# What does a 50 ohm coax "look like" when neither the source nor the load are matched to it, and the center conductor and shield are used for signals?

Say I have a π-network filter for which the inductor is on a separate board. For convenience, I use a 50 ohm coax (4" long, frequency is 13.56 MHz) to connect to the inductor, using the center conductor and the shield to connect to the leads on the inductor. What does this 50 ohm cable look like in this circuit? How will it impact the tuning of the filter?

Since the signals through the cable are neither differential, nor referenced to ground, I don't know how to model this connection done this way (I guess I don't fully understand how/when a 50 ohm cable "is" a 50 ohm cable -- Is using this coax in this way better, the same, or worse than using individual wires to connect to the inductor?

In another example (in an effort to better understand the meaning of cables with characteristic impedances), what does a 50 ohm cable look like in a circuit where the source and load impedances are not matched to 50 ohms (assume signal in center conductor, shield grounded). I assume I'd get reflections at both the source and the load ---or do I not care if the length of the cable is less than $$\\frac{\lambda}{10}\$$?

• if you have the cable, measure resistance on each of the leads separately and then measure capacitance. model it as such - two resistors and a cap connecting them.
– Abel
Aug 5, 2021 at 3:17
• I think it would degenerate in two transmission lines, capacitively coupled. So some kind of capacitive transformer. Abel missed out all the RF behaviour :P Aug 5, 2021 at 8:31
• The capacitance is key to the reduction in LC resonance dependent on filter C value and cable C value. R is less important and C is predictable by type. Thus a slight reduction in fo due to sqrt of C change, Ignore if low Q, but may affect if high Q. [Edited - non-technical - RM] Aug 6, 2021 at 1:03

13.5 MHz has a wavelength of about 22 m, so your 4" (physical length, electrical length around 6") is very 'short', well below λ/20.

With a short length of transmission line, you can model it quite well as a lumped series L, series R and parallel C.

Increasing the effective L and the C of your inductor will drop its tuning slightly, and decrease the SRF of the inductor, compared to using shorter wires.

Individual wires of the same length will have a higher characteristic impedance than 50 Ω so will have an effective smaller C and larger L. They would also be more likely to couple magnetically into adjacent wiring loops.

In the more general case, yes, you will get reflections when a line is mismatched.

For short lines operated with an actual sine wave meeting λ/20 or λ/10, see the first part of this answer. The reflections add up to give the effect of a small series L and a small parallel C. This effectively makes the group delay element of a low pass filter's passband. We don't get to stopband frequencies as the line is 'short'.

For short lines operated with wideband or digital signals meeting λ/20 or λ/10, equivalent maximum frequency of the signal risetimes (not the pulse repetition frequency), exactly the same can be said, but with a time domain rather than frequency domain viewpoint. As the round-trip time for a reflection is much less than the transit time of a signal edge along the line, the first reflection will arrive back at the sending end before a small fraction of the transition has been made. The effect on the small signals is to slow the risetime slightly just like the effect of a lowpass filter.

In contrast, when a line is 'long', most or all of the transition will have been executed before the reflection arrives back. If you like, the line is 'storing' most of the edge waveform. In this case, the reflections cause additional discrete transitions to the pulse shape, which is usually bad if you're not in control of it.

The inductance of signal and return coax would tend to cancel out while the capacitance per inch would tune the filter perhaps less than 1%, depending on the LC values. E.g., 222 nH, C1,C2 = 1300 pF, and Ccoax = 10 pF.

The resonant frequency becomes sensitive to (C1 + C2)/Ccoax, so removing C2 and make C1 twice the value now makes the coax length a filter BPF tuner.

• thank you. How does the inductance of signal and return coax cancel each other out? and this capacitance per inch would be in "parallel" with the physical inductor, right? Aug 5, 2021 at 11:30
• The coax was short when compared to wavelength. The currents in the coax center and shield are opposite and their magnetic field partially cancel each other. 2 separate wires would not be that close to easch other, I presume. The C of the cable is in parallel with the inductor. Try simulation. At least Micro-Cap (=freeware) has a model for coaxial cable.
– user136077
Aug 5, 2021 at 12:29
• currents are not opposite....it is not the return--it is the same current through the inductor that is coming out through the shield. Aug 5, 2021 at 22:10
• Yes the coax parallel capacitance of ~ 30pF /ft will tune the LC resonant frequency dependent on the C ratio for deviation regardless of wavelength being much longer than cable length. Aug 6, 2021 at 0:58