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      <text>20.6.2.2 Engineering Example - System response</text>
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      <text><![CDATA[<h4 class="HELM_example">Example</h4>
<p>An engineering system is modelled by the block diagram in Figure 29: </p>
<p></p><center><img src="@@PLUGINFILE@@/20-6-fig-29-eng_example_feedback_loop.svg" alt="Figure showing a block diagram for a given feedback loop." width="555" height="145" class="img-responsive atto_image_button_text-bottom"></center><p></p>
<p>Determine the system response \(v_0(t)\) when the input function is a unit step function when \(K=2.5\) and \(a=0\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>If the system has an overall transfer function \(H(s)\) then \(V_0(s)=H(s)V_1(s)\). But this particular system is the negative feedback loop described earlier and so \[
H(s)={G(s)\over {1+G(s)}}={\displaystyle{{K\over{1+as}}}\over \displaystyle{{1+{K\over{1+as}}}}} =\displaystyle{{K\over {K+1+as}}}\] In this particular case \[\displaystyle H(s)={2.5\over {3.5+0.5s}}={5\over {7+s}}\] Thus the impulse response \(h(t)\) is \[\displaystyle h(t)={\cal L}^{-1}\{H(s)\}={\cal L}^{-1}\left\{{5\over
{(7+s)}}\right\}=5\text{e}^{-7t}u(t)\] and so the response to a step input \(u(t)\) is given by the convolution of \(h(t)\) with \(u(t)\) \[\begin{eqnarray*}
v_0(t)&amp;=&amp;\int_0^tu(t-x)5\text{e}^{-7x}u(t)\,dx\\
&amp;=&amp;\int_0^t5\text{e}^{-7x}\,dx\qquad t&gt;0\\
&amp;=&amp;\left[\phantom{\dfrac{1}{1}}\!\!\!-\dfrac{5}{7}\text{e}^{-7x}\right]_0^t=-\dfrac{5}{7}[\text{e}^{-7t}-1].
\end{eqnarray*}\]</p>]]></text>
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