Signal reflections in open and closed circuits

In summary, the conversation discusses the behavior of signals in coaxial cables and the formation of standing waves. It is mentioned that a circuit with an earthed end will force a node at that end, while an open end will create an antinode. The question is posed about what would happen if a switch were to be added to the circuit, and the response is that closing the switch would create a short at the far end and enforce v(t)=0. The conversation also touches on the use of a string analogy to understand the concept and the importance of matching the transmission line impedance with the source impedance.
  • #1
mrneglect
11
0
I'll try to make this as concise as possible. I'm trying to get my head around how signals behave in coaxial cables. If you set up a circuit like this:

Exp4.png


Then the earthed end will force a node at that end, right? So you'll end up with your reflections having a phase shift of pi and looking like this:

Sine.png


And you find that a standing wave arises with a node at the earthed end. I also understand that if that end were not earthed but left open (infinite resistance) then the signal would reflect without any phase shift, and a similar standing wave would arise but with an antinode at the open end of the circuit. But what if one were to construct a circuit like the one below?

Exp3.png


With the switch open, it should be just the same as an open circuit: a standing wave should arise with an antinode at the open end.
With the switch closed, the sine waves should hit one another in phase and cause an identical standing wave to occur.

However, I have measured the electric field strength of the parallel wires in the circuit above with both an open and closed switch, but the standing waves produced are not only of different amplitudes, but also out of phase:

Exp3Graph1.png


Exp3Graph2.png


Could somebody explain to me why this happens? I feel I'm missing something blindingly obvious.

Many thanks in advance for any help! :smile:
 
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  • #2
Thread moved to Homework Help. Welcome to the PF!

Why do you say this?

With the switch open, it should be just the same as an open circuit: a standing wave should arise with an antinode at the open end.
With the switch closed, the sine waves should hit one another in phase and cause an identical standing wave to occur.

You seem to have gotten much of it right up until you say that you still have an antinode when you close the right-hand switch. The v(t) condition changes from an anti-node to a node when you close the switch.
 
  • #3
Ah, sorry, I wondered if I had posted in the wrong place.

Well the reason is that I don't see how closing the switch will force a node. I can understand than an earthed end will force zero volts and thus a node (like wiggling a string which is nailed to a wall), and an open circuit will create an antinode (like wiggling a free string), but I don't understand what happens when the switch is closed.

Maybe I should get a long piece of string and a friend to help me wiggle it from both ends at the same frequency.
 
  • #4
mrneglect said:
Ah, sorry, I wondered if I had posted in the wrong place.

Well the reason is that I don't see how closing the switch will force a node. I can understand than an earthed end will force zero volts and thus a node (like wiggling a string which is nailed to a wall), and an open circuit will create an antinode (like wiggling a free string), but I don't understand what happens when the switch is closed.

Maybe I should get a long piece of string and a friend to help me wiggle it from both ends at the same frequency.

By closing the switch, you create a short at the far end, which envorces v(t)=0.

Don't think of the bottom wire as being "ground" -- as you get away from the signal generator, it is just part of the transmission line. The TL can be unbalanced (like with coax), or balanced (like with twisted pair).

The string analogy works pretty well. To force a node, tie the string to a wall at the far end. To force an anti-node, you need a pretty low-friction ring on a rod, so that it can move freely up and down (or side-to-side, if you excite it that way).
 
  • #5
mrneglect said:
Ah, sorry, I wondered if I had posted in the wrong place.

Well the reason is that I don't see how closing the switch will force a node. I can understand than an earthed end will force zero volts and thus a node (like wiggling a string which is nailed to a wall), and an open circuit will create an antinode (like wiggling a free string), but I don't understand what happens when the switch is closed.

Maybe I should get a long piece of string and a friend to help me wiggle it from both ends at the same frequency.

When the switch closed, the end shorted to ground and there cannot be any voltage swing therefore the end becomes a note.

I still don't see why your scope traces have different amplitude between open and closed. Did you match the transmission line impedance with the generator source impedance? I guess you just roll two wires parallel onto a spool, no matching, no shielding to get the strange looking wave!.

If you match the Tx line Z to source without rolling into a roll, it should be a sine wave and the amplitude should be the same except they are out of phase between open and close of the switch if the generator drive a sine wave.
 
  • #6
But how can there be an Earth when the switch is closed when there isn't an Earth anywhere in the circuit?

And I don't know why the two waves have different amplitudes; they were held taut and parallel (not on a spool). The equipment looked like this:

Exp3Equipment.jpg
 
  • #7
From the wave form, the wave length is about 40cm. Assume speed of signal travel at light speed ( which should be close). The freq is 3EE8m/0.4m = 7.5 GHz! Is this true? That is too high a freq to run in the setup. Be careful of the setup. At RF, setup will very likely cause more error on the result than the circuit itself.

Your scope probe ground lead hook up to the wire will change the circuit. Without a proper ground plane, the scope probe ground literally change the length of your wire between switch open and close. I can't even start to name all the parasitic effect also. The only way to do this is either Fab a long micro strip with ground along side but not close to the line with many VIAs stitch to the ground plane below for the prob return...Or coax cut at certain segment for measuring and need to be very short at the cut... Hard!

BTW I asked a stupid question. Of cause the amplitude should be different. That is the whole thing about STANDING WAVE. The amplitude is the sum of the forward and reflected wave. At different point you have different amplitude. The amplitude also different at the same point if the switch is open or close. I wasn't thinking when I asked, sorry.
 
Last edited:

FAQ: Signal reflections in open and closed circuits

What are signal reflections in open and closed circuits?

Signal reflections occur when an electrical signal encounters a change in impedance in a circuit, causing some of the signal to be reflected back towards the source. In open circuits, the impedance change is caused by a break in the circuit, while in closed circuits, it can be caused by a mismatch in impedance between different components.

How do signal reflections affect circuit performance?

Signal reflections can cause interference and distortions in the original signal, leading to errors and malfunctions in the circuit. This can result in decreased performance or even complete failure of the circuit.

How can signal reflections be minimized in open and closed circuits?

In open circuits, signal reflections can be minimized by ensuring the circuit is properly closed and there are no breaks or gaps. In closed circuits, impedance mismatches can be reduced by using components with matching impedance values and proper circuit design.

What are some common causes of signal reflections in open and closed circuits?

In open circuits, signal reflections are commonly caused by faulty connections, damaged wires, or breaks in the circuit. In closed circuits, they can be caused by using components with different impedance values, improperly matched components, or inadequate circuit design.

How can signal reflections be measured and analyzed in open and closed circuits?

Signal reflections can be measured and analyzed using specialized equipment such as an oscilloscope or a network analyzer. These tools can capture and display the reflected signals, allowing for analysis and troubleshooting of the circuit.

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