Kurt's Assignment: Difference between revisions

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=Common Synthesizer Waveforms=
=Common Synthesizer Waveforms=
<math>\begin{align}
<math>\begin{align}
         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==


<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]\\
         &=\frac{H}{2}-\frac{H}{2}\\
         &=\frac{H}{2}-\frac{H}{2}\\
        &=0
\end{align}</math>
<math>\begin{align}
        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\\
        &=\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\\
        &=\frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\sin(n\pi)\right]
       
         &=0
         &=0
\end{align}</math>
\end{align}</math>


TODO: finish
TODO: finish

Revision as of 14:44, 1 November 2010

Common Synthesizer Waveforms

x(t)=x(t+T)=a0+∑n=1∞ancos⁡(nω0t)+bnsin⁡(nω0t)a0=1T∫0Tf(t)dtan=2T∫0Tf(t)cos⁡(nω0t)dtbn=2T∫0Tf(t)sin⁡(nω0t)dt

Square Wave

a0=1T∫012THdt+1T∫12TT−Hdt=1T[Ht]|t=012T−1T[Ht]|t=12TT=1TH12T−0−[1THT−1TH12T]=H2−[H−12H]=H2−H2=0


an=2T∫012THcos⁡(nω0t)dt+2T∫12TT−Hcos⁡(nω0t)dt=2T[Hnω0sin⁡(nω0t)]012T+2T[−Hnω0sin⁡(nω0t)]12TT=2T[Hn2πTsin⁡(nπ)]=0

TODO: finish