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

  1. Apr 8, 2007 #1
    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?
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 8, 2007 #2

    mjsd

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    note [tex]I_0=V_2[/tex]
    alternatively, you can use current divider formula a few times to solve this problem.
     
  4. Apr 9, 2007 #3
    Ok, let's try again...

    I am going to use s in place of jw.

    [tex]I_S\,=\,V_1\,+\,\frac{V_1\,-\,V_2}{s}[/tex]

    [tex]I_o\,=\,\frac{V_1\,-\,V_2}{s}\,-\,\frac{V_2}{s}[/tex]

    What do I do now? This problem really sucks, and it's the first one in the assignment! Please help.
     
  5. Apr 9, 2007 #4

    mjsd

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

    while I haven't check your final answer, but methods are all good. from here as I said before [tex]I_0=V_2[/tex]... so sub in and solve for [tex]I_0/I_s[/tex]
    will get same answer as your latest trial once you set [tex]I_0=V_2[/tex]
     
  6. Apr 10, 2007 #5
    [tex]I_S(\omega)\,=\,I_0\left(2\,+\,j\omega\right)\,+\, \frac{I_0\left(1\,+\,j\omega\right)}{j\omega}[/tex]

    [tex]I_S(s)\,=\,I_0\left(2\,+\,s\right)\,+\,\frac{I_0\left(1\,+\,s\right)}{s}[/tex]

    [tex]I_S(s)\,=\,I_0\left(3\,+\,s\,+\,\frac{1}{s}\right)[/tex]

    Finally, I get:

    [tex]\frac{I_0(s)}{I_S(s)}\,=\,\frac{1}{3}\,+\,\frac{1}{s}\,+\,s[/tex]

    Is that the correct transfer function?
     
    Last edited: Apr 10, 2007
  7. Apr 10, 2007 #6

    mjsd

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    one should be confident with one's work...again I don't see a problem with the methods... to avoid silly mistakes like 2+3=6, just check your workings....
     
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