Ejemplo 3

Encontrar la serie trigonométrica de Fourier para la señal triangular que se muestra en la siguiente figura:

Figura 5. Señal triangular. La figura muestra la señal triangular y los intervalos de tiempo definidos para aplicar la serie trigonométrica.

(Para ampliar la imagen haga clic sobre ella)

Para definir la función que describe esta señal, se analiza cada uno de los intervalos y, por medio de la ecuación de la recta, se establece:


f\left( t \right)=\left\{ \begin{matrix} ~~1+\frac{4t}{T},~~~-\frac{T}{2} \textless t\le 0 \\ \\ 1-\frac{4t}{T},~~~~0 \textless t\le \frac{T}{2} \\ \end{matrix} \right.


{{a}_{0}}=\frac{2}{T}\left[ \underset{-\frac{T}{2}}{\overset{0}{\mathop \int }}\,\left( 1+\frac{4t}{T} \right)dt+\underset{0}{\overset{\frac{T}{2}}{\mathop \int }}\,\left( 1-\frac{4t}{T} \right)dt \right]=\frac{2}{T}\left[ \left. t+\frac{2{{t}^{2}}}{T} \right|\begin{matrix} 0 \\ -\frac{T}{2} \\ \end{matrix}+\left. (t-\frac{2{{t}^{2}}}{T} \right|\begin{matrix} \frac{T}{2} \\ 0 \\ \end{matrix} \right]

=\frac{2}{T}\left[ -\frac{T}{2}+\frac{T}{2}+\frac{T}{2}-\frac{T}{2} \right]=0

Para desarrollar la integral
\underset{-\frac{T}{2}}{\overset{0}{\mathop \int }}\,\underline{\left( 1+\frac{4t}{T} \right)Cos\text{ }\!\!~\!\!\text{ }\left( n{{\omega }_{0}}t \right)dt}

En la integral definimos {{u}_{1}}=1+\frac{4t}{T} ~~~~~,~~~~~ d{{u}_{1}}=\frac{4}{T}dt

y
d{{v}_{1}}=Cos\text{ }\!\!~\!\!\text{ }\left( n{{\omega }_{0}}t \right)dt ~~~~~,~~~~~ {{v}_{1}}=\text{ }\!\!~\!\!\text{ }\frac{1}{n{{\omega }_{0}}}Sen\left( n{{\omega }_{0}}t \right)
(Integral por partes)
u\text{*}v-\int v\text{*}du)

{{a}_{n}}=\frac{2}{T}\left[ \underset{-\frac{T}{2}}{\overset{0}{\mathop \int }}\,\underline{\left( 1+\frac{4t}{T} \right)Cos\text{ }\!\!~\!\!\text{ }\left( n{{\omega }_{0}}t \right)dt}+\underset{0}{\overset{\frac{T}{2}}{\mathop \int }}\,\underline{\left( 1-\frac{4t}{T} \right)Cos\text{ }\!\!~\!\!\text{ }\left( n{{\omega }_{0}}t \right)dt} \right]


\underset{\frac{T}{2}}{\overset{0}{\mathop \int }}\,\left( 1+\frac{4t}{T} \right)Cos\text{ }\!\!~\!\!\text{ }\left( n{{\omega }_{0}}t \right)dt=\left. \left( 1+\frac{4t}{T} \right)\frac{1}{n{{\omega }_{0}}}Sen\left( n{{\omega }_{0}}t \right)\text{ }\!\!~\!\!\text{ } \right|\begin{matrix} 0 \\ -\frac{T}{2} \\ \end{matrix}-\underset{-\frac{T}{2}}{\overset{0}{\mathop \int }}\,\text{ }\!\!~\!\!\text{ }\frac{1}{n{{\omega }_{0}}}Sen\left( n{{\omega }_{0}}t \right)\frac{4}{T}dt

Con {{\omega }_{0}}=\frac{2\pi }{T} entonces n{{\omega }_{0}}=n\frac{2\pi }{T}.


=\left. \left( 1+\frac{4t}{T} \right)\frac{1}{n{{\omega }_{0}}}Sen\left( n{{\omega }_{0}}t \right)\text{ }\!\!~\!\!\text{ } \right|\begin{matrix} 0 \\ -\frac{T}{2} \\ \end{matrix}=\text{ }\!\!~\!\!\text{ }0-\left( 1+\frac{4.\left( -\frac{T}{2} \right)}{T} \right)Sen\left( n\frac{2\pi }{T}.\left( -\frac{T}{2} \right) \right)

=0-\left( 1-2 \right)Sen\left( -n\pi \right)=0

Ahora bien,

\underset{-\frac{T}{2}}{\overset{0}{\mathop \int }}\,\text{ }\!\!~\!\!\text{ }\frac{1}{n{{\omega }_{0}}}Sen\left( n{{\omega }_{0}}t \right)\frac{4}{T}dt=-\frac{4}{T}{{\left( \frac{1}{n{{\omega }_{0}}} \right)}^{2}}\left. Cos\left( n{{\omega }_{0}}t \right) \right|\begin{matrix} 0 \\ -\frac{T}{2} \\ \end{matrix}=-\frac{4}{T}\left( \frac{{{T}^{2}}}{{{n}^{2}}4{{\pi }^{2}}} \right)\left( 1-Cos\left( n\pi \right) \right)

