Serie exponencial de Fourier

Es común que para algunas aplicaciones sea conveniente expresar las series en términos de los exponenciales complejos {{e}^{\pm jn{{\omega }_{0}}t}}.

Sean:     Cos\left( n{{\omega }_{0}}t \right)=\frac{1}{2}\left( {{e}^{jn{{\omega }_{0}}t}}+{{e}^{-jn{{\omega }_{0}}t}} \right) ~~~ y ~~~ Sen\left( n{{\omega }_{0}}t \right)=\frac{1}{j2}\left( {{e}^{jn{{\omega }_{0}}t}}-{{e}^{-jn{{\omega }_{0}}t}} \right)


Si sabemos que:     f\left( t \right)=\frac{{{a}_{0}}}{2}+\underset{n=1}{\overset{\infty }{\mathop \sum }}\,({{a}_{n}}Cos\left( n{{\omega }_{0}}t \right)+~{{b}_{n}}Sen\left( n{{\omega }_{0}}t \right))


Se puede decir que:

f\left( t \right)=\frac{{{a}_{0}}}{2}+\underset{n=1}{\overset{\infty }{\mathop \sum }}\,({{a}_{n}}\frac{1}{2}\left( {{e}^{jn{{\omega }_{0}}t}}+{{e}^{-jn{{\omega }_{0}}t}} \right)+~{{b}_{n}}\frac{1}{j2}\left( {{e}^{jn{{\omega }_{0}}t}}-{{e}^{-jn{{\omega }_{0}}t}} \right))

f\left( t \right)=\frac{{{a}_{0}}}{2}+\underset{n=1}{\overset{\infty }{\mathop \sum }}\,\left( \frac{{{a}_{n}}}{2}{{e}^{jn{{\omega }_{0}}t}}+\frac{{{a}_{n}}}{2}{{e}^{-jn{{\omega }_{0}}t}} \right)+~\left( \frac{{{b}_{n}}}{j2}{{e}^{jn{{\omega }_{0}}t}}-\frac{{{b}_{n}}}{j2}{{e}^{-jn{{\omega }_{0}}t}} \right)

f\left( t \right)=\frac{{{a}_{0}}}{2}+\underset{n=1}{\overset{\infty }{\mathop \sum }}\,\left( \frac{{{a}_{n}}}{2}{{e}^{jn{{\omega }_{0}}t}}-\frac{j{{b}_{n}}}{2}{{e}^{jn{{\omega }_{0}}t}} \right)+~\left( \frac{{{a}_{n}}}{2}{{e}^{-jn{{\omega }_{0}}t}}+\frac{j{{b}_{n}}}{2}{{e}^{-jn{{\omega }_{0}}t}} \right)

f\left( t \right)=\frac{{{a}_{0}}}{2}+\underset{n=1}{\overset{\infty }{\mathop \sum }}\,{{e}^{jn{{\omega }_{0}}t}}\left( \frac{{{a}_{n}}}{2}-\frac{j{{b}_{n}}}{2} \right)+~{{e}^{-jn{{\omega }_{0}}t}}\left( \frac{{{a}_{n}}}{2}+\frac{j{{b}_{n}}}{2} \right)

f\left( t \right)=\frac{{{a}_{0}}}{2}+\underset{n=1}{\overset{\infty }{\mathop \sum }}\,\frac{1}{2}\left( {{a}_{n}}-j{{b}_{n}} \right){{e}^{jn{{\omega }_{0}}t}}+~\frac{1}{2}\left( {{a}_{n}}+j{{b}_{n}} \right){{e}^{-jn{{\omega }_{0}}t}}

Si se hace:

{{C}_{0=~}}\frac{{{a}_{0}}}{2},~~{{C}_{n}}=\frac{1}{2}\left( {{a}_{n}}-j{{b}_{n}} \right)~~{{C}_{-n}}=\frac{1}{2}\left( {{a}_{n}}+j{{b}_{n}} \right)

Entonces:

f\left( t \right)=~{{C}_{0}}+\underset{n=1}{\overset{\infty }{\mathop \sum }}\,{{C}_{n~}}{{e}^{jn{{\omega }_{0}}t}}+{{C}_{-n}}{{e}^{-jn{{\omega }_{0}}t}}

f\left( t \right)=~{{C}_{0}}+\underset{n=1}{\overset{\infty }{\mathop \sum }}\,{{C}_{n~}}{{e}^{jn{{\omega }_{0}}t}}+\underset{n=-1}{\overset{-\infty }{\mathop \sum }}\,{{C}_{n}}{{e}^{jn{{\omega }_{0}}t}}

Es decir:     f\left( t \right)=\underset{n=-\infty }{\overset{\infty }{\mathop \sum }}\,{{C}_{n~}}{{e}^{jn{{\omega }_{0}}t}}


Los coeficientes de la serie compleja de Fourier se calculan así:

{{C}_{0}}=\frac{1}{T}\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\mathop \int }}\,f\left( t \right)dt=\frac{1}{T}\underset{0}{\overset{T}{\mathop \int }}\,f\left( t \right)dt

{{C}_{n}}=\frac{1}{T}\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\mathop \int }}\,f\left( t \right){{e}^{-jn{{\omega }_{0}}t}}dt=\frac{1}{T}\underset{0}{\overset{T}{\mathop \int }}\,f\left( t \right){{e}^{-jn{{\omega }_{0}}t}}dt

{{C}_{n}}=\left| {{C}_{n}} \right|{{e}^{j{{\text{ }\!\!\Phi\!\!\text{ }}_{n}}}} ~~~ , ~~~ {{C}_{-n}}={{C}^{*}}=\left| {{C}_{n}} \right|{{e}^{-j{{\text{ }\!\!\Phi\!\!\text{ }}_{n}}}}

{{C}_{n}}=\frac{1}{2}\sqrt{{{a}_{n}}^{2}+{{b}_{n}}^{2}}

{{\phi }_{n}}=Ta{{n}^{-1}}\left( -\frac{{{b}_{n}}}{{{a}_{n}}} \right)

Para todos los valores n, excepto para n=0. En este caso {{C}_{0}} es real y


{{C}_{0}}=\frac{1}{2}{{a}_{0}}