All About Circuits Forum Solving Transfer Function
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#1
10-29-2013, 08:08 PM
 Sgt.Incontro Junior Member Join Date: Dec 2012 Location: (GMT + 0) Posts: 47
Solving Transfer Function

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 by Sgt.Incontro; 11-02-2013 at 05:38 PM.
#2
10-29-2013, 08:25 PM
 Jony130 Senior Member Join Date: Feb 2009 Location: Poland/Wroclaw Posts: 2,496

Try to use a current divider rule
http://en.wikipedia.org/wiki/Current_divider
 The Following User Says Thank You to Jony130 For This Useful Post: anhnha (10-31-2013)
#3
10-30-2013, 03:50 AM
 WBahn Senior Member Join Date: Apr 2012 Location: Larkspur, Colorado Posts: 8,077 Blog Entries: 9

Quote:
 Originally Posted by Sgt.Incontro 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.

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?
 The Following User Says Thank You to WBahn For This Useful Post: anhnha (10-31-2013)
#4
10-30-2013, 07:06 AM
 Sgt.Incontro Junior Member Join Date: Dec 2012 Location: (GMT + 0) Posts: 47

Quote:
 Originally Posted by WBahn Using your answer you have: 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?
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
10-30-2013, 07:33 AM
 WBahn Senior Member Join Date: Apr 2012 Location: Larkspur, Colorado Posts: 8,077 Blog Entries: 9

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?
 The Following User Says Thank You to WBahn For This Useful Post: anhnha (10-31-2013)
#6
10-30-2013, 03:55 PM
 Sgt.Incontro Junior Member Join Date: Dec 2012 Location: (GMT + 0) Posts: 47

Quote:
 Originally Posted by WBahn If it was a voltage source, then the transfer function would be exactly 1. Think about it. 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?
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
10-30-2013, 04:07 PM
 WBahn Senior Member Join Date: Apr 2012 Location: Larkspur, Colorado Posts: 8,077 Blog Entries: 9

Quote:
 Originally Posted by Sgt.Incontro 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?
Sure.

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

Quote:
 and then from there?
And then write i(c) and i(RL) in terms of the output voltage.
#8
10-30-2013, 04:39 PM
 Sgt.Incontro Junior Member Join Date: Dec 2012 Location: (GMT + 0) Posts: 47

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?)

Quote:
 Originally Posted by WBahn 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.
I considered both currents facing downwards.

Quote:
 Originally Posted by WBahn And then write i(c) and i(RL) in terms of the output voltage.
How exactly?
#9
10-31-2013, 09:17 AM
 anhnha Senior Member Join Date: Apr 2012 Posts: 456

Hi,
Maybe, I can help a bit.
Quote:
 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 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?
Attached Images
 Transfer Function.PNG (20.7 KB, 18 views)
 The Following User Says Thank You to anhnha For This Useful Post: Sgt.Incontro (10-31-2013)
#10
10-31-2013, 03:54 PM
 WBahn Senior Member Join Date: Apr 2012 Location: Larkspur, Colorado Posts: 8,077 Blog Entries: 9

Quote:
 Originally Posted by Sgt.Incontro 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?
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?
 The Following User Says Thank You to WBahn For This Useful Post: Sgt.Incontro (10-31-2013)

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