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<h2 class="hd hd-2 unit-title">6.1. Impedance of components in parallel.</h2>
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<p><p><b class="bfseries">Example 1.1 </b> Suppose instead that we put a capacitor and an inductor in parallel. </p><center><img src="/assets/courseware/v1/6e8f046072bd1168625bbb5d5ccb4057/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c3_CLparallel.svg" width="300px" style="margin: 0px 10px 10px 10px"/></center><p>
What is the complex gain of this system? What is the sinusoidal response to the input signal [mathjaxinline]\cos (\omega t)[/mathjaxinline]? Is there a frequency for which the gain is 0? </p></p><p><b class="bfseries"><span style="color:#FF7800">Solution:</span></b> The impedances of components in parallel do not add, but rather their reciprocals do. So the net impedance [mathjaxinline]Z[/mathjaxinline] in the circuit is </p><table id="a0000000714" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{1}{Z} = \frac{1}{Ls} + Cs.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The system function [mathjaxinline]H(s)[/mathjaxinline] is the reciprocal of [mathjaxinline]Z[/mathjaxinline], so </p><table id="a0000000715" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]H(s)=\frac{1}{Ls}+Cs=\frac{1+LCs^2}{Ls}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The complex gain is </p><table id="a0000000716" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]G(\omega )=H(i\omega )=\frac{1-LC\omega ^2}{L\omega i}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The requested system response is then </p><table id="a0000000717" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x_ p(t)=\mathrm{Re}\, (G(\omega )e^{i\omega t})= \frac{1-LC\omega ^2}{L\omega }\sin (\omega t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The gain is [mathjaxinline]\displaystyle {\frac{|1-LC\omega ^2|}{L\omega }}[/mathjaxinline]. </p><p>
Note the interesting fact that if we drive the system at the angular frequency [mathjaxinline]\omega _0=1/\sqrt {LC}[/mathjaxinline], the gain is 0: no current passes through the system. The transfer function has <b class="bf">zeros</b> at [mathjaxinline]s=\pm \omega _0i[/mathjaxinline]. </p>
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Activity with impedance.
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What is the transfer function of the system illustrated below? </p>
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Given [mathjaxinline]R[/mathjaxinline] and [mathjaxinline]C[/mathjaxinline], for what values of [mathjaxinline]L[/mathjaxinline] does this system exhibit oscillatory transients? </p>
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<p style="display:inline">[mathjaxinline]L&lt;[/mathjaxinline]</p>
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<h3 class="hd hd-3 problem-header">Summary</h3><ul class="itemize"><li><p><b class="bfseries"><span style="color:#0000FF">The complex impedance of a system is the reciprocal of its system function.</span></b></p></li><li><p><b class="bfseries"><span style="color:#0000FF">Impedances add in series; reciprocals of impedances add in parallel.</span></b></p></li></ul><p><b class="bfseries"><span style="color:#FF7800">Warning (again):</span></b> Circuit diagrams and block diagrams have a superficial resemblance. But they are different! </p><p>
A “series" relationship in a block diagram is a cascade; <b class="bf">the transfer functions multiply</b>. In contrast, in a series circuit diagram <b class="bf">the impedances add</b>. </p><p>
The “parallel" block diagram </p><center><img src="/assets/courseware/v1/9b1e8b93cff74755acc4100cd237cf65/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c3_inparallelC.svg" width="400px" style="margin: 0px 10px 10px 10px"/></center><p>
indicates that the same signal enters both blocks, and the output signals of those blocks are added together: it is equivalent to </p><center><img src="/assets/courseware/v1/719b42e9bc20d18be6aa99c61b6453a7/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c3_inparallelD.svg" width="400px" style="margin: 0px 10px 10px 10px"/></center><p>
In a circuit containing this configuration, if we declare the voltage drop from left to right as the input signal and the current through the whole configuration as the system response, then the same relationship obtains: the transfer functions add. This should be viewed as a coincidence. </p>
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