Martinez's Fourier Assignment: Difference between revisions

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Solve for a<sub>3</sub> for the
Prove that a<sub>3</sub> = 0 for the waveform below:

[[Image:Circuit8.png|thumb|400px|center]]



<math>\begin{align}
T &= 6 seconds\\
a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\
b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\
\frac{2\pi}{\omega_0}\ &= 6\\
\because \!\, T &= 6\\
\therefore \!\,\omega_o &= \frac{pi}{3}\\
a_3 &= \frac{2}{6}\int_0^6 f(t)\cos(3\omega_0t)\, dt\\
a_3 &= \frac{1}{3}[\int_2^3 \cos(\pi*t)\, dt + \int_3^4 \cos(3\omega_ot)\, dt\\
\because \!\, \omega_o &= \frac{pi}{3}\\
a_3 &= \frac{1}{3}[\frac{10}{pi}(sin(3\pi)-sin(2\pi)+\frac{5}{pi}(sin(4\pi)-sin(3\pi)]\\
a_3 &= \frac{1}{3}[\frac{10}{pi}(0)+\frac{5}{pi}(0)]\\
a_3 &= 0\\
\end{align}

Latest revision as of 01:03, 13 December 2010

Prove that a3 = 0 for the waveform below:

Circuit8.png


<math>\begin{align} T &= 6 seconds\\ a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\ b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\ \frac{2\pi}{\omega_0}\ &= 6\\ \because \!\, T &= 6\\ \therefore \!\,\omega_o &= \frac{pi}{3}\\ a_3 &= \frac{2}{6}\int_0^6 f(t)\cos(3\omega_0t)\, dt\\ a_3 &= \frac{1}{3}[\int_2^3 \cos(\pi*t)\, dt + \int_3^4 \cos(3\omega_ot)\, dt\\ \because \!\, \omega_o &= \frac{pi}{3}\\ a_3 &= \frac{1}{3}[\frac{10}{pi}(sin(3\pi)-sin(2\pi)+\frac{5}{pi}(sin(4\pi)-sin(3\pi)]\\ a_3 &= \frac{1}{3}[\frac{10}{pi}(0)+\frac{5}{pi}(0)]\\ a_3 &= 0\\ \end{align}