S-Matrix calculation for H-Plane Tee

Updated on 2017/07/29 20:22

H-plane Tee

 

We know that,
When TE10 mode is made to propagate into port 3 the two outputs at port 1 & port 2 will have a output equal in phase

Since it is a three port junction the s-matrix can be derived as follows.

\begin{vmatrix} S \end{vmatrix}=\begin{bmatrix} S11 &S12 &S13 \\ S21 &S22 &S23 \\ S31 &S32 &S33 \end{bmatrix}

Step 1: Scattering coefficient (plane of symmetry)
S23= S13 ------- (1)
the outputs at ports 1 and port 2  with an input at port 3 will be equal in phase

Step 2: If port 3 is perfectly matched to the junction
S33 = 0 -------- (2)

Step 3: From symmetric property

Sij = Sji
S12 = S21
S13 = S31
S23 = S32=S13 --------- (3)

Using values of equation 1, 2, 3 in S matrix, it becomes:

\begin{bmatrix} S_{11} &S_{12} &S_{13} \\ S_{12}& S_{22} &S_{13} \\ S_{13} &S_{13} & 0 \end{bmatrix}--- (4)

Step 4: From unitary property \begin{bmatrix} S \end{bmatrix}.\begin{bmatrix} S^{*} \end{bmatrix}=I

i.e\begin{bmatrix} S_{11} &S_{12} &S_{13} \\ S_{12}& S_{22} &S_{13} \\ S_{13} &S_{13} & 0 \end{bmatrix}.\begin{bmatrix} S^{*}_{11} &S^{*}_{12} &S^{*}_{13} \\ S^{*}_{12}& S^{*}_{22} &S^{*}_{13} \\ S^{*}_{13} &S^{*}_{13} & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0& 0\\ 0& 1 &0 \\ 0 & 0&1 \end{bmatrix}

  • R1C1: |S11 |2 + |S12|2 + |S13|= 1 -------- (5)
    R2C2: |S12|2 + |S22|2 + |S13|2 = 1 --------- (6)
    R3C3: |S13|2 + |S13|= 1 --------- (7)
    R3C1: S13.S11* + S13.S12* = 0 ---------- (8)
     
  • Equating (5) & (6), we get,
  • |S11 |2 + |S12|2 + |S13|2=|S12|2 + |S22|2 + |S13|2

               S11 = S22 ---------- (9)

  • From equation (7) S13 = \frac{1}{\sqrt{2}} --------- (10)
  • From equation (8), S13 (S11* + S12*) = 0
    Or S11 = -S12  or S12=-S11----------from (9) ------- (11)
  • Using equation (9), (10) & (11) in equation (5) |S11|2 + |S11|2 + \frac{1}{2} = 1
    Or 2|S11|= \frac{1}{2} or S11 = \frac{1}{2} --------- (12)
  • S12=(-1/2)
    S22=(1/2) ---------- (13)
  • Substitute all these values in S-matrix

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

We know that [b] = [S] [a]

\begin{bmatrix} b1\\ b2\\ b3 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} &\frac{-1}{2} &\frac{1}{\sqrt{2}} \\ \frac{-1}{2} &\frac{1}{2} &\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} & 0 \end{bmatrix}.\begin{bmatrix} a1\\ a2\\ a3 \end{bmatrix} ---------- (14)

b1=\frac{1}{2}a1-\frac{1}{2}a2+\frac{1}{\sqrt{2}}a3---------- (15)

b2=\frac{-1}{2}a1+\frac{1}{2}a2+\frac{1}{\sqrt{2}}a3---------- (16)

b3=\frac{1}{\sqrt{2}}a1+\frac{1}{2}a2---------- (17)

Case 1: a1=a2=0, a3≠ 0
i.e Input at Port 3 and no input at port1 and port 2

Put case 1 values in equation 15, 16, 17:

b1=\frac{a3}{\sqrt{2}}

b3=\frac{a3}{\sqrt{2}}

b3=0

Conclusion: An input at port 3 equally divides between port 1 and port 2 

Case 2: a1=a2=a; a3=0;

Put case 2 values in equation 15, 16, 17:

b1=\frac{a}{2}-\frac{a}{2}=0

b2=\frac{-a}{2}+\frac{a}{2}=0

b3=\frac{a}{\sqrt{2}}+\frac{a}{\sqrt{2}}

Conclusion: Hence the output at port 3 is addition of  the two Equal inputs at port 1 and port 2.

Case 3: a1≠0, a2 = 0, a3 = 0

Put case 3 values in equation 15, 16, 17:

b1=\frac{a1}{2}

b2=-\frac{a1}{2}

b3=\frac{a1}{\sqrt{2}}

Conclusion

 

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Created by Sujit Wagh on 2017/07/27 11:22