Branchline Coupler – Theory and Design

Alright, let’s move on and talk about 90-Degree Divider and Combiner. Sometimes, we may need the output signals to have different phases as well as being equal in amplitude. And a great example is we’ve got an IQ modulator. If you remember, we need to generate two signals; one is generated directly from the oscillator and then 90 degrees out-of-phase. So, a first attempt in doing this might be using the in-phase divider and introduce an additional phase shift to one of the paths, so this is conceptionally how you might do that. So this might be a resistor or a Wilkinson Combiner and we’re using the transmission line as a phase shifter. And here’s the picture of what it may look like as a phase shifter. The problem here is that the two paths are going to have a different loss, and therefore this could be a imbalance at the output powers. And also, the phase shift is narrow-band because we have this 90-degree shift. So, there are downsides in addition to this could be a little large if we’re trying to do a transmission line and it’s difficult to lay-out.

So, let’s take a look at a particular coupler that is able to provide us with a plus or minus 90-degree phase shift and also split our signal into two pieces. It’s called the Branch-Line Coupler and it is shown here. It’s a basic design. It has four quarter-wavelength transmission lines; so here’s one, here’s two, here’s three, and here’s four. And you can already see from the thickness of the diagram that the characteristic impedance of the top two are different from the characteristic impedance of the sides. And our output impedance here that we’re assuming is at all of the ports.

And what we’re going to do is we’re going to use Even and Odd Mode Analysis in order to analyze this circuit. And this is one of our motivations to go ahead and learn Even and Odd Mode Analysis because circuits like these can be much easier to analyze if we do an Even and Odd Mode Analysis. So, let’s take a look and split this up into Even and Odd Modes. So, if you think about mode coming in from the left ports that are in-phase, which is even, we have a virtual open. And so, that means we can actually split the circuit right here and say that’s an open. And if you remember the reason why it is an open is if I have a circuit element between two signals that are exactly in-phase and the same amplitude, no current flows because there’s no potential drop and that’s why I consider this midpoint simply being an open. Likewise for the Odd Mode, where we have an in-phase signal on one port, and an out-of-phase signal on the other port, I can use symmetry, and now I can make this into a virtual ground because I can split up any impedance between this into two pieces and call this ground because it’s not going to move because the two signals are symmetric. Alright, so now what we’re going to do is to analyze what happens in the Even/Odd Mode, and then I’m going to use ABCD Matrices to figure out what the total transfer function of a Branchline coupler is, and then I’ll convert that to S Parameters. But don’t worry; we’re going to go through that step by step.

So, let’s start with the Even Mode, where I’ve gone ahead and removed the lower half because all we need to do is to analyze what happens at the top half because they’re symmetric. And we can write an equivalent circuit for this, where we have our quarter -wavelength transmission line here. And remember that this is half our quarter-wavelength. So, they’re, and I’ve put in the correct characteristic impedances. And so, what we’re going to do is we’re going to analyze what does this look like for the Even Mode and also the Odd Mode. And if you remember, the only difference between the Even Mode is that this is open, and for the Odd Mode, this is simply a ground.

Alright, so for the Even Mode, let’s take a look at what’s happening with these two transmission lines here at the end. And this is just a simple telegraph equation, and you should remember this from previous slides, and another thing we did in the lecture where we talked about stubs and transmission line transformers was that if we have a transmission line, that means and and so, that means this is 1, and this is 1. And this is basically infinite impedance, so I can set my load to infinity. And it’s clear from this fraction that these two will dominate, and the Z_Ls will cancel out, and I’ll be left simply with. And I’m going to write that in terms of admittance and I’ll show you in a minute why I want to do that. So, if I write and submit it, I just flipped it over and I can write it simply as this.

Before I go on and use this high result, I want to talk a little bit about the Odd Mode. For the Odd Mode, what I would do is ground this because it’s a virtual ground and certainly, this is zero, and this is zero. And that means the dominating terms are going to be these, and I end up with that in terms of admittance, I end up with. So, the only difference between the Even/Odd Mode Circuit Analysis is the virtual ground or virtual open and all it does is the result in the change of the minus sign. That’s going to be helpful because when we do our calculations, we can use the same result, which is the change of minus sign so I would consider the Odd Mode. So, the only difference is the negation.

So, we’re going to look back at the Even Mode, and here we have our admittance. And below, what I’ve done is I’ve gone through a micro-wave textbook or you can search it online and I found out that the ABCD Matrix by a shunt-by-element. So, this is just something that people have figured out analytically. We did our two-port network. Remember, we did some voltage and currents at certain ports were open or shorted and we’re able to find different parameters. Somebody has gone ahead and done this for a shunt element and this is the ABCD. And now, it’s clear why I wanted everything in terms of Y because I basically calculated before with the telegraphs equation what Y_IN is and this could be represented simply as a shunt element. I know that it is not touching anything here but remember, we calculated with the input impedance was, and that’s the same input impedance is here if we consider this route. So, the two are equivalent, and I could just plug in my Y and I’m done from my Even Mode now I found the ABCD matrix for my Even Mode. And I’ll just point that out that it’s the same ABCD Matrix for this. And what we’re going to do now is we’re going to look and see if the ABCD Matrix of our line is.

And once again; first-off, I’ve changed these to the circuit picture just to make things a little bit easier on how we are progressing. And then, I have gone into the textbook and I have found the ABCD Matrix for a transmission line and for, this means and we can see the cosine is equal to 0, sine is equal to 1. So, this goes to 0, this goes to 0, this goes to 1, this goes to 1, and you can see that my 0 is here, my 0 is here, and this is the textbook for Z_Line. The Z_Line that we’re looking at is this one and that characteristic impedance is . So, I plugged that in here and now I have the ABCD Matrix of this set or item. And all I have to do now is cascade this with matrix multiplication and figure out the ABCD Matrix of the entire circuit.

