Two-port network z-parameters

[acw260]

Homework Statement

Homework Equations

As above.

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!

 

[BvU]

Hello acx, welcome to PF :)

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

 

[acw260]

Thanks BvU, I shall do just that!

 

[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 can't seem to work out and my notes tell me virtually nothing. Appreciated thanks

 

[NascentOxygen]

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 can't seem to work out and my notes tell me virtually nothing. Appreciated thanks

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.

 

[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.

 

[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

 

[NascentOxygen]

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 $Z11,V1=R8 + R6$

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

Does this answer your question?

 

[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

 

[rude man]

₁₂!

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).

 

[Jason-Li]

Homework Statement

View attachment 112127

Homework Equations

As above.

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!

Hi, sorry to be this a pain as I know this thread is a couple of years old but I'm now working through the same question, was just wondering if anyone could explain why for z21 V2 was changed to V1 x (R6 / (R8+R6) ) x ( (R8+R6) / V1) and how that equals 6 ohms? I'm looking to understand this as I don't want to 'just know the answer'. Is this because the R8 & R6 are effectively in parallel like the attached drawing?

Also is this approach also acceptable?
Using the formulas V1 = z11*I1+z12*I2 and also V2 = z21*I1+Z22*I2
For inpute open:
Z22 = V2 / I2 = (Z*I2) / I2 = z22 = 6 ohms
For output open:
z11= V1 / I1 = (Z*I1) / I1 = 8+6 = Z11 = 14 ohms

I am happy with the above as the the formula rearranges nice and easily however with the following two I end up with the 'wrong V over the wrong I'. Unsure of how to work around this without values for V or I
Z12 = V1 / I2 =
Z21= V2 / I1 =

Appreciate any help!

 

[rude man]

Using your two equations:
z₁₂ = v₁/i₂ |_i1=0
z₂₁ = v₂/i₁ |_i2=0

what do you compute?

(Fact: in any passive 2-port network, z₁₂ = z₂₁). If you have any active element in your network like a transistor, this does not necessarily hold).

 

[Jason-Li]

Okay so if z12 = z21
V1 / I2 = V2 / I1

which I could rearrange to
V1 * I1 = V2 * I2
V1 = (V2*i2) / I1 & V2 = (V1*I1) / I2
and
I1 = (V2*I2) / V1 & I2 = (V1*I1) / V2
so
I1 = (V2*I2) / ((V2*I2) / I1)
I1*I1 = (V2*I2) / ((V2*I2)
I1*I1 = (Z*I2*I2) / ((Z*I2*I2)
I1*I1 = 0 ?

V1 = (V2*i2) / I1 & V2 = (V1*I1) / I2
V1 = (((V1*I1) / I2)*i2) / I1 = (V1*I1) / I1 = V1 ?

z12 = ((V2*i2) / I1) / ((V1*I1) / V2)
z12 = (V2*i2) / I1) * (V2 / (V1*I1))
z12 = (I2*V2^2) / (V1*I1^2)

Safe to say I'm confused haha

 

[Jason-Li]

To try and understand the current flow path, I think I understand why Z12 = 6ohms as it passes only through the 6ohm resistor ( first crude drawing )
I would've though that Z21 = 14ohms as well as it would take the red path ( second drawing), can't quite understand why it is 6ohms unless it takes the blue path, meaning that the 8 & 6 are in parallel?

 

[rude man]

To try and understand the current flow path, I think I understand why Z12 = 6ohms as it passes only through the 6ohm resistor ( first crude drawing )
I would've though that Z21 = 14ohms as well as it would take the red path ( second drawing), can't quite understand why it is 6ohms unless it takes the blue path, meaning that the 8 & 6 are in parallel?
View attachment 237875

Right, the red path. Let's look at the red path.
So z₂₁ = v2/i1 with i2=0. So I run a current i1 into the input, what's the output voltage with the output open?

 

[Jason-Li]

Sorry, I assume you mean the red path in the second drawing? If so current would flow from top terminal 1 to bottom terminal one, so with both terminal 2s open the output voltage would equal the voltage across the 6 ohm resistor?

Wait, so would:

Z21 = V2 / I1 = (6 * I1) / I1 = 6ohms as I1 cancels out?

 

[Jason-Li]

Then Z12 = V1/ I2 , the current would act as in the first drawing so V1 = I2 * Z

So:
Z12 = (Z*I2)/ I2 = (6 * I2) / I2 = 6 ohms

If this is the case then I now don't understand why Z11 is 14 ohms rather than the same as Z21 with 6, unless it is because from the perspective from Z21 the 8 ohm resistor is downstream of the 6ohm?

 

[rude man]

Then Z12 = V1/ I2 , the current would act as in the first drawing so V1 = I2 * Z
So:
Z12 = (Z*I2)/ I2 = (6 * I2) / I2 = 6 ohms
If this is the case then I now don't understand why Z11 is 14 ohms rather than the same as Z21 with 6, unless it is because from the perspective from Z21 the 8 ohm resistor is downstream of the 6ohm?

Well, you got z₁₂ and z₂₁ right.
So, let's re-examine z₁₁: z₁₁ = v₁/i₁ with i₂ zero (nothing connected to port 2). You put in v₁ volts, what do you get for the current?
BTW although I told you about z₁₂ = z₂₁, you shouldn't use that fact until you're comfortable with solving for them with the basic equations. It takes a while and you'd benefit from the double-check. And also BTW you may know there are several other two-port parameters, like y_ij, h_ij, a_ij. You should be able to find those from the respective equations also.

 

[Jason-Li]

Okay so Z11 = V1 /I1 with V1 volts would mean that current = I1. I think I maybe understand now.

So Z11 = V1 / I1, voltage will run from top terminal 1 to bottom terminal 1 so will. = 14 ohms

However for Z21 = V2 / I1, V2 contacts will be in parallel with the 6 ohm resistor, so Z21 = (Z6+I1)/ I1 = 6 ohms


Okay noted I'll try learn to use the basic ones first.

 

[rude man]

Okay so Z11 = V1 /I1 with V1 volts would mean that current = I1. I think I maybe understand now.
Z21 = (Z6+I1)/ I1 = 6 ohms

What kind of equation is this? But your answer is right. So are z12 and z21.

 

[Jason-Li]

Sorry I'm not sure what you mean by type of equation? Basic ohms law or?

 

[rude man]

Sorry I'm not sure what you mean by type of equation? Basic ohms law or?

Different two-port parameters have different equations.
For example, the y parameters are defined by the equations
i1 = y11 v1 + y12 v2
i2 = y21 v1 + y22 v2

etc. You can look upthe ones for the h or a or other parameters. There is even a set (the "s" parameters) based on incident, transmitted and reflected waves. They're particularly appropriate for microwave and other high-frequency circuits. You won't run into those unless you're going for an EE degree, probably.

I should also mention that all parameters can be complex transforms, i.e. have real and imaginary parts, if there are inductive and/or capacitive components in the network. For example, jωL for an inductor's impedance etc.

 

[Jason-Li]

What kind of equation is this? But your answer is right. So are z12 and z21.

Yeah I was using V1 = z11*I1+z12*I2 with I2=0 so that V1 = z11*I1 which I rearranged to z11 = V1/I1

Just wondering if you could explain in algebraic terms what the original post of:

z21 = V2 / I1 = V1 x (R6 / (R8+R6) ) x ( (R8+R6) / V1) = R6 = 6ohms;

is?