Mark all as read

Understanding the s11 and s21 coefficients of a microstrip line with resistor to ground


I thought I understood the theory of scattering parameters, smith charts etc. until I tried a practical example.

The following stupid example is supposed to be a microstrip line shunted by two 100 Ohm resistors to ground, that is, an equivalent resistance of 50 Ohm (also checked at DC with a voltmeter). microstrip 50ohm

Note: the bottom of the plate is copper plated.

Now, I cannot afford a good network analyzer, so I bought a nanoVNA network analyzer (image below). Admittedly, this is a toy, but it is a well regarded toy, especially for frequencies below 300MHz. So, I assume it gives not so bad results for this experiment. nanovna

Note: the analyzer has been calibrated.

Testing the example above, here are the results returned by the nanoVNA:

vnaResults1 The yellow line is s11, approximately -9db, and the blue line is s21, approximately -3.53 db, measured and constant in the range 10kHz - 1GHz.

Now, -9 dB is equivalent to a ratio of 0.34 approximately, and -3.53 db is equivalent to a ratio of 0.66 approximately. I expected S11 to be near 0. Regarding s21, I didn't know exactly what to expect.

Notice that these measurements begin at 10 kHz, where everything should behave like the theory, due to the very low frequency.

Trying to understand these values, I finally found an excellent article, full of practical data.

Now, according to fig. 1, it seems that the stimulus has an output impedance of 50 ohm. This is my first question:

Question 1: Should we assume that network analyzers are built in such a way the stimulus (signal toward the port) has an output impedance of 50 ohm?

Question 2: How to understand the above results ?

Why should this post be closed?


What's the impedance of cold solder blobs? Probably not 50 ohm... also, looks like a potential short to the left on the first pic. These are some seriously cold joints. My advise is to throw this in the garbage, then make a new attempt with larger 1210 or 2512 etc, as large as you can fit. Heat the part like there's no tomorrow and use the thickest tip you got. Lundin‭ 16 days ago

Q1A) We shouldn't assume anything about the network analyser that isn't documented in the manual/data sheet. Andy aka‭ 15 days ago

@Lundin. Admittedly, the "thing" looks bad, worse than it really is, because there are reflections and other image artifacts. But I am sure there is no short. I may do a more pretty circuit and post it this week when I am in my lab. On the other hand, if you already know how the s11 and s21 SHOULD look in theory, please, just tell that in an answer. coquelicot‭ 14 days ago

2 answers


In the absence of VNA product documentation......

Now, -9 dB is equivalent to a ratio of 0.34 approximately, and -3.53 db is equivalent to a ratio of 0.66 approximately. I expected S11 to be near 0.

A 9 dB return loss is a reflection coefficient of $\pm$0.355. Accounting for slight discrepancies in the measurement, I suspect that the reflection coefficient ($\Gamma$) might in fact be -0.3333 (somewhere between 9 dB and 10 dB of return loss). And, a $\Gamma$ of -0.3333 would be obtained from a 25 Ω resistor relative to a measurement reference of 50 Ω: -

$$\Gamma = \dfrac{R_x- 50}{R_x + 50}$$

Plugging 25 Ω into $R_x$ means $\Gamma$ is -0.3333 and return loss is $-20 log_{10}{|\Gamma|}\text{ dB}$ = -9.54 dB.

So, why 25 Ω? It's the input impedance of the VNA in parallel with two 100 Ω resistors in parallel inside the soldered blob i.e. 25 Ω.

Regarding s21, I didn't know exactly what to expect.

If you used a straight cable with no 100 Ω resistors fitted then I would expect 0 dB but, now that the power is shared between the two parallel 100 Ω resistors and the input of the VNA, the signal power level measured by the VNA is half or -3.01 dB. OK the cable and other errors makes this -3.53 dB.