And I’ve gone ahead and done that and you can see that I have a plus and minus. And the plus and the minus is tracking the Even and Odd Mode where plus is Even, and minus is Odd. I should just caution you here that you might want to take the wavelength where plus is always even and the minus is always the odd. It just happened to be in this particular circuit that they had the exact same values, except for when the even had a plus and the odd had the minus. I could do a different circuit that it might have it backwards, where the odd had the plus, and the even had the minus. So, don’t take this as an explicit; it’s a specific result of our circuit.

And if I go through that multiplication, I end up with a final ABCD Matrix of the Even and Odd Modes, as shown here. And now that we have this, it’s not particularly useful in this form because we want to examine our Branchline Coupler. And what we really want to do is to talk about how signals propagate through, how do we get our 90-degrees phase shift, and typically, what we’re going to do is in terms of S-Parameters. But luckily as we’ve talked about in class, it’s really easy to convert ABCD Parameters to S-Parameters. And you can do this analytically by going online or in a microwave textbook and you can find cables that will relay ABCD Matrix to S-Parameters and they will simply say that S11 is equal to some combination of these components, or if you have numbers for these, you can go into math lab and it will numerically convert for you. I think it pretty much does all of the parameters you can ever deal with. So, I went to Pozar and converted these two, the Even and Odd Mode, as the S-Parameters and basically, what I did is just took the S, the Even Mode Matrix, calculated the S-Parameters for that and the Odd Mode Matrix, calculated the S-Parameters for that. And it turns out that once again, we just have a minus and a plus and that does the Even and the Odd Modes. And in some point, I’ve got to actually combine these Even and Odd Modes into a complete signal and that’s what I’m going to do next.

[Slide] So, let’s put this all together and try to calculate S₁₁ and this is for the waves going out of Port 1 due to waves going into Port 1 and all of these just mean I’ve terminated everything in its characteristic impedance. So, there are no waves coming in and there is no reflection. Alright, so my singular wave is V+V+, that’s these two, and then I’m going to have my Even Mode reflection and my Odd Mode reflection. And then, I can rewrite this here, here, and it’s pretty easy, and it just comes into my S₁₁ at my even S₁₁, my odd half; both of these are 0, and I end up with S₁₁ = 0, and I could go into the same mathematics due to symmetry for S₂₂, S₃₃, S₄₄, and they all are 0. And what that means is that if I’ve terminated all of these un-impedance matched with all of my terminals, which is one of the big goals we want to do in our engineering is to make sure we don’t have any reflections because our next transmission line and maybe taking this signal someplace else. I don’t want any reflection, which is off of that interface. And if this is Z₀, and if the input impedance is Z₀, I’m all set.

Right, let’s look at Z₂₁. And so, these are the waves coming out of 2, and coming in, and just remember the minus here is telling me out of the port. And so, once again my “in” is V+ and V+ and I’m basically going to have the wave coming out of port 2 due to the Even wave, due to the Odd Mode in port 2, and I could do the same kind of math and ends up that I just have . I plugged in the correct equations for those and I end up with this value () and now I have S₂₁ and I go through the same mathematics to solve S₂₁ and S₁₂ and are always equal to the passive system and the passive reciprocal system. And S₄₃, if I do the same math, I get the same thing for S₃₄ and I have the answer for those, and it’s great. And if I go look at S₃₁, that’s what is coming out of 3 due to what’s coming into 3, and I go ahead and split it up just like I did before, but because of symmetry, what’s coming out of 3 is the same coming out of 4, except that on this side, I would have a minus sign because it’s the Odd Mode and they can write that simply as and I go through the mathematics and I end up with . These two are equal because we have a passive symmetric system. These two are the same because its symmetrical system, if I went through the analysis, I get the same answer. And I end up with the final result here: .

Last, but not least is S₄₁. What is coming out of 41, from what’s coming into 1; same mathematics, and I do the same trick here, where I realized that what’s come out of here due to this can be mirrored to what comes out of , yes, what comes out of here. Due to that, and I just put the minus on the Odd Mode, and I end up with S₁₁. By the way, this is what we’re converting 4s and 1s, just so we’re clear why we get back to S₁₁ and they’re both 0 anyway. And so, we end up with all of our odds has been, all of our 411, 411, and 4323 as being 0.

Alright, so here we are. So, I basically calculated the full S-Parameters. And just as a reminder, this is 11, 12, 13, 14, 22, 33, 44, and so there are a couple of things that we see right away. One is that as we’ve said, all of the ports are matched because the S-Parameter, S₁₁, is just gamma or is just 0. And so then, let’s look at some other things. We noticed that S₁₂ has a 90-degree phase shift compared to S₁₃, and that means what is coming out of here and what’s coming out of here, we’ll have 90-degrees of phase shift between the two because one is a j and the other is a 1. And then, last but not least, if I look at port 4, you’d notice that I have a zero diagonal here, but most importantly, I’ve got a zero here and a zero here. And that tells me that anything coming into here is never coming out of port 4. So, port 4 is considered isolated. So, what we’ve done is we’re able to build a system that is impedance-matched. And that takes the input signal and it splits it between two ports with even magnitude – notice the magnitude of this as 1. And remember this gets squared as a power; remember S is in terms of voltage. So in squared, we get the power in half and that this through here had a 90-degrees phase shift. So, this is a really nice way to take your local oscillator and generate sine and cosine for your quadrature amplitude modulator signals…