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Square and symmetric triangle waves

Figure 10.6: Symmetric triangle wave, obtained by superposing parabolic waves with $(M, c)$ pairs equal to $(0, 8)$ and $(N/2, -8)$.
\begin{figure}\psfig{file=figs/fig10.06.ps}\end{figure}

To see how to obtain Fourier series for classical waveforms in general, consider first the square wave,

\begin{displaymath} x[n] = s[n] - s[n-{N \over 2}] \end{displaymath}

equal to $1/2$ for the first half cycle ($0 <= n < N/2$) and $-1/2$ for the rest. We get the Fourier series by plugging in the Fourier series for $s[n]$ twice:
\begin{displaymath} x[n] \approx {1 \over \pi} \left [ {\sin ( \omega n )} + ... ...\over 2} + {{\sin ( 3 \omega n)} \over 3} + \cdots \right . \end{displaymath}


\begin{displaymath} \left . -{\sin ( \omega n )} + {{\sin ( 2 \omega n)} \over 2} - {{\sin ( 3 \omega n)} \over 3} \pm \cdots \right ] \end{displaymath}


\begin{displaymath} = {2 \over \pi} \left [ {\sin ( \omega n )} + {{\sin ( 3 ... ...\over 3} + {{\sin ( 5 \omega n)} \over 5} + \cdots \right ] \end{displaymath}

The symmetric triangle wave (Figure 10.6) given by

\begin{displaymath} x[n] = 8 p[n] - 8 p[n-{N \over 2}] \end{displaymath}

similarly comes to
\begin{displaymath} x[n] \approx {8 \over {{\pi^2}}} \left [ {\cos ( \omega n ... ...over 9} + {{\cos ( 5 \omega n)} \over 25} + \cdots \right ] \end{displaymath}



Miller Puckette 2006-12-30