Find the values of $R, R_1, L$ and $C$.

Solution:

$ \begin{aligned} \\ &\begin{aligned} \\ &\text { Given } \\\\ &\text { poles } \Rightarrow-1+j \sqrt{5} ;-1-j \sqrt{5} \\ \end{aligned}\\\\ &\therefore \text { polefactors }=(s+1-j \sqrt{5})(s+1+j \sqrt{5})\\\\ &=(5+1)^2-(j \sqrt{5})^2\\\\ &=s^2+2 s+1+5\\\\ &=s^2+2 s+6\\\\ &\text { zeras } \Rightarrow-1,-3\\\\ &\text { Zero factors }=(s+1)(s+3) \\ &=(s^2+4 s+3)\\\\ &T \cdot F \cdot=z(s)=\frac{(\text { scale factor) (zerofactor) }}{(\text { polo factor) }}\\\\ &\therefore z(s)=\frac{k\left(s^2+4 s+3\right)}{\left(s^2+2 s+6\right)}...(1) \\ \end{aligned} \\ $

$ \begin{aligned} \\ \text { Bit } z(0)&=1 \\\\ \text { Hence put s }&=0 \text { in (1) } \\\\ \therefore z(0)&=\frac{k(0+0+3)}{(0+0+6)} \\\\ 1&=\frac{3 \cdot k}{6} \\\\ \therefore \quad k&=2 \\\\ \text { from eqn (1) } \therefore z(s)&=\frac{2 s^2+8 s+6}{s^2+2 s+6} ....(2)\\ \end{aligned} \\ $

$ \begin{aligned} \\ &\text { Now from CKt. diagram } \\\\ &\begin{aligned} \\ Z_1(S) &=R_1+L S \cdot Y_1(S)=\frac{1}{L S+R_1} \\\\ \& Z_2(S) &=R+\frac{1}{C S} \\\\ &=\frac{R C S+1}{C S} \therefore Y_2(S)=\frac{C S}{R C S+1} \\\\ \therefore Y(S) &=Y_1(S)+Y_2(S) \\\\ &=\frac{1}{L S+R 1}+\frac{C S}{R C S+1} \\\\ &=\frac{R C S+1+L C S^2+R_1 C S}{\left(L S+R_1\right)(R C S+1)} \\\\ &=\frac{L C S^2+\left(R C+R_1 C\right) S+1}{R L C S^2+L S+R R_1 C S+R_1} \\\\ &=\frac{L C S^2+\left(R C+R_1 C\right) S+1}{R L C S^2+\left(L+R R_1 C\right) S+R_1}....(3) \\ \end{aligned} \\\\ \end{aligned} \\ $

$ \therefore z(S)=\frac{1}{y(S)}=\frac{R L C S^2+\left(L+R R_1 C\right) S+R_1}{L C S^2+\left(R C+R_1 C\right) S+1}...(4) \\ $

compalring (4) whth (2) and rearranging (2) we get

$ \begin{aligned} Z(s) &=\frac{2 s^2+8 s+6}{6\left(0.166 s^2+0.333 s+1\right)} \\\\ &=\frac{0.333 s^2+1.333 s+1}{0.166 s^2+0.333 s+1}....(5)\\ \end{aligned}\\ $

Now comparing (5) with (4) \& rearranging (4) we get,

$ Z(S)=\frac{\left(\frac{R L C}{R_1}\right) S^2+\left(\frac{L+R R_1 C}{R_1}\right) S+1}{L C S^2+\left(R C+R_1 C\right) S+1}.....(6)\\ $

equating (5) \& (6) we can write,

$ \frac{R L C}{R_1}=0.333....(7)\\ $

$ L C=0.166...(8) =\frac{L+R_1 C}{R_1}=1.333...(9)\\ $

$ \quad R C+R_1 C=0.333...(10)\\ $

$ \frac{R(0.166)}{R_1}=0.333\\ $

$ R=2 R_1 ...(11)\\ $

put (11) in (10)

$ \begin{array}{r} 2 R_1 C+R_1 C=0.333 \\\\ 3 R_1 C=0.333 \\\\ R_1 C=0.111...(12) \\ \end{array} \\ $

$ \begin{aligned} \\ &\begin{gathered}\\\\ \text { Put (12) In (8) } \\ \frac{L+R(0.111)}{0.5 R}=1.333 \\\\ L+0.111 R=0.666 R \\ \end{gathered}\\\\ &L=0.555 \mathrm{R}...(13)\\ \end{aligned}\\ $