On 7/10/2015 1:39 PM, wrote:

standing waves are
a consequence of a VSWR greater than 1:1 on a transmission line.

Did you really write that? The standing waves are a consequence of a
standing wave ratio of greater than 1:1?

Ok, I've got my horse, now I just need a cart to pull it with.

--

Rick

Jeff wrote:

That is correct, but not the situation that we are discussing, we are
talking about matching a load to a 50 ohm transmission line. In that
case changing the length of line will NEVER result in a match. Using a
*different impedance* length of coax as a transmission line transformer
is a totally different case, and as you say will result is a standing
wave on the line and associated losses.

Jeff

So you are only interested in special cases?


No. I am commenting on the original proposal that changing the length of
a line of the same impedance as the system impedance can result in a
good match when one did not exist to start with.

Jeff

OK, but that has been settled long ago.

BTW, for a real world transmission line transformer, the transformer
section where the impedance mismatch occurs will be of a very short
length compared to the total transmission line and thus will have low
loss.


--
Jim Pennino

Jeff wrote:

How about a section of transmission line with one impedance of some
length attached to a section of transmission line with a different
impedance of random length?


That is not the same situation as changing the length of a transmission
line of the same impedance which is what we are discussing.

Jeff

OK; I thought that was totally settled.

--
Jim Pennino

On 7/10/2015 12:29 PM, wrote:
John S wrote:
On 7/10/2015 12:29 AM, wrote:
John S wrote:
On 7/9/2015 12:32 AM, wrote:
John S wrote:
On 7/8/2015 4:48 PM, wrote:
John S wrote:
On 7/8/2015 12:47 PM, wrote:
John S wrote:

So, at 1Hz the law has changed, eh? What new law do I need to use?

To be pendatic, there is only one set of physical laws that govern
electromagnetics.

However for DC all the complex parts of those laws have no effect and
all the equations can be simplified to remove the complex parts.

In the real, practical world people look upon this as two sets of
laws, one for AC and one for DC.

A good example of this is the transmission line which does not exist
at DC; at DC a transmission line is nothing more than two wires with
some resistance that is totally and only due to the ohmic resistance
of the material that makes up the wires.

So, is .01Hz AC or DC, Jim? How about 1Hz? 10Hz? Where does AC begin and
DC end?

It is called a limit.

If there is NO time varying component, it is DC, otherwise it is AC.

Are you playing devil's advocate or are you really that ignorant?

Then there is no such thing as DC because even a battery looses voltage
over a period of time. DC voltage sources have noise.

An ideal battery doesn't.

Where can one be purchased?

At the ideal battery store.

C'mon, jimp, what concession from me will it take to get us back on
track so we can discuss this topic in an adult and gentlemanly manner?

When one analyzes circuits, it is done with ideal components.


Yes, until you throw in the requirement that one must go get an
off-the-shelf ham transmitter for measurements.

If the real world properties are important, they are in turn modeled
with additional ideal components.

Yes.

For example, an ideal voltage source has constant voltage and zero
source resistance forever.

Yes.

If the source resistance is important to the circuit, then it is modeled
by putting an ideal resistor in series with the voltage source.

Yes.

Your statement:

"Then there is no such thing as DC because even a battery looses voltage
over a period of time. DC voltage sources have noise."

is either just a childish strawman or you have no real clue how circuits,
including electromagnetic circuits, are modeled.

This was in response to your asinine statement "At the ideal battery
store." And, the whole thing started when I asked you to define the
difference between AC and DC. There is no breakpoint, jimp, it is
dependent on which tool we use to analyze the problem at hand.

The only time where the fact that a real battery discharges would be of
any significance is if you were analyzing a circuit for it's performance
over a voltage range, in which casee one would step the voltage source.


Okay, jimp. You did not answer my direct question, so I must assume you
have no further interest in continuing this childish argument. I am
tired of your nit-picky responses that we can all see through. They are
designed exclusively to satisfy your ego and we don't need that in spite
of the fact that you can be a valuable source of information. Yeah, I
needled you right back because you refused to continue in a reasonable
manner.

From now on, I will not respond to you. I'm not saying that I will
killfile you. I will just not respond to any of your childish pranks. I
really don't have time to argue with your ego. That's too bad.

John S wrote:
On 7/10/2015 12:29 PM, wrote:
John S wrote:
On 7/10/2015 12:29 AM, wrote:
John S wrote:
On 7/9/2015 12:32 AM, wrote:
John S wrote:
On 7/8/2015 4:48 PM, wrote:
John S wrote:
On 7/8/2015 12:47 PM, wrote:
John S wrote:

So, at 1Hz the law has changed, eh? What new law do I need to use?

To be pendatic, there is only one set of physical laws that govern
electromagnetics.

