Hi everyone,
I'm new to this forum. I'd really appreciate some help with a circuit analysis exercise in preparation for a Circuit Theory exam.

Given the following circuit:

Calculate the NETWORK FUNCTION F(s) between I_L and I_g i.e.
I_L(s)=F(s) I_g(s)

Where do I start from?
Thanx in advance.

 

exam_help,

Where do I start from?

Well, first of all, it is a nonlinear circuit. So you have to evaluate it twice for t<=0 and t>0. Are you going to give us the inductance of the primary and secondary of the transformer? Designate which coils of the transformer are 2:1 . Is "Y" an admittance matrix? How is it applied across its three terminals within the circuit. What does 1[Y]2 mean?

This looks like a loop mesh problem. Could you take a stab at it first?

Ratch

Ratch

 

exam_help,

Well, first of all, it is a nonlinear circuit.

I have to take it back. The excitation is discontinuous, but the circuit is linear.

Ratch

 

The "2" of 2:1 refers to the vertical coil of the ideal transformer (the one on the lower left side).

"Y" is a matrix measured in Ω\(^{-1} \)

As to " 1[Y]2 ", I don't really know what that means. I guess it's just a label indicating it's a 2-port-network (?). Anyway, I'm just reporting what the text of the exercise says.

I know I'll have to work in Laplace's domain (that's what I've been doing in the latest exercises). Do I have to apply any theorem to make things easier like say Superposition Th, Thevenin or Norton? I'm not very thorough with these theorems so I don't know where to apply them and how to work it out.

For now I just translated all the circuit elements to their equivalents in Laplace's domain:
\( C \rightarrow \frac{1}{sC} \)

\( L \rightarrow sL \)

\( R \rightarrow R \)

Ideal transformer:
\(
\left \{
\begin{array}{ll}
V_3(s)= 2V_4(s) & \\
I_3(s)= -\frac{1}{2}I_4(s) &
\end{array}
\right.
\)

(V3 and I3 are the transformer's vertical coil's voltage and current. I4 and V4 are current and voltage of the horizontal coil)

"Y" matrix:
\(
\left[
\begin{array}{c}
I_1 \\
I_2
\end{array} \right]

=
\left[
\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right]

\left[
\begin{array}{c}
V_1 \\
V_2
\end{array}
\right]
\)

I1 and I2 are the currents flowing into Y from the lateral terminals. V1 and V2 are the voltage drops between the central terminal and the lateral ones. (left side is I1, V1)

 

Use node method? If so, which nodes do I have to consider?

 

"Y" matrix:
\(
\left[
\begin{array}{c}
I_1 \\
I_2
\end{array} \right]

=
\left[
\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right]

\left[
\begin{array}{c}
V_1 \\
V_2
\end{array}
\right]
\)

I1=Ig. but I must resolve for \(I_L(s) \). How to continue?

 

exam_help,

You did a good job of explaining the circuit. I must say, I have never seen an admittance matrix used within a circuit before.

Use node method? If so, which nodes do I have to consider?

Of course you can try it, but I would use the loop method.

I1=Ig. but I must resolve for . How to continue?

As shown below, I consider Ig to be (Isub1 - Isub3) .

Looks to me like you need to find the division of Isub2/Ig which equals Isub2/(Isub1 - Isub3)

Let's gather up all the equations.

Vsub3 + Vsub2 - Vsub1 = 0 >>>>>>>>> loop 1
Vsub2 - (Ls)Isub2 = 0 >>>>>>>>>>>>loop 2
-R(Isub2)-Isub3/Cs +Vsub4 - Vsub2 +Vsub1 - Vg = 0 >>>loop3
Vsub3 = -2Vsub4 >>>>>>>>>>>from transformer specifications
Ig = Iy1 = Isub1 - Isub3
Iy2 = Isub2 - I sub3 >>>>>>>>from admittance matrix
Ig = 2V1 - Vsub2 >>>>>>>>>>from admittance matrix
Isub2-Isub3 = -Vsub1+2Vsub2

We have unknowns :
Isub1
Isub2
Isub3
Ig
Vsub1
Vsub2
Vsub3
Vsub4
Vg

So now we have 9 equations and 9 unknowns. We should be able to find Isub1, Isub2, and Isub3 and plug into Isub2/(Isub1 - Isub3) to get the transfer function. Is there any question as to the accuracy of any of these equations? That should be resolved first.

