[SOLVED] Need help to solve S-matrix problem from Pozar; Microwave Engineering

I don't get how Pozar used the voltage division rule twice to solve for S21. Can anyone explain in detail? Please!

Can anybody Help?

First you have voltage division 8.56 and 141.8||8.56 and then you have voltage division between 8.56 and 50 Ohm.

for al_dddd:
you just forgot the 50 ohm load in your answer. I rewrite it here with the correction:
First you have voltage division 8.56 and 141.8||(8.56+50) and then you have voltage division between 8.56 and 50 Ohm.

Elaborating a little bit more:

let's call "x" the node where the two 8.56 ohm and the 141.8 ohm resistors are connected.
The voltage Vx will be given by the voltage divider formed by the "input" 8.56 ohm series resistor and the 141.8 resistor in parallel with the sum of the "output" 8.56 ohm resistor with the 50 ohm load.
The sum of the "output" 8.56 ohm resistor with the 50 ohm load will be obviously 58.56 ohm. This value in parallel with 141.8 ohm leads to 41.44 ohm.
Then:
Vx=V1*41.44/(41.44+8.56)

Now to go from Vx to V2 we will have the voltage divider formed by the "output" 8.56 ohm resistor and the 50 ohm load, thus

V2=Vx*50/(8.56+50)

from which:

V2=V1*41.44/(41.44+8.56)*50/(8.56+50)

Thanks a lot albbg.

So, what about this question???

A load of 50 Ω is connected in shunt in a 2-wire transmission line of Zo = 50 Ω as shown in the figure. The 2-port scattering parameter matrix (s-matrix) of the shunt element is?

https://obrazki.elektroda.pl/8252618700_1431863907.jpg

We know that S11=(Zin-Zo)/(Zin+Zo) where Zo=50 ohm and Zin=50/2=25 ohm (you have the load in parallel with the 50 ohm shunt) so:

S11=(25-50)/(25+50)=-0.33
due to the symmetry S22=s11

about S21 take into account that V2=V1*25/(50+25)=0.33. The same to calculate S12

But in the answer S21 is given as 2/3 and not 1/3!

Also, S21 = (2*Z21*Zo) / (((Z11 + Z0)(Z11 + Z0)) - Z12*Z21) = 2/3 ! But how to do it directly without conversion? I don't get it!

cannibol_90 said:

But in the answer S21 is given as 2/3 and not 1/3!

Click to expand...

Not sure if I'm looking a the correct figure, but in the one I'm looking at, V1 (the left hand side) = V2 (the right hand side).

The input power is divided equally between the resistor and the output TL. Since (1/9) of the power reflects, the remaining power (8/9) is divided between the two loads, such that each receives (4/9) of the power. The insertion loss is related to the power which is received by the output TL, which is the square root of (4/9) - 2/3.

What I said is that to calculate S21 you have to take into account the relationship: V2=V1*25/(50+25)=0.33, not that S21=V2/V1

In fact S21=2*(V2/V1) that is 2*0.33=0.66

albbg said:

In fact S21=2*(V2/V1) that is 2*0.33=0.66

Click to expand...

We must not be looking at the same image.

If you prefer working in terms of voltages, V1 = V2.

V1 is the sum of an incident wave a1 and a reflected wave b1. => V1 = a1 + b1.

We know through the reflection coefficient that => b1 = -(1/3)*a1.

Therefore, V1 = (2/3)*a1.

Since at the output port, only the b-wave exists (there is no excitation and the impedance is matched), a2 = 0, b2 ≠ 0, and => V2 = a2 + b2 = b2.

Since V1 = V2, V2 = (2/3)*a1, and therefore, b2 = (2/3)*a1.

Since S21 = b2/a1, we have in conclusion that S21 = (2/3).