=\frac{-T}{{{n}^{2}}{{\pi }^{2}}}\left( 1-Cos\left( n\pi \right) \right)

\underset{0}{\overset{\frac{T}{2}}{\mathop \int }}\,\left( 1-\frac{4t}{T} \right)Cos~\left( n{{\omega }_{0}}t \right)dt=\left. \left( 1-\frac{4t}{T} \right)\frac{1}{n{{\omega }_{0}}}Sen\left( n{{\omega }_{0}}t \right)~ \right|\begin{matrix} \frac{2}{T} \\ 0 \\ \end{matrix}-\underset{0}{\overset{\frac{2}{T}}{\mathop \int }}\,~\frac{1}{n{{\omega }_{0}}}Sen\left( n{{\omega }_{0}}t \right)\left( -\frac{4}{T} \right)dt

Para desarrollar la integral
\underset{0}{\overset{\frac{T}{2}}{\mathop \int }}\,\underline{\left( 1-\frac{4t}{T} \right)Cos\text{ }\!\!~\!\!\text{ }\left( n{{\omega }_{0}}t \right)dt}~

En la integral definimos {{u}_{1}}=1-\frac{4t}{T} ~~~~~,~~~~~ d{{u}_{1}}=-\frac{4}{T}dt

y
~~d{{v}_{1}}=Cos~\left( n{{\omega }_{0}}t \right)dt ~~~~~,~~~~~ {{v}_{1}}=~\frac{1}{n{{\omega }_{0}}}Sen\left( n{{\omega }_{0}}t \right)
(Integral por partes)
u\text{*}v-\int v\text{*}du)

Con {{\omega }_{0}}=\frac{2\pi }{T} entonces n{{\omega }_{0}}=n\frac{2\pi }{T}


=0\left. \left( 1-\frac{4t}{T} \right)\frac{1}{n{{\omega }_{0}}}Sen\left( n{{\omega }_{0}}t \right)~ \right|\begin{matrix} \frac{T}{2} \\ 0 \\ \end{matrix}=~\left( 1-\frac{4.\left( \frac{T}{2} \right)}{T} \right)Sen\left( n\frac{2\pi }{T}.\frac{T}{2} \right)-0

=\left( 1-2 \right)Sen\left( n\pi \right)-0=0

Ahora bien,

-\underset{0}{\overset{\frac{T}{2}}{\mathop \int }}\,~\frac{1}{n{{\omega }_{0}}}Sen\left( n{{\omega }_{0}}t \right)\frac{4}{T}dt=\frac{4}{T}{{\left( \frac{1}{n{{\omega }_{0}}} \right)}^{2}}\left. Cos\left( n{{\omega }_{0}}t \right) \right|\begin{matrix} \frac{T}{2} \\ 0 \\ \end{matrix}=\frac{4}{T}\left( \frac{{{T}^{2}}}{{{n}^{2}}4{{\pi }^{2}}} \right)\left( Cos\left( n\pi \right)-1 \right)

=\frac{T}{{{n}^{2}}{{\pi }^{2}}}\left( Cos\left( n\pi \right)-1 \right)

Entonces:

{{a}_{n}}=\frac{2}{T}\left[ \frac{T}{{{n}^{2}}{{\pi }^{2}}}\left( 1-Cosn\pi \right)-\frac{T}{{{n}^{2}}{{\pi }^{2}}}\left( Cosn\pi -1 \right) \right]=~~~\frac{4}{T}.\frac{T}{{{n}^{2}}{{\pi }^{2}}}\left( 1-Cosn\pi \right),~~

{{a}_{n~~=}}~~\frac{4}{{{n}^{2}}{{\pi }^{2}}}\left( 1-Cosn\pi \right)

Como    Cosn\pi =~{{\left( -1 \right)}^{n}} ~~~,~~~ {{a}_{n}}=\left\{ \begin{matrix} 0,~~si~n~~es~par~ \\ \frac{8}{{{n}^{2}}{{\pi }^{2}}},~si~n~es~impar \\ \end{matrix} \right.


{{b}_{n}}=\frac{2}{T}\left[ \underset{-\frac{T}{2}}{\overset{0}{\mathop \int }}\,\underline{\left( 1+\frac{4t}{T} \right)Sen~\left( n{{\omega }_{0}}t \right)dt}+\underset{0}{\overset{\frac{T}{2}}{\mathop \int }}\,\underline{\left( 1-\frac{4t}{T} \right)Sen~\left( n{{\omega }_{0}}t \right)dt} \right]=0

Entonces podemos interpretar la solución como f\left( t \right)=~\frac{8}{{{\pi }^{2}}}\underset{n~impar}{\overset{\infty }{\mathop \sum }}\,\frac{1}{{{n}^{2}}}Cos\left( n{{\omega }_{0}}t \right)

Para el análisis de señales por medio de Fourier es bueno tener en cuenta: