E&M Question
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Born2bwire
Diamond Member
- Oct 28, 2005
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Reflections only occur when there are impedance mismatches. As long as the short wire is matched to its terminations, then there will not be any reflections from the short wire's load. A quick refresher: http://www.amanogawa.com/archive/docs/D-tutorial.pdf .
- Sep 10, 2005
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My confusion comes from the fact that I never used to think about waves when I think about circuits. I was told that if you had two matched parallel cables where one was terminated and the other was open, the wave would travel down both cables as if they were both parallel with the same resistance (aka, the power would initially be split half and half between each cable). But when the wave reaches the open circuit, it reflects back creating a standing wave on the open cable and dividing in two again between the source and the terminated cable. I feel like the shorter cable would effectively have less resistance, but since the wave takes time to travel, it sends equal amount of power down each cable because it only 'sees' the characteristic impedance of each wire. When it finally realizes that the resistance of one cable is less than the other, I feel like reflection should occur to compensate for this. If this is one of those cases where Ohm's law does not hold, let me know as I am probably making wrong assumptions. Also, an open circuit is not supposed to have any current. But I get the feeling that the standing wave has some sort of current thing going.
DominionSeraph
Diamond Member
- Jul 22, 2009
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I'm by no means qualified to answer this question but I thought the impedence of a cable determined the velocity of the signal. Going from a medium with one velocity to another is what causes reflections. Having one cable longer than the other won't change this but when the 2 signals arrive at the other end, they will be out of phase unless the cables lengths are one wavelength different. I would also think the only other thing the different length affects is signal strength at the receiver end. As for an open circuit not having current, it does have a small level of capacitance so current can travel in for a short period of time.
Modelworks
Lifer
- Feb 22, 2007
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The easiest way to understand reflections and standing waves is a simple experiment.
A box with a fan at one end and a vent at the other. Fill the box with smoke and the vent fully open and you will see the smoke exit in almost a straight line with no turbulence. Close the vent partially and now the smoke will exit but some of it will curl back and start to churn round and round waiting for its chance to exit, standing waves. Close the vent completely and the smoke will hit the vent, create a standing wave which takes in more smoke until it finally pushes back at the fan, reflections.
In your cables the shorter cable is just a smaller box. Change the length of the cables and it still doesn't matter because the fans air is still contained in the box and will have the same effect on the smoke.
Where you get into issues is changing the signal, that would be like changing the fan on the box without changing the vent size. The vent that worked before half open would have issues now if you changed the fan to one twice the size.
A box with a fan at one end and a vent at the other. Fill the box with smoke and the vent fully open and you will see the smoke exit in almost a straight line with no turbulence. Close the vent partially and now the smoke will exit but some of it will curl back and start to churn round and round waiting for its chance to exit, standing waves. Close the vent completely and the smoke will hit the vent, create a standing wave which takes in more smoke until it finally pushes back at the fan, reflections.
In your cables the shorter cable is just a smaller box. Change the length of the cables and it still doesn't matter because the fans air is still contained in the box and will have the same effect on the smoke.
Where you get into issues is changing the signal, that would be like changing the fan on the box without changing the vent size. The vent that worked before half open would have issues now if you changed the fan to one twice the size.
I think the example should almost be thought of as thinking of impedence as a resistance to induction of a current (not actual resistance to the current itself).
If there is a difference between them, one of the ones (My logic would say the one with the lower impedence) would not be able to force its way through (fully) to the other. You would get a reflection according to the relative difference in EM properties.... I could be wrong (and tell me if I am).
The only thing I believe would be a problem is if one was substantially shorter. I believe that the EM effect travels at the speed of light through wire (not the actual electrons). Again, that may not be correct, but there is a speed.
If the wires are significantly different in length you will get a lag/shadowing of the signal at the receiving end.
If there is a difference between them, one of the ones (My logic would say the one with the lower impedence) would not be able to force its way through (fully) to the other. You would get a reflection according to the relative difference in EM properties.... I could be wrong (and tell me if I am).
