# S-matrix calculation for Directional Coupler

Updated on 2017/08/04 13:05

## Syllabus

• Microwave Network Analysis: S-Matrix calculations for Directional coupler

## Directional Coupler

• A portion of power travelling from Port 1 to Port 2 is coupled to Port 4 but not to Port 3
• A portion of power travelling from Port 2 to Port 1 is coupled to Port 3 but not to Port 4
• Also Port 1 and Port 3 are decoupled as are Port 2 and Port 4

We will use the above properties of D.C. as follows:

1. Directional coupler is 4-Port network. Hence [S] is a 4*4 matrix

$\begin{bmatrix} S \end{bmatrix}=\begin{bmatrix} S_{11} &S_{12} &S_{13} &S_{14} \\ S_{21} &S_{22} &S_{23} &S_{24} \\ S_{31}& S_{32} &S_{33} &S_{34} \\ S_{41}& S_{42} &S_{43} & S_{44} \end{bmatrix}$----(1)

2. In a DC all four ports are perfectly matched to the junction Hence the diagonal elements are zero.

$S11=S22=S33=S44=0$ ---(2)

3. From symmetric property Sij = Sji
$S23=S32$
$S13=S31$
$S24=S42$
$S34=S43$
$S41=S14$ -------(3)
Ideally back power is zero (Pb=0) i.e. There is no coupling between port 1 and Port 3.

Hence, $S13=S31=0$ ------(4)

4. There is no coupling between port 2 and port 4.
Hence, $S24=S42=0$ --------(5)

• Substitute the above values, we get:

$[S]=\begin{bmatrix} 0 &S_{12} &0 &S_{14} \\ S_{12}& 0 &S_{23} &0 \\ 0 &S_{23} & 0 & S_{34}\\ S_{14}& 0 &S_{34} &0 \end{bmatrix}$ -----------(6)
• From unity matrix property: [s].[s]*=[I]

$\begin{bmatrix} 0 &S_{12} &0 &S_{14} \\ S_{12}& 0 &S_{23} &0 \\ 0 &S_{23} & 0 & S_{34}\\ S_{14}& 0 &S_{34} &0 \end{bmatrix}.\begin{bmatrix} 0 &S^{*}_{12} &0 &S^{*}_{14} \\ S^{*}_{12}& 0 &S^{*}_{23} &0 \\ 0 &S^{*}_{23} & 0 & S^{*}_{34}\\ S^{*}_{14}& 0 &S^{*}_{34} &0 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 &0 \\ 0& 1 & 0 &0 \\ 0&0 &1 &0 \\ 0& 0 & 0 &1 \end{bmatrix}$
• R1c1:- |s12|2+|s14|2=1 –-------------(7)
R2c2:- |s12|2 +|s23|2=1 —-----------(8)
R3c3:- |s23|2 +|s34|2=1 --------------(9)
R1c3:-s12. s*23+ S14 s*34=0 --------(10)

• Comparing equation 7 and 8
S14=s23 -------(11)

• Comparing equation 8 and 9
S12=S34 -------(12)

• Let us assume that S12 is real and positive = 'P'
Therfore s12=s34=P=S*34 .........(13)

• Put equation 11 value in equation 10 i.e s14=s23

• From equation 10 & 13
P.s*23+s23.P=0
P[s23+s*23]=0

• Since P$\neq$0, s23+s*23=0
s23=jy
s*23=-jy
i.e. s23 must be imaginary

• Let s23=jq=s14 .........(14)
Therfore, s12=s34=P and s23=s14=jq also  $p^{2}+q^{2}=1$

• Substituting these values, we get:
$[S]=\begin{bmatrix} 0 & p & 0&jq \\ p&0 &jq &0 \\ 0& jq & 0 &p \\ jq & 0 &p &0 \end{bmatrix}$

## Scattering matrix for 3-port Circulator

As it is 3-port device [S] matrix will be of order 3x3:
$[S]=\begin{bmatrix} S_{11}& S_{12} &S_{13} \\ S_{21}& S_{22} &S_{23} \\ S_{31}&S_{32} &S_{33} \end{bmatrix}$

As we have seen properties of circulator so only s21,s32,s13 parameters exist. So the equation becomes:
$[S]=\begin{bmatrix} 0& 0 &S_{13} \\ S_{21}& 0 &0 \\ 0&S_{32} &0 \end{bmatrix}$ ---------(1)

The above S-matrix is for perfectly matched, lossless non-reciprocal 3-port circulator:

• $S_{13}=$ output at port 1/input at port 3
• $S_{21}=$ output at port 2/input at port 1
• $S_{32}=$ output at port 3/input at port

## References

• WikiNote Foundation
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Created by Vishal E on 2017/08/04 13:05

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