Ratch

 

WOw that's an amazing writing job though I'm sorry I haven't understood much of it.
I decided to proceed with node method (see picture attached).

\(
\left[
\begin{array}{ c c c }
(\frac{1}{R} + sC) & 0 & -(\frac{1}{R} + sC) \\
0 & 0 & 0\\
-(\frac{1}{R} + sC) & 0 & (\frac{1}{R} + sC +\frac{1}{sL})
\end{array} \right]

\left[
\begin{array}{ c }
E_1 \\
E_2 \\
E_3
\end{array} \right]
=

\left[
\begin{array}{ c }
-I_3-I_4-I_g \\
I_1 \\
I_2-I_L+I_4
\end{array} \right]

\)

however there's a mistake somewhere since the second line is all "0" and then equals to I1. This cannot be. Please check the updated drawing.
Since I3=-½ I4, the currents' matrix becomes
\(
\left[
\begin{array}{ c }
-\frac{1}{2}I_4-I_g \\
I_1 \\
I_2-I_L+I_4
\end{array} \right]

\)

Could you please help me with my method? Where did I go wrong?

 

exam_help,

there's a mistake somewhere ....

We both made mistakes. I just realized that there will be initial conditions at t=0, such as voltage across the capacitor and current through inductance L. This is because Ig has output 1 amp since the Jurassic Period. After t = 0, then Ig does a step function to 2 amps, so its transform at that time is 2/s.

I will try to spend some time tonight on a solution and test it out. Right now I have other things to do.

Ratch

 

I guess we have to first of all find the initial conditions of the capacitor and the inductance and then transform the circuit in Laplace's domain, adding some indipendent voltage/current generators if necessary, and finally solving with node/mesh method for \(I_L \).

I know we should get something like

\(I_L= (\frac{A}{s+s_0}) \cdot I_g\)

So the first problem is: finding the initial conditions of C & L.

 

exam_help,

I sent a message similiar to this one last night, but I it appears to have gotten lost. I think a loop analysis should be done for current, because that is what you eventually want for your transfer function. I changed the "Y" admitance matrix to a "Z" impedance matrix by simply inverting it. That makes a resistance of 1 ohm on the left and right side of the Y junction. And a resistance of -1/3 ohms on the center leg. Notice that this is a negative resistance. I can explain it better for you if you don't understand what I did. It makes it easier to do a loop equation with resistance instead on conductance. It appears to me that there will be no initial voltage on the capacitor, because it has had a long time to deenergize through the transformer coils since during t<0. And current will be blocked by the capacitor in the upper loop. So the current division rule could be applied to get the initial current in L

Now, I need a good way to make a schematic that can be sent along with my solution response. Preferable in *.png format. How did you do yours? I hope it can be done with freely available software.

Ratch

 

exam_help,

I changed the "Y" admitance matrix to a "Z" impedance matrix by simply inverting it. That makes a resistance of 1 ohm on the left and right side of the Y junction. And a resistance of -1/3 ohms on the center leg. Notice that this is a negative resistance.
Ratch

When I invert the Y matrix, I get:

Z =
[ 2/3 1/3 ]
[ 1/3 2/3 ]

which is equivalent to a T network of 3 resistors, all 1/3 ohm.

Or, a PI network of 3 resistors, all 1 ohm.

The convention for Y matrices is that the current enters at each node. I think you may have taken the current at node 2 to be exiting the node, and that's how you got a negative value for the center leg resistance.

 

I first calculate Z

\( Z=Y^{-1}=\left [ \begin{array}{cr}
\frac{2}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{2}{3} \end{array} \right ]
\)

This means

\(
\left { \begin{array}{ll}
\frac{2}{3}I_g + \frac{1}{3}I_2 &= E_3\\
\frac{1}{3}I_g + \frac{2}{3}I_2 &= E_2
\end{array}
\right.
\)

now I need to know the initial conditions \(i(0^-) \) of the capacitor and the inductors (both transformer and the single coil down right).

Somehow I need to get a set of equations which can be solved for I_L, so that I_L is a function of "s" and I_g. I have to solve this problem with pen and pencil, without the aid of computer programmes as they're not allowed in exam.
Guys, thank you for helping me. Keep posting, please.

 

exam_help,

Sorry I did not get back to you sooner, but my computer died dead. I now have a new computer with the Vista O/S. I read up on 2 port networks and believe the Electrician has it it correct. If you need any more assistance with this problem, let us know.

Ratch