Figure: Determine values of R, L, C. If Z(0)= 1.

Solution:

From circuit diagram:

$\begin{aligned} \\ z(S) &=(L S+R) \|\left(\frac{1}{C S}\right) \\\\ &=\frac{(L S+R)(1 / C S)}{L S+R+1 / C S} \\\\ z(S) &=\frac{L S+R}{L C S^2+R C S+1}....(1) \\ \end{aligned} \\ $

From pole, zero diagram,

zeros $=-3 \\ $

$\therefore$ $zero factors,$ $=(s+3)(s+3) \\ $

$poles,$

$\Rightarrow-1.5+j \frac{\sqrt{111}}{2} ;-1.5-j \frac{\sqrt{111}}{2} \\ $

$-1.5+j 5.267 \quad-1.5-j 5.267 \\ $

pole factors,

$=(s+1.5-j 5.267)(s+1.5+j 5.267) \\ $

$=(S+1.5)^2-(j 5.267)^2 \\ $

$=s^2+3 s+2.25+27.741 \\ $

$=s^2+3 s+30 \\ $

$\therefore T \cdot F . Z(S)=\frac{H(z e r o \text { factors })}{\text { (pole factors) }} \\ $

$z(s)=\frac{H(s+3)}{\left(s^2+3 s+30\right)}...(2) \\ $

$To calculate,$

$H$, given $z(0)=1 \ $ $z(0)=\frac{H(0+3)}{0+0+30}=\frac{3 H}{30} \\ $ $1=H / 10 \quad \therefore \quad H=10 \\ $ from (2) $Z(s)=\frac{10 s+30}{s^2+3 s+30}....(3) \\ $ compairing (3) with (1) and rearranging (3) we get $\begin{aligned} \\ &z(s)=\frac{10 s+30}{30\left(\frac{1}{30} s^2+\frac{3}{30} s+1\right)} \\ &z(s)=\frac{0.333 s+1}{0.0333 s^2+0.1 s+1} ....(4) \\ \end{aligned} \\ $ Now equating (1) and (4), we get $\begin{array}{ll} \\ L=0.333 \mathrm{H} & R=1 \Omega \\ L C=0.0333 & R C=0.1 \\ C=\frac{0.0333}{0.333} & C=0.1 \mathrm{~F} \\ C=0.1 \mathrm{~F} & \\ \end{array} \\ $