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      <text>3.6.4.5 Engineering Example</text>
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      <text><![CDATA[<h4>Engineering Example</h4>

<h5>Admittance</h5>

<p>Admittance, \(Y\), is a quantity which is used in analysing electronic circuits. A typical expression for admittance is \[ Y(s) = {s^2+4s+5\over (s^2+2s+4)(s+3)}\] where \(s\) can be thought of as representing frequency. To predict the behaviour of the
    circuit it is often necessary to express the admittance as the sum of its partial fractions and find the effect of each part separately. Express \(Y(s)\) in partial fractions.</p>
<p>The fraction is proper. The denominator contains an irreducible quadratic factor, which cannot be factorised further, and also a linear factor. Thus \[{s^2+4s+5\over (s^2+2s+4)(s+3)} = {As+B\over s^2+2s+4} + {C\over s+3} \qquad \qquad \qquad (1) \] Multiplying
    both sides of Equation (1) by \((s^2+2s+4)(s+3)\) we obtain \[ s^2+4s+5 = (As+B)(s+3) + C(s^2+2s+4) \qquad \qquad \qquad (2) \] To find the constant \(C\) we let \(s=-3\) in Equation (2) to eliminate \(A\) and \(B\). Thus \[ (-3)^2+4(-3)+5 = C((-3)^2+2(-3)+4)\]
    so that \[ 2 = 7C\qquad \text{and so}\qquad C={2\over 7}\] Equating coefficients of \(s^2\) in Equation (2) we find \( 1= A+C\) so that \(A=1-C= 1-\displaystyle{2\over 7} = \displaystyle{5\over 7}\). Equating constant terms in Equation (2) gives \(
    5=3B+4C\) so that \[{3B=5-4C=5-4\left({2\over 7}\right)={27\over 7}}\] so \[ B=\displaystyle{9\over 7}\] Finally\[{Y(s) = {s^2+4s+5\over (s^2+2s+4)(s+3)} = {{5\over 7}s + {9\over 7} \over s^2+2s+4} + {{2\over 7}\over s+3}}\] which can be written as
    \[{Y(s) = {5s+9\over 7(s^2+2s+4)} + {2\over 7(s+3)}}\]</p>]]></text>
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