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I do not intuitively understand this. A short circuit simply means there's no impedance. If there's much less impedance than the transmission line, the waves should simply move through the short-circuit, uninhibited. The voltage and current waves being reflected back do not make sense to me.

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  • \$\begingroup\$ Traveling waves don't behave like simple circuits -- they have different rules! \$\endgroup\$ – Daniel Jan 20 '17 at 3:49
  • \$\begingroup\$ @Daniel But it is just common sense to me for something to travel where there is less resistance. \$\endgroup\$ – JobHunter69 Jan 20 '17 at 5:05
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    \$\begingroup\$ Have you seen this related question? \$\endgroup\$ – Dave Tweed Jan 20 '17 at 5:28
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    \$\begingroup\$ Light hitting a mirror is a similar situation. The mirror acts very much like a short-circuit. The light is not absorbed, but reflects. \$\endgroup\$ – glen_geek Jan 20 '17 at 7:09
  • \$\begingroup\$ @glen_geek that argument is flawed because a mirror has much more impedance than air. Moving through air is easy but moving through a mirror is not. \$\endgroup\$ – JobHunter69 Nov 12 '20 at 19:12
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In a transmission line, you have an electromagnetic wave traveling along. This is a time varying electric and magnetic field. When the wave reaches a short circuit, the short circuit enforces the rule that V=0 at that location. This destroys the conditions necessary for the wave to continue traveling. Because the electric field can no longer vary with time at that location. Without this time variation, the wave cannot continue to travel.

And, as it happens, it also creates the conditions needed for the wave to reflect.

You could also consider this from a conservation of energy perspective. An electromagnetic wave has energy. It is actually a form of traveling energy. The short circuit cannot dissipate energy (when V=0, power=0). BUT, the wave cannot continue to travel, either, as previously mentioned. So, really, there is no other thing that can happen other than reflection.

You could say that when a wave in a transmission line encounters a load, any energy which is not delivered to the load MUST be reflected in order to satisfy conservation of energy. Of course, if the load is an antenna, some of the energy will be radiated into space, but that does not really change anything. The antenna is modeled as some kind of load, and the energy that is radiated into space is accounted for by a resistor in the model.

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  • \$\begingroup\$ What is the relationship between current and EM waves? \$\endgroup\$ – JobHunter69 Jan 20 '17 at 5:06
  • \$\begingroup\$ In space, there is no current associated with an EM wave. However, if the wave encounters an antenna, it will cause current to flow in the antenna. And in the type of transmission lines we are considering here, the kind made with two conductors separated by a dielectric, the traveling wave causes current to flow in the conductors due to the time varying electric field. \$\endgroup\$ – mkeith Jan 20 '17 at 5:12
  • \$\begingroup\$ What I mean is, if you take a snapshot at any instant, you will see that the electric potential varies as you move along the transmission line. This is hypothetical, if you could freeze time. Since there is an instantaneous electric field along the transmission line, there will be current flowing along the electric field gradient. \$\endgroup\$ – mkeith Jan 20 '17 at 5:15
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    \$\begingroup\$ You might also find some enlightenment if you read about how time domain reflectometry works. When a source applies a voltage to a transmission line, the impedance that source feels is the characteristic impedance of the line. For example, it could be 50 Ohms. So, at the source, the current will be V/50, even if the end of the line has a short circuit. Only when the reflection comes back from the short circuit will the source "feel" the 0 Ohm load. If you observe the voltage at the source with an oscilloscope, you can calculate the round trip time of the EM wave to the short. \$\endgroup\$ – mkeith Jan 20 '17 at 5:22
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A transmission line is sometimes modeled as a series of lumped elements. Each tiny imcrement of line has an equally tiny inductance and capacitance associated with it.

schematic

simulate this circuit – Schematic created using CircuitLab

A narrow pulse traveling down the line--say it's passing L4 at the moment--would be stored in C4 and is now passing through L4, on its way to charging C5. As it passes through L4, it is stored there as well. The inductor current creates a magnetic field, which builds and then collapses, producing the voltage that charges C5. The pulse continues down the line in this fashion.

When the short is encountered, the voltage goes nearly to zero, and the current to a high value. The current through inductance of the short stores the energy in a magnetic field, and then as the field collapses, the current continues on around (per Lenz's law) but the voltage across the inductor has reversed.

Having attempted an explanation without ropes, equations, or Star Trek terminology, I must say that the rope analogies are very good. And there's similar question to this one, and in an answer, a link to this film from Bell Labs showing a very interesting wave demonstrator. I encourage readers to watch it.

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Let's imagine waves moving down a transmission line. A current wave, and a voltage wave. Their ratio is the line impedance, V/I. For instance, a 50v wave and a 1A wave moving down a 50ohm line. They always move together, and V=50*I at every point.

If they hit an open circuit, no current can flow there. So they generate a reverse travelling wave so the 1A reverse cancels out the 1A forward to produce zero amps at the open circuit (don't ask me 'how' this reverse wave happens, but it is observed to happen). This reverse travelling wave must have a voltage component as well, which must be 50v as well, and these add up to make 100v at the far end. The 50v 1A wave travels back along the line to the source.

If they hit a short ciruit, then no voltage can exist there. So they generate a reverse travelling wave so the 50v cancels out to produce 0v at the short circuit. The 50v 1A wave travels back along the line to the source.

