Back in Lecture 23, we began our discussion of dividers and couplers by considering important general properties of three and four-port networks. This was followed by an analysis of three types of three-port networks in Lectures 24 and 25. we will now move on to (reciprocal) directional couplers, which are four-port networks. As in the text, we will consider these specific types of directional couplers:
2. 180 ° Hybrid,
3. Coupled Line, and
4. Lange Coupler.
We will begin with the quadrature (90 °) hybrid. Shows this coupler implemented with microstrip as a 1:1 power divider:

Because of symmetry, we can simplify the analysis of this circuit considerably using even-odd mode analysis. This process is similar to what we did in the last lecture with the Wilkinson power divider.
Even-Odd Mode Analysis of the Quadrature Hybrid
The normalized (wrt Z0 ) TL circuit is shown in minus the return lines:

A symmetric (even mode) excitation of this circuit is shown in

and an anti-symmetric (odd mode) excitation is shown in

Observe that the circuit and its boundary conditions remain the same in both the even and odd mode configurations. It is only the excitation that changes. Because of this and the circuit being linear, the total solution is simply the sum of the even and odd mode solutions. each solution (even and odd) is simpler to determine than the complete circuit, which is why we employ this technique.
• Even mode. Because the voltages and currents must be the same above and below the line of symmetry (LOS) in then I = 0 at the LOS Þ open circuit loads at the ends of l/8 stubs, as shown.
Referring to the definition of Bi (i =1,…,4) in we can write from that for the even mode excitation:
Be1 = te Ae1 , Be2 = Te Ae1
Be3 = Be2 = Te Ae1 , Be4 = Be1 =te Ae1
where Ae1 =1/2, and te and Te are the reflection and transmission coefficients for the even mode configuration.
• Odd mode. Because the voltages and currents must have opposite values above and below the LOS in then V = 0 along the LOS Þ short circuit loads at the ends of l/8 stubs, as shown.
B01 = t0 A01 , B02 = T0 A01
Then,
B03 =-B02 =-T0 A01; B04 =-B01 =-t0A01
where Ao1 =1/2 and t0 and T0 are reflection and transmission coefficients for the odd mode configuration.
• Total solution. The total solution is the sum of the voltages in both circuits. From this fact, we can deduce that the total Bi coefficients will be the sum of (1) and (2):
B1 =Be1 +B01 =(1/2) te +(1/2) t0
B<sub2< sub=""> =Be2 +B02 =(1/2)Te +(1/2)T0
B<sub3< sub=""> =Be3 +B03 =(1/2)Te +(1/2)T0
B4 =Be4 +B04 =(1/2) te +(1/2) t0

Likewise, the incident wave coefficients are
A1 =Ae1+ A01 =1/2+1/2 =1
A4 =Ae4+ A04 =1/2-1/2 =0
These match the assumed excitation in the original circuit on
p. 2. To finish the calculation of the S parameters for the quadrature hybrid, we need to determine the reflection and transmission coefficients for the even- and odd-mode configurations. Your text shows that the solutions for te and Te are
te =0 and Te -1/Ö(1+j)
Here we’ll derive solutions for t0 and T0 .From

We have three cascaded elements, so we’ll use ABCD parameters to solve for the overall S parameters of this circuit.
. Elements 1 and 3. These are short circuit stubs of length l/8, which appear as the shunt impedance
Zin=jZ0 tanbl where bl =(2p/l).(/8)= p/4)
There fore
Zin/Z0 =j,or YN =-j
From the inside flap of your text:

. Element 2. This is a l/4-length of TL where
bl =(2p/l).(p/4) =(p/2)
From the inside flap of your text:

Cascading these three ABCD matrices we find the overall ABCD matrix for odd mode excitation:

Using Table 4.2, we can convert these to S parameters:

=Ö2(2/2+2j)= Ö2/1+j
Since the ports are matched, then:
t0 =S11 =0 T0 =S21 =(Ö2/1+j).(1-j/1-j)=1/Ö2(1-j)
Finally, using (7), (8), (13) and (14) in (3)-(6) we find • B1 = 0
. B2 =-1/2 Ö2(1+j)+1/2 Ö2(1-j)=-j/ Ö2
. B3 =-1/2(1/ Ö2)(1+j)-(1/2 Ö2)(1-j)=-1/ Ö2
. B4 =1/2 . 0 -1/2 .0 = 0
When properly interpreted, these results tell us much about the circuit. In particular, when port 1 is excited and all other ports terminated in matched loads, then:
• B1 = 0 Þ port 1 is matched.
• B2 =-j2 Ö -90 ° phase shift from port 1 to 2, and 1/2 of input power is delivered to port 2.
• B3 =-1/ Ö2 Þ -180 ° phase shift from port 1 to 3 (90 ° phase shift between ports 3 and 2), and one half of the input power is delivered to port 3.
• B4 = 0 Þ no power output to port 4.
Because of the high degree of symmetry, we can treat any port as the input port. Then, the isolation is the other port on the same “side” as the input and the outputs are the two ports on the other “side” of the circuit.
Employing this concept and the results above, we can construct the other three columns in the full S matrix for the quadrature (90 °) hybrid by simply transposing rows of the first column:

That is, the first column in (19) is the results from (15)-(18) when the input was assumed at port 1. In the second column, we can directly deduce that the outputs are at ports 1 and 4, the input is at port 2 and the isolation is at port 4. Further transposition of the rows in column 1 produces columns 3 and 4.
Example N26.1. Design a branch line hybrid coupler using 100- ohm microstrip on 32-mil RO4003C for a center frequency of 2.5 GHz. Include the effects of copper and substrate losses.
Because there are two different characteristic impedances needed for the 90 ° hybrid device, two different widths of microstrip must be computed (because W/d depends on Z0) and two different l/4-lengths must be determined (because er,e depends on W/d).
• Z0 =100 ohm sections. Using LineCalc, W =18.02 mil and er,e= 2.424 The guide wavelength at this frequency is then
l=c/f =(2.998´108)/(2.5´109.Ö2.424) =7.702 cm
Hence, this branch line coupler should have 100-ohm lines with length = l/4 =1.93 cm.
• Z0 Ö2 = 70.71 ohm sections. Using LineCalc, W = 39.62 mil and er,e=2.545. The guide wavelength is then
l=c/f =(2.998´108)/(2.5´109.Ö2.545) =7.517 cm
Hence, this branch line coupler should have 70.71-ohm lines with length = l 4 =1.88 cm. The following S parameter results were obtained for this design using ADS.