Two-port network z-parameters

  • #1
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Homework Statement



Z_PARAMETERS.JPG


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!
 

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  • #2
BvU
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Hello acx, welcome to PF :)

Some things are easy :) enjoy while it lasts...
 
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  • #3
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Thanks BvU, I shall do just that!
 
  • #4
David J
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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
NascentOxygen
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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
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.
 
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  • #6
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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
David J
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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
NascentOxygen
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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.

Z_PARAMS.JPG


Does this answer your question?
 
  • #9
David J
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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
 
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  • #10
rude man
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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).
 
  • #11
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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!!
Untitled.png
 

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  • #12
rude man
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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).
 
  • #13
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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
 
  • #14
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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?

2.png
 

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  • #15
rude man
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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?
 
  • #16
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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?
 
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  • #17
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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?
 
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  • #18
rude man
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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 yij, hij, aij. You should be able to find those from the respective equations also.
 
  • #19
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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.
 
  • #20
rude man
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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.
 
  • #21
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Sorry I'm not sure what you mean by type of equation? Basic ohms law or?
 
  • #22
rude man
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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.
 
  • #23
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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?
 
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