Martinez's Fourier Assignment - Revision history

Jonathanmartinez321 at 09:03, 13 December 2010

2010-12-13T09:03:16Z

← Older revision		Revision as of 02:03, 13 December 2010
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	~~Solve for~~ a<sub>3</sub> for the waveform below:		Prove that a<sub>3</sub> = 0 for the waveform below:

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


	~~T=6 seconds~~

	<math>\begin{align}		<math>\begin{align}
	a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\		T &= 6 seconds\\
	b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\		a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\
	\because \T=6, 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}		\end{align}

2010-12-13T08:31:02Z

← Older revision		Revision as of 01:31, 13 December 2010
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	a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\		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\\		b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\
	\because \!\T=6, dt\\		\because \T=6, dt\\
	\end{align}		\end{align}

2010-12-13T08:30:52Z

← Older revision		Revision as of 01:30, 13 December 2010
Line 9:		Line 9:
	a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\		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\\		b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\
			\because \!\T=6, dt\\
	\end{align}		\end{align}

2010-12-13T08:27:46Z

← Older revision		Revision as of 01:27, 13 December 2010
Line 8:		Line 8:
	<math>\begin{align}		<math>\begin{align}
	a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\		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\		b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\
	\end{align}		\end{align}

2010-12-13T08:27:33Z

← Older revision		Revision as of 01:27, 13 December 2010
Line 8:		Line 8:
	<math>\begin{align}		<math>\begin{align}
	a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\		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\
	\end{align}		\end{align}

2010-12-13T08:26:23Z

← Older revision		Revision as of 01:26, 13 December 2010
Line 5:		Line 5:

	T=6 seconds		T=6 seconds

	<math>\begin{align}		<math>\begin{align}
	a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\		a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\

	\end{align}		\end{align}

2010-12-13T08:26:02Z

← Older revision		Revision as of 01:26, 13 December 2010
Line 5:		Line 5:

	T=6 seconds		T=6 seconds
			<math>\begin{align}
	a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\		a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\

			\end{align}

2010-12-13T08:25:13Z

← Older revision		Revision as of 01:25, 13 December 2010
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	[[Image:Circuit8.png\|thumb\|400px\|center]]		[[Image:Circuit8.png\|thumb\|400px\|center]]


			T=6 seconds
			a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\

2010-12-13T08:23:39Z

← Older revision		Revision as of 01:23, 13 December 2010
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	Solve for a<sub>3</sub> for the waveform below:		Solve for a<sub>3</sub> for the waveform below:

	[[Image:Circuit8.png\|thumb\|~~200px~~\|center]]		[[Image:Circuit8.png\|thumb\|400px\|center]]

2010-12-13T08:22:47Z

← Older revision		Revision as of 01:22, 13 December 2010
Line 1:		Line 1:
	Solve for a<sub>3</sub> for the waveform below:		Solve for a<sub>3</sub> for the waveform below:

	[[Image:Circuit8.~~jpeg~~\|thumb\|200px\|center]]		[[Image:Circuit8.png\|thumb\|200px\|center]]