Mains interference: Digital TV problems


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The errors will be symbols with the wrong values. They come from any part of the circuit. For example it might be due to low input SNR, it might be due to faulty components, or whatever.
(I suppose it might be fun here to inject the fact that not all digital systems actually work with "bits". A bit is a *binary* information block, and many digital systems are m-ary, where "m" is greater than 2. No actual bits, just symbols, some with values larger than 2.)

You appear to think that "digital" means it is some kind of a binary voltage pulse. Something like you'd see for TTL, RS-232, T1, where a "pulse" of the appropriate amplitude is a 1 or a 0 value.
First, that is *not* what defines "digital", and secondly not a one of those is necessarily edge triggered vs. level triggered. Which is to say that rise time is absolutely insignificant for some applications.
T1 is an example. Just where do you claim the signal for a T1 carrier system is sensitive to rise time?

Yes, the advantage of digital transmission systems is the fact that a new signal is regenerated at each node.

As I previously noted, noise is cumulative in analog systems, and errors are cumulative in digital systems.

*Errored* bits! No bits are added. No bits are dropped. They just have the *wrong* *value*. A symbol that was supposed to have the value 64 ends up with a value of 2.

Wrong. I just gave you an example, and you will *not* find *any* place in a T1 digital carrier system where the signal is rise time dependent for correct decoding.
In fact, the rise time will be *very* significantly changed between the transmitter's output and a receiver's input, depending on the length of cable. Hence the rise times expected if the cable is 10 feet long are very short, and voltage pulses look like square waves, but if there is a mile of cable the voltage pulses are nothing close to a square wave and have very long rise times (a sine wave!).

Sine waves, at any given frequency, have significantly slow rise times compared to, for example, a square wave pulse at the same frequency.

That's a wonderful set of words. Do you know what you said? Was it supposed to means something specific? I particularly liked the "rather unit function" group! But "a sinusoid on its leading edge" ain't bad either.
I've got a program that generates nonsense phrases from real words somewhere, but it doesn't do that good.

I've seen *many* amplifiers used for audio that could and would reproduce RF.
Regardless, all you are saying is that amplitude distortion is inherent in electronic circuitry. That is true, but digital amplifiers to not necessarily make any inherent use of that characteristic; in fact they often go to extremes to avoid it! (Which is to say, digital is commonly very broadband... :-)

What is a "100 G sinusoid"? What is a "1M amplifier"?
I take it that was meant to be some sort of frequency specification; and if so you are *wrong*. It might not be able to respond to it, but in fact it might. Excess bandwidth is not inherently detrimental...

But it is not part of the functional design for digital amplifiers.

No response?

No response? I clearly said that you were making false statements, because you don't even realize that the input to the tuner is analog, not digital.

No response? What's the matter, did you suddenly realize how far off base you were with those statements?

The definition of "Insertion loss" negates everything you've said about it.
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Floyd L. Davidson <http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska) snipped-for-privacy@apaflo.com
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On Wed, 14 Jun 2006 08:53:38 -0800, snipped-for-privacy@apaflo.com (Floyd L. Davidson) Gave us:

And we won't even mention systems which use CML! HA!
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So just when are you going to answer that question?
...

How to you account for the above?

No response?

*You* are the one who wrote "not the LeGrange/LePlace pulse", thus excluding those transforms. Why didn't you also mention Fourier, and we might as well comment on Maxwell and Newton too.
But the point is that what *I* questioned was what *you said* it was, not what you said it wasn't:
"That's a wonderful set of words. Do you know what you said? Was it supposed to means something specific? I particularly liked the "rather unit function" group! But "a sinusoid on its leading edge" ain't bad either.
I've got a program that generates nonsense phrases from real words somewhere, but it doesn't do that good."
It seems the only way you can defend what you've written is to edit what I said, and talk about what you didn't say. But, alas, everyone can look at the Google archives and see what was actually posted.

So tell us again about how the input to the tuner is a digital amplifier! Tell us more about how digital systems all use rise time to differentiate between symbol values.

Your point?

Like, for example, the analog input to a tuner?

All digital signals are made of a finite set of discrete *symbols*.

Bits are *binary* information units. Binary is only one particular type of digital signal. Not all digital symbols are binary in value. For example, a single symbol might have a set of values that range from 0 to 255. If converted to a binary digital system, that would require 8 symbols to hold the same data.
Do *not* confuse "bits" with defining "digital". There are many many digital systems that do not use bits to carry information.

