(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For the circuit below, find H(s) = [tex]\frac{I_0(s)}{I_S(s)}[/tex]

http://img108.imageshack.us/img108/7136/problem146td5.jpg [Broken]

2. Relevant equations

Inductor in the s-domain: [itex]j\omega[/itex](Inductor Value)

3. The attempt at a solution

I made a new circuit diagram for the s-domain:

http://img293.imageshack.us/img293/7284/problem146part2gg0.jpg [Broken]

[tex]i_1\,=\,\frac{V_1}{1\Omega}[/tex] [tex]i_2\,=\,\frac{V_1\,-\,V_2}{j\omega}[/tex] [tex]i_3\,=\,\frac{V_2}{j\omega}[/tex] [tex]I_0(\omega)\,=\,\frac{V_2}{1\Omega}[/tex]

KCL @ [itex]V_1[/itex]:

[tex]I_S(\omega)\,=\,i_1\,+\,i_2[/tex]

[tex]I_S(\omega)\,=\,V_1\,+\,\frac{V_1\,-\,V_2}{j\omega}[/tex]

KCL @ [itex]V_2[/itex]:

[tex]i_2\,=\,i_3\,+\,I_0(\omega)[/tex]

[tex]\frac{V_1\,-\,V_2}{j\omega}\,=\,\frac{V_2}{j\omega}\,+\,V_2[/tex]

After a couple of algebra moves(is it safe to multiply both sides by jw?), I get for [itex]V_1[/itex]:

[tex]V_1\,=\,V_2\left(2\,+\,j\omega\right)[/tex]

Substituting into the [itex]V_1[/itex] KCL equation:

[tex]I_S(\omega)\,=\,V_2\left(2\,+\,j\omega\right)\,+\,\frac{V_2\left(1\,+\,j\omega\right)}{j\omega}[/tex]

I also have this:

[tex]I_S(\omega)\,=\,i_1\,+\,i_3\,+\,I_0(\omega)[/tex]

[tex]\frac{I_0(\omega)}{I_S(\omega)}\,=\,1\,-\,\frac{i_1}{I_S(\omega)}\,-\,\frac{i_3}{I_S(\omega)}[/tex]

But how do I get rid of the i1 and i3? That would require ridding the equations of the V1 and V2 right? How to do that? I bet there is a much easier way of going about this. Does anyone know of an easier way?

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# Homework Help: Frequency response: Find H(s) for a circuit w/ 2 res. - 2 Inductors and an I - source

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