Encontrar la representación en serie de fourier de la siguiente señal \[ x(t) = cos(2t + 3) + sin(6t) + 1 \]

Desarrollo

1) Convertir a forma de Euler

Es caso 2 (expresado por coseno y seno) Usar relación de Euler \[ \sin(\theta) = - \frac{1}{2}i \left( e^{i\theta} - e^{-i\theta} \right) \] \[ \cos(\theta) = \frac{1}{2} \left( e^{i\theta} + e^{-i\theta} \right) \] La ecuación nos queda de la siguiente manera: \[ x(t) = \frac{1}{2} \left( e^{i(2t + 3)} + e^{-i(2t + 3)} \right) - \frac{1}{2}i \left( e^{i6t} - e^{-i6t} \right) + 1 e^0 \]

2) Encontrar la frecuencia fundamental.


\[ x(t) = \frac{1}{2} \left( e^{i( \color{purple} 2 \color{black}t + 3)} + e^{-i( \color{purple} 2 \color{black}t + 3)} \right) - \frac{1}{2}i \left( e^{i \color{purple} 6 \color{black} t} - e^{-i \color{purple} 6 \color{black} t} \right) + 1 e^{\color{purple}0 \color{black}} \] Ahora, calculamos el máximo común divisor.
2 6
2 1 3
3 1 1
\[ \text{MCM} = 2 \cdot 3 = 6 \]
\[ MCD = \frac {2 \cdot 6} {6} \] \[ MCD = 2 \] \[ \omega_0 = 2 \] Entonces, reemplazamos en \(e^{ik\omega_0t}\) \[ e^{ik2t} \]

3) Encontrar los coeficientes \(a_k\)


Reordenamos la ecuación para identificar los coeficientes más rápidamente.
\[ x(t) = \frac{1}{2} \left( e^{i( \color{purple} 2 \color{black}t + 3)} + e^{-i( \color{purple} 2 \color{black}t + 3)} \right) - \frac{1}{2}i \left( e^{i \color{purple} 6 \color{black} t} - e^{-i \color{purple} 6 \color{black} t} \right) + 1 e^{\color{purple}0 \color{black}} \] \[ x(t) = \frac{1}{2} e^{i( \color{purple} 2 \color{black}t + 3)} + \frac{1}{2} e^{-i( \color{purple} 2 \color{black}t + 3)} -\frac{1}{2}i e^{i \color{purple} 6 \color{black} t} + \frac{1}{2}i e^{-i \color{purple} 6 \color{black} t} + 1 e^{\color{purple}0 \color{black}} \] \[ x(t) = \frac{1}{2} e^{i\color{purple} 2 \color{black}t} e^{i3} + \frac{1}{2} e^{-i\color{purple} 2 \color{black}t} e^{-i3} -\frac{1}{2}i e^{i \color{purple} 6 \color{black} t} + \frac{1}{2}i e^{-i \color{purple} 6 \color{black} t} + 1 e^{\color{purple}0 \color{black}} \] \[ x(t) = \color{orange} \frac{1}{2} \color{black} e^{i\color{purple} 2 \color{black}t} \color{orange} e^{i3} \color{black} + \color{orange} \frac{1}{2} \color{black} e^{-i\color{purple} 2 \color{black}t} \color{orange} e^{-i3} \color{black} \color{orange} -\frac{1}{2}i \color{black} e^{i \color{purple} 6 \color{black} t} + \color{orange} \frac{1}{2}i \color{black} e^{-i \color{purple} 6 \color{black} t} + \color{orange} 1 \color{black} e^{\color{purple}0 \color{black}} \] \[ x(t) = \color{orange} \frac{1}{2} \color{black} \color{orange} e^{i3} \color{black} e^{i\color{purple} 2 \color{black}t} + \color{orange} \frac{1}{2} \color{black} \color{orange} e^{-i3} \color{black} e^{-i\color{purple} 2 \color{black}t} \color{orange} -\frac{1}{2}i \color{black} e^{i \color{purple} 6 \color{black} t} + \color{orange} \frac{1}{2}i \color{black} e^{-i \color{purple} 6 \color{black} t} + \color{orange} 1 \color{black} e^{\color{purple}0 \color{black}} \] Ahora, identificamos a los \(k\) evaluando \(\omega_0\).
Exponente en \(x(t)\) Relación Valor de \(k\)
\[ e^{i2t} \] \[ \color{purple} k \color{black} \omega_0 = \color{purple} 2 \color{black} \] \[ \color{purple} 1 \color{black} \cdot 2 = \color{purple} 2 \color{black} \] \(1\)
\[ e^{-i2t} \] \[ \color{purple} k \color{black} \omega_0 = \color{purple} -2 \color{black} \] \[ \color{purple} -1 \color{black} \cdot 2 = \color{purple} -2 \color{black} \] \(-1\)
\[ e^{i6t} \] \[ \color{purple} k \color{black} \omega_0 = \color{purple} 6 \color{black} \] \[ \color{purple} 3 \color{black} \cdot 2 = \color{purple} 6 \color{black} \] \(3\)
\[ e^{-i6t} \] \[ \color{purple} k \color{black} \omega_0 = \color{purple} -6 \color{black} \] \[ \color{purple} -3 \color{black} \cdot 2 = \color{purple} -6 \color{black} \] \(-3\)
\[ 1 e^{\color{purple}0 \color{black}} \] \[ \color{purple} k \color{black} \omega_0 = \color{purple} 0 \color{black} \] \[ \color{purple} 0 \color{black} \cdot 2 = \color{purple} 0 \color{black} \] \(0\)
Entonces, los coeficientes \(a_k\) son:
\[ a_{\color{purple} 1 \color{black}} = \color{orange} \frac{1}{2} e^{i3} \color{black} \] \[ a_{\color{purple} -1 \color{black}} = \color{orange} \frac{1}{2} e^{-i3} \color{black} \]
\[ a_{\color{purple} 3 \color{black}} = \color{orange} -\frac{1}{2}i \color{black} \] \[ a_{\color{purple} -3 \color{black}} = \color{orange} \frac{1}{2}i \color{black} \]
\[ a_{\color{purple} 0 \color{black}} = \color{orange} 1 \color{black} \]

4) Finalmente, construimos la sumatoria de Fourier.


\[ x(t) = \sum_{k=-\infty}^{k=\infty} a_k e^{ik\omega_0 t} \]
\[ x(t) = \sum_{\color{purple} k=-3\color{black}}^{\color{purple} k=3 \color{black}} \color{orange} a_{\color{purple} k }\color{black} e^{i \color{purple} k \color{black} 2 t} \] Con \(k \neq 2, -2\)