Kurt's Assignment: Difference between revisions
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=Common Synthesizer Waveforms= |
=Common Synthesizer Waveforms= |
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<math>\begin{align} |
<math>\begin{align} |
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x(t) &= x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)\\ |
x(t) &= x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)\\ |
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==Square Wave== |
==Square Wave== |
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<math>\begin{align} |
<math>\begin{align} |
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&=\frac{H}{2} - \left[H-\frac{1}{2}H\right]\\ |
&=\frac{H}{2} - \left[H-\frac{1}{2}H\right]\\ |
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&=\frac{H}{2}-\frac{H}{2}\\ |
&=\frac{H}{2}-\frac{H}{2}\\ |
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&=0 |
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\end{align}</math> |
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<math>\begin{align} |
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a_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\cos(n\omega_0t) dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\cos(n\omega_0t) dt\\ |
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&=\frac{2}{T}\left[\frac{H}{n\omega_0}\sin(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{-H}{n\omega_0}\sin(n\omega_0t)\right]_{\frac{1}{2}T}^T\\ |
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&=\frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\sin(n\pi)\right] |
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&=0 |
&=0 |
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\end{align}</math> |
\end{align}</math> |
Revision as of 13:44, 1 November 2010
Common Synthesizer Waveforms
Square Wave
TODO: finish