However for DC all the complex parts of those laws have no effect and
all the equations can be simplified to remove the complex parts.

In the real, practical world people look upon this as two sets of
laws, one for AC and one for DC.

A good example of this is the transmission line which does not exist
at DC; at DC a transmission line is nothing more than two wires with
some resistance that is totally and only due to the ohmic resistance
of the material that makes up the wires.

So, is .01Hz AC or DC, Jim? How about 1Hz? 10Hz? Where does AC begin and
DC end?

It is called a limit.

If there is NO time varying component, it is DC, otherwise it is AC.

Are you playing devil's advocate or are you really that ignorant?

Then there is no such thing as DC because even a battery looses voltage
over a period of time. DC voltage sources have noise.

An ideal battery doesn't.

Where can one be purchased?

At the ideal battery store.

C'mon, jimp, what concession from me will it take to get us back on
track so we can discuss this topic in an adult and gentlemanly manner?

When one analyzes circuits, it is done with ideal components.


Yes, until you throw in the requirement that one must go get an
off-the-shelf ham transmitter for measurements.

Babbling nonsense.

Amateur radio equipment is designed by the same engineering tools that
design everything else.

If the real world properties are important, they are in turn modeled
with additional ideal components.

Yes.

For example, an ideal voltage source has constant voltage and zero
source resistance forever.

Yes.

If the source resistance is important to the circuit, then it is modeled
by putting an ideal resistor in series with the voltage source.

Yes.

Your statement:

"Then there is no such thing as DC because even a battery looses voltage
over a period of time. DC voltage sources have noise."

is either just a childish strawman or you have no real clue how circuits,
including electromagnetic circuits, are modeled.

This was in response to your asinine statement "At the ideal battery
store." And, the whole thing started when I asked you to define the
difference between AC and DC. There is no breakpoint, jimp, it is
dependent on which tool we use to analyze the problem at hand.

Nope; In mathematics it is called a lower limit.

My "asinine statement" was appropriate for your asinine question.

The tools used at DC may have been derived from that same base principals
as used at AC, but they are much simplified and many things that exist
at AC simply do not exist at DC, e.g. capacitors, inductors, transmission
lines, propagating fields, etc.


The only time where the fact that a real battery discharges would be of
any significance is if you were analyzing a circuit for it's performance
over a voltage range, in which casee one would step the voltage source.


Okay, jimp. You did not answer my direct question, so I must assume you
have no further interest in continuing this childish argument. I am
tired of your nit-picky responses that we can all see through. They are
designed exclusively to satisfy your ego and we don't need that in spite
of the fact that you can be a valuable source of information. Yeah, I
needled you right back because you refused to continue in a reasonable
manner.

I am responding as an engineer would respond.

An engineer doesn't arm wave crap about battery discharge disproving
DC theory.


--
Jim Pennino

rickman wrote:
On 7/10/2015 1:39 PM, wrote:

standing waves are
a consequence of a VSWR greater than 1:1 on a transmission line.

Did you really write that? The standing waves are a consequence of a
standing wave ratio of greater than 1:1?

As VSWR is a consequence of an impedance mismatch, yes.

You keep ignoring or failing to understand that part.

No matter what you call it, a value of greater than 1:1 indicates an
IMPEDANCE mismatch.

If the equations that describe VSWR were discovered by Dr. Snagpuss and
the effect were named after him, then the statement would be that
a Snagpuss greater than 1:1 causes standing waves on a transmission line.

An IMPEDANCE mismatch on a TRANSMISSION LINE results in standing waves.

It is irrelevant what you call the value, only the value itself is of
any importance.

--
Jim Pennino

In message ,
writes
Jeff wrote:
On 09/07/2015 18:35, wrote:
Jeff wrote:

Can you measure VSWR on a 1 meter long Lecher line at 1 MHz?

VSWR is not meaningful in such a situation, however, you can measure
return loss and Reflection Coefficient etc.. Of course that in not to
say that VSWR is not used in situations where it is not appropriate in
order to indicate how good a match is, when RL or Reflection Coefficient
would be more appropriate.

Jeff

Jeff

Are you trying to say that VSWR is not meaningfull at 160M (to put it
in an Amateur context)?

For those that don't know, a Lecher wire is just a carefully contructed,
rigid parallel transmission line upon which one would slide a high
impedance sensor to find voltage minimum, maximum, and where they
occured. That and a Smith chart were used to solve transmission line
and impedance matching problems and were often home built by Amateurs
in the early VHF days.

Today you would use a VNA (Vector Network Analyzer).

Unless you have a very long feeder at 160m you cannot have a complete
voltage maxima and minima from the standing wave on the line so VSWR is
meaningless. That is not to say that you cannot calculate an 'effective'
VSWR from other quantities such as return loss, S11, by measuring the
forward and reflected signals as you would with a Network Analyser or
SWR bridge.

