Question:

Published on: 24 July, 2024

**With the neat sketch, explain working principle of a two – hole directional coupler. Derive the scattering matrix of magic Tee. A multi hole directional coupler is fed with single power of 2.8 mW at 10 GHz. The coupling factor is 3 dB and the directivity is better than 40 dB over X-band range. Find the distribution of power at all other ports.**

Answer:

**Two holes directional coupler: **

The ideal two holes coupler is depicted in Fig. 5. It consists of two guides the main and the auxiliary with two tiny holes between them as shown in Fig. 5. The two holes are separated from each other by the distance \(\lambda_g/4\), wher \(\lambda_g\) is the guide wavelength.

Fig. 5 Two hole directional coupler

The power is incident at port 1 and wave is traveling towards port 3 through hole A and B simultaneously. Both the waves which is passing through hole 1 and hole 2 are covered same distance from port 1 to port 3. Hence power is added up constructively at port 3 by the two holes. When power propagating in reverse way, then wave covers \(\lambda_g/2\) more distance by hole B than the coming from hole A. Therefore power added at port 4 is destructively because both wavers are in out of phase each other. Hence no power gets coupled to this port. In this coupler the directivity as a very sensitive function of frequency.

The number of holes can be one (as in Brthe cross guide coupler) or more than two (as in a Multi – holes coupler). The degree of coupling is determined by size and location of the holes in the waveguide walls. Although a high degree of directivity can be achieved as a fixed frequency, it is quite difficult over a band of frequencies. In this connection, it should be realized that the frequency determines the separation of the two holes as a fraction of the wavelength.

**Magic Tee: **

Any two port networks are represented by a special type of parameter by which reflection co- efficient or transmission co – efficient of any network is measured. This special type of parameter is called scattering parameter of S - parameter.

Characteristics of microwave devices can be represented by its scattering parameters (S- parameters) instead of conventional Z, Y or hybrid parameter for following reason

- No equipments are readily available to measure total voltage and total currents at the ports of the network.
- Over a broad band of frequencies it is difficult to achieve short and open circuits.
- Active devices like tunnel diode and power transistors frequently will not have stability for a short or open circuit. The scatter parameters are the reflection co- efficient of a port and transmission co - efficient between different ports with other ports matched terminated. These are also the frequently measured quantities and can be measured readily with the help of a network analyzer.

**Scattering matrix of Magic Tee**

Fig. 6 Hybrid Tee or Magic Tee

Hybrid means mixture of two different types of things. In case of microwave the hybrid junctions are four port networks and are of different types, like hybrid T junction, ratrace coupler, branchline hybrid etc.

Fig. 6 shows the construction of hybrid Tee or magic Tee. Here, the collinear arms are the side arms while the other two are called E – arm (series) and H- arm (shunt). The E and H – arm are crossed polarized or are out of phase and hence there is no coupling. Power entering from the H- arm is equally divided in phase in the two side arms. Similarly, the power entering from the E-arm is equally divided in side arms but is out of phase.

Further, magic tee being combination of E and H plane tees, if power is feed through port 1 and 2, it is added in h- arm, while it is subtracted in the E – arm.

As it is four ports network, thus the S- matrix can be written as

\(\left[S\right]=\left[\begin{matrix}\begin{matrix}\begin{matrix}S_{11}\\S_{21}\\\end{matrix}&\begin{matrix}S_{12}\\S_{22}\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}S_{13}\\S_{23}\\\end{matrix}&\begin{matrix}S_{14}\\S_{24}\\\end{matrix}\\\end{matrix}\\\begin{matrix}\begin{matrix}S_{31}\\S_{41}\\\end{matrix}&\begin{matrix}S_{32}\\S_{42}\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}S_{33}\\S_{43}\\\end{matrix}&\begin{matrix}S_{34}\\S_{44}\\\end{matrix}\\\end{matrix}\\\end{matrix}\right]\)

For H- plane Tee

S_{23} = S_{13}

For E – plane tee

S_{24} = S_{14}

Because of geometrical shape input of port 3 cannot come out through port 4.

