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Lets say I have a lossless coaxial cable with a characteristic impedance of 50 ohms. The length of the cable is very long compared to the the wavelength of the relevant signals on the coaxial cable. This is an obvious transmission line.

Now, lets say I attach an ordinary 50 ohm through-hole resistor to the end of the transmission line by attaching one lead of the resistor to the inner conductor of the coaxial line and the other lead of the resistor to the outer conductor. From what I have learned, this should result in no reflections in the transmission line when a source generates a signal at the other end of the coaxial cable because the transmission line is matched to the load.

In this situation, what is happening at the junction between the coaxial line and the leads of the resistor? Would the leads of the resistor (although very short) not make up a sort of transmission line with its own characteristic impedance? I know that characteristic impedance is not dependent on length of the transmission line, but if characteristic impedances do not match, there will be reflections. Why would there be no reflections?

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Would the leads of the resistor (although very short) not make up a sort of transmission line with its own characteristic impedance?

The leads of the resistor will (mainly) introduce parasitic inductance. This will make the resistor not behave exactly like a resistor, but like an LR series combination.

Of course this means that when the inductive term becomes significant, the termination will no longer be matched to the characteristic impedance of the transmission line, and you will get reflections.

If you increase the frequency further (to try to make these leads look like transmission lines) you'll probably get into a regime where you must do a full electromagnetic analysis (to find how much they radiate, like little antennas) before they ever act like transmission lines.

Why would there be no reflections?

There's no reflection when you terminate a transmission line with a matched load because the ratio of I to V in the load is equal to the ratio of I to V in the forward travelling wave on the transmission line. Therefore, there's no impedance mismatch at the connection of the line to the load.

Looked at more mathematically, you can say the solution to the boundary value problem for the transmission line with a matched load is to have no reflection.

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  • \$\begingroup\$ A more intuitive way to see it is that the source sees the 50 ohm impedance of the transmission line until the signal has had a chance to get to the other end and back...then the source sees the impedance of the termination. If that's also 50 ohms, no change (therefore no reflection) is necessary. \$\endgroup\$ – Cristobol Polychronopolis Feb 10 at 13:46
  • \$\begingroup\$ @Cristobol Polychronopolis that makes sense to me but at the same time I don't understand why the leads of the resistor don't make up a sort of "twisted pair" transmission line. Using an online impedance calculator, a twisted pair transmission line with conductor diameter of 0.5 mm and a wire separation of 5 mm with a substrate dielectric of 1 has an impedance of 359 ohms. Why would there not be reflections due to this difference from the 50 ohm coaxial line? \$\endgroup\$ – Nick Feb 10 at 15:41
  • \$\begingroup\$ @Nick, what are you supposing is the length of those leads? If it's, say, 5 mm, then about 17 ps after the reflection from the 50 ohm to 359 ohm interface you get another reflection from the 359 to 50 ohm interface which nearly cancels it out. If you're working below 300 MHz, the net reflection will likely be negligible. \$\endgroup\$ – The Photon Feb 10 at 16:21
  • \$\begingroup\$ At the same time, there is a reason why we generally don't use axial leaded component for RF. An SMT component will act more like an ideal component (resistor, capacitor, whatever) up to a higher frequency. So now put an SMT resistor on the end of your coaxial cable. How long are the "leads" in that scenario? \$\endgroup\$ – The Photon Feb 10 at 16:23
  • \$\begingroup\$ @The Photon I was assuming small (5mm) lead length like you described. Your first comment cleared up my confusion. \$\endgroup\$ – Nick Feb 10 at 18:25

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