x} dx

Normalizando

Cn=1T∫0Tf(x)e−i2πxn/TdxC_n = \frac{1}{T} \int_0^T f(x) e^{-i 2\pi x n /T} dx

"Continuizando" la serie de Fourier

Válido cuando T→∞T \to \infty

f(t)=∑n=−∞∞Cneintf(t) = \sum_{n=-\infty}^{\infty} C_n e^{i n t}f(t)=∑n=−∞∞Cnei2πtn/Tf(t) = \sum_{n=-\infty}^{\infty} C_n e^{i 2 \pi t n / T}
w=n/Tw = n/T
f(t)=∑n=−∞∞Cwei2πtwf(t) = \sum_{n=-\infty}^{\infty} C_w e^{i 2 \pi t w}f(t)=∫−∞∞C(w)ei2πtwdwf(t) = \int_{-\infty}^{\infty} C(w) e^{i 2\pi t w} dwf(t)e−i2πtw′=∫−∞∞C(w)ei2πtwe−i2πtw′dwf(t) \textcolor{#FF90FF}{e^{ - i 2\pi t \textcolor{#FF9055}{w'}}} = \int_{-\infty}^{\infty} C(w) e^{i 2\pi t w} \textcolor{#FF90FF}{e^{ - i 2\pi t \textcolor{#FF9055}{w'}}} dw∫−∞+∞f(t)e−i2πtw′dt=∫−∞∞∫−∞∞C(w)ei2πtwe−i2πtw′dwdt\int_{-\infty}^{+\infty} f(t)\textcolor{#FF90FF}{e^{ - i 2\pi t \textcolor{#FF9055}{w'}}} dt= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} C(w) e^{i 2\pi t w}\textcolor{#FF90FF}{e^{ - i 2\pi t \textcolor{#FF9055}{w'}}} dw dtei2πtw∗e−i2πtw′dt=∫−∞∞ei2πtwe−i2πtw′dt e^{ i 2\pi t w} *\textcolor{#FF90FF}{e^{ - i 2\pi t \textcolor{#FF9055}{w'}}} dt= \int_{-\infty}^{\infty} e^{i 2\pi t w}\textcolor{#FF90FF}{e^{ - i 2\pi t \textcolor{#FF9055}{w'}}} dtei2πtw∗e−i2πtw′dt=δ(w−w′) e^{ i 2\pi t w} *\textcolor{#FF90FF}{e^{ - i 2\pi t \textcolor{#FF9055}{w'}}} dt= \delta(w-\textcolor{#FF9055}{w'})∫−∞+∞f(t)e−i2πtw′dt=∫−∞∞C(w)δ(w−w′)dw\int_{-\infty}^{+\infty} f(t)\textcolor{#FF90FF}{e^{ - i 2\pi t \textcolor{#FF9055}{w'}}} dt= \int_{-\infty}^{\infty} C(w) \delta(w-\textcolor{#FF9055}{w'}) dw ∫−∞+∞f(t)e−i2πtw′dt=C(w′)\int_{-\infty}^{+\infty} f(t)\textcolor{#FF90FF}{e^{ - i 2\pi t \textcolor{#FF9055}{w'}}} dt= C(\textcolor{#FF9055}{w'})
F(w)=∫−∞+∞f(t)e−i2πtwdtF(w) = \int_{-\infty}^{+\infty} f(t)e^{ - i 2\pi t w} dt