S_{34} = S_{43} = 0

Now for symmetric property

S_{12} = S_{21}, S_{13} = S_{31, S23 = S32, S34 = S43, S24 = S42, S41 = S14}

Let port 3 and 4 are perfectly matched with respect to source.

Hence no reflection

S_{44} = S_{33} = 0

Thus S – matrix of a magic tee can be rewritten with the help of above consideration

\(\left[S\right]=\left[\begin{matrix}\begin{matrix}\begin{matrix}S_{11}\\S_{21}\\\end{matrix}&\begin{matrix}S_{12}\\S_{22}\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}S_{13}\\S_{13}\\\end{matrix}&\begin{matrix}S_{14}\\{-S}_{14}\\\end{matrix}\\\end{matrix}\\\begin{matrix}\begin{matrix}S_{13}\\S_{14}\\\end{matrix}&\begin{matrix}S_{13}\\{-S}_{14}\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}0\\0\\\end{matrix}&\begin{matrix}0\\0\\\end{matrix}\\\end{matrix}\\\end{matrix}\right]\) … (1)

Using unitary property

\(\left[\begin{matrix}\begin{matrix}\begin{matrix}S_{11}\\S_{21}\\\end{matrix}&\begin{matrix}S_{12}\\S_{22}\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}S_{13}\\S_{13}\\\end{matrix}&\begin{matrix}S_{14}\\{-S}_{14}\\\end{matrix}\\\end{matrix}\\\begin{matrix}\begin{matrix}S_{13}\\S_{14}\\\end{matrix}&\begin{matrix}S_{13}\\{-S}_{14}\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}0\\0\\\end{matrix}&\begin{matrix}0\\0\\\end{matrix}\\\end{matrix}\\\end{matrix}\right]\left[\begin{matrix}\begin{matrix}\begin{matrix}{S_{11}}^\ast\\{S_{21}}^\ast\\\end{matrix}&\begin{matrix}{S_{12}}^\ast\\{S_{22}}^\ast\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}{S_{13}}^\ast\\{S_{13}}^\ast\\\end{matrix}&\begin{matrix}{S_{14}}^\ast\\{{-S}_{14}}^\ast\\\end{matrix}\\\end{matrix}\\\begin{matrix}\begin{matrix}{S_{13}}^\ast\\{S_{14}}^\ast\\\end{matrix}&\begin{matrix}{S_{13}}^\ast\\{{-S}_{14}}^\ast\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}0\\0\\\end{matrix}&\begin{matrix}0\\0\\\end{matrix}\\\end{matrix}\\\end{matrix}\right]=\left[\begin{matrix}\begin{matrix}\begin{matrix}1\\0\\\end{matrix}&\begin{matrix}0\\1\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}0\\0\\\end{matrix}&\begin{matrix}0\\0\\\end{matrix}\\\end{matrix}\\\begin{matrix}\begin{matrix}0\\0\\\end{matrix}&\begin{matrix}0\\0\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}1\\0\\\end{matrix}&\begin{matrix}0\\1\\\end{matrix}\\\end{matrix}\\\end{matrix}\right]\)

\(\left|S_{11}\right|^2+\left|S_{12}\right|^2+\left|S_{13}\right|^2+\left|S_{14}\right|^2=1\) … (2)

\(\left|S_{12}\right|^2+\left|S_{22}\right|^2+\left|S_{13}\right|^2+\left|S_{14}\right|^2=1\) … (3)

\(\left|S_{13}\right|^2+\left|S_{13}\right|^2=1\) … (4)

\(\left|S_{14}\right|^2+\left|S_{14}\right|^2=1\) … (5)

From equation (4) and (5), we have

\(S_{13}=S_{14}=\frac{1}{\sqrt2}\) … (6)

Comparing equation (2) and (3), we have

\(S_{11}=S_{22}\) … (7)

