# Solving Transfer Function

Discussion in 'Homework Help' started by Sgt.Incontro, Oct 29, 2013.

1. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
Hi,

I am stuck on this question...

The current source is throwing me off this, and I do not know exactly how to deal with it.

My answer seems very wrong -> V(0)/i(g) = Zr+Zl, so a hand would be greatly appreciated, as I am really stuck.

Thanks.

Last edited: Nov 2, 2013
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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3. ### WBahn Moderator

Mar 31, 2012
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V(0) = i(g)(Zr+Zl)

But this would require that ALL of the i(g) goes through the Zr and Zl. What about the portion that goes through Zc?

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4. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
I knew my answer is incorrect, as it does not make any sense.

If it was an ordinary voltage source instead of current source, you would have:

V(o)/V(i)= (Zl+Zr)/( (Zl+Zr) || Zc )

wouldn't you? So how would deal with the current source?

Sorry if I am heading in the wrong direction.

5. ### WBahn Moderator

Mar 31, 2012
18,093
4,918
If it was a voltage source, then the transfer function would be exactly 1.

Going back to the original circuit:

Just analyze the circuit. For instance, sum up the currents going into the top node and set that equal to the currents leaving the top node. You know that the voltage on the top node is Vo. What do you get?

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6. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
You are totally correct about the transfer function being 1 if it was a voltage source. (lol) But is it not possible to do a source transformation?

For the currents:
i(g) = i(c) + i(RL)

and then from there?

7. ### WBahn Moderator

Mar 31, 2012
18,093
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Sure.

I don't know. You haven't indicated what the reference direction for i(c) and i(RL) and I'm not a mind reader.

And then write i(c) and i(RL) in terms of the output voltage.

8. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
I tried a source transformation, am I correct in saying the source transformation would result in a voltage source V(g) followed by a series source resistance of Z(c)||( Z(r) + Z(l) )? (And then the rest of the circuit attached to this?)

I considered both currents facing downwards.

How exactly?

9. ### anhnha Active Member

Apr 19, 2012
776
48
Hi,
Maybe, I can help a bit.
I think it is right. However, what is the purpose for this?

1.What is relation between ig, I1, I2? In other words,write the equation that relates ig, I1, I2 through Kirchhoff's Current Law for node A.
2.Express I1 in terms of Vo?
3 Express I2 in terms of Vo?

Substitute I1, I2 expressions you got in #1 and #2 into #1.
Now can you derive Vo/ig?

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10. ### WBahn Moderator

Mar 31, 2012
18,093
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I don't know what happen to the response I thought I had posted. Oh well.

1) What is the voltage across the capacitor?
2) What is the impedance of the capacitor?
3) What is the current through the capactor?

4) What is the voltage across the series combination of R & L?
5) What is the impedance of the series combination of R & L?
6) What is the current through the series combination of R & L?

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11. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
1. i(g) = i(1) + i(2)
2. i(1) = V(o)/Z(c)
3. i(2) = V(o)/( Z(r)+Z(l) )

Substituting 2& 3 into equation #1 leads me to:

V(o)/i(g) = (R+sL)/(sCR + LC(s^2) + 1)

Is this correct?

12. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
1) V(o)
2) Z(c) = 1/sC
3) Current = V(o)/Z(c)

4) V(o)
5) Z(c)+Z(l) = 1/SC + sL
6) V(o)/(part 5)

Thank you everyone, and especially anhnha & WBahn for your help, much appreciated. I am glad I understand this problem now, it seems much more simple than I initially imagined it to be.