What is "an AM riding on a digit"???
Note that "levels of digital" is hardly BS. For example, take PCM code, which has 256 values per symbol, not 2 as with a bit. Or try ISDN which is commonly encoded as 2B1Q (Two Binary One Quaternary), where each pulse has 4 possible values, not 2.

No response?

No response?

No response?

Well, which is it that we are going to talk about, digital amplifiers or the input stage to a tuner?
I don't blame you for not wanting to read and not wanting to attempt a response. You've got a pretty good sized hole dug with this troll effort of yours, and standing your ground in any one place just won't get it.

The input to a tuner is not digital. It is analog. The signals are analog. The *information* carried is digital, but there are no digital signals until the analog modulation is demodulated.
Your claims about the input signals to a tuner are ridiculous, because you don't recognize where you are working with digital and where you are working with analog.
And to add to that, you don't seem to realize that "digital" is not always "binary". A single pulse is not necessarily either 0 or 1. And a 1 is not necessarily a pulse...

I said you were wrong, and did not agree with you at all. You said RCA cables are not coaxial, and that insertion loss is not the total loss. Are you now retracting what you said?

Where you said the opposite? You were wrong.
--
Floyd L. Davidson <http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska) snipped-for-privacy@apaflo.com
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I'm claiming *exactly* the opposite. You are the one who says it is in some way sensitive to rise time.
But it is also clear that you have no perception of what that means, and cannot support your claim with anything that makes an ounce of sense.
The *facts* are that *nothing* in a T1 carrier system is sensitive to rise time relative to encoding or decoding digital information.
In fact the only part that is sensitive to rise time is the *analog* phase locked loop used for clock recovery!
It is a very nice demonstration of how incorrect your statements about rise time are and digital signals are.

No response? You don't seem to understand much about this topic, as you cannot explain either your own statements or the most obvious rebutals.

Still no response. Not just to that, but to *anything*!
--
Floyd L. Davidson <http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska) snipped-for-privacy@apaflo.com
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...
All of the above is fine.

That statement is absurd. Learn how to articulate, because the words you used do *not* mean what you were should have been saying.

What you meant was that a sine wave can be analyzed as if it were an equivalent series of pulses with linear rise times.

Hee hee, that *is* funny. It doesn't "create an impedance" during the first 10% or during the last 10%?
So you are blowing blue smoke, trying to obscure some fairly simple principles. Input circuits are frequency sensitive, which manifests itself as an input impedance that varies with frequency, and to whatever degree that occurs and is significant, the amplifier will exhibit amplitude distortion.

Really? Who'd a thunk, eh?
Of course in practice, if broadband characteristics are desired, the circuit is designed to make a short enough rise time to cause that effect shorter than will in fact exist during operation. Generally speaking, with modern technologies that is fairly easy to do too!

That statement is meaningless. In particular it is invalid in an argument claiming that rise time has some significance in a digital circuit in regard to decoding the symbol values in a digital signal.
The fastest possible rise time might well *limit* the symbols rate that a particular channel can use, but it is *not* necessarily used to decode symbol values. Which is to say, if the transient response is at least sufficient to allow a rise time higher than the requirements for any specified symbols rate, it makes *no* difference how much faster than that the actual rise time is, or how much slower the actual symbols rate is.
For any given symbol to symbol change in values, the fact that rise time might be virtually non-existent (they may have the same pulse amplitudes) means that lower rise times equally do not affect decoding values of the symbols.
As I've said, rise time is not a factor in decoding symbol values for a T1 system, nor is it necessarily the case for other digital systems.

Insignificant. A T1 system has a broadband input, it does *not* have an impedance mismatch sufficient to cause amplitude distortion and *any* of the requisite frequencies (DC to 775KHz).

Unless of course the input is broadband, compared to the signal frequency spectrum.

And if the slope does *not* "create" such a mismatched impedance, then everything you are saying can be ignored!
All you have demonstrated is that a 1.544 Mbps T1 circuit might not work if clocked at 10 Mbps. Whoop dee doo!

