What is E-plane Tee? Derive its S-matrix.
E-plane tee is a three port network. Here the auxiliary or side arm axis is parallel t the plane of the electric field of the main guide and hence this tee is known as E –plane Tee.
Fig. 15 Structure of E\ - plane tee.
It is a three port network. The load connected to port 1 and port 2 are in series with the input power, hence it is also known as series tee. When the port 1 and 2 are terminated with identical loads, then the power will be equally divided between the loads but power flowing in port 1 is 180° out of phase as the flowing in port 2 and hence this tee can be used as a subtractor or differentiator.
As it is three ports network, thus the S- matrix can be written as
\(\left[S\right]=\left[\begin{matrix}S_{11}&S_{12}&S_{13}\\S_{21}&S_{22}&S_{23}\\S_{31}&S_{32}&S_{33}\\\end{matrix}\right]\)
Port 3 is the E - arm through which TE10 mode is made to propagate and if this port is ideally matched with the source then the reflection coefficient S33=0
The S – matrix is modified as
\(\left[S\right]=\left[\begin{matrix}S_{11}&S_{12}&S_{13}\\S_{21}&S_{22}&S_{23}\\S_{31}&S_{32}&0\\\end{matrix}\right]\)
Since the outputs at port 1 and 2 are out of phase by 180° with the input port 3, thus the scattering matrix
S23=-S13
For E – plane tee
Now for symmetric property
S12=S21, S13=S31, S23=S32
Thus finally the S – matrix of a three port E – plane Tee becomes
\(\left[S\right]=\left[\begin{matrix}S_{11}&S_{12}&S_{13}\\S_{12}&S_{22}&S_{13}\\S_{13}&S_{13}&0\\\end{matrix}\right]\) … (1)
Using unitary property
\(\left[\begin{matrix}S_{11}&S_{12}&S_{13}\\S_{12}&S_{22}&S_{13}\\S_{13}&S_{13}&0\\\end{matrix}\right]\left[\begin{matrix}{S_{11}}^\ast&{S_{12}}^\ast&{S_{13}}^\ast\\{S_{12}}^\ast&{S_{22}}^\ast&{S_{13}}^\ast\\{S_{13}}^\ast&{S_{13}}^\ast&0\\\end{matrix}\right]=\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right]\)
\(\left|S_{11}\right|^2+\left|S_{12}\right|^2+\left|S_{13}\right|^2=1\) … (2)
\(\left|S_{12}\right|^2+\left|S_{22}\right|^2+\left|S_{13}\right|^2=1\) … (3)
\(\left|S_{13}\right|^2+\left|S_{13}\right|^2=1\) … (4)
\(S_{13}=\frac{1}{\sqrt{2\ }}\) … (5)
Now again, \(S_{13}{S_{11}}^\ast-\ S_{13}{S_{12}}^\ast=0\) … (6)
and equating equations (2) and (3), we have
S11=S22 … (7)
Thus from equation (6), we have
\(S_{13}{(S_{11}}^\ast-\ {S_{12}}^\ast)=0\) … (8)
S11=S12=S21 … (9)
Using the values of equations (5), (7) and (9), we have
\(\left|S_{11}\right|^2+\left|S_{11}\right|^2+\frac{1}{2}=1\)
\(2\left|S_{11}\right|^2=\frac{1}{2}\)
\(S_{11}=\frac{1}{2}\) … (8)
Hence the S – matrix of a four port magic Tee becomes
\(\left[S\right]=\left[\begin{matrix}\frac{1}{2}&\frac{1}{2}&\frac{1}{\sqrt2}\\\frac{1}{2}&\frac{1}{2}&-\frac{1}{\sqrt2}\\\frac{1}{\sqrt2}&-\frac{1}{\sqrt{2\ }}&1\\\end{matrix}\right]\)
Now, we can write
\(\left[b_i\right]=\left[S_{ij}\right]\left[a_j\right] \left[\begin{matrix}\begin{matrix}b_1\\b_2\\\end{matrix}\\b_3\\\end{matrix}\right]=\left[\begin{matrix}\frac{1}{2}&\frac{1}{2}&\frac{1}{\sqrt2}\\\frac{1}{2}&\frac{1}{2}&-\frac{1}{\sqrt2}\\\frac{1}{\sqrt2}&-\frac{1}{\sqrt{2\ }}&1\\\end{matrix}\right]\left[\begin{matrix}\begin{matrix}a_1\\a_2\\\end{matrix}\\a_3\\\end{matrix}\right]\)
Therefore
\(b_1=\frac{1}{2}a_1+\frac{1}{2}a_2+\frac{1}{2}a_3\)
\(b_2=-\frac{1}{2}a_1+\frac{1}{2}a_2-\frac{1}{2}a_3\)
\(b_3=\frac{1}{\sqrt2}a_1-\frac{1}{2}a_2\)
If \(a_3\neq0,\ a_1=a_2=0\) then
\(b_1=\frac{1}{\sqrt2}a_3\)
\(b_2=-\frac{1}{\sqrt2}a_3\)
b3=0
Thus an input power in port 3 divided equally between port 1 and 2 but 180° out of phase with each other. Thus E – plane tee acts as a 3 dB splitter.
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