EEGR$\mathrm{.}331-$ Project $3$: Simple RLC Circuit$\%$ Part $-$ I: Simulation and Data Generation$\%\%$ Values for R$,$ L & C used for generating the dataset $\%\{($w$\_$n$,$G$\_$maes$($w$\_$n$))\},$ n$=1,2,...,$N $\%+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++\%$ Assumption: r $=0$ for this simulationR$=50$; $\%$ R $=50$ OhmsL$=0\mathrm{.}0023$; $\%$ L $=2\mathrm{.}3$ mHC$=0\mathrm{.}0000015$; $\%$ C $=1\mathrm{.}5$ micro$-$Faradf$\_$n$=1$:$50$:$100001$;w$\_$n$=2*$pi$*$f$\_$n;h$\_$n$=$abs$(($R$/$L$).*($j$*$w$\_$n$)./(-$ w$\_$n$.^2+($R$/$L$)*($j$*$w$\_$n$)+(1/($L$*$C$))))$;subplot $(2,1,1)$;semilogx$($w$\_$n$,$h$\_$n$)$;grid on; title$(\text{'}|$H$($j$\backslash$omega$\_$n$)|\text{'})$;xlabel $(\text{'}\backslash$omega$\_$n$\text{'})$;ylabel $(\text{'}|$H$($j$\backslash$omega$\_$n$)|\text{'})$;subplot $(2,1,2)$;theta$\_$n$=$angle$(($R$/$L$).*($j$*$w$\_$n$)./(-$ w$\_$n$.^2+($R$/$L$)*($j$*$w$\_$n$)+(1/($L$*$C$))))$; degree$\_$n$=$theta$\_$n$*180/$pi;semilogx$($w$\_$n$,$degree$\_$n$)$;grid on;title$(\text{'}\backslash$theta$($j$\backslash$omega$\_$n$)\text{'})$;xlabel$(\text{'}\backslash$omega$\_$n$\text{'})$;ylabel$(\text{'}\backslash$theta$($j$\backslash$omega$\_$n$)\text{'})$;Gmeas$\_$n $=1./($h$\_$n$.^2)$;temp $=[$w$\_$n$\text{'},$ h$\_$n$\text{'},$ Gmeas$\_$n$\text{'}]$; save$(\text{'}$RLC$.$mat','temp'$)$; $\%$ Saving the data in an ascii file$\%+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++\%$ EEGR$\mathrm{.}331-$ Project $3$: Simple RLC Circuit$\%$ Part $-$ II: From the supplementary document for Project $3$: Solving the $\%$ linear algebraic equation $\%\%$ A $*$ alpha $=$ b $\%\%$ with unknown parameters aplha $=[$alpha$\_1,$ aplha$\_2,$ aplha$\_3]\text{'}\%+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++$clear all;load RLC$.$mat;w$\_$n $=$ temp$($:$,1)$; h$\_$n $=$ temp$($:$,2)$; Gmeas$\_$n $=$ temp$($:$,3)$;A $=$ zeros$(3,3)$; $\%$ Initialization of the matrix A and column vector bb $=$ zeros$(3,1)$;N $=$ length$($w$\_$n$)$; $\%$ Computing the system matrix A and column vector bfor i$=1$:$1$:N $\%$ A$(1,1)=$ A$(1,1)+$ w$\_$n$($i$)^(-4)$; A$(1,2)=$ A$(1,2)+$ w$\_$n$($i$)^(-2)$; A$(1,3)=$ A$(1,3)+1$; A$(2,1)=$ A$(2,1)+$ w$\_$n$($i$)^(-2)$; A$(2,2)=$ A$(2,2)+1$; A$(2,3)=$ A$(2,3)+$ w$\_$n$($i$)^(2)$; A$(3,1)=$ A$(3,1)+1$; A$(3,2)=$ A$(3,2)+$ w$\_$n$($i$)^(2)$; A$(3,3)=$ A$(3,3)+$ w$\_$n$($i$)^(4)$; b$(1)=$ b$(1)+$ w$\_$n$($i$)^(-2)*$Gmeas$\_$n$($i$)$; b$(2)=$ b$(2)+$ Gmeas$\_$n$($i$)$; b$(3)=$ b$(3)+$ w$\_$n$($i$)^(2)*$Gmeas$\_$n$($i$)$;endalpha $=$ A$\backslash$b; $\%$ Solving the linear algebraic equations A $*$ alpha $=$ b; $\%++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++\%$ Note that we can also use the Moore$-$Penrose Pseudoinverse of matrix A$,\%$ that is$,$ alpha$=$pinv$($A$)*$b$,$ to solve the algebraic equation, where the norm$\%$ for the solution is smaller than for any other solution including A$\backslash$b$.\%\%$ REMARK: Your approximate solution for alpha must close to the following $\%$ values: alpha$\_1=1/($RC$)^2,$ alpha$\_2=[1-2$L$/($R$^2$ C$)],\%$ with r $=0,$ & alpha$\_3=$ L$^2/$R$^2*$End of Code$*$

Note that $(0\mathrm{.}3)$ is a frequency$-$domain model for the system with parameters ${}_0,{}_1$ and ${}_2.$ It is easy to verify that the theoretical transfer function for the $RLC$ circuit has the form in $(0\mathrm{.}3),$ that is$,$

$|H_{\text{t}\text{h}\text{e}\text{o}\text{r}}(){|}^2=\frac{{}^2}{(\frac{1}{RC}{)}^2+[(\frac{R+r}{R}{)}^2-\frac{2L}{R^{2}C}]{}^2+(\frac{L}{R}{)}^2{}^4}$

Here, the objective is to develop a procedure for choosing the parameters ${}_0,{}_1$ and ${}_2$ in $(0\mathrm{.}1),$ so that the model $|H_{\text{t}\text{h}\text{e}\text{o}\text{r}}(){|}^2$ is a good approximation to the noisy frequency response $|H_{\text{m}\text{e}\text{a}\text{s}}(){|}^2$ that would be obtained through laboratory measurements. Then, once ${}_0,{}_1$ and ${}_2$ are known, we should be able to relate $(0\mathrm{.}3)$ and $(0\mathrm{.}4),$ and derive ${}_0,$ and $Q$ in $(0\mathrm{.}2)$ to compute ${}_0,{}_1$ and ${}_2.$ Note that solving for the optimum ${}_0,{}_1$ and ${}_2$ directly from $(0\mathrm{.}3)$ is difficult, because $|H_{\text{m}\text{o}\text{d}\text{e}\text{l}}(){|}^2$ is a nonlinear function of ${}_0,{}_1$ and ${}_2.$ Note, however, that the reciprocal is a linear function of the parameters:

$G_{\text{m}\text{o}\text{d}\text{e}\text{l}}()=\frac{1}{|H_{\text{m}\text{o}\text{d}\text{e}\text{l}}(){|}^2}$

$={}_0{}^{-2}+{}_1+{}_2{}^2$

Thus, it is easy to find ${}_0,{}_1$ and ${}_2$ that make $G_{\text{m}\text{o}\text{d}\text{e}\text{l}}()$ the "best approximation" to the reciprocal of the measured data

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