# Two Port Network - Impedances

Discussion in 'Homework Help' started by Sgt.Incontro, Jan 7, 2014.

1. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
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Last edited: Jan 7, 2014
2. ### studiot AAC Fanatic!

Nov 9, 2007
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b) (i)What does the negative sign imply for the ratio of a voltage across a resistor to the current through it?

3. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
Oops, wrong link, haha. Will fix it shortly.

Jan 3, 2011
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5. ### studiot AAC Fanatic!

Nov 9, 2007
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Why not just post it here?

What about my question on the negative sign?

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6. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
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I wrote: The magnitude of the current is solely given by i(2), as there is no current entering the second loop. The negative sign is due to the passive sign convention?

About the solutions that I posted up - am I barking up the completely wrong tree, or am I headed in the right direction?

7. ### rsashwinkumar New Member

Feb 15, 2011
23
2
Hi,
I think instead of using KVL and crunching through the math as you had done, you could very well use the result from part a.
Take Z1=1/(j*w*C) ; Z2=j*w*L; Z3=Z1;
Expand the matrix form equation as
V1=Z11*I1 + Z12*I2;
V2=Z21*I1 + Z22*I2;
Also use the fact that V2=-I2*R;
With this express I2 as a function of I1, and from expression of V1 you can get Zin as V1/I1.

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8. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
Hmm, that actually sounds like a better idea. I will try this tomorrow, and report back.

So does what I have done hold true, or is it complete rubbish though?

9. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
Turns out this is indeed the best way to solve it IMO, worked a treat. Thanks.

10. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
Hmm, this is what I got? Not sure if it is correct??

I have a feeling it ISN'T because when the time comes to use it for part C, the solution becomes excruciatingly complicated.

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11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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You're making finding Zin much harder than it needs to be.

Forget about two-ports; it's just a simple ladder network. To find Zin, start at the right side. You've got R in series with 1/jωC. Then that combination is in parallel with jωL, and finally add 1/jωC in series, simplify and you have Zin.

12. ### Sgt.Incontro Thread Starter Member

Dec 5, 2012
50
1
Hi,

I tried this method the very first time I attempted this question, but it starts getting VERY complicated. My math teacher told the class that it wasn't meant to be attempted this way too, if I remember correctly.

Do you know if my previous answer is correct?

13. ### The Electrician AAC Fanatic!

Oct 9, 2007
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No matter what you do, a fair amount of complex algebra is involved. Using a computer program such as Matlab or Mathematica, or a modern calculator can help you avoid the mistakes that are regular occurrences when you do a lot of complex algebra. I'm going to show the correct result and leave it to you to determine where you made your mistake.

Using the method I described, starting on the right and building the impedance Zin a step at a time:

You can also do it directly from the Z matrix. Adding the load resistor R to the Z matrix is trivial. Then invert the Z matrix to get the Y matrix. The impedance Zin is then the reciprocal of the 1,1 element of the Y matrix. This method looks as easy as it does because the huge amount of complex algebra involved in the inversion of the Z matrix is done by the computer, and is thus hidden from the user:

The format of the final result shown here, while correct, is most likely not the result a human being (you, in other words) would get. This is typical of what happens when a computer program is used to get results of this sort. Some additional manipulation of the result may be required to get it in the form most pleasing to your instructor.

Something that can help in trying to get a better looking result would be to leave the impedances Z1, Z2 and Z3 as such, compute the result, then substitute the C and L components at the end.

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• ###### DoZinMat.png
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14. ### The Electrician AAC Fanatic!

Oct 9, 2007
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What happened to part C? Even more algebra is involved, but the final answer is amazing simple.

15. ### Tesla23 Active Member

May 10, 2009
323
67
Hi Electrician, the original poster has vanished and I suspect his assignment is past due now, but part C was interesting, thought you may be interested in an alternative approach (of no use to the original poster - he was set an algebra grind).

Every symmetric T (i.e. Z1 = Z3) and where the impedances are reactive and Z1 is of opposite sign to Z3, at a given frequency is equivalent to a piece of transmission line, impedance Z0 and length θ, for some Z0 and θ.

For this circuit:

$jZ_0 tan(\theta/2) = \frac{1}{j\omega C}$

$jY_0 sin(\theta) = \frac{1}{j\omega L}$

So for a real input impedance, either the line is acting as a λ/4 transformer or Z0 = ±R.

1. λ/4 transformer
θ = 90 hence

$jZ_0 = \frac{1}{j\omega C}$

$jY_0 = \frac{1}{j\omega L}$

and using $Z_0 Y_0 = 1$ you immediately get

$\omega^2 LC = 1$ and

$Z_0^2 = \frac{L}{C}$ and so $Z_{in} = \frac{L}{RC}$

2. $Z_0 = -R$
A couple of lines of trig to eliminate θ shows that this happens if

$L = \frac{1+R^2\omega^2C^2}{2\omega^2C}$

which if you check, results in $Z_{in} = R$

So the input impedance is real if

$\omega^2 LC = 1$ or

$L = \frac{1+R^2\omega^2C^2}{2\omega^2C}$

16. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Those are the two results I got.