| written 4.6 years ago by |
$Z\left(s\right)=\dfrac{\left(s^2+1\right)\left(s^2+3\right)}{s\left(s^2+2\right)\left(s^2+4\right)} $ By partial fraction expansion $Y\left(s\right)=\dfrac{k_1}{s+j+1}+\dfrac{k_1}{s-j1}+\dfrac{k_2}{s+j\sqrt{3}}+\dfrac{k_2}{s-j\sqrt{3}} $ $\therefore{}Y\left(s\right)=\dfrac{2k_1s}{s^2+1}+\dfrac{2k_2s}{s^2+3\ } $ $where \ \ k_{1} =\dfrac{s^{2}+1}{2s}Y(s)\Bigg|_{s^2=-1}=\dfrac{(s^{2}+2)(s^2+4)}{2(s^2+3)}\Bigg|_{s^2=-1}=\dfrac{3}{4}$ $where \ \ k_{2} =\dfrac{s^{2}+3}{2s}Y(s)\Bigg|_{s^2=-3}=\dfrac{(s^{2}+2)(s^2+4)}{2(s^2+1)}\Bigg|_{s^2=-3}=\dfrac{1}{4}$ $\therefore{}Y\left(s\right)=\dfrac{\left(\frac{3}{2}\right)s}{s^2+1}+\dfrac{\left(\frac{1}{2}\right)s}{s^2+3} $ The 2 terms represent admittance of series LC N/W $Y_{LC}\left(s\right)=\dfrac{\left(\dfrac{1}{L}\right)s}{s^2+\dfrac{1}{LC}} $ $By\ comparison\ \ \ \ L_1=\frac{2}{3}H,\ L_2=2H,\ C_1=\frac{3}{2}F,\ C_2=\frac{1}{6}F$


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