CÁLCULO GRACELI SINTÉTICO.

$T^{\nu }{}_{\alpha }$ $\lambda$ ω ${\mathcal {T}}$ , $\sigma$ , $\psi$

A(z)=\sum _{{n=-\infty }}^{\infty }a_{n}z^{n}\,

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

B(z)=\sum _{n=-\infty }^{\infty }b_{n}z^{n}

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

\sum _{{n=-\infty }}^{\infty }a_{n}b_{n}^{*}={\frac {1}{2\pi }}\int _{{-\pi }}^{\pi }A(e^{{i\phi }})B(e^{{i\phi }})^{*}d\phi \,

/

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

\sum _{{n=-\infty }}^{\infty }|a_{n}|^{2}={\frac {1}{2\pi }}\int _{{-\pi }}^{\pi }|A(e^{{i\phi }})|^{2}d\phi \,

/

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

a_{0}b_{0}+2\Re \sum _{{n=1}}^{\infty }a_{n}b_{n}^{*}={\frac {1}{2\pi }}\int _{{-\pi }}^{\pi }A(e^{{i\phi }})B(e^{{i\phi }})d\phi \,

/

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

\int _{-\infty }^{\infty }|x(t)|^{2}dt=\int _{-\infty }^{\infty }|X(2\pi f)|^{2}df

/

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

\sum _{{n=-\infty }}^{{\infty }}|x[n]|^{2}={\frac {1}{2\pi }}\int _{{-\pi }}^{{\pi }}|X(e^{{j\phi }})|^{2}d\phi

/

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

\sum _{{n=0}}^{{N-1}}|x[n]|^{2}={\frac {1}{N}}\sum _{{k=0}}^{{N-1}}|X[k]|^{2}

/

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

X(\omega )=\sum _{n=-\infty }^{\infty }x[n]\,e^{-i\omega n}.

/

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

e^{-i(\omega +2\pi k)n}=e^{-i\omega n}\,

/

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

X(\omega +2\pi k)=X(\omega )\,

/

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

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