Finding H(s) for Circuit in s-Domain

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Homework Help Overview

The discussion revolves around finding the transfer function H(s) = I_0(s)/I_S(s) for a given circuit in the s-domain. Participants are exploring the relationships between currents and voltages in the circuit using Kirchhoff's laws and the properties of inductors in the s-domain.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to derive expressions for I_S and I_0 using circuit analysis techniques, including KCL and algebraic manipulation. Some express uncertainty about the correctness of their algebraic steps and seek alternative methods, such as the current divider formula.

Discussion Status

Several participants have provided insights and alternative approaches, with some confirming the validity of previous methods. There is ongoing exploration of different expressions for I_S and I_0, and while some participants express frustration, they continue to engage with the problem collaboratively.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may impose specific methods or formats for the solution. There is a noted transition from using jω to s in the analysis, indicating a shift in the approach to the problem.

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Homework Statement



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

http://img108.imageshack.us/img108/7136/problem146td5.jpg

Homework Equations



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

The Attempt at a Solution



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

http://img293.imageshack.us/img293/7284/problem146part2gg0.jpg

[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:
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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.
 
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.
 
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]
 
[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:
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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