Jeff

Nope, VSWR is always meaningful and you have the cart before the horse.

VSWR is a consequence of an impedance match and standing waves are
a consequence of a VSWR greater than 1:1 on a transmission line.

Attach a SWR meter directly to the output of YOUR transmitter and a
1 Ohm resistor directly to the other end of the SWR meter.

The meter reading will be the same as the calculated value, there will
be no standing waves as there is no transmission line, but the results
WILL be meaninful to your transmitter.

Even when the only transmission line consists the output connector of
the SWR meter, and maybe an inch of internal coax, there will still BE a
standing wave - but it will only be a tiny portion of longer one.

However, surely an SWR meter is really measuring the ratio of the go and
return signals - ie the RLR? If so, would it not end all this
essentially esoteric argument if we called it an RLR meter?




--
Ian

Ian Jackson wrote:
In message ,
writes
Jeff wrote:
On 09/07/2015 18:35, wrote:
Jeff wrote:

Can you measure VSWR on a 1 meter long Lecher line at 1 MHz?

VSWR is not meaningful in such a situation, however, you can measure
return loss and Reflection Coefficient etc.. Of course that in not to
say that VSWR is not used in situations where it is not appropriate in
order to indicate how good a match is, when RL or Reflection Coefficient
would be more appropriate.

Jeff

Jeff

Are you trying to say that VSWR is not meaningfull at 160M (to put it
in an Amateur context)?

For those that don't know, a Lecher wire is just a carefully contructed,
rigid parallel transmission line upon which one would slide a high
impedance sensor to find voltage minimum, maximum, and where they
occured. That and a Smith chart were used to solve transmission line
and impedance matching problems and were often home built by Amateurs
in the early VHF days.

Today you would use a VNA (Vector Network Analyzer).

Unless you have a very long feeder at 160m you cannot have a complete
voltage maxima and minima from the standing wave on the line so VSWR is
meaningless. That is not to say that you cannot calculate an 'effective'
VSWR from other quantities such as return loss, S11, by measuring the
forward and reflected signals as you would with a Network Analyser or
SWR bridge.

Jeff

Nope, VSWR is always meaningful and you have the cart before the horse.

VSWR is a consequence of an impedance match and standing waves are
a consequence of a VSWR greater than 1:1 on a transmission line.

Attach a SWR meter directly to the output of YOUR transmitter and a
1 Ohm resistor directly to the other end of the SWR meter.

The meter reading will be the same as the calculated value, there will
be no standing waves as there is no transmission line, but the results
WILL be meaninful to your transmitter.

Even when the only transmission line consists the output connector of
the SWR meter, and maybe an inch of internal coax, there will still BE a
standing wave - but it will only be a tiny portion of longer one.

There will NOT be standing waves and there will not be a voltage
maximum and a voltage minimum unless there is a transmission line.

However, that is irrelevant to the point.

From the transmitter point of view, there is no difference between
a directly connected 1 Ohm resistor and a 1 Ohm resistor at the end
of a lossless 1 Ohm transmission line 1,000,000 meters long.

However, surely an SWR meter is really measuring the ratio of the go and
return signals - ie the RLR? If so, would it not end all this
essentially esoteric argument if we called it an RLR meter?

The SWR meter will read the IMPEDANCE MISMATCH.

Here's a hot flash; people in the engineering world have been using
the consept and equations of VSWR for many decades for many things
that do NOT include transmission lines with no problems whatsoever.

It seems only some Amateur radio operators are hung up on terminology
like "standing waves".


--
Jim Pennino

On 7/10/2015 2:34 PM, wrote:
rickman wrote:
On 7/10/2015 1:39 PM, wrote:

standing waves are
a consequence of a VSWR greater than 1:1 on a transmission line.

Did you really write that? The standing waves are a consequence of a
standing wave ratio of greater than 1:1?


An IMPEDANCE mismatch on a TRANSMISSION LINE results in standing waves.

First you say the standing waves are a result of the SWR being greater
than 1:1, now you say it is a result of the impedance mismatch. I'm
*so* confused....

I'm getting the impression you are being water boarded and will say
anything you think will make it end! Give us a location and we will
save you!

--

Rick

In message ,
writes
Ian Jackson wrote:


Even when the only transmission line consists the output connector of
the SWR meter, and maybe an inch of internal coax, there will still BE a
standing wave - but it will only be a tiny portion of longer one.

There will NOT be standing waves and there will not be a voltage
maximum and a voltage minimum unless there is a transmission line.

Are you saying that for a standing wave to qualify as a standing wave,
the transmission line needs to be long enough for there to be a voltage
maximum a voltage minimum?





--
Ian