# Transmission Lines and Standing Waves, the Effect of Short Circuit and Open Circuit

#### Like To Learn

Joined Jul 11, 2015
5
Hello, this is my first post to the All About Circuits forums and I'm not sure I've posted in the correct forum.

I am learning about Transmission Lines and have been reading the article Google found on the All About Circuits website, Home/Textbooks/Vol II Alternating Current (AC)/Transmission Lines/Standing Waves and Resonance. Chapter 14 Transmission Lines

This article is very good and the illustrated waveforms demonstrate the effect of reflection very well but perhaps I have misunderstood the terminology used in this article. I am a bit confused about the impact of open circuit and short circuit conditions at the end of the transmission line. That is, the info in this article seems to differ from other reference sources.

The 2nd para says "The following illustration shows how a triangle-shaped incident waveform turns into a mirror-image reflection upon reaching the line’s unterminated end."

For unterminated I have understood this to mean open circuit which I thought does not produce a mirror image reflection wave, which I have interpreted to mean inverted, but an erect reflection wave which algebraically adds to the incident wave as they each travel in opposite directions. So for the waveforms illustrated in this article where the blue reflected wave is shown inverted relative to the green incident wave, is that illustration actually showing the status for a short circuit at the end of the transmission line? The illustrations also say unterminated line but should it read short circuited line for these illustrations?

I will be grateful if any expert on this topic could clarify this for me please. Thanks

#### Lool

Joined May 8, 2013
116
Mirror image does not mean inverted in amplitude. Imagine the waveform flips around a vertical axis. So open circuit has reflection coefficient of 1, which does produce a mirror image reflected wave. A shorted line has reflection coefficient of -1 which means an inverted mirror image.

So, yes you just got the terminology confused, and it is confusing terminology by the way. But, the math/equations leave no ambiguity.

#### nsaspook

Joined Aug 27, 2009
6,256
This old video gives a good visual of the subject:

#### Like To Learn

Joined Jul 11, 2015
5
Mirror image does not mean inverted in amplitude. Imagine the waveform flips around a vertical axis. So open circuit has reflection coefficient of 1, which does produce a mirror image reflected wave. A shorted line has reflection coefficient of -1 which means an inverted mirror image.

So, yes you just got the terminology confused, and it is confusing terminology by the way. But, the math/equations leave no ambiguity.
Thanks for this reply. I'm studying this as well as the video below. I understand your point about mirror image being about the vertical axis. But I'm now unclear as to whether the ilustrations of waveforms in the article are those of current or voltage? The standing wave has a node at the unterminated end so I suppose the waveforms must be of current? Still studying

#### Like To Learn

Joined Jul 11, 2015
5
This old video gives a good visual of the subject:
Thank you for this video which is excellent at explaining wave phenomena. I note that the presenter describes the mechanical clamping at the ned of the mechanical transmission line as being analogous to an open circuit transmission line in which case I assume the mechanical wave motion is analogous to electrical current wave motion? So at the moment I am assuming that the waveforms illustrated in the article are those of current. But I understood that the standing wave is a voltage standing wave as described in the article. I need to study this some more to get this clear.

#### Like To Learn

Joined Jul 11, 2015
5
First : The Textbook, Chapter 14

I've looked at this many times together with other educational source material and it still seems to me that the transmisson line demonstrating total reflection of the triangular waveform in Chapter 14 of the All About Circuits textbook mentioned in my first post is showing the case of a short circuit termination and not an unterminated line as the text and illustrations say. I interpret unterminated to mean open circuit.

In the illustrations the blue reflected wave is clearly inverted and as Lool has pointed out an inverted reflection wave results from a reflection coefficient of -1 which is the status for a short circuit termination.

Second : The Video

I have watched the video sent by nsaspook a few times which is very good at explaining the wave behaviours but unless I have completely misunderstood there seems to be a mistake in the section where the presenter compares the mechanical situation with the electrical transmission line situation

Starting at about 4.19 into the video the presenter uses a blackboard to make comparisons.