That range might be from DC to 10 times the fastest rise time necessary. And in fact, with a T1 system that is true. It has essentially a broadband input. The rise time is *not* affected by the amplifier input circuitry *at all*, if for no other reason than the fact that all practical transmission lines (i.e., cable pairs) have a dramatically greater affect on the slope of the signal.

Still no ability to respond to the obvious.

No response...
--
Floyd L. Davidson <http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska) snipped-for-privacy@apaflo.com
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It is total nonsense. A sine wave is *not* "identical to the rise time of a pulse" if the pulse is a '("square wave")'.

You therefore meant nothing, because what you said does not make sense.

The waveform does not create an impedance. The amplifier does not see impedance. You are not making sense. (The amplifier is what creates an impedance, and the waveform is what will see it. Not the twist...)

The above addresses "ONLY the rise time". If the input circuit is designed to be sufficiently broadband, the rise time of the signals seen in actual operation will be longer than the minimum rise time acceptable to the circuit.
Again, you are not making sense.

I'm obviously far more versed in digital amplifiers than you, and realize that there are other forms of "digital" signals than just binary pulse trains. Regardless, the input to the tuner that was being discussed when you originally made that sort of statement, is *analog*.

You do realize that the pulse, in the systems that you are discussing, *is* nothing but a "symbol", and that the pulse rate is the symbols rate.
The difference is that your terminology is very restrictive, while what I'm saying is very generic. You are limiting what you say to binary digital systems, and I'm not.

Rise time is not necessarily related to decoding/encoding digital signals. And worse yet you are again talking strictly about binary systems with pulses of voltage or current. Such systems are very common, but there are many others that are different.
And, as I noted, sometimes there simply is *no* rise time between two adjacent symbols, even in a binary encoded digital system. Consider an RS-232 circuit, with the bit pattern 101. Clearly the second bit is encoded in a pulse that has some significant rise time relative to the value of the preceding 0 bit. Now think of the bit pattern 1011111. There are 3 bits there which have a pulse that exhibits *no* rise time at all.
1 0 1 1 1 1 1
+ +-----+ | | | 0 == | === | === | =========================== | =====> | | | | - +-----+ +-----+-----+-----+-----+-----+
The same is true of a pattern like 1010000, where again there is no rise time at all on the pulses for the last 3 bits.
1 0 1 0 0 0 0
+ +-----+ +-----+-----+-----+-----+ | | | | | 0 == | === | === | === | ===================== | =====> | | | | - +-----+ +-----+
Of course, if we are talking about a T1... The last 4 bits in that sequence result not only in no rise time, but not even a pulse!
1 0 1 0 0 0 0
+ x---x | | 0 == +-x x-+---+-x x-+-x---x-+-x---x-+-x---x-+-x---x-+ ====> | | - x---x
How can you say that rise time is the determining factor in decoding a sequence digital symbols?

Clue: rise time of the individual pulses is *not* significant.
Take a look at a diagram I drew a few years ago:
http://www.apaflo.com/floyd_davidson/t1pulse.jpg
That shows the specifications for the waveform of a T1 pulse as seen at the receiver input. Note that rise time is not specified. Pulse width, overshoot, amplitude and wavelength are all specified... because *that* is what determines the value of the pulse.
The only frequences that make any difference at all are the above mentioned range from DC to 775KHz. Give it some thought for a moment, and try to see how a 1.544 Mbps signal can be decoded using *no* frequency higher than 775KHz. Obviously rise time for pulses *cannot* be significant!
If what you are claiming were correct, a T1 circuit's bandwidth would necessarily have to be specified at some frequency greater than 775KHz, which is indeed the pulse rate frequency. In fact, it would have to be at least 3 times that frequency in order to retain sufficient information in the pulse to indicate rise time.
But... it is *not* required. Hmmm...

What, you think "ONE" pulse requires less bandwidth than a series of pulses???? You do realize that in order to have your "square wave" pulse, the bandwidth has to be *much* higher than that required to simply pass a series of pulses...
Ahem, a square wave results from odd *harmonics* of the basic pulse rate frequency...

The rise time is *directly* related to the pulse rate.

You are so confused that you have yet to understand what we are even talking about. You do realize that in a T1 half of the information bits are sent without sending any pulse at all??? No pulse, no rise time. How can you claim rise time is used to distinguish the value of a symbol???

