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    <name>
      <text>23.2.4.2 FS of square wave</text>
    </name>
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      <text><![CDATA[<p class="HELM_exercise">Exercise</p>
Obtain the Fourier series of the square wave one period of which is shown below:<p></p><p style="text-align: center;"><img src="@@PLUGINFILE@@/23-2-4%20task%20square%20pulse.svg" alt="square pulse" width="330" height="113" class="img-responsive atto_image_button_text-bottom"></p>&nbsp;

Fill in the boxes below. The layout may serve as a hint&nbsp;
<p></p>
\(a_0 = \) [[input:ans1]] [[validation:ans1]] [[feedback:prt1]]
<p></p>
For even values of \(n\)
<p></p>
\(a_n = \) [[input:ans2]][[validation:ans2]] [[feedback:prt2]]
<p></p>
For \(n = 1, 5, 9, \dots \)
<p></p>
\(a_n = \) [[input:ans3]][[validation:ans3]] [[feedback:prt3]]
<p></p>
For \(n = 3, 7, 11, \dots \)
<p></p>
\(a_n = \) [[input:ans4]][[validation:ans4]] [[feedback:prt4]]
<p></p>
Finally, 
\( b_n =\) [[input:ans5]][[validation:ans5]] [[feedback:prt5]]]]></text>
<file name="23-2-4 task square pulse.svg" path="/" encoding="base64">[encoded data, 5904 characters]</file>
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    <generalfeedback format="html">
      <text><![CDATA[<p>We have, since the function is non-zero only for \( -\dfrac{\pi}{2} &lt; t &lt; \dfrac{\pi}{2}\),
 \[ a_0 = \dfrac{1}{\pi} \displaystyle{ \int _{-\frac{\pi}{2}} ^{\frac{\pi}{2}} } 4 dt = 4\]
\[ \therefore \dfrac{a_0}{2} = 2  \]
is the constant term as we would expect. Also
\[
\begin{align}
a_n &amp;= \dfrac{1}{\pi} \displaystyle{ \int _{-\frac{\pi}{2}} ^{\frac{\pi}{2}} } 4 \cos nt \hspace{0.15cm}dt = \left[ \dfrac{\sin nt}{n} \right] _{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \\
&amp;= \dfrac{4}{n\pi} \left\{ \sin \left( \dfrac{n\pi}{2} \right) - \sin \left( - \dfrac{n\pi}{2} \right) \right\} = \dfrac{8}{n\pi} \sin \left( \dfrac{n\pi}{2} \right) \hspace{1cm} n = 1, 2, 3, \dots 
\end{align}
\]</p>
It follows from a knowledge of the sine function that
\[
a_n = 
\begin{cases}
\hspace{0.5cm}0 \hspace{1.5cm} n = 2, 4, 6, \dots \\
\hspace{0.3cm}\dfrac{8}{n\pi} \hspace{1cm} n = 1, 5, 9, \dots \\
-\dfrac{8}{n\pi} \hspace{1cm} n = 3, 7, 11, \dots 
\end{cases}
\]
Also
\[ b_n = \dfrac{1}{\pi} \displaystyle{ \int _{-\frac{\pi}{2}} ^{\frac{\pi}{2}} } 4 \sin nt \hspace{0.15cm}dt = \dfrac{4}{\pi} \left[ - \dfrac{\cos nt}{n} \right] _{-\frac{\pi}{2}}^{\frac{\pi}{2}} =
 - \dfrac{4}{n\pi} \left\{  \cos \left( \dfrac{n\pi}{2} \right) - \cos \left( - \dfrac{n\pi}{2} \right)  \right\} = 0 \]
Hence the Fourier series would be
\[ f(t) = 2 + \dfrac{8}{\pi} \left( \cos t - \dfrac{1}{3}\cos 3t + \dfrac{1}{5} \cos 5t - \dfrac{1}{7} \cos 7t + \dots \right) \]]]></text>
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    <stackversion>
      <text>2021120900</text>
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      <text>ta1:4;
ta2:0;
ta3:8/(n*pi);
ta4:-8/(n*pi);
ta5:0;</text>
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      <text><![CDATA[<span style="font-size: 1.5em; color:green;"><i class="fa fa-check"></i></span> Correct answer, well done.]]></text>
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