This script determines the initial zero crossing of the

$\%$ underdamped response of a parallel RLC circuit. The

$\%$ underdamped response is:

$\%$ v$($t$)=$exp$(-$t$/(2$RC$))*($A$*$cos$($sqrt$(1/$LC$-(1/2$RC$)^2)$t$)$

$\%+$ B$*$sin$($sqrt$(1/$LC$-(1/2$RC$)^2)$t$))$

$\%$ The circuit component values and initial conditions

$\%$ are:

$\%$ R$=100$ ohms; L$=1\mathrm{.}0$e$-3$ henry; C$=1\mathrm{.}0$e$-6$ farads;

$\%$ A$=6\mathrm{.}0$ volt, B$=-9\mathrm{.}0$ volt

$\%$ Solve over the interval $0<=$ t $<=0\mathrm{.}5$e$-3$ sec

clear; clc;

R$=100$; L$=1\mathrm{.}0$e$-3$; C$=1\mathrm{.}0$e$-6$; A$=6\mathrm{.}0$; B$=-9$;

fprintf$(\text{'}$Example $4\mathrm{.}1$: Find first zero crossing of $\text{'})$;

fprintf$(\text{'}$underdamped RLC circuit

$\text{'})$;

tmin$=0\mathrm{.}0$; tmax$=0\mathrm{.}5$e$-3$;

$\%$ split up timespan into $100$ intervals:

N$=100$;

dt$=($tmax$-$tmin$)/$N;

$\%$ First, calculate t and v$($t$)$ at each timestep

for n$=1$:N$+1$

t$($n$)=$ tmin$+($n$-1)*$dt;

v$($n$)=$ func$\_$RLC$($t$($n$),$A$,$B$,$R$,$L$,$C$)$;

fprintf$(\text{'}\%10\mathrm{.}4$e $\%10\mathrm{.}3$f

$,$t$($n$),$v$($n$)$;

end

plot$($t$,$v$),$ xlabel$(\text{'}$t$\text{'}),$ ylabel$(\text{'}$v$\text{'}),$

title$(\text{'}$v vs t for a RLC circuit'$),$ grid;

$\%$ Next, use the search method to find the first timestep

$\%$ where the sign of v$($t$)$ changes. When found, use fzero

$\%$ to solve.

for n$=1$:N

sign $=$ v$($n$)*$v$($n$+1)$;

if sign $<=0\mathrm{.}0$

root$\_$interval$(1)=$t$($n$)$;

root$\_$interval$(2)=$t$($n$+1)$;

$\%$ Solve for the zero of the multi$-$argument

$\%$ function func$\_$RLC$.$m by invoking func$\_$RLC as an

$\%$ anonymous function of one variable.

root $=$ fzero$($@$($t$)$ func$\_$RLC$($t$,$A$,$B$,$R$,$L$,$C$),...$

root$\_$interval$)$;

break;

end

end

fprintf$(\text{'}$

$\text{'})$;

if n $<$ N

fprintf$(\text{'}$First zero crossing is at t$=\%10\mathrm{.}4$e s

$\text{'},...$

root$)$;

else

fprintf$(\text{'}$No roots lie within $\%$g $<=$ t $<=\%$g s

$\text{'},...$

tmin,tmax$)$;

end

$\%$ Check solution. Putting root into func$\_$RLC should

$\%$ produce zero.

v $=$ func$\_$RLC$($root$,$A$,$B$,$R$,$L$,$C$)$;

fprintf$(\text{'}$v$=\%8\mathrm{.}6$e

$\text{'},$v$)$;

$----------------------------------------------------------------------------$

$\%$ func$\_$RLC$.$m

$\%$ This function works with Example$\_4\_1.$m

function v$=$func$\_$RLC$($t$,$A$,$B$,$R$,$L$,$C$)$

arg$1=1/(2*$R$*$C$)$;

arg$3=$sqrt$(1/($L$*$C$)-1/(2*$R$*$C$)^2)$;

$\%$ Equation $(4\mathrm{.}5)$:

v$=$exp$(-$arg$1*$t$)*($A$*$cos$($arg$3*$t$)+$B$*$sin$($arg$3*$t$))$;

Look at the above programme and find out what is wrong. Discuss as

students and correct if necessary.