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      <text>23.2.4.1 Examples of Fourier series</text>
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      <text><![CDATA[<h4>Examples of Fourier series</h4>
We shall obtain the Fourier series of the \half-rectified" square wave shown in the Figure below.<p><br></p><p style="text-align: center;"><img src="@@PLUGINFILE@@/23-2-4-fig-10-square-pulse.svg" alt="half rectified square pulse" width="494" height="122" class="img-responsive atto_image_button_text-bottom"><br></p><p style="text-align: center;"><br></p>
<p></p>
We have
\[
f(t) = 
\begin{cases}
1 \hspace{1cm} 0 &lt; t &lt; \pi \\
0 \hspace{1cm} \pi &lt; t &lt; 2\pi
\end{cases}
\]
\[f(t + 2\pi) = f(t) \]
The calculation of the Fourier coefficients is merely a straightforward integration using the results
already obtained:
\[ a_n = \dfrac{1}{\pi} \displaystyle{ \int_{-\pi}^\pi } f(t) \cos nt \hspace{0.15cm}dt \]
in general. Hence, for our square wave
\[ a_n = \dfrac{1}{\pi} \displaystyle{ \int_{0}^\pi } (1) \cos nt \hspace{0.15cm}dt = \dfrac{1}{\pi} \left[ \dfrac{\sin nt}{n} \right] _0 ^\pi = 0 \]
provided \( n \neq 0\).
<p></p>
But \( a_0 = \dfrac{1}{\pi}\displaystyle{ \int_{0}^\pi } (1)  \hspace{0.15cm}dt = 1\), so the constant term is \( \dfrac{a_0}{2} = \dfrac{1}{2} \).
<p></p>
(The square wave takes on values \(1\) and \(0\) over equal length intervals of \(t\) so \( \dfrac{1}{2}\) is clearly the mean value.)
<p></p>
Similarly
\[ b_n = \dfrac{1}{\pi} \displaystyle{ \int_{0}^\pi } (1) \sin nt \hspace{0.15cm}dt = \dfrac{1}{\pi} \left[ \dfrac{-\cos nt}{n} \right] _0 ^\pi  \]
Some care is needed now!
\[ b_n = \dfrac{1}{n\pi}(1 - \cos n\pi)\]
But 
\[ 
\begin{align} 
\cos n\pi = +1 \hspace{1cm} \text{ for }  \hspace{1cm} &amp;n = 2, 4, 6, \dots \\
\therefore b_n = 0 \hspace{1cm} \text{ for } \hspace{1cm} &amp;n = 2, 4, 6, \dots
\end{align}
\]
However
\[ 
\begin{align} 
\cos n\pi = -1 \hspace{1cm} \text{ for }  \hspace{1cm} &amp;n = 1, 3, 5, \dots \\
\therefore b_n = \dfrac{1}{n\pi}(1 - (-1)) = \dfrac{2}{n\pi} \hspace{1cm} \text{ for } \hspace{1cm} &amp;n = 1, 3, 5, \dots
\end{align}
\]
i.e \(b_1 = \dfrac{2}{\pi}\), \(b_3 = \dfrac{2}{3\pi}\), \(b_5 = \dfrac{2}{5\pi}\), \(\dots\)
<p></p>
Hence the required Fourier series is
\[
\begin{align}
f(t) &amp;= \dfrac{a_0}{2} + \sum _{n = 1} ^\infty (a_n\cos nt + b_n \sin nt ) \hspace{1cm} \text{in general} \\
f(t) &amp;= \dfrac{1}{2} + \dfrac{2}{\pi} \left(\sin t + \dfrac{1}{3}\sin 3t + \dfrac{1}{5}\sin 5t + \dots \right) \hspace{1cm} \text{in this case}
\end{align}
\]
Note that the Fourier series for this particular form of the square wave contains a constant term and
odd harmonic sine terms. We already know why the constant term arises (because of the non-zero
mean value of the functions) and will explain later why the presence of any odd harmonic sine terms
could have been predicted without integration.
<p></p>
The Fourier series we have found can be written in summation notation in various ways:
<p></p>
For odd values of \(n\) we have
\[ \dfrac{1}{2} + \dfrac{2}{\pi} \sum _{n=1}^\infty \dfrac{1}{n} \sin nt \]
or, since \(n\) is odd, we may rewrite \( n = 2k - 1\), \(k = 1,2 \dots \) and write the Fourier series as 
\[ \dfrac{1}{2} + \dfrac{2}{\pi} \sum_{k = 1}^\infty \dfrac{1}{2k - 1} \sin (2k -1) t \]]]></text>
<file name="23-2-4-fig-10-square-pulse.svg" path="/" encoding="base64">[encoded data, 11672 characters]</file>
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