lumanauw said:The graphs are interesting
They are also incorrect. They show a system with matched inputs and outputs having 5 Vp-p at the source and 4 Vp-p at the receiver. Then they show a system with a high impedance receiver with 8 Vp-p at the receiver.
That's a pretty good trick -- a "straight wire with gain", eh?
Charles Hansen said:
Thank you, Charles.
Here's a decent article on that subject:
Characteristic Impedance of Cables at High and Low Frequencies
se
This article is confusing me. If Rsource=50ohm, Rload=50ohm, then this will make a 1/2 voltage divider.
But in this article, the original signal and reflected signal can make input~output?
http://www.ecircuitcenter.com/Circuits/tline1/tline1.htm
But in this article, the original signal and reflected signal can make input~output?
http://www.ecircuitcenter.com/Circuits/tline1/tline1.htm
lumanauw said:
Yes, which is why the first article you linked had silly graphs. They showed 5 volts in and 4 volts out. It should have been 5 volts in and 2.5 volts out.
lumanauw said:
Don't worry, just keep re-reading it. I had to read Howard Johnson's book 3 times before it started to make sense to me.
Hi, Steve Eddy,
I got a question regarding termination, say the source has 50ohm series resistor, the receiver has 50ohm+100pf termination.
Is this law still works if we put isolation transformer in the middle of the cable? 10k/10k isolation transfomer needs the receiver impedance >10k, 50ohm is way too low.
I got a question regarding termination, say the source has 50ohm series resistor, the receiver has 50ohm+100pf termination.
Is this law still works if we put isolation transformer in the middle of the cable? 10k/10k isolation transfomer needs the receiver impedance >10k, 50ohm is way too low.
Lumanauw --
Go read the book I recommended. This will answer all of your questions.
It is a very complex topic and one or two posts probably will not clarify things for you.
One problem is that audio guys are normally thinking about things in a static way -- either DC or constant sine waves. But with this, you have to think about what is happening at each instant. It is a very different way to think about things, which is why I had to re-read the book 3 times.
As far as your question about the quote, the only thing I can say in a short posting is that you have to remember that there can be both positive and negative reflections. When the characteristic impedance of the receiver is lower than the transmission line, then the reflection will be of the opposite polarity. When the impedance of the receiver is higher than the transmission line, the reflection will be of the same polarity.
In the example given, the pulse traveling down the transmission line is only 0.5 because of the voltage divider action. When the pulse reaches the end and the characteristic impedance is extremely high, the reflection is of the same polarity and the mis-match is so great that the reflection is essentially the same amplitude.
So then the 0.5 incident pulse and the 0.5 reflected pulse superimpose to form the 1.0 volt pulse at the receiver. The reflected pulse keeps going back "upstream", but when it hits the (properly terminated) source, it is fully absorbed and "disappears".
This is why (in most situations) it is only necessary to terminate one end of a transmission line to avoid problems.
Go read the book I recommended. This will answer all of your questions.
It is a very complex topic and one or two posts probably will not clarify things for you.
One problem is that audio guys are normally thinking about things in a static way -- either DC or constant sine waves. But with this, you have to think about what is happening at each instant. It is a very different way to think about things, which is why I had to re-read the book 3 times.
As far as your question about the quote, the only thing I can say in a short posting is that you have to remember that there can be both positive and negative reflections. When the characteristic impedance of the receiver is lower than the transmission line, then the reflection will be of the opposite polarity. When the impedance of the receiver is higher than the transmission line, the reflection will be of the same polarity.
In the example given, the pulse traveling down the transmission line is only 0.5 because of the voltage divider action. When the pulse reaches the end and the characteristic impedance is extremely high, the reflection is of the same polarity and the mis-match is so great that the reflection is essentially the same amplitude.
So then the 0.5 incident pulse and the 0.5 reflected pulse superimpose to form the 1.0 volt pulse at the receiver. The reflected pulse keeps going back "upstream", but when it hits the (properly terminated) source, it is fully absorbed and "disappears".
This is why (in most situations) it is only necessary to terminate one end of a transmission line to avoid problems.
syn08 said:
Yes. If you take an ideal transmission line with a 50-ohm characteristic impedance and terminate it at one end with a 50-ohm resistor, you will see a resistive 50-ohm impedance looking into it at the other end, regardless of its length. This is the beauty of transmission lines.
