Correct model for interference from an unterminated transmission line branch

Question

Question about the correct mathematical model for a branched transmission line

Baseline Setup

• Waveform generator: Output 10-60MHz sine wave, 8Vp-p, Output impedance 50Ohms
• RG58 50Ohm coax: propagation speed 0.659c (length ~0.5m, but that should not be relevant)
• 50Ohm "feed through" terminator connected to
• 200Mhz digital oscilloscope

Result: 4Vp-p sine wave. Amplitude flat across frequency range. 50% attenuation, as expected, due to "voltage divider" of the 50Ohm source with 50Ohm terminator

Interference Setup

• Add a coax-T before 50Ohm feed through terminator
• Connect additional ~1.6m piece of RG58 coax to the 3rd leg of the T.
• Leave the end of the ~1.6m piece unterminated.

Result: As expected, we get reflection at unterminated end. This results in interference being seen by oscilloscope. Amplitude measured by oscilloscope collapses to near zero (about 200mVp-p in reality) at about 31Mhz. Given prop-speed of \$1.98e8m/s\$ this is at \$\lambda = 6.38m\$, which is a good match for \$\lambda / 4 = 1.63m\$.

At 10Mhz \$\lambda \approx 20m\$ and interference almost perfectly "constructive" resulting in an observed amplitude of ~4Vp-p, very similar to the baseline. (note there is no "doubling" due to constructive interference, see below)

Question

What is the correct relationship for observed amplitude in the interference setup as a function of frequency?

My attempt was to just model sinusoidal addition from here: https://dspguru.com/files/Sum_of_Two_Sinusoids.pdf

gives us:

$$ A \cos(\omega t + \alpha) + B \cos(ωt) = \sqrt{A^2 + B^2 + 2AB \cos(\alpha)} \cos(\omega t + \arctan(\frac{A \sin(\alpha)}{A cos(\alpha) + B})) $$

And if we assume near perfect reflection (we nearly have that), then A and B are equal. Just focusing on the amplitude of resultant we have:

$$ Resultant Amplitude = \sqrt{2 A^2 + 2 A^2 \cos(\alpha)} $$

This does indeed give the rough shape of the frequency response however the scale is wrong. Expression above gives 2A at \$\alpha = 0\$ which is clearly wrong by factor of 2 (no doubling). It correctly shrinks to zero at \$\alpha = \pi\$. The amplitudes at the intermediate frequencies are also too high.

A simple linear scale factor to push the low frequency amplitudes down to the correct, observed value, fails to align the frequencies around 20Mhz.

I suspect this is because we are not modelling the actual network of impedances here, and the reflected wave is not a "second source" (hence no doubling).

There is also the potential issue that the reflected wave may(!) see two 50Ohm loads in parallel, ie 25Ohms and that there will therefore be a secondary reflection back to the open ended stub?

I am ideally looking for a simple electrical equivalent circuit, for example:

Z will be changing from an "open circuit" at low frequencies to a "short circuit" ~ 31Mhz.

So, in summary, what I am looking for is an expression for the signal seen by the oscilloscope across the 50Ohm feed through terminator which incorporates a good model for the changing Z.

Answer 1

Answering my own question with help from EEVBlog and Wikipedia:

The impedance looking into an unterminated piece of transmission line is:

$$ Z_{in} = -j Z_0 \cot(\beta l) $$

Where \$\beta = 2 \pi / \lambda\$ is the wavenumber, \$Z_0\$ is the characteristic impedance of the coax, and \$l\$ is the length of unterminated transmission line.

This reduces the problem of calculating the voltage seen by the oscilloscope to a simple voltage divider with bottom half being a parallel pair of impedances:

$$ V_o = \frac{V_{in} Z_l || Z_{in}}{Z_l || Z_{in} + Z_s} $$

Below is some python code implementing the above equation and normalising to the base (non-interference) case.

import numpy as np

def interference_attenuation(frequency, length):
    c = 3e8
    v_rel = 0.659  # velocity in coax relative to c - from RG58 spec sheet
    v = c * v_rel

    wavelength = v / frequency

    Zs = 50 # source impedance
    Z0 = 50 # characteristic impedance of the coax (for both pieces)
    Zl = 50 # terminator impedance
    
    wavenumber = 2 * np.pi / wavelength

    # impedance of an open ended transmission line
    # https://en.wikipedia.org/wiki/Transmission_line#Open
    Zin = 0 + 1j * (-Z0 / np.tan(wavenumber * length)) 

    ZlparZin = Zin * Zl / (Zin + Zl) # normal parallel formula
    non_inteference_amplitude = Zl / (Zs + Zl)

    # voltage divider in complex domain, then take magnitude with abs()
    # and make it relatrive to the base case
    # web calculator here:
    # http://hyperphysics.phy-astr.gsu.edu/hbase/electric/vdivac.html
    return abs(ZlparZin / (Zs + ZlparZin)) / non_inteference_amplitude

This model agrees very well with experimental results (for a 50cm unterminated stub here):

It was also pointed out that the reflected wave will see \$50\Omega || 50\Omega = 25\Omega\$ and therefore a secondary reflection will occur. In practice this is normally avoided by not using a terminator at the oscilloscope and relying purely on the waveform generator source impedance to dampen all reflections.

In that case the transfer function would simply be:

$$ V_o = \frac{V_{in} Z_{in}}{Z_{in} + Z_s} $$