Microwave Coupled Lines

1.1 Introduction

We consider the lines of propagation in which length and coupling are about the wavelength λ. We use to say that the effect is distributed. The electromagnetic (EM) field of the line (1) induces an EM field of the line (2) and reciprocally.

1.2 Description

The phenomenon is induced on a wavelength λ that is around some Giga Hertz = GHz = 109 Hz in microwaves.

=cf=300,000km/s109=3×108109=0.3m

1.2.1 Proximity coupler

We like to say that going from 1 to 2 is the direct way and going from 3 to 4 is the coupling way. Then, we define:

The same definitions occur to the other type of coupler as the hybrid ring or the guide coupler.

1.2.2 Hybrid ring

Suppose that 1 is the input of the waves. From 1 to 3, there is a path of λ/4. Through the other way, 1, 4, 2, 3, there is a path of 5λ/4. And a difference of way of:

l=5λ4−λ4=λ

Then, 3 is in phase with 1 and this recombines two waves in 3.

Now, consider 1 as the input and 2 as the output. The way 1, 3, 2 is 4λ/4 and the way 1, 4, 2 is 2λ/4. This induces a difference in way of:

l=4λ4−2λ4=λ2

And nothing appears in 2. We have constructed a coupler.

1.2.3 Guide coupler

Coupling is made by the two irises I1 and I2 upon a length of λg/4.

On iris I1, a small part of the energy (10−3) goes through the secondary guide and induces B′4 and B′3.

On iris I2, a small part of the energy (10−3) goes through the secondary guide and induces B″4 and B″3.

– In branch 3, if waves B′3 and B″3 have the same phase, then B3 ≠ 0.

– But in branch 4, if waves B′4 and B″4 are in opposition, then B4 ≠ 0.

The difference in way is 2l = λg/2. Then, we have a diphase of π.

1.2.4 Use of couplers

This system is used to take a previous small part of the microwave energy. It can be used:

– to measure the frequency, power, etc.

– to make a feedback technique and then equalize the output power.

1.3 Lossless equivalent circuit

1.3.1 Mesh law

We give the lossless equivalent circuit on a dz length. We know the equivalent circuit in the case of alone line and we suppose two lines are coupled by:

– a magnetic coupling M; and

– a capacitive coupling Cm.

The mesh law of the equivalent circuit is now:

1(z)=Ldz∂I1∂t+Mdz∂I2∂t+V1(z+dz)V2(z)=Ldz∂I2∂t+Mdz∂I1∂t+V2(z+dz)

But by definition:

1(z)−V1(z+dz)=−∂V1∂zdzV2(z)−V2(z+dz)=−∂V2∂zdz

And we get:

∂V1∂z=L∂I1∂t+M∂I2∂t−∂V2∂z=M∂I1∂t+L∂I2∂t

1.3.2 Notch law

Now, for the notch law, we have to consider the currents:

1(z)=C0dz∂V1∂t+Cmdz∂(V1−V2)∂t+I1(z+dz)I2(z)=C0dz∂V2∂t+Cmdz∂(V2−V1)∂t+I2(z+dz)

Using:

1(z)−I1(z+dz)=−∂I1∂zdzI2(z)−I2(z+dz)=−∂I2∂zdz

We also get:

∂I1∂z=(C0+Cm)∂V1∂t−Cm∂V2∂t−∂I2∂z=−Cm∂V1∂t+(C0+Cm)∂V2∂t

1.3.3 Coupled differential system of the first order

Considering the whole capacity:

=C0+Cm

We have to find a solution of the coupled differential system of four equations:

∂I1∂z=C∂V1∂t−Cm∂V2∂t[1.1]−∂I2∂z=−Cm∂V1∂t+C∂V2∂t[1.2]−∂V1∂z=L∂I1∂t+M∂I2∂t[1.3]−∂V2∂z=M∂I1∂t+L∂I2∂t[1.4]

1.3.4 Non-coupled differential system of the second order

The resolution of these four coupled equations [1.1] – [1.4] is easy.

First, we transform to non-coupled equations. We consider the sums of equations [1.1] + [1.2] and [1.3]+ [1.4] and have:

∂Ie∂z=(C−Cm)∂Ve∂t−∂Ve∂z=(L+M)∂Ie∂t

With the even (sum) voltages and currents:

e=I1+I2Ve=V1+V2

It is the same as if there is only one line of capacity (C − Cm) and inductance (L + M).

We also consider the differences of equations [1.1] – [1.2] and [1.3] – [1.4] and then we have:

∂Io∂z=(C+Cm)∂Vo∂t−∂Vo∂z=(L−M)∂Io∂t

With the odd (differences) voltages and currents:

o=I1−I2Vo=V1−V2

It is the same as if there is only one line of capacity (C + Cm) and inductance (L − M).

Combining these two groups of two equations, we obtain equations on the even modes and on the odd modes (non-coupled telegraph equations).

2∂z2|IeVe−LC(1+kL)(1−kC)∂2∂t2|IeVe=0∂2∂z2|IoVo−LC(1−kL)(1+kC)∂2∂t2|IoVo=0

where we have defined the magnetic and capacitive coefficients and the even and odd speed of propagation:

L=MLandkC=CmCve=1LC(1+kL)(1−kC)vo=1LC(1−kL)(1+kC)

EM state on the two lines results from the superposition of two Transversal Electro Magnetic (TEM) modes. These two modes are orthogonal and they are the normal modes of the coupler.

1.4 Homogeneous medium of permittivity ε

If the medium is homogeneous with a permittivity ε (this is not the case of the microstrip) and the TEM waves are propagating with the speed of the light in this medium, then:

=ve=vo=1με

And the coupling coefficient k will be:

=kL=kC=CmC=ML

We also have:

=1/LC1−k2

where /LC is the propagation speed without coupling. Now it is possible to have an expression of the coupling coefficient:

=1−μεLC

1.5 Sinusoidal excitation, even and odd modes

In the case of a sinusoidal excitation (i = e and o):

i(z,t)=ReVi(z)ejωtIi(z,t)=ReIi(z)ejωt

The equations to be satisfied are:

2dz2|IeVe+ω2LC(1−k2)|IeVe=0∂2dz2|IoVo+ω2LC(1−k2)|IoVo=0

The propagation constant is:

=ωLC(1−k2)

The general solutions of these two last equations with...