6.15 Double stub matching

There is one last technique we can look at which is somewhat more flexible than the single stub matching which we just lookedat. This is called double stub matching! Suppose we have the following situation, as depicted in the figure . There is a load of $\frac{Z_L}{Z_0}=0.2+1.3i$ located at the end of the line, and then some arbitrary distance away ( $0.11$ ) an adjustable stub. Another (arbitrary) $0.11$ from the first stub, there is a second one. Let's plot $\frac{Y_L}{Y_0}$ on the Smith Chart , and then, since the stubs are in shunt across the line, switch toadmittance, and find $\frac{Y_L}{Y_0}$ . It is easy to see that $\frac{Y_L}{Y_0}=1.5+2.3i$ .

This is tricky now, so you have to pay attention and think. If I want to find a place which, when moved from A toB, 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 Irotate the matching circle around $0.175$ towards the load , then any place on that rotated matching circle is guaranteed to end up on the real matching circle, when we go $0.175$ back towards the generator.

OK, so here's what we do. First, we rotate the matching circle 0.175 around towards the load (gocounterclockwise) . 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 imaginarypart of $Y_A$ with the stub. Suppose we move as shown in . 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 $-(i\times 0.6)$ to $i\times 0.05$ , thus we have added $i\times 0.65$ 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 $$∞ , $()$∞ (the short at the end of the stub) and go around the outside of the Smith Chart until we find $i\times 0.05$ . To get from one place to the next we went $(0.25+0.09)=0.34$ and so the length of the first stub, $L_1$ should be $0.09$ . Now we are at $\frac{Y_A}{Y_0}$ with stub. The next thing we have to do is to rotate another $0.175$ towards the generator so that we can get to stub B. As we do this rotation, we again stay on acircle 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 $0.175()$ towards the load from the matching circle. As shown here , we are now at the point $\frac{Y_B}{Y_0}$ without stub $1.0+1.6i$ . Thus we need to adjust the length $L_2$ of the second stub to give us $-(i\times 1.6)$ of reactance, so we can move (along a circle of constant real part = 1.0) into the center of the Smith Chart . We have to find the length $L_2$ for the second stub, but that is now easy! ( )