The only thing I believe would be a problem is if one was substantially shorter. I believe that the EM effect travels at the speed of light through wire (not the actual electrons). Again, that may not be correct, but there is a speed.
If the wires are significantly different in length you will get a lag/shadowing of the signal at the receiving end.
- Sep 10, 2005
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The reflection stuff makes sense to me. The fact that the wave does not realize that the lumped resistance of one cable is greater than the other until it gets to the end slightly baffles me because I almost expect a reflection to occur to compensate for this. To make this clearer, if you had two resistors in parallel with a voltage source, you expect the current to be the same in both resistors. If you double the length of one of the resistors, the wave is supposed to travel down both resistors just as if they were parallel. But the amount of current in the smaller resistance is supposed to be greater than the larger resistance. I expect the wave to double back and compensate for this somehow. I'm confused so this may not make sense at all. I'm definitely not accounting for something.
Yeah, you're not accounting for capacitance and inductance in a simple voltage source & resistor model.
The characteristic impedance of a transmission line is not the same as the "impedance" of the line measured from point a to point b.
Forgive my rusty EE, but is impedence the resistance to the change in voltage, not to the current?
(Me lazy not Googling...)
(Me lazy not Googling...)
- Sep 10, 2005
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I think I accounted for this when I said the wave will split evenly between the two resistors (initially).
Then why would the wave "split evenly" when one cable is properly terminated and the other has infinite impedance?
- Sep 10, 2005
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Not sure if OP understands "properly terminated". His / her second post talks about creating a reflection back from an open-ended cable. But "properly terminated" means that at the end of the cable there is a load with the same characteristic impedance of the line, and all the signal energy arriving at the line end is absorbed in the load. That is why there is no reflection. (The load could be an amplifier input circuit, or an antenna, or just a small resistor from center to outer conductor of the cable.) The length of the cable does not change that - there is no reflection if the cable truly is, as specified in the question, "properly terminated". Now, it is true that with different cable lengths and thus different total signal attenuation in the two cables, the signal that reaches the end of the shorter cable will be stronger and hence the energy absorption rate in the load on its end will be higher that in the load at the end of the longer cable. But that still does NOT cause any difference in reflections (that do not exist in either cable).
- Sep 10, 2005
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You missed the part where I said there were two cables in parallel, where one is terminated and the other is open circuited.
Another way to think of being properly terminated is that the line appears to be infinitely long.
OP is mixing 2 "fields" ... weak pun. Transmission line theory and traveling wave theory ... with slightly more on traveling waves than the tx lines.
Also, to be explicit, separate DC resistance from characteristic impedance. The characteristic impedance is really just Ohms law, but using complex math, aka the Telegraphers Equations.
And uh ... a transmission line can be modeled as an infinite series of a RLGC circuit. Series R & L with shunt G & C. A computer model of a transmission line might have as many as 100 of these little 2 ports connected in series. Simpler models maybe only 10 and may only use the series L & shunt C (the lossless tx line). It depends on how much time it takes to crunch and what you are willing to give up in terms of accuracy. Ten circuits may be accurate enough to represent propagation delay, but there will be a large phase error ... if you don't care about the phase error, then no biggy. Depends on the application.
Traveling waves ... hmmm, do not actually need a medium. Thinking about EM fields in space. So the velocity of propagation is at the speed of light. But send in impulse (the wave) down a coaxial tx line such as a TV cable, then the propagation time is slowed by the dielectric surrounding the center conductor. Different dielectrics have different dielectric constants. So 2 lines with different dielectrics but with identical physical length will have different propagation delays ... the electrical length is different. Another point of view is that the dielectric makes the line appear electrically longer.
What you haven't touched on is that the wave is reflected by 100% (thinking of a lossless line) for either an open or a short circuited tx line, but reflects with opposite polarity.
