# S-matrix calculations for E-plane Tee

Updated on 2017/07/27 23:26

## E-plane Tee

We know that,
When TE10 mod is made to propagate into port 3 the two outputs at port 1 & port 2 will have a phase shift of 180°

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)
Since outputs at ports 1 and port 2 are out of phase by 180° with an input at port 3.

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

Step 3: From symmetric property Sij = SjiS12 = S21
S13 = S31
S23 = S32 --------- (3)

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

$\begin{bmatrix} S11 &S12 &S13 \\ S21 &S22 &-S13 \\ S13&-S32 &0 \end{bmatrix}$-------- (4)

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

i.e. $\begin{bmatrix} S11 &S12 &S13 \\ S21 &S22 &-S13 \\ S13 & -S32 &0 \end{bmatrix}.\begin{bmatrix} S11^{*} &S12^{*} &S13^{*} \\ S21^{*} &S22^{*} &-S13^{*} \\ S13^{*} & -S32^{*} &0 \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

• 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 = S22 ---------- (9)
• From equation (7) S13 = $\frac{1}{\sqrt{2}}$ --------- (10)
• From equation (8), S13 (S11- S12*) = 0
Or S11 = S12 = S22 ----------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)
• $\dpi{150} \begin{bmatrix} S \end{bmatrix}=\begin{bmatrix} \frac{1}{2} &\frac{1}{2} &\frac{1}{\sqrt{2}} \\ \frac{1}{2} & \frac{1}{2} &\frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}} & 0 \end{bmatrix}$ ---------- (13)

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

$\dpi{150} \begin{bmatrix} b1\\ b2\\ b3 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} &\frac{1}{2} &\frac{1}{\sqrt{2}} \\ \frac{1}{2} & \frac{1}{2} &\frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}} & 0 \end{bmatrix}\begin{bmatrix} a1\\ a2\\ a3 \end{bmatrix}$ ---------- (14)

$\dpi{150} b1=\frac{1}{2}a1+\frac{1}{2}a2+\frac{1}{\sqrt{2}}a3$---------- (15)

$\dpi{150} b2=\frac{1}{2}a1+\frac{1}{2}a2-\frac{1}{\sqrt{2}}a3$---------- (16)

$\dpi{150} b3=\frac{1}{\sqrt{2}}a1-\frac{1}{\sqrt{2}}a2$---------- (17)

### Case 1: a1=a2=0, a3≠ 0

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

$\dpi{150} b1=\frac{a3}{\sqrt{2}}$

$\dpi{150} b2=(-\frac{a3}{\sqrt{2}})$

$\dpi{150} b3=0$

Conclusion: An input at port 3 equally divides between port 1 and port 2 but introduces phase shift of 180°.

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

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

$\dpi{150} b1=\frac{a}{2}+\frac{a}{2}$

$\dpi{150} b2=\frac{a}{2}+\frac{a}{2}$

$\dpi{150} b3=\frac{a}{\sqrt{2}}-\frac{a}{\sqrt{2}}=0$

Conclusion: Equal inputs at port 1 and port 2 results in no output at port 3

### Case 3: a1≠0, a2 = 0, a3 = 0

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

$\dpi{150} b1=\frac{a1}{2}$

$\dpi{150} b2=\frac{a1}{2}$

$\dpi{150} b3=(-\frac{a1}{\sqrt{2}})$

Conclusion

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Created by Vishal E on 2017/07/26 11:00

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