- #1
Xyius
- 508
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I am having some issues with this problem I am doing. I will show what I have done so far and where I am getting confused.
1. Homework Statement
The 2-port shown in Figure 2 is characterized by its Z-parameters, where ##Z_{11}=Z_{12}=Z_{21}=Z_{22}=R##. Find the input impedance ##Z_{in}##. Express your result in terms of ##R## only. Hint: The input impedance of a 2-port terminated with load impedance ##Z_L## is ##Z_{in}=Z_{11}-\frac{Z_{12}Z_{21}}{Z_{22}+Z_L}##
[tex]V_1=Z_{11}I_1+Z_{12}I_2[/tex]
[tex]V_1=Z_{21}I_1+Z_{22}I_2[/tex]
So the first thing I notice is that this is two 2-port networks in series which is then terminated by a load resistor ##R## on the RHS. The top one I will call ##A##, and the bottom one I will call ##B##. Therefore the Z matrix is simply the sum of the two matrices from each 2-port network.
[tex]Z=\left( \begin{array}{c,c} R+Z^B_{11} & R+Z^B_{12} \\ R+Z^B_{21} & R+Z^B_{22} \end{array} \right)[/tex]
If I find this matrix, I can easily plug in the values to find ##Z_{in}##. So first I would like to find the first element in the matrix. I can do this by setting ##I_2=0##. In order to do this, the right hand side must be turned into an open circuit. Using the first equation in section 2 above gives me,
[tex]Z_{11}^B=\frac{V_1}{I_1}-R[/tex]
This is where I am not sure. So I need to determine the term ##\frac{V_1}{I_1}##, which I know is just equal to the resistance across the terminals for ##V_1##. So if I follow the circuit starting from the top left, it goes through the first 2-port, and then through the second 2-port. Is the contribution from the first 2-port just ##Z_{11}^A##? If so, then ##\frac{V_1}{I_1}=R+R## meaning ##Z_{11}^B=R## which makes perfect sense.
Likewise, the same can be done for ##Z_{22}## and I get ##Z_{22}^B=R##.
I am confused about ##Z_{21}## and ##Z_{12}##. In general, I do not know what these terms mean physically which makes it hard for me to know what to do mathematically. If I can get these two terms then I can solve the problem. I know that..
[tex]Z_{21}^B=\frac{V_2}{I_1}-R[/tex]
and
[tex]Z_{12}^B=\frac{V_1}{I_2}-R[/tex]
But I do not know where to go from here.
1. Homework Statement
The 2-port shown in Figure 2 is characterized by its Z-parameters, where ##Z_{11}=Z_{12}=Z_{21}=Z_{22}=R##. Find the input impedance ##Z_{in}##. Express your result in terms of ##R## only. Hint: The input impedance of a 2-port terminated with load impedance ##Z_L## is ##Z_{in}=Z_{11}-\frac{Z_{12}Z_{21}}{Z_{22}+Z_L}##
Homework Equations
[tex]V_1=Z_{11}I_1+Z_{12}I_2[/tex]
[tex]V_1=Z_{21}I_1+Z_{22}I_2[/tex]
The Attempt at a Solution
So the first thing I notice is that this is two 2-port networks in series which is then terminated by a load resistor ##R## on the RHS. The top one I will call ##A##, and the bottom one I will call ##B##. Therefore the Z matrix is simply the sum of the two matrices from each 2-port network.
[tex]Z=\left( \begin{array}{c,c} R+Z^B_{11} & R+Z^B_{12} \\ R+Z^B_{21} & R+Z^B_{22} \end{array} \right)[/tex]
If I find this matrix, I can easily plug in the values to find ##Z_{in}##. So first I would like to find the first element in the matrix. I can do this by setting ##I_2=0##. In order to do this, the right hand side must be turned into an open circuit. Using the first equation in section 2 above gives me,
[tex]Z_{11}^B=\frac{V_1}{I_1}-R[/tex]
This is where I am not sure. So I need to determine the term ##\frac{V_1}{I_1}##, which I know is just equal to the resistance across the terminals for ##V_1##. So if I follow the circuit starting from the top left, it goes through the first 2-port, and then through the second 2-port. Is the contribution from the first 2-port just ##Z_{11}^A##? If so, then ##\frac{V_1}{I_1}=R+R## meaning ##Z_{11}^B=R## which makes perfect sense.
Likewise, the same can be done for ##Z_{22}## and I get ##Z_{22}^B=R##.
I am confused about ##Z_{21}## and ##Z_{12}##. In general, I do not know what these terms mean physically which makes it hard for me to know what to do mathematically. If I can get these two terms then I can solve the problem. I know that..
[tex]Z_{21}^B=\frac{V_2}{I_1}-R[/tex]
and
[tex]Z_{12}^B=\frac{V_1}{I_2}-R[/tex]
But I do not know where to go from here.
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