Standing waves occur with a constant wave source. The reflected signal canceling & adding to the forward signal.
OP is mixing 2 "fields" ... weak pun. Transmission line theory and traveling wave theory ... with slightly more on traveling waves than the tx lines.
Also, to be explicit, separate DC resistance from characteristic impedance. The characteristic impedance is really just Ohms law, but using complex math, aka the Telegraphers Equations.
And uh ... a transmission line can be modeled as an infinite series of a RLGC circuit. Series R & L with shunt G & C. A computer model of a transmission line might have as many as 100 of these little 2 ports connected in series. Simpler models maybe only 10 and may only use the series L & shunt C (the lossless tx line). It depends on how much time it takes to crunch and what you are willing to give up in terms of accuracy. Ten circuits may be accurate enough to represent propagation delay, but there will be a large phase error ... if you don't care about the phase error, then no biggy. Depends on the application.
Traveling waves ... hmmm, do not actually need a medium. Thinking about EM fields in space. So the velocity of propagation is at the speed of light. But send in impulse (the wave) down a coaxial tx line such as a TV cable, then the propagation time is slowed by the dielectric surrounding the center conductor. Different dielectrics have different dielectric constants. So 2 lines with different dielectrics but with identical physical length will have different propagation delays ... the electrical length is different. Another point of view is that the dielectric makes the line appear electrically longer.
What you haven't touched on is that the wave is reflected by 100% (thinking of a lossless line) for either an open or a short circuited tx line, but reflects with opposite polarity.
Standing waves occur with a constant wave source. The reflected signal canceling & adding to the forward signal.
William Gaatjes
Lifer
- May 11, 2008
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I am a bit confused now too.
This may be a bit of topic but i had to think about it when the 2 different lengths where mentioned.
I remember : That when i take an signal in the shape of an impulse(although any signal with a short enough wavelength will do) as used with a clock signal and i take 2 wires of different length, i get an phase shift if the wavelength of the signal is short enough when i measure the time between sending the signal and measuring the signal at both wires at the same time. Even with the proper termination. If i would send databits over, some bits would still be traveling while the clock is already there creating data corruption.
Now that i think about it, perhaps because those 2 wires are different in length, this does not alters the characteristic impedance of the cable but it does affect the signal when seen as an inductor. The signal transmission takes time to travel. copper wire was about 2/3 of the speed of light in vacuum i think, but i am not sure. But again, this has nothing to do with the impedance. It does however affect the standing wave in the wire when the wire is short enough to come in the realm of the wavelength of the signal through the wire.
Can you refresh me a bit ?
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Comments in RED & I could write pages about grammar & the science in this 1 post.
Reflection will not occur in either cable if "properly" terminated (characteristic impedance of the port each cable is connecting to is the same as that of the corresponding cable).
What I expect would happen is the magnitude of the voltage at the receiving end would be somewhere between the magnitudes at the end of each cable. ie if at one instant the voltage on the longer cable is V1 and the magnitude of the voltage on the shorter cable is V2, the voltage magnitude seen at the receiving port would be V2 + [ (V1 - V2) * L2/(L1 + L2) ] where L1 is the length of the longer cable and L2 is the length of the shorter one. What you have is a voltage divider since each cable exhibits a finite, lumped resistance.
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- Sep 10, 2005
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Okay, I think this is what I'm trying to say. Say you have one co-ax cable. You turn on the generator, and the wave begins to propagate down the cable. As the wavefront moves, the capacitance along the cable gets charged little by little. Imagine a bajillion capacitors in parallel along the cable. If there is no resistance, they each get charged up as the wave propagates across them. If there is resistance, the rate at which all the little caps get charged isn't dependent only on the speed of the wave. So now as you charge the caps further and further down the line, your current decreases and so does the rate of charge. In fact, the rate of charge may not even keep up with the speed that the wave propagates and you may end up charging several caps in parallel rather than just the one nearest the wavefront. As the length over which the wave occupies increases, so does the resistance. So as the wave moves across the line, your overall current is dropping. This causes a change in the circular H-field. Would this not cause transmission all over the place?
