📡 RF Power & S-Parameters

S-parameters are a compact language for that question. Instead of measuring open-circuit voltage and short-circuit current, which are hard or destructive at RF, you measure waves entering and leaving ports. The most famous one is S11: the reflected wave at port 1 divided by the incident wave at port 1. In one-port language, S11 is the same thing as the reflection coefficient $\Gamma$.

By the end, you can

1. Compute the reflection coefficient $\Gamma = (Z_L - Z_0)/(Z_L + Z_0)$ for a load
2. Connect $|\Gamma|$ to reflected power, return loss and VSWR
3. Interpret S11 in dB as a measure of input match
4. Explain why impedance matching matters for RF power delivery and measurement

Intuition first

Imagine a pulse running down a transmission line. The line has a characteristic impedance $Z_0$, the ratio of voltage wave to current wave it naturally supports. If the load has the same impedance, the pulse flows into it cleanly. The load looks like more line, so there is nothing to echo from.

If the load is different, the boundary cannot absorb the wave as-is. Some energy enters the load and some reflects back toward the source. The reflection coefficient is the complex number that says how large that echo is and what phase it has:

$$\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}$$

The magnitude $|\Gamma|$ tells how much voltage wave reflects. Reflected power is $|\Gamma|^2$. A $|\Gamma|$ of 0.1 reflects one percent of the power. A $|\Gamma|$ of 0.5 reflects 25 percent of the power.

Return loss is the same idea written in dB:

$$\text{return loss} = -20 \log_{10}|\Gamma|$$

Higher return loss is better. S11 in dB is $20\log_{10}|\Gamma|$, so it is negative: S11 = -20 dB means $|\Gamma| = 0.1$, one percent reflected power.

Drive it

Move load resistance, load reactance and system impedance. The point is the complex reflection coefficient plotted inside the unit circle. Near the centre is a good match. Far from the centre means a loud echo.

Try setting the load to 50 Ω with zero reactance on a 50 Ω system. The dot collapses to the centre, S11 falls toward negative infinity, and VSWR becomes 1. Then add reactance. The resistance did not change, but the load stopped matching because RF cares about the full complex impedance.

?A 50 Ω transmission line is terminated in 50 Ω with no reactance. What is Γ?

• A 0, because the load equals the line impedance
• B 1, because all power reaches the load
• C -1, because the line is shorted
• D 0.5, because 50 Ω is half of 100 Ω

?If |Γ| = 0.2, about what fraction of incident power is reflected?

• A 2%
• B 4%
• C 20%
• D 40%

↘ What a vector network analyzer is really measuring

A vector network analyzer launches a known wave into a port and samples the outgoing waves after directional couplers separate incident and reflected energy. Calibration moves the reference plane to the end of your test cable, so the S11 you see belongs to the device, not the cable and adapters. The word "vector" matters: the instrument keeps phase, so it can distinguish a capacitive mismatch from an inductive one even when their magnitudes look similar.

Key takeaways

• ▸S11 is the reflected wave at port 1 divided by the incident wave at port 1.
• ▸$\Gamma = (Z_L - Z_0)/(Z_L + Z_0)$, and S11 is $\Gamma$ for a one-port load.
• ▸Reflected power fraction is $|\Gamma|^2$.
• ▸Return loss is positive and higher is better; S11 in dB is negative and more negative is better.
• ▸Complex impedance matters: resistance alone does not guarantee a match.

Practice 1 core

A load has S11 = -10 dB. What is $|\Gamma|$, and what fraction of incident power is reflected?

↘ Show worked solution

S11 in dB is $20\log_{10}|\Gamma|$. So

$|\Gamma| = 10^{-10/20} \approx 0.316.$

Reflected power fraction is $|\Gamma|^2 \approx 0.316^2 \approx 0.10$, or about 10 percent. This is not catastrophic for every system, but it is a visible mismatch.