@Andy, we have written answers in parallel. I'm happy to see that your condensed answer coincides with mine. coquelicot‭ 13 days ago

And as always, where I have to spend hours to understand the math, you just feel things, and you feel them right. coquelicot‭ 13 days ago

Note: Actually, the analyzer displayed -9.28db at the marker position, as can be hardly seen (a ratio of 0.3435). I've incorrectly approximated with -9db in the question. coquelicot‭ 13 days ago

@Andy. Actually, our answers do not coincide for s21. You say that the expected s21 should be 0.5, but I've computed it should be 0.66... as indicated by the analyzer. Is there some mistake in my answer or yours? coquelicot‭ 13 days ago

@coquelicot‭ I think S21 would be 0 dB for a straight through wire with no added resistors. But, without documentation, it's just a guess. Andy aka‭ 13 days ago

Show 1 more comments

Eureka! thanks to the article cited in the question, I've finally understood the matter.

I will describe here the general procedure for computing the network parameters of a grounded two port device (DUT), without entering into the intuition behind the math. The generalization to more ports is easy.

Let me call "active terminal" the terminal of the analyzer that sends the stimulus, that is, the S11 terminal. Similarly, let me call "passive terminal" the S21 terminal of the analyzer.

The first point to understand is, as vaguely suggested by the cited article, that the signal exiting from the active terminal of the network analyzer has a given output impedance. I believe that in general, most analyzers are built in such a way that this impedance is equal to the characteristic impedance of the usual coaxial cables, that is, 50 ohm. This will be assumed here for the sake of completeness, but be aware that this impedance plays no role in the computations below.

At the other end, the passive terminal entering into the analyzer is terminated by a resistance equal to the characteristic impedance of the cables, that is, 50 ohm again. Thus, we have the following diagram:


Notice that the analyzer termination impedance seen from the S21 ref plane is 50 ohm at low frequencies, but also at high frequencies because the characteristic impedance of the cable is just 50 ohm. That means that this diagram is valid at all frequencies.

This being said, according to the article cited in the question, with $$a_i = \frac{U_i + Z_0I_i}{2\sqrt{Z_0}}\ \ \text{and} \ \
b_i = \frac{U_i - Z_0I_i}{2\sqrt{Z_0}}, $$ we have, for $(i, j) = (1,1)$ and $(i, j) = (2, 1)$, $$ S_{ij} = \frac{b_i}{a_j}.$$

The rest consists in computing $U_i$ and $I_i$ with complex analysis, using the usual electrical circuit laws. Let us apply that to our problem:


We have $$I_1 = U_1 / (50\Omega || 50\Omega) = U_1/25;$$ $$I_2 = -U_2/50 = -U_1/50;$$

Omitting the $1/2\sqrt{Z_0}$ factor that vanishes in the fractions below, we then have $$a_1 = U_1+ \frac{50}{25}U_1 = 3U_1, \ \ b_1 = U_1- \frac{50}{25}U_1= -U_1,\ \ b_2 = U_2 +\frac{50}{50}U_1 = U_1+U_1 = 2U_1.$$

We deduce $$S_{11} = \frac{b_1}{a_1} = -\frac{1}{3} = -0.33\ldots \ \ \text{and}\ \ S_{21} = \frac{b_2}{a_1} = \frac{2}{3} = 0.66\ldots$$ BINGO!!!


Now that you grasp the basics, the geometry is a big factor moving towards 5% of the wavelength and beyond. Although video SA's are 75 Ohms the std versions are 50 Ohms. But the only difference in coax is the ratio of the conductor radius for outer/inner. In your example the small solder gap adds C and thus lowers Zo with rising f while L typ. ranges from 5 to 10 nH/cm typ. depending of the log of length/width ratio TonyStewart‭ 12 days ago

Then s21 losses in the connector threads of 0.1dB is good 0.5dB is bad, so calibrated torque is important. TonyStewart‭ 12 days ago

@TonyStewart. What is calibrated torque? coquelicot‭ 11 days ago

That’s the spec on the tightening of the connectors to achieve desired low loss without removal of excessive gold plating. TonyStewart‭ 9 days ago

Sign up to answer this question »