S-Matrix Calculation For E-H Plane Tee (Magic Tee)

Updated on 2017/07/30 01:23

E-H Plane Tee (Magic Tee)

MagicTee.jpg

  • [S]= 412089720.png ----(1)
  • Because of H-plane Tee section, s23=s13 -----(2)
  • Because of E-plane Tee section, s24=-s14 ----(3)
  • Because of geometry of the junction on inputs at port 3 can’t come out of port 4 since they are isolated ports and vice-versa
    s34=s43=0 -------(4)
  • From symmetric property sij = sji:
    s12=s21; s13=s31;
    s23=s32; s34=s43;
    s24=s42; s41=s14; --------(5)
  • If ports 3 and 4 are perfectly matched to the junction:
    s33=s44=0 ---(6)
     
  • Substitute above properties in equation (1)
    [s]= 1619085221.png -------(7)
     
  • From unity matrix property, [s].[s]*=[I]
    1619085221.png . 2032323542.png = 1716666570.png
     
  • R1c1:- |s11|2 +|s12|2+|s13|2 +|s14|2=1 -----(8)
    R2c2:- |s12|2 +|s22|2+|s13|2 +|s14|2=1 -----(9)
    R3c3:- |s13|2 +|s13|2=1 ------(10)
    R4c4:- |s14|2 +|s14|2=1 ------(11)
     
  • From equation 10 and 11,
    S13= 1/√2 — (12)
    S14= 1/√2 — (13)
     
  • Comparing equation (8) and (9) we get,
    s11=s22 ------(14)
     
  • Using these values from equation (12) and (13) in equation (8), we get:
    |s11|2 + |s12|2 + \frac{1}{2} + \frac{1}{2} =1
    i.e |s11|2 + |s12|2=0
     
  • We know that
    s11=0 hence s12=0 ------------------------------(15)
    So from equation (9) s22=0 --------------------(16)

This means port 1 and port 2 are also perfectly matched to the junction.
Hence in any four ports junction, if any two ports are perfectly matched to the junction then the remaining two are automatically matched to the junction such a junction where in all the four ports are perfectly matched to the junction is a Magic Tee.

[S] = \begin{bmatrix} 0& 0 &\frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\ 0& 0 &\frac{1}{\sqrt{2}} &\frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} &0 & 0\\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} &0 & 0 \end{bmatrix} -----(17)

  • We know that, [b]=[s] . [a]
    \begin{bmatrix} b1\\ b2\\ b3\\ b4 \end{bmatrix}=\begin{bmatrix} 0& 0 &\frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\ 0& 0 &\frac{1}{\sqrt{2}} &\frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} &0 & 0\\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} &0 & 0 \end{bmatrix}\begin{bmatrix} a1\\ a2\\ a3\\ a4 \end{bmatrix}
     
  • b1= 1/√2(a3+a4)
    b2= 1/√2(a3-a4) -----(18)
    b3= 1/√2(a1+a2)
    b4= 1/√2(a1-a2)
     

From equation (18):

Case 1: a3\neq 0, a1=a2=a4=0

b1= a3/√2  
b2= a3/√2 
b3= 0 
b4= 0 --- satisfies property of H-plane Tee

Case 2: a4\neq 0 , a1=a2=a3=0

b1= a4/√2 
b2= -a4/√2
b3= 0 
b4= 0 --- satisfies property of E-plane Tee

Case 3: a1\neq 0, a2=a3=a4=0

b1= 0;
b2= 0; 
b3=  a1/√2;
b4= a1/√2;

That is, 
When power is fed into the port 1 nothing comes out of port 2 and vice-versa even though they are collinear ports (MAGIC)Hence they are isolated ports.

Case 4: a3=a4, a1=a2=0

So we have,
b1=1/√2(2a3);
b2= 0; 
b3= 0; 
b4= 0;  

Case 5: a1=a2 , a3=a4=0 (ADDITIVE PROPERTY)

So we have,
b1=0;
b2= 0; 
b3= 1/√2(2a3);
b4= 0;  

Equal inputs at port (1) and (2) results in on outputs at port (3) and no output at 1,2 & 4.

References

  • Notes by Prof. Sujit Wagh
  • WikiNote Foundation
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Created by Sujit Wagh on 2017/07/26 09:54