Two hole directional Coupler:-

• It consists of two guides the main and the auxiliary with two tiny holes common between them as shown in figure.
• The two holes are at a distance of λ9/4 where λ9 is the guide wavelength.

• The two leakages are out of phase by 1800 at the position of second hole and hence they add up contributing to Pf.
• But the two leakages are out of phase by 1800 at the position of the first hole and therefore they cancel each other making Pb = 0
• Since the distance between holes is λ9/4, Pb is made ‘O’.

• [S] matrix for two hole directional coupler.

  Directional coupler is four port.

  Hence, [S] is 4 x 4 matrix.

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

Since, all ports are perfectly matched.

Hence, $S_{11}= S_{22} = S_{33} = S_{44} = 0$

Port 1 and 3 are decoupled as well as port 2 and 4

$S_{13} = S_{31}= S_{24} = S_{42} = 0$

Also, [s] is symmetric matrix i.e. Sij = Sji

$\begin{bmatrix}0 & S_{12} & 0 & S_{14} \\ S_{21} & 0 & S_{23} & 0 \\ 0 & S_{32} & 0 & S_{34} \\ S_{41} & 0 & S_{43} & 0 \end{bmatrix}$

From unitary property of [S] matrix.

[S] [S]* = I

$\begin{bmatrix}0 & S_{12} & 0 & S_{14} \\ S_{21} & 0 & S_{23} & 0 \\ 0 & S_{32} & 0 & S_{34} \\ S_{41} & 0 & S_{43} & 0 \end{bmatrix} \begin{bmatrix}0 & S_{12} * & 0 & S_{14}* \\ S_{21}* & 0 & S_{23}* & 0 \\ 0 & S_{32}* & 0 & S_{34}* \\ S_{41}* & 0 & S_{43}* & 0 \end{bmatrix}= \begin{bmatrix}1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}$

$R_1C_1 = |S_{12}|^2 + |S_{14}|^2 = 1 ----- 1) \\ R_2C_2 = |S_{12}|^2 + |S_{23}|^2 = 1 ----- 2) \\ R_3C_3 = |S_{23}|^2 + |S_{34}|^2 = 1 ----- 3) \\ R_4C_4 = |S_{14}|^2 + |S_{34}|^2 = 1 ----- 4) \\ R_1C_3 = S_{12}.S_{23}* + S_{14}.S_{34}*= 0 ----- 5)$

From equation 1) and 2)

$S_{14} = S_{23}$

From equation 1) and 4)

$S_{12} = S_{34}$

Let $S_{12} = S_{34} – P$ (Real no.)

Hence, $S_{12}* = S_{34}* = P$

Equation 5) becomes

$P. S_{23}* + S_{14}P = 0 \\ P (S_{23}* + S_{14}) = 0 \\ But, S_{14} = S_{23} \\ P (S_{23}* + S_{23}) = 0 \\ S_{23}* = - S_{23} \\ Let S_{23} = jq = S_{23}* = -jq$

$[S]=\begin{bmatrix}0&P&0&jq \\ P&0&jq&0 \\ 0&jq&0&P \\ jq&0&P&0 \end{bmatrix}$