Electronic – Transmission Line Impedance Matching Basics

impedance-matchingtransmission line

From what I understand, matching a load to a transmission line means that no signal is reflected from the load, which means that at the two sides of the connection point between the load and the transmission line the impedance is the same, the load impedance. I tried to test my understanding with the following example:

If we have a \$50\Omega\$ transmission line connected at one end to a \$50\Omega\$ source and at the other end to a \$100+50j\$ load. We would like to match this load to the transmission line, using a shunt short stub. I used the Smith chart to calculate the point where we connect the stub to be \$0.199\lambda\$ away from the load. I also used the Smith chart to calculate the length of the stub to be \$0.125\lambda\$.

Now for the load to matched, the impedance to the left of the connection point should also be \$100 + 50j\$. But I am getting \$100 – 50j\$. I used the equation
$$Z_{in} = Zo \frac{Z_L + jZ_0\tan\beta l}{Z_0 + jZ_L\tan\beta l}$$

with \$Z_o = 50\Omega\$, \$Z_L =\$ parallel impedance of the stub and the source : $$\beta l = 2\pi \times 0.199 =0.398\pi $$

$$Z_L = \frac{j50*50}{50+j50} =25+25j $$

What am I doing wrong?

Best Answer

When you match to complex valued loads the matching for zero power reflection states that the impedance seen from your complex-valued load (\$100+50j\$) has to be its complex conjugate (\$100-50j\$).

This is because that way we would be satisfying the max. power theorem, and, at the same time, getting rid of the imaginary part of your load.