Using the values of equations (6), (7) and (2), we have

\(\left|S_{11}\right|^2+\left|S_{12}\right|^2+\frac{1}{2}+\frac{1}{2}=1\)

\(\left|S_{11}\right|^2+\left|S_{12}\right|^2=0\) … (8)

\(\left|S_{12}\right|^2+\left|S_{22}\right|^2+\frac{1}{2}+\frac{1}{2}=1\)

\(\left|S_{12}\right|^2+\left|S_{22}\right|^2=0\) … (9)

Therefore S_{11} = S_{12} = S_{22} = 0

Hence the S – matrix of a four port magic Tee becomes

\(\left[S\right]=\left[\begin{matrix}\begin{matrix}\begin{matrix}0\\0\\\end{matrix}\ \ \ \ \ &\begin{matrix}0\\0\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}\frac{1}{\sqrt{2\ }}\\\frac{1}{\sqrt{2\ }}\\\end{matrix}&\begin{matrix}\frac{1}{\sqrt{2\ }}\\-\frac{1}{\sqrt{2\ }}\ \\\end{matrix}\\\end{matrix}\\\begin{matrix}\begin{matrix}\frac{1}{\sqrt{2\ }}\\\frac{1}{\sqrt{2\ }}\\\end{matrix}&\begin{matrix}\frac{1}{\sqrt{2\ }}\\-\frac{1}{\sqrt{2\ }}\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}0\\0\\\end{matrix}\ \ \ \ &\begin{matrix}0\\0\\\end{matrix}\\\end{matrix}\\\end{matrix}\right]\)

Now, we can write

\(\left[\begin{matrix}\begin{matrix}b_1\\b_2\\\end{matrix}\\\begin{matrix}b_3\\b_4\\\end{matrix}\\\end{matrix}\right]=\left[\begin{matrix}\begin{matrix}\begin{matrix}0\\0\\\end{matrix}\ \ \ \ \ &\begin{matrix}0\\0\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}\frac{1}{\sqrt{2\ }}\\\frac{1}{\sqrt{2\ }}\\\end{matrix}&\begin{matrix}\frac{1}{\sqrt{2\ }}\\-\frac{1}{\sqrt{2\ }}\ \\\end{matrix}\\\end{matrix}\\\begin{matrix}\begin{matrix}\frac{1}{\sqrt{2\ }}\\\frac{1}{\sqrt{2\ }}\\\end{matrix}&\begin{matrix}\frac{1}{\sqrt{2\ }}\\-\frac{1}{\sqrt{2\ }}\\\end{matrix}\\\end{matrix}&\begin{matrix}\begin{matrix}0\\0\\\end{matrix}\ \ \ \ &\begin{matrix}0\\0\\\end{matrix}\\\end{matrix}\\\end{matrix}\right]\left[\begin{matrix}\begin{matrix}a_1\\a_2\\\end{matrix}\\\begin{matrix}a_3\\a_4\\\end{matrix}\\\end{matrix}\right]\)

Therefore

\(b_1=\frac{1}{\sqrt2}(a_3+a_4)\), \(b_3=\frac{1}{\sqrt2}(a_1+a_2)\)

\(b_2=\frac{1}{\sqrt2}(a_3-a_4)\), \(b_4=\frac{1}{\sqrt2}(a_1-a_2)\)

If \(a_3\neq0,\ a_1=a_2=a_4=0\)

Then

\(b_1=\frac{1}{\sqrt2}a_3\), \(b_3=0\)

\(b_2=\frac{1}{\sqrt2}a_3\), \(b_4=0\)

Thus when port – 3 is the input port, then power will be divided equally in port 1 and 2.

In the second case when \(a_4\neq0,\ a_1=a_2=a_3=0\) then we can write

\(b_1=\frac{1}{\sqrt2}a_4\) , \(b_3=0\)

\(b_2=-\frac{1}{\sqrt2}a_4\) , \(b_4=0\)

Thus when port 4 is the input port then power will be divided equally but in out of phase between ports 1 and 2.

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