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

1. Apr 8, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

For the circuit below, find H(s) = $$\frac{I_0(s)}{I_S(s)}$$

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

2. Relevant equations

Inductor in the s-domain: $j\omega$(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]

$$i_1\,=\,\frac{V_1}{1\Omega}$$ $$i_2\,=\,\frac{V_1\,-\,V_2}{j\omega}$$ $$i_3\,=\,\frac{V_2}{j\omega}$$ $$I_0(\omega)\,=\,\frac{V_2}{1\Omega}$$

KCL @ $V_1$:

$$I_S(\omega)\,=\,i_1\,+\,i_2$$

$$I_S(\omega)\,=\,V_1\,+\,\frac{V_1\,-\,V_2}{j\omega}$$

KCL @ $V_2$:

$$i_2\,=\,i_3\,+\,I_0(\omega)$$

$$\frac{V_1\,-\,V_2}{j\omega}\,=\,\frac{V_2}{j\omega}\,+\,V_2$$

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

$$V_1\,=\,V_2\left(2\,+\,j\omega\right)$$

Substituting into the $V_1$ KCL equation:

$$I_S(\omega)\,=\,V_2\left(2\,+\,j\omega\right)\,+\,\frac{V_2\left(1\,+\,j\omega\right)}{j\omega}$$

I also have this:

$$I_S(\omega)\,=\,i_1\,+\,i_3\,+\,I_0(\omega)$$

$$\frac{I_0(\omega)}{I_S(\omega)}\,=\,1\,-\,\frac{i_1}{I_S(\omega)}\,-\,\frac{i_3}{I_S(\omega)}$$

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. Apr 8, 2007

### mjsd

note $$I_0=V_2$$
alternatively, you can use current divider formula a few times to solve this problem.

3. Apr 9, 2007

### VinnyCee

Ok, let's try again...

I am going to use s in place of jw.

$$I_S\,=\,V_1\,+\,\frac{V_1\,-\,V_2}{s}$$

$$I_o\,=\,\frac{V_1\,-\,V_2}{s}\,-\,\frac{V_2}{s}$$

What do I do now? This problem really sucks, and it's the first one in the assignment! Please help.

4. Apr 9, 2007

### mjsd

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

while I haven't check your final answer, but methods are all good. from here as I said before $$I_0=V_2$$... so sub in and solve for $$I_0/I_s$$
will get same answer as your latest trial once you set $$I_0=V_2$$

5. Apr 10, 2007

### VinnyCee

$$I_S(\omega)\,=\,I_0\left(2\,+\,j\omega\right)\,+\, \frac{I_0\left(1\,+\,j\omega\right)}{j\omega}$$

$$I_S(s)\,=\,I_0\left(2\,+\,s\right)\,+\,\frac{I_0\left(1\,+\,s\right)}{s}$$

$$I_S(s)\,=\,I_0\left(3\,+\,s\,+\,\frac{1}{s}\right)$$

Finally, I get:

$$\frac{I_0(s)}{I_S(s)}\,=\,\frac{1}{3}\,+\,\frac{1}{s}\,+\,s$$

Is that the correct transfer function?

Last edited: Apr 10, 2007
6. Apr 10, 2007

### mjsd

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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