6.16: Double Stub Matching

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There is one last technique we can look at, which is somewhat more flexible than the single stub matching which we just looked at. This is called double stub matching! Suppose we have the following situation, as depicted in Figure $\PageIndex{1}$. There is a load of $\frac{Z_{L}}{Z_{0}} = 0.2 + 1.3 i$ located at the end of the line, and then some arbitrary distance away ( $0.11 λ$ ) an adjustable stub. Another (arbitrary) $0.11 λ$ from the first stub, there is a second one. Let's plot $\frac{Y_{L}}{Y_{0}}$ on the Smith Chart in Figure $\PageIndex{2}$, and then, since the stubs are in shunt across the line, switch to admittance, and find $\frac{Y_{L}}{Y_{0}}$ . It is easy to see that $\frac{Y_{L}}{Y_{0}} = 1.5 + 2.3 i$ .

The right end of a transmission line contains a load impedance Z_L/Z_0 of 0.2 + 0.3j. A distance of 0.11 lambda along the line, to the left of the load impedance, is a stub labeled Stub #1 or A, of length L_1. A distance 0.175 lambda to the left of Stub #1 is a second stub labeled Stub #2, or B, or length L_2.

Mini Smith Chart labeled with the point Z_L/Z_0 with a value of 0.2 + j0.3. Directly across from this point is the point corresponding to Y_L/Y_0 = 1.5 - j2.3.

The first thing we might as well do is move down to the first stub, and see what admittance we have there, as shown in Figure $\PageIndex{3}$. We go from the load, to the first stub by rotating on a circle of constant radius (constant $| r (s) |$ )since all we are doing is going from one place on the line to another. If we call the location on the line of the first stub $A$, then we can see that $\frac{Y_{A}}{Y_{0}} = 0.25 + 0.6 i$ .

On a Smith Chart, moving clockwise from the point Y_L/Y_0 from Figure 2 above on a circle of constant radius takes a distance of 0.11 lambda to arrive at the location corresponding to the value of Y_A/Y_0 without the stub.

Now, what can the first stub accomplish? A shorted stub can create any imaginary admittance we want, but can not change the real part of the admittance. Thus, by adjusting the first stub, we can move around on a circle of constant real part $0.25 Y_{0}$ , and have any imaginary part we want. This is shown schematically below in Figure $\PageIndex{4}$.

From the point corresponding to the value of Y_A/Y_0 without the stub, established in Figure 3, we can travel on the Smith Chart along a circle of constant real component 0.25.

Now, where do we want to go? Well, we would like to end up someplace so that, after we have moved from $A$ to $B$ on the line (gone from the first stub to the second), we are on the matching circle. If this were so, then, since we are on the matching circle, we could use the second stub to match the whole line and we would be done.

This is tricky now, so you have to pay attention and think. If I want to find a place which, when moved from $A$ to $B$, ends up on the matching circle, then what I should do is take the matching circle and move it from $B$ to $A$. That is, if I rotate the matching circle around $0.175 λ$ towards the load, then any place on that rotated matching circle is guaranteed to end up on the real matching circle, when we go $0.175 λ$ back towards the generator.

