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sandy.bridge
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Homework Statement
Hello all,
Representing RLC circuits via differential equations is rather new to me, in fact differentials as an entirety are new for me. Essentially I am looking for clarification regarding this type of question and whether or not I am doing it right. Thanks in advance.
[tex]I=i_C+i_L+i_R\rightarrow{I}=C\frac{dv}{dt}+\frac{1}{L}\int_0^tvdt+\frac{v}{R}[/tex]
Differentiating both sides, we have
[tex]I=C\frac{d^2v}{dt^2}+\frac{v}{L}+R\frac{dv}{dt} \rightarrow{0}=v''+\frac{v'}{CR}+\frac{v}{LC}[/tex]
let
[tex]v=e^{st}, v'=se^{st}, v''=s^2e^{st}[/tex]
[tex]{0}=s^2e^{st}+\frac{se^{st}}{CR}+\frac{e^{st}}{LC}[/tex]
dividing out e^{st} and solving for the roots we have:
[tex]0=s^2+s/(CR)+1/(LC)\rightarrow{s=(-1/(CR)+/-\sqrt{(1/(CR))^2-4(1/LC)})/2}[/tex]
Let C=1, R=1, L=0.5, also note that for t<0 the current source is grounded and there is absolutely no current or voltage in the network.
Thus,
[tex](s_1, s_2)=((-1+i\sqrt{7})/2, (-1-i\sqrt{7})/2)[/tex]
The general solution for second order differential can be expressed as
[tex]v(t)=C_1e^{s_1t}+C_2e^{s_2t}=C_1e^{-0.5t}e^{i\sqrt{7}/2t}+C_2e^{-0.5t}e^{-i\sqrt{7}/2t}[/tex]
[tex]v(t)=C_1e^{-0.5t}(cos(\sqrt{7}/2t)+sin(\sqrt{7}/2t)i)+C_2e^{-0.5t}(cos(-\sqrt{7}/2t)-sin(-\sqrt{7}/2t)i)[/tex]
at t=0-, v(0)=0
thus,
[tex]0=C_1(cos(0)+sin(0)i)+C_2(cos(0)-sin(0)i)\rightarrow{C_1+C_2=0}[/tex]
however, I am a little unsure of how to determine the other equation for determining the constants..
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