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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?

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  • \$\begingroup\$ FYI, EE uses \$ instead of just $ for inline Mathjax. \$\endgroup\$
    – The Photon
    Dec 5, 2014 at 18:41
  • \$\begingroup\$ I have learned that only noob users say Thank you but I have been wondering how to do that inline thing, so what the hell, Thank you. \$\endgroup\$
    – user120404
    Dec 5, 2014 at 19:02
  • \$\begingroup\$ Complex conjugate, not equal impedance. \$\endgroup\$
    – MarkU
    Dec 6, 2014 at 0:54
  • \$\begingroup\$ So the reflection coefficient formula \$ \gamma = \frac{Z_{L1} - Z_{L2}}{Z_{L1} + Z_{L2}}\$ is not correct ? \$\endgroup\$
    – user120404
    Dec 6, 2014 at 5:24

2 Answers 2

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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.

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Your calculations look good and your confusion is justified. It's matched, so why isn't the reflection coef. zero?

Most of the time, the reference impedance is 50ohms or at least positive real. Most books assume Zo is postive and purely real, but don't specifically say so. This leads to the common expression for reflection coef.

$$\Gamma = \frac{Z_L-Z_o}{Z_L+Z_o}$$

The expression for reflection coef. when working with generalized s-parameters or complex reference impedances is...

$$\Gamma = \frac{Z_L-Z_o^*}{Z_L+Z_o}$$

I wish it wasn't so common to assume Zo is positive and real. I don't think Pozar mentions this at all. Gonzalez talks about this difference, but only in passing. The only book I know that properly explains/derives general forms of these equations is by Max Medley, called Microwave and RF Circuits: Analysis, Synthesis and Design

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