OK, so here's what we do. First, we rotate the matching circle 0.175 around towards the load (go counterclockwise), as in Figure $\PageIndex{5}$. Now what we have to do is somehow get from $\frac{Y_{A}}{Y_{0}}$ without stub to someplace on the rotated matching circle. The only way we can do this is to change the imaginary part of $Y_{A}$ with the stub. Suppose we move as shown in Figure $\PageIndex{6}$. In going from $\frac{Y_{A}}{Y_{0}}$ without stub to $\frac{Y_{A}}{Y_{0}}$ with stub we have changed the imaginary part from $- (i 0.6)$ to $i 0.05$ , thus we have added $i 0.65$ to the imaginary part of $\frac{Y_{A}}{Y_{0}}$ . Thus using our standard method for finding the length of the first stub, we start at $\infty$ , $- \infty$ (the short at the end of the stub) and go around the outside of the Smith Chart until we find $i 0.05$ . To get from one place to the next we went $(0.25 + 0.09) λ = 0.34 λ$ and so the length of the first stub, $L_{1}$ should be $0.09 λ$ . Now we are at $\frac{Y_{A}}{Y_{0}}$ with stub. The next thing we have to do is to rotate another $0.175 λ$ towards the generator so that we can get to stub B. As we do this rotation, we again stay on a circle of constant radius, because now we are moving down the transmission line not adding reactance by using a stub! This rotation is guaranteed to end us up on the matching circle because every point on the rotated circle (the one we start from) is exactly $0.175 (λ)$ towards the load from the matching circle. As shown in Figure $\PageIndex{8}$, we are now at the point $\frac{Y_{B}}{Y_{0}}$ without stub $1.0 + 1.6 i$ . Thus we need to adjust the length $L_{2}$ of the second stub to give us $- (i 1.6)$ of reactance, so we can move (along a circle of constant real part = 1.0) into the center of the Smith Chart in Figure $\PageIndex{9}$. We have to find the length $L_{2}$ for the second stub, but that is now easy! (Figure $\PageIndex{10}$)

A Smith Chart containing a point for Y_A/Y_0 evaluated without stub has its matching circle rotated a distance of 0.175 lambda about the center point of the chart, towards the load (meaning counterclockwise).

Smith Chart showing movement in a clockwise direction along a circle of constant real part from Y_A/Y_0, evaluated without stub, to the rotated matching circle from Figure 5 above. The intersection of the rotated matching circle with the circle of constant real part from Y_A/Y_0 without stub represents the value of Y_A/Y_0 evaluated with stub.

Rotating clockwise along the outermost circle of the Smith Chart by a distance of 0.34 lambda, from the rightmost point of the outermost circle to the point representing a real component 0 and imaginary component 0.65j.

Moving on the Smith Chart from the point representing Y_A/Y_0 evaluated without stub, by a clockwise distance of 0.175 lambda along a circle of constant radius, to reach the point representing the value of Y_B/Y_0 evaluated without stub.

Moving counterclockwise on the Smith Chart along the matching circle, from the point representing Y_B/Y_0 evaluated without stub to the center point of the chart, which represents Y_B/Y_0 evaluated with stub.

Moving clockwise on the Smith Chart along the outermost chart circle by a distance of 0.09 lambda, from the chart's rightmost point to the point representing a real value of 0 and a imaginary component of -j1.6.

Thus, by doing double stub matching, we are able, by adding the additional degree of freedom of two adjustable stubs, not to have to specify exactly where the stubs have to be placed, so they can be in the line before the matching is attempted. Figure $\PageIndex{11}$ below shows the whole sequence of changes that we made. See if you can begin at "Start" and go through the numbers $0 \to 5$ and get from $\frac{Z_{L}}{Z_{0}}$ to the matching point at the center of the Smith Chart. Remember, when we move from one place to another on the line, we must stay on a circle of constant radius. When we change reactance by adjusting a stub, we must move along circles of constant real part. If you do that, it's easy!

Smith Chart showing the 5 labeled steps of the double stub matching process.

There's just one little problem. What if $\frac{Y_{A}}{Y_{0}}$ without stub had ended up as shown in Figure $\PageIndex{12}$. We are on the $ℜ (\frac{Y_{A}}{Y_{0}} | without stub) = 2.0$ circle. No matter how hard I try, and no matter where I set $L_{1}$ , all I can do is spin around on the little circle as shown, and I will never end up on the rotated matching circle, and I won't be able to make a match! Well, if I add a third stub...I'll let you work it out!

A Smith Chart where the point representing Y_A/Y_0 evaluated without stub falls on the chart circle representing a real value of 2. This chart circle, the locus of places we can get to by adjusting the stub length L_1, does not intersect with the rotated matching circle.

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