# Branchline Coupler Theory

# 3dB Branchline Coupler

The Branchline coupler is a signal divider that can separate an incoming signal into two equal power, but 90 degree phase shifted signals. It consists of 4 ports: Input (*P _{1}*), Output 1 (

*P*) and Output 2 (

_{2}*P*). The fourth port (

_{3}*P*) is terminated at 50 ohms for the signal splitter and is isolated from the input.

_{4}The branchline coupler is used to generate quadrature signals for our QAM radio.

# Theory of Operation

The branchline coupler consists of four transmission lines, each a quarter-wavelength long at the target frequency. Two have the characteristic impedance of the terminations (Ports P_{1}-P_{4}) and two have one that is reduced by 1/√2 or is 35.35 Ohms.

It should be obvious that the impedance of the transmission lines do not match with the port, i.e. 50 Ohm || 35.35 Ohm is not 50 Ohm. This tells us that the operation is a little more complicated than just a phase shift through a few transmission lines. We will use even-odd mode analysis to determine the operation of the branchline coupler.

Every signal can be split into an even and odd mode, or a common mode and differential mode. Drawn here are the two modes: even (left) and odd (right). We will use the fact that in the **even mode**, the dashed line is an **open** – no current flows through the transmission lines as the signals are in phase. For the **odd mode** the dashed line is a **virtual ground**, or short.

We will examine the even mode first. We have split the coupler into two parts, just looking at the top half (they are symmetrical). We treat the dashed-line plane as a virtual open.

We then calculate the input impedance into the /8 transmission lines using the standard transmission line impedance calculation. For /8, *l*=/4 and tan(/4)=1. Since *Z _{L}*=∞, the input impedance can be written as

We can do the same analysis for the odd mode, here simply replacing *Z _{L}*=∞ with

*Z*=0.

_{L}Note that the only difference is the negation.

We are now going to make an ABCD matrix of each element: first /8 transmission line, the middle /4 transmission line and the final /8 transmission line. The /8 transmission line can be easily modeled from its admittance:

Likewise, we can find the ABCD for the /4 transmission line from standard microwave textbooks.

The cascade to ABCD can now easily be done, noting the difference between even and odd modes is just a negation. The ABCD once multiplied, can be converted to *S * parameters using equations from standard textbooks. The final *S* matrix is, where + is for even mode and – is for the odd mode.

We find the *S* parameters for the system through the standard definition of incident and reflected waves. *S _{11}* is the ratio of the reflected wave to incident. We assume there is no incident waves on any other ports, i.e.

*a*=

_{2}*a*=

_{3}*a*=0. We decompose both into the even and odd modes. For

_{4}*S*the input is componesed of two equal input,

_{11}*V*

^{+}_{.}The reflected wave has an even and odd component (also the same for this example), however we leave them in terms of even and odd so that we can utilize the

*S*-matrix we found before. From our

*S-*matrix for even and odd, we see that both ratios are zero. In addition, due to symmetry, we would find using the same analysis that all input ports are matched, i.e.

*S*=

_{11}*S*=

_{22}*S*=

_{33}*S*=0

_{44}*.*

We can now follow the same procedure for *S _{21}* and use the

*S*-matrix as before. Due to symmetry, we also can solve for the corresponding

*S*parameters for the other through signals. We see that the results is that half of the power (remember

*S*is the voltage, so we square for the power) comes out of

_{31}*P*. In addition it is phase shifted by 90 degrees.

_{3}For *S _{31}*, we apply an even and odd mode input to

*P*and measure the output at

_{1}*P*. Now we are going to do a trick here. Having an even and odd mode at

_{3}*P*is the same terminating

_{1}*P*with 50 Ohms and having the same even mode at

_{1}*P*and the opposite polarity odd mode at

_{4}*P*. This is useful, as we already know what the

_{4}*S*=

_{21}*S*is. So we simply replace the even and odd mode with the even and (negative) odd mode, and use the solution we had found for

_{34}*S*and

_{21}*S*. The only difference is the negative sign. Once again, we can use this to solve for the corresponding permutations of

_{34}*S*parameters as before. We see that the results is that half of the power (remember

*S*is the voltage, so we square for the power) comes out of

_{31}*P*. In addition it is phase shifted by 180 degrees.

_{3}To solve for *S _{41}* we use the same trick of swapping the even/odd modes from

*P*to

_{1}*P*and reusing the solution from

_{3}*S*. As before, we can use the same procedure to solve for the permutations. We see that

_{11}*P*is isolated from

_{4}*P*, a feature we leverage in our design. We want the input to split into two equal power signals. We do not want any power to be lost in

_{1}*P*.

_{4}Here is our final result. The input at *P _{1}* is divided into two equal power signals at

*P*and

_{3}*P*. The phase shifts are 180 and 90 degrees respectively, creating the desired 90 degree phase shift or quadrature output we will use for our radio.

_{2}