I doubt you understood the signficance of that statement. The input to the T1 equipment is broadband, and can accept a very high frequency (compared to either the pulse rate or the rise time of the fastest pulse that will be seen). And that makes little difference simply because the cable connecting a T1 receiver to the output from a T1 transmitter is in fact a low pass filter which removes virtually all high frequency components. 775KHz is the highest frequency component required to allow a T1 to function. Lines are specified by the amount of loss at 775KHz, and no higher frequency is ever measured.
-- Floyd L. Davidson <http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) snipped-for-privacy@apaflo.com
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Not too bad, but you failed to actually *read* it. It does *not* say the trans-hybrid loss will be 3 dB. What it actually says is that it will be "less than a 3dB loss". As I noted, it is usually about 0.5 dB. That is a *long* ways from 3 dB.
Typical insertion loss for a hybrid is between 3.5 and 4.5 dB.

Me thinks you are blowing blue smoke. What you describe might work for certain applications, but it *won't* work in the one under discussion.
And I would be *very* surprised if it can be made to work with "multimegawatt transmitters".
It appears that you don't understand what ground loops are or how they work, not to mention having not a clue what a transmission line does or does not do.

That is a really dumb idea. I realize that a lot of people use that for audio, where the "transmission line" is very very very short compared to the wavelength of the signal. It isn't the best way to do it there either, but it actually can work. But doing that at RF is absurd.

Now I'm *positive* that you are blowing blue smoke. That's meaningless verbiage.

Nothing.
--
Floyd L. Davidson <http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska) snipped-for-privacy@apaflo.com
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us:

Nearly ALL audio RCA type patch cables are a pair of coaxial runs.
I cannot think off any that are not. Can you? Give the model and brand name here:
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Gave

vs
far
I can think of no RCA patch cables that are coaxial conductors.
coaxial cables are concentric conductors (co-axial conductors , hence the name) that use fields to transfer energy near the conductors. Both the outer and inner axial metal parts of a coaxial cable are ("same-direction") conductors.
RCA audio cables are shielded conductors that use the center conductor to carry energy. The outer metal part is not intended to carry the signal, just to shield the conductor and sometimes provide a ground path.
( Some people undoubtedly sell coax for audio hookup and some fools buy it, but some people also buy "super-pure" copper to carry a signal to a speaker that has more inherent distortion than a thousand miles of romex.)
Yes, coax will conduct audio signal in its center conductor if you want to waste your money, but an RCA audio cable is not coax and doesn't work like it.
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us:

Even the tiny feeds up to my headphones are coax.
Co... axial. That's all that is required. It is a physical structure, and YES, the audio uses the shield of the coax as a return.
It is typically referred to as a completed circuit.
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Gave

to
like
A completed circuit on a pair of wires does not make a coaxial cable - co-axial conductors of unidirectional transmission makes a coaxial cable.
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Wrong.
"Unidirectional transmission" has nothing to do with defining a coaxial cable. A coaxial cable is a *physical* description, it has *nothing* to do with what kind of signal it might have on it.
A coaxial cable that has a DC voltage on it is still just a coaxial cable. Cut it out of that circuit and splice it into a circuit with bidirectional audio signals, and it is *still* just a coaxial cable. Plug it into a circuit with microwave RF on it... and it is *still* just a coaxial cable.
Ground the inner conductor and the outer conductor at to the same point, and it will still be nothing more or less than a coaxial cable.
--
Floyd L. Davidson <http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska) snipped-for-privacy@apaflo.com
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Coaxial cable (coax):
A cable consisting of a center conductor surrounded by an insulating material and a concentric outer conductor. (188) Note: Coaxial cable is used primarily for wideband, video, or rf applications.
From Federal Standard 1037C, Telecommunications: Glossary of Telecommunication Terms. Note that the definition lacks any reference to the circuit a cable is used in, or any parameters of that use. Coax is defined by the physical construction of the cable, and *nothing* else.
As I noted correctly, the inner conductor and the outer conductors at both ends of a length of coax can all be grounded at a single point, it is *still* defined as a coaxial cable.
Your suggestions otherwise, and particularly concerning the requirement that it be used as a transmission line to be properly called "coax", is hilarious, but very indicative of virtually everything you've been saying in this thread: abject stupidity.
--
Floyd L. Davidson <http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska) snipped-for-privacy@apaflo.com
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