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Square and symmetric triangle waves

Figure 10.6: Symmetric triangle wave, obtained by superposing parabolic waves with $(B, c)$ pairs equal to $(0, 4/N)$ and $(N/2, -4/N)$.
\begin{figure}\psfig{file=figs/fig10.06.ps}\end{figure}

To demonstrate 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] = {{4} \over {N}} p[n] - {{4} \over {N}} p[n-{N \over 2}]
\end{displaymath}

similarly comes to

\begin{displaymath}
x[n] \approx {4 \over {{\pi^2}}} \left [
{\cos ( \omega n ...
...over 9}
+ {{\cos ( 5 \omega n)} \over 25}
+ \cdots
\right ]
\end{displaymath}



Miller Puckette 2006-03-03