# Two-port network z-parameters

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1. Feb 10, 2015

### acw260

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

2. Relevant equations

As above.

3. The attempt at a solution

(z11) With the output open circuit (and therefore I2=0) z11 = V1 / I1 = (R8 + R6)I1 / I1 = 14ohms;

(z12) With the input open circuit (and therefore I1=0) z12 = V1 / I2 = R6 = 6ohms;

(z21) With the output open circuit (and therefore I2=0) z21 = V2 / I1 = V1 x (R6 / (R8+R6) ) x ( (R8+R6) / V1) = R6 = 6ohms;

(z22) With the input open circuit (and therefore I1=0) z22 = V2 / I2 = R6xI2 / I2 = R6 = 6ohms

Is the above correct? It seems so easy that I'm worried I've overlooked something and made a really basic mistake!

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Last edited by a moderator: Jan 26, 2017
2. Feb 10, 2015

### BvU

Hello acx, welcome to PF :)

Some things are easy :) enjoy while it lasts...

3. Feb 11, 2015

### acw260

Thanks BvU, I shall do just that!

4. Jan 25, 2017

### David J

Where does the value for $I1$ come from? In the first equation for $z11$ where $I2=0$ there appears to be a value of 1 for $I1$. Then in the next equation for $z12$ and $I1 =0$ there appears to be a value of 1 for $I2$. How are these values derived please as I cant seem to work out and my notes tell me virtually nothing. Appreciated thanks

5. Jan 25, 2017

### Staff: Mentor

Can you enlarge your screen or change the default font to see print more clearly?

z11 = V1 / I1 = (R8 + R6) I1 / I1 = 14ohms

The formulae come from applying the z-parameter equations characterising a 2 port network.

6. Jan 25, 2017

### acw260

For z11 assume that the input is not open circuit. Therefore I1 is not equal to zero. I1 then cancels out from the equation leaving z11 = R8+R6. You don't need a value for I1, just that it doesn't equal zero.

For z12 again it doesn't matter what value you have for I2. You show that z12 = R6 = 6ohms. The value of I2 becomes irrelevant.

7. Jan 26, 2017

### David J

OK thanks for the info.

So as I understand this $Z11$ is the impedance seen looking into port 1 when port 2 (output) is open
.

Output is open circuit so $I2=0$

$Z11=\frac{V1}{I1}=\frac{(R8+R6)(I1)}{I1}$ The $I1$ cancels leaving $R8+R6=14\Omega$

So for $Z22$ is the impedance looking into port 2 when port 1 (input) is open

Input is open circuit so $I1=0$

$Z22=\frac{V2}{I2}=\frac{(R6)(I2)}{I2}$ The $I2$ cancels leaving $R6=6\Omega$

The reason $Z22$ only has $R6$ to deal with is because looking into the network from port 2 we can only see the $6\Omega$ resistance

So for $Z12$ i understand this is the ratio of the voltage at port 1 to the current at port 2 when port 1 is open. If port 1 is open the $I1=0$

You have used the equation $Z12 =\frac{V1}{I2} = R6 = 6\Omega$

My question is why does $V1=6\Omega$ this time whereas when for $Z11,V1=R8 + R6$

Thanks

8. Jan 26, 2017

### Staff: Mentor

Inject current at top terminal 2 and it exits at lower terminal 2. With port 1 open, determine V1.

9. Jan 26, 2017

### David J

So current will flow through $R6$ and back out via the lower terminal 2 because port 1 is open. So R8 does not come into the equation so to speak??

I have a better understanding of this now, yes. Thanks for the help

10. Jan 28, 2017

### rude man

Sneak extra info: for a passive network, x12 = x21. x can be either z or y parameters. (The rules for h, a, s parameters differ slightly).