If you mis-terminate it by some amount, as with, say, 40 or 60 ohms, then you will see non-idealities at the other end which are length-dependent.
Obviously, the best scenario is for a transmission line to be properly terminated at both ends. Under those conditions, a reflection will not occur at either end.
When unable to terminate properly at both ends, one can get good transmission line behavior by series terminating at the source or by shunt terminating at the receiver.
Cheers,
Bob
lumanauw said:
What about this do you not understand? This is very standard transmission line theory as applied to unterminated digital electronics such as TTL, or ACT, etc. The impedances and voltages are different but the theory is the same.
Keep in mind that the Zo of the line is an AC characteristic, a coax or transmission line typically has nearly no DC leakage so why would it be surprising that an unterminated net does not settle to the input voltage? What is special about this case is that it is (match) source terminated so that while there is a strong reflection from the end of the line, it is matched and properly terminated at the sending end which stops further reflections. What is fascinating is that the far end receives the "full transition" signal before the sending end. If this was a bus with receivers along the line, timing analysis would have to account for prop delay down the line and back for the receiver closest to the source. And the receiver *near* the source would see a full strength transition *last*! This is the nature of transmission lines.
Consider the DC case where you apply a step input, and let's say that it is not match source, or end terminated, there would be multiple reflections but it would still settle to the input voltage at all points on the line.
I have debugged hacked together TTL/CMOS systems for a client, operating on 5V where there were nearly double strength (8-9V) signals on some of the nets which burnt out many of the CMOS parts. This was not my design. I was called on to debug and fix it.
Anyone want to explain how you might find 8-9V in a 5 V only system?
Is this off topic?
Pete B.
Edmond Stuart said:
Hi Bob,
So the other end needs to be improperly terminated?
And how, with 0 Ohm, 100 Ohms, 1kOhm 10kOhm?
Please explain.
Cheers,
Edmond.
Hi Edmond,
The driving end does not need to be mis-terminated. In the frequency range where the far-end termination is valid and accurate, the transmit end can be terminated with whatever impedance you want, since it will see 50 ohms resistive looking into the cable. However, in this same frequency range, the amplitude of the signal that will be delivered to the other end will depend on the voltage divider formed by the output impedance of the source and the impedance of the terminated transmission line.
Bear in mind that the approach I described only provides a valid resistive far-end termination of the cable at very high frequencies.
At much lower frequencies, the transmission line largely behaves like a lumped element. If I recall correctly, the transition frequency from lumped behavior to transmission line effects is at a frequency where the round-trip distance is about 1/6 of a wavelength (I could be off by a factor of two).
Cheers,
Bob
Bob Cordell said:
At this point I am confused: what is the purpose of terminating the "line" with 50ohm? The "line" does not have 50ohm characteristic impedance and therefore it will be barely matched at the receiver's end by a 50 ohm termination. OTOH, even if the line would be 50 ohm and properly terminated by 50 ohm at both ends, you would get the maximum power transmission which I somehow doubt it's good for the purpose of the whole setup (protecting against HF?).
I think a current drive schema (high output impedance preamp and low input impedance amp) would be better in terms of protecting the link against HF but I haven't closely analyzed this scenario.
Charles Hansen said:
Low power systems are typically source terminated if possible since there is no static power dissipated. High speed systems are typically driven by low impedance sources, to provide incident wave switching, and are terminated at the far end to avoid the prop delay round trip effect.
High speed TTL (S, AS, FAST) which was often unterminated, typically had clamp diodes on the inputs to provide a loss path for overshoot in the waveform. This sometimes shunted so much current into the ground that it caused ground or Vcc bounce. I have seen RAMs lose data, due to spurious writes, and PALs go into program verify mode due to ground/Vcc bounce.
These effects are highly important for systems operating in the 10-1000+ MHz range, really it is the edge speed and component bandwidth that matters. Not really a factor in audio with the much lower bandwidth/edge speeds involved.
Transmission line effects are important in high bandwidth video. Such video signals are typically source and end match terminated.
The source and end termination do form a voltage divider and therefore the termination (75 ohms typically) can usually be switched on or off in pro equipment. This allows it to be daisy-chained with the termination turned on at the end of the line only.
Pete B.
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