Characteristic Impedance for a (∏) two port network

[agata78]

Homework Statement

Calculate the Characteristic Impedance (Z0) of the two port network?

Homework Equations

When port 2 open-circuit, I2 = 0

So, Z11 = V1 / I1

= 40(60) / 100 = 24 Ω

When I2 = 0, thus V2

V2 = 40 / 20+40 V1 = 0.666 V1

Z21 = V2 / I1

= 0.666 V1 / (V1 / 40-20)

= 0.666 V1 / (V1/20) = 0.666 V1 : V1/20

= 0.666 V1 x 20/v1

= 0.666 V1 x 20 / V1

= 13.32 V1 / V1

= 13.32 Ω

Now, when I1 = 0, so:

Z22 = V2 / I2

= 40(60) / 100

= 2400 / 100

= 24 Ω

When port 2 open circuit, I1 - 0, so:

V1 = 40 / (20+40) V2

= 40 / 60 V2

= 0.666 V2

I2 = V2 / 24

Z12 = V1 / I2

= 0.666 V2 / (V2 / 24)

= 0.666 V2 : (V2 / 24)

= 0.666 V2 (24/ V2)

= 15.984 Ω

AM I CORRECT?

Can someone please help?

 

[rude man]

Write the equations defining Z parameters, then set i2 = 0.

I have never heard the term 'characteristic impedance' apply to z parameter networks. It's used in transmission lines.

 

[The Electrician]

When your network is symmetrical, Z12 has to equal Z21.

Why are you going to all the trouble of calculating the Z parameters, and the voltage transfer ratios, when (apparently) all you want is the characteristic impedance?

And, if you want the characteristic impedance, why did you stop your calculations before you calculated it?

 

[agata78]

Z0 = √ R1 x R2² / R1 + (2 x R2)

Z0 = √ 20 x (40)² / 20 + (2 x 40)

Z0 = √ 20 x 1600 / 20 + 80

Z0 = √ 32000 / 100

Z0 = √ 320

Z0 = 17.8885

Z0 = 17.88 Ω

Can someone please confirm that I am correct? Much appreciated!

 

[The Electrician]

Your final answer is correct, but your arithmetic procedures leading to it are not unambiguous.