Impedance matching a purely resistive load

[timnswede]

Homework Statement

A 25-Ω antenna is connected to a 75-Ω lossless transmission line. Reflections back toward the generator can be eliminated by placing a shunt reactance Z at a distance l from the load (Fig. 1). Determine the values of Z and l.

Homework Equations

Z(l)=Zo((ZL+jZotan(Bl))/(Zo+jZLtan(Bl))

The Attempt at a Solution

My first intuition was to put a transmission line of length lambda/4 since the load impedance is just resistive, but the question asks for reactance which has me a bit stumped. I used a Smith chart and normalized the load to .33 ohms and found the conductance to be 3. Since I want a real part 1, I found the intersection of the constant vswr circle and the conductance equal to 1 circle. That point was 1+j1.15 and then I found the unnormalized reactance to be 65.2 ohms. I also got the length of l to be .083 lambda.

It's probably confusing to follow what I did on the Smith chart, but I just did the lumped element impedance matching. Apparently this is the right answer, but I'm having a hard time understanding why. The point of impedance matching is so that what the input impedance looks like (for this problem) at point B is 75 ohms. How can putting a complex component in shunt cause it to be 75 ohms? Is it because the transmission line from B to A is complex and it cancels that? I feel like I'm missing something fundamental.

 

[tech99]

I seem to obtain different figures to you. I think you are have an Impedance Smith Chart but you are using it to transform Admittances. This is my attempt - I may be wrong but no one has answered yet.
The 25 Ohm load can be entered on the chart as 0.33, then rotate clockwise (towards generator) until you hit the R=1 circle. This gives a line length of about 0.17 Lambda and a series reactance of about +j1.2. As we now want to connect a shunt capacitor, we need the equivalent admittance value. So cross to the diametrically opposite point and we obtain an admittance of 0.4-j0.5. So a capacitor having a susceptance of +j0.5 will cancel this out. Such a capacitor has an X of 2 and an actual reactance of 150 Ohms.
The reason the circuit works is that the line AB is operated with standing waves, so the impedance will rise and fall with distance. But at positions other than the nodes and antinodes, there is always a reactive component which we have to cancel out if we wish to obtain a resistive match at that point. It is of no consequence that the conductance (or shunt resistance) at point B is an odd value as we never use this in this circuit as we use the series value.
It is also worth mentioning that if we convert between series R and X into shunt R and X, both values change, not just the reactance. This is because if you place an R and an X in a black box, from the outside it must always exhibit the same Q, whether they are connected in series or parallel, so both values change. For parallel connection, Q = R/X and for series connection, Q = X/R.

 

[rude man]

Assume Z = jX or Y = 1/Z = jB. Then the transformed admittance will have to look like Y₁ = Y₀ - jB to terminate the xmsn line in its characteristic impedance of 75Ω, where Y₀ = 1/Z₀.

So solve Y₁ for βl and B in terms of your "relevant equation" except I suggest working with Y's instead of Z's so your equation changes to an equivalent admittance equation.
Setting real and imaginary parts of that equation = Y₀ - jB gives you the needed tan(βl) and B.
The math is mostly changing a fraction (a+jb)/(c+jd) to the form e + jf which is not too bad.
I'm willing to compare results if you post your work & results.
(I believe the phase velocity v is needed in adition to λ, or the frequency f = v/λ. You can assume v = c/2 or whatever.)

 

[tech99]

I now agree with your working! It is OK to do transmission line transformations with the ordinary chart, and you are right to start out with admittance as we will want to add a shunt reactance at point B.