You can now see that if the line suddenly changes impedance to 75ohms, we can no longer have a 50v 1A wave propagating forward by itself. It generates a small reverse wave such that when you add up the voltage and current at the junction point, the forward wave now obeys V=75*I, and the reverse wave makes up the difference.

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A simple analogy

Consider a long rope tied to a 'wall'. The end of the rope is flicked to produce a 'pulse'. The pulse travels along the rope until it meets the 'wall' or NODE. (a travelling transverse wave)

At the wall the rope is prevented from moving (i.e. a point of NO DEviation).

What happens is that an equal and opposite reaction at the 'wall' creates an anti-phase pulse travelling in the opposite direction.

The superposition of the forward and reflected pulse creates the unique solution that the deviation at the wall (or NODE) is zero.

enter image description here

This is a general property of waves. In a transmission line the 'wall' is a short circuit and the 'wave' is the electric field component of the electromagnetic wave.

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    \$\begingroup\$ No, the wall is analogous to am open circuit. \$\endgroup\$ – JobHunter69 Jan 20 '17 at 18:22
  • \$\begingroup\$ @Goldname the 'wall' is a node - The simple analogy only shows one type of transverse wave whereas an EM wave has two mutually combined waves - the electric field and the magnetic field. From a voltage perspective the wave is reflected by a short circuit, from a current perspective it is reflected by an open circuit. The reflection coefficient (r) can vary between +1 and -1 (= (Zl-Zc)/(Zl+Zc)) . The interesting value is when r = 0 - then there is no reflection and the transmission line is impedance matched. \$\endgroup\$ – JIm Dearden Jan 20 '17 at 18:55
  • \$\begingroup\$ I can see that the wall is a node in this case, but I am not understanding how open circuits create a node. Also, this analogy would be parallel if the wall represented an open circuit. That would make more sense. \$\endgroup\$ – JobHunter69 Jan 20 '17 at 22:09
  • \$\begingroup\$ From a current perspective its what would stop any current flowing - R is 'infinite' , I = V/R. From a voltage perspective its what would stop a voltage from changing - R is 'zero' - V = IR. The way I look at it for an EM wave is that the electric field component produces a changing voltage, the magnetic field component produces a changing current (see Maxwell's equations). So, for both a short circuit or an open circuit termination you get the same effect - a total reflection. Any other value will absorb the wave energy and reduce reflection. At characteristic impedance, no reflection \$\endgroup\$ – JIm Dearden Jan 21 '17 at 20:59
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You try to apply your basic "Kirchoff's intuition" to a much more complex phenomenon.

Waves in a circuit can be correctly described only using Maxwell's equations. Kirchhoff's laws are only an approximation of Maxwell's equations that work when the waves have a wavelength that's much bigger than the physical dimensions of the circuit, i.e. only if the highest frequency component in a signal is much smaller than c/d, where c is the speed of light and d is the dimension of the circuit.

That is, KVL and KCL hold only if the circuit is working in so-called quasi-static conditions. In these conditions you can approximate Maxwell's equations and eliminate some terms from them and treat the problem as it were an electrostatics problem. In electrostatics you can define an electrostatic potential directly related to the (static) electric field, and this gives the definition of voltage. Also current becomes easily defined in term of current density.

When those simplifying assumptions no longer hold, you can't even define voltage properly. In the context of transmission lines voltage can only be meaningfully defined between two points on the same plane perpendicular to the propagation direction. I.e. you can't measure a voltage between a point in the upper wire and another point, say, two centimeters away on the other wire (as you can easily do when KVL holds).

Bottom line: drop your "Kirchhoff's mindset" when you try to understand transmission lines and try to develop a more "Maxwellian mindset".

Anyway, this thread might give you more insight in that direction.

EDIT (to answer a comment)

Common sense is not applicable in this field (pun intended). You could get away with some, very imperfect, analogy (and for that, see the link I posted above or the other answers in this thread), but that's all. There is a reason why it took mankind millennia to understand and explain electro-magnetic phenomena (static electricity was known experimentally since ancient Greece, let alone lightnings and their consequences, for instance). If these phenomena had been easy to grasp with common-sense, it wouldn't have required sages and scientists ages to understand them.

Impedance/resistance, as an element in which current flows, is a concept tightly related to Kirchhoff's laws and lumped element circuit analysis.

In EM theory impedance has a whole different meaning: it doesn't relate voltage with current, but, more or less (...hand-waving...), E field with H field.

Fields don't flow like currents or exists across two points like voltages. Understanding exactly how, why and when fields propagate requires tons of advanced math. Maxwell's equations is, IMO, the conceptually toughest topic in Physics that an Electrical Engineer has to grok during his/her university education (bar quantum mechanics and solid-state physics).

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  • \$\begingroup\$ I understand what you mean, but I am not thinking in terms of kirchkoff. I'm thinking in terms of common sense. it simply would be easier for something to go where there's less impedance/resistance, so it should go there, am I wrong? \$\endgroup\$ – JobHunter69 Jan 20 '17 at 4:59
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    \$\begingroup\$ @Goldname Unfortunately 'common sense' only repeats to us what we have already learnt from experience. When we move into a new field, that experience is irrelevant, and common sense often leads us astray. \$\endgroup\$ – Neil_UK Jan 20 '17 at 6:58
  • \$\begingroup\$ @Goldname see my edit. \$\endgroup\$ – Lorenzo Donati -- Codidact.com Jan 21 '17 at 4:56

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