On the lefthand side there are the mechanical, electrical and acoustical versions of transmission medium with a free end at the end of the transmission medium.

On the right hand side there are the mechanical, electrical and acoustical versions with a restrained end of the transmission medium.

The blackboard RHS versions are said to be those that produce an inverted reflected wave. The example for the electrical transmission line shows an open circuit. But an open circuit gives a reflection coefficient of +1 and hence an upright reflected wave.

Similarly the blackboard LHS versions are said to be those that produce a reflection "right side up" in other words not inverted. But the diagram on the board shows a short circuit at the end of the electrical transmission line which gives a reflection coefficient of -1 and hence an inverted reflected wave.

So its seems to to me that the LHS and RHS electrical transmission lines should be open circuit and short circuit respectively.

I will be grateful to know from those more expert if I have interpreted these observations of the textbook and the video correctly and if I am way off the mark for clarification of where I have gone wrong

Thank you

#### nsaspook

Joined Aug 27, 2009
6,256
Pressed for time for a road trip:
Yes, you are correct about the electrical transmission line reflections in the old video. #### Like To Learn

Joined Jul 11, 2015
5
Thanks for your reply. I am grateful for your confirmation about the transposed examples shown in the old video for electrical transmission lines with a short circuit and open circuit termination.
I am continuing to study this topic and have again been looking at the AAC textbook Chapter 14 but this time focussing on the articles on Characteristic Impedance and Finite-Length Transmission Lines and I now have a supplementary question related to this. Here's what I have understood so far.
These sections of the textbook show a hypothetical infinitely long, loss free transmission line made up from an infinite number of (imaginary?) "small chunks" of line connected in succession. Each "chunk" has series inductance in each of the conductors and parallel capacitance across the conductors. It describes the way the inductance limits the current which flows to charge the capacitors progressively down the line.
In this infinite length line the current is limited by the characteristic impedance of the line, Zo. So for an applied voltage of V volts the current will be V/Zo. For this hypothetical line Zo will be purely resistive.
It also shows that for a finite length loss free line terminated with a resistor equal to the characteristic impedance of the line, then the impedance seen from the source end will equal the characteristic impedance.
So if a source signal with zero internal impedance provides a perfect dc voltage step V and this is applied to such a line the wavefront (i.e. the perfect rising edge of the dc voltage step) will travel down the line at the propagation velocity of the line ( in the hypothetical perfect line at the speed of light ?).
I have difficulty in visualising the waveforms of the current in the line for that short period of time when the wavefront is still travelling down the line and has not reached the load end. I'll try and explain that.
The current leaving the source = V/Zo. But if some of that current has to follow a transverse path through a capacitance and some of it continues to flow down the line through the next series inductance/parallel capacitance (or "small chunk"), what are the relative waveforms of these components of current at different propagation times, which together equal V/Zo ?
Again I will be grateful if you or any reader can offer clarification about this or reference to other educational material. Thank you

#### nsaspook

Joined Aug 27, 2009
6,256
So if a source signal with zero internal impedance provides a perfect dc voltage step V and this is applied to such a line the wavefront (i.e. the perfect rising edge of the dc voltage step) will travel down the line at the propagation velocity of the line ( in the hypothetical perfect line at the speed of light ?).
I have difficulty in visualising the waveforms of the current in the line for that short period of time when the wavefront is still travelling down the line and has not reached the load end. I'll try and explain that.
The current leaving the source = V/Zo. But if some of that current has to follow a transverse path through a capacitance and some of it continues to flow down the line through the next series inductance/parallel capacitance (or "small chunk"), what are the relative waveforms of these components of current at different propagation times, which together equal V/Zo ?
Again I will be grateful if you or any reader can offer clarification about this or reference to other educational material. Thank you
The textbook answer is to use the Telegrapher equations but perfection is generally not possible in this universe.