# Z Parameter Questions

1. Sep 1, 2015

### Xyius

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. The problem statement, all variables and given/known data

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 $Z_{in}=Z_{11}-\frac{Z_{12}Z_{21}}{Z_{22}+Z_L}$

2. Relevant equations
$$V_1=Z_{11}I_1+Z_{12}I_2$$
$$V_1=Z_{21}I_1+Z_{22}I_2$$

3. 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.

$$Z=\left( \begin{array}{c,c} R+Z^B_{11} & R+Z^B_{12} \\ R+Z^B_{21} & R+Z^B_{22} \end{array} \right)$$

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,

$$Z_{11}^B=\frac{V_1}{I_1}-R$$

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..

$$Z_{21}^B=\frac{V_2}{I_1}-R$$
and
$$Z_{12}^B=\frac{V_1}{I_2}-R$$

But I do not know where to go from here.

Last edited: Sep 1, 2015
2. Sep 1, 2015

### Xyius

I don't know what you mean. The circuit must be valid since this is my homework problem and not a trick question. Is this issue a simple assumption could fix?

At any rate, I believe I can write $V_2$ in terms of $I_1$ to have some cancelation, but I haven't tried to work out the details yet. Too late in the evening!

3. Sep 2, 2015

### rude man

Oops, I redrew the circuit wrong.
I din't go the two 2-port network route.
Call the low side of input and output of the 2-port ground (zero volts).
Call the floating input voltage source V.
Then, use the relation between v1 and v2 to simplify the usual two equations relating v1 and v2 to i1 and i2.
Then, find equations relating V, v1 and i1; also V to i2. It helps to redraw the circuit, but don't goof like I did ...
Finally, Zin = V/i1.
I have to say I don't see any help in the given hint. Maybe with the 2-network approach. Let me know what you got & we can compare.

4. Sep 2, 2015

### Xyius

Okay so in order to get an expression for $Z^B_{21}$ I choose to let $I_2=0$. To do this, the circuit must be open at the output of the two 2-port networks. I then get (as before)

$$Z_{21}^B=\frac{V_2}{I_1}-R$$

So now I need to find $V_2$. This is where I am having confusion but I believe I may be on the right track.

Since this is a series connection of two 2-port networks, we have $V_2=V_2^A+V_2^B$. $V_2^A$ is the voltage across the two terminals on the RHS of 2-port A and $V_2^B$ is the voltage across the two terminals on the RHS of 2-port B. Both of these I believe are simply $I_1 R$ because the top voltage $V_2^A$ is the voltage across both terminals on the RHS, which corresponds to $Z_{22}^A=R$, and the bottom voltage $V_2^B$ is simply the voltage across the resistor $R$. This gives me the following.
$$V_2=V_2^A+V_2^B=I_1R+I_1R=2I_1R$$
$$Z_{21}=\frac{V_2}{I_1}-R=\frac{2I_1R}{I_1}-R=R$$

In the same manner, $Z_{12}$ would also be equal to $R$.

Thus, the Z matrix for the resistor is equal to $R$ for each element in the matrix, same as the first and the total Z matrix is simply 2R for each element. Which means this problem is nothing more than 2 resistors in series (very cool!).

So now for the load resistor. I think the hint may be there so I can use what I have found and simply plug it in as follows.

$$Z_{11}=Z_{12}=Z_{21}=Z_{22}=2R$$
$$Z_{in}=2R-\frac{(2R)(2R)}{2R+2R}=2R-\frac{4R^2}{4R}=2R-R=R$$

I am not sure if this makes sense though.

5. Sep 2, 2015

### rude man

OK as I said I didn't go the two-network route. I combined one-network equations with KVL/KCL. I didn't get Zin = R; mine was < R. But tell you what, I will look at it the way you did & see howa two-network approach goes. Stay tuned, I just got up.

6. Sep 2, 2015

### Xyius

Thanks I appreciate it!

7. Sep 2, 2015

### rude man

I'm having a rough time redrawing the circuit as two 2-port networks. Could you draw something up showing that configuration?
Otherwise I suggest looking again at my post # 3. It was really pretty straightforward.

8. Sep 2, 2015

### Xyius

This is how I drew it. Excuse the crudeness.

This configuration looks identical to what is in the book for a series of 2-port networks. I will go through it again to double check my work.

9. Sep 2, 2015

### rude man

OK, I think that looks good. I suggest:
1. combine your two 2-port Z networks into one; then
2. use the hint for a loaded Z network.

I did it that way & got the same answer as before, halleluia!

10. Sep 2, 2015

### Xyius

Ha! Well I must be doing something wrong then. It was my understanding that I did indeed combine the two networks into one. The Z matrix for each 2-port each have all elements equal to $R$. So that would mean the combined Z matrix would have all elements equal to $2R$ would it not? When plugging into the equation for the hint I get R as I have shown. What am I doing wrong?? :(

11. Sep 2, 2015

### rude man

You have the right combined Z matrix. Could you show your math step-by-step?

12. Sep 2, 2015

### Xyius

Oh! I plugged in the wrong value for $Z_L$! I did it again and got $(2/3)R$.

13. Sep 2, 2015

### rude man

Bingo!

14. Sep 2, 2015

### Xyius

Awesome! Hey thanks a lot for your help. I very much appreciate it!

15. Sep 2, 2015

My pleasure.