I am having trouble understanding what is happening physically during a transmission line reflection, in particular the short circuit and open circuit case. I understand that when when a transmission line is terminated with a resistance equal to the transmission line impedance, then

$$ \Gamma = \frac{R_T - Z_o}{R_T + Z_o} = 0 $$

which means that there is no reflection, and physically it makes sense because the resistor is dissipating power through it, and so therefore the wave cannot propagate back. However, I don't understand what is happening in the closed and open circuit cases.

In this post On a transmission line why are voltage and current waves reflected at a short circuit? the answer makes sense in a power perspective for the short circuit case, but I don't understand why the reflected wave is inverted. This post Transmission Line reflection. I would like a non-mathematical explanation seems to answer the question with

But now we have another unstable situation: That end of the line is at 0V, but the rest of the line is still charged to Vs/2. Therefore, additional current flows into the short, and this current is equal to VS/2 divided by Z0 (which happens to be equal to the original current flowing into the line). A voltage wave (stepping from VS/2 down to 0V) propogates in the reverse direction, and the current behind this wave is double the original current ahead of it.

but I don't understand why additional current flows into the short. Is this some application of Len's Law? Or is there something that I am missing? Also, what happens when the reflected wave reaches the voltage source? There would be an enforcement of some voltage so would it be inverted again?

For the open circuit case, I understand that voltage "Piles up" at the end of the transmission line since there is no path for it to continue traveling, but why is it twice as much voltage as what is started with? Mathematically, it makes sense, but I still don't understand this physically.

Any help would be great, I am very confused about this topic and would like to get a firm physical understanding before accepting the equations. Thank you!

edit: I walked through the equations a few hundred times, and was able to figure out what is happening physically.

  • 1
    \$\begingroup\$ The open circuit enforces the rule that I = 0. Voltage is a free variable. The short circuit enforces the rule that V = 0. In that case I is a free variable. In both cases, though, the energy delivered to the load is zero. Therefore 100% of the energy must be reflected back toward the source. \$\endgroup\$
    – user57037
    Commented May 10, 2020 at 23:16
  • \$\begingroup\$ That makes sense, but why does it get reflected with a 180º phase shift in the closed circuit case, and why does the voltage become twice that of the initial voltage at the end of the line in the open circuit case? \$\endgroup\$
    – Rico
    Commented May 10, 2020 at 23:19
  • \$\begingroup\$ Before the reflection gets back to the source, the source acts as if it is connected to a resistor whose value is the line impedance. Let's say we have a source with 5V open circuit voltage. Series resistance is 50 Ohms. Line impedance is 50 Ohms. Initially, voltage will be 2.5V. But when the reflection gets back to the source, the source will now see the end-of-line impedance and the voltage will drop to zero. \$\endgroup\$
    – user57037
    Commented May 10, 2020 at 23:27
  • \$\begingroup\$ This is the best intuitive explanation of transmission lines that I've seen: electronics.stackexchange.com/a/93254 \$\endgroup\$
    – Drew
    Commented May 10, 2020 at 23:28
  • \$\begingroup\$ Possible dup of electronics.stackexchange.com/questions/484269/… \$\endgroup\$
    – hotpaw2
    Commented May 11, 2020 at 1:57

1 Answer 1


This video explains it so well for being so old. Bell Labs experiment of wave propagation


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