I realize this is not reflection, but my real question was actually the mechanism which alters the field such that it reflects the current that should be in the line. I thought some kind of reflection was the source of the field correction. Cuz ultimately, I should be able to apply ampere's law around the cable and find the current. But with there being no reflection, it almost didn't seem like the field was changing. I don't even want to think about the transient behavior of the field in the conductor itself
I realize this is not reflection, but my real question was actually the mechanism which alters the field such that it reflects the current that should be in the line. I thought some kind of reflection was the source of the field correction. Cuz ultimately, I should be able to apply ampere's law around the cable and find the current. But with there being no reflection, it almost didn't seem like the field was changing. I don't even want to think about the transient behavior of the field in the conductor itself
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Okay, I think this is what I'm trying to say. Say you have one co-ax cable. You turn on the generator, and the wave begins to propagate down the cable. As the wavefront moves, the capacitance along the cable gets charged little by little. Imagine a bajillion capacitors in parallel along the cable. If there is no resistance, they each get charged up as the wave propagates across them. If there is resistance, the rate at which all the little caps get charged isn't dependent only on the speed of the wave. So now as you charge the caps further and further down the line, your current decreases and so does the rate of charge. In fact, the rate of charge may not even keep up with the speed that the wave propagates and you may end up charging several caps in parallel rather than just the one nearest the wavefront. As the length over which the wave occupies increases, so does the resistance. So as the wave moves across the line, your overall current is dropping. This causes a change in the circular H-field. Would this not cause transmission all over the place?
I realize this is not reflection, but my real question was actually the mechanism which alters the field such that it reflects the current that should be in the line. I thought some kind of reflection was the source of the field correction. Cuz ultimately, I should be able to apply ampere's law around the cable and find the current. But with there being no reflection, it almost didn't seem like the field was changing. I don't even want to think about the transient behavior of the field in the conductor itself
As for your top paragraph, if the resistance of your central conductor and/or your outer (shield) conductor is high then yes, the signal will be attenuated. Since I think you have just been introduced to this subject, (I mean no offence) assume a lossless case since the math is easier to work with.
Also, remember this is a transmission line only for frequencies which are high enough. Consult a textbook for why. The point is that KVL/KCL break down at those frequencies. We treat t-lines as multiple RLGC segments in series (see PsiStar's post) and you may think of those little series inductances maintaining the "momentum" of the wave throughout the line.
- Sep 10, 2005
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Well I keep stating some DC voltage because we all know from intro circuits lab how two resistors in parallel with a DC source behave. So you have that expectation. Then there are people who tell me that the currents will be the same because the characteristic impedance in both cables are the same (despite length). So to combine the expectations from intro circuits and wave theory, I went with a step function (lots of high frequency content). That essentially combines both wave propagation stuff and circuits 101 expectations.
William Gaatjes
Lifer
- May 11, 2008
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Technoob I think you should take a look at an electromagnetics textbook since you seem very interested about this. But don't start with transmission lines -- instead look up em wave propagation in free space and pay particular attention to normal incidence at boundaries between regions of different relative permittivity and permeability. They use continuity equations at the boundaries for current and voltage which can help you understand what's going on.
Note that the math for em waves is a little more...complicated than for t-lines but most courses on em seem to cover waves then t-lines (as they always seem to go from more complicated stuff to the less complicated.
Note that the math for em waves is a little more...complicated than for t-lines but most courses on em seem to cover waves then t-lines (as they always seem to go from more complicated stuff to the less complicated.
William Gaatjes
Lifer
- May 11, 2008
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Although i think you deserve such name calling, i will apologize anyway for my rude outburst. I do not want to lower myself to an individual as yourself. Therefor i shall file it under not yet enough coffee this dreaded morning and these dreaded last 10 days.
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