John S wrote:
On 7/9/2015 1:01 PM, wrote:
Jeff wrote:
you may get a 50 ohm match at that point.

https://en.wikipedia.org/wiki/Standi...dance_matching

"if there is a perfect match between the load impedance Zload and the
source impedance Zsource=Z*load, that perfect match will remain if the
source and load are connected through a transmission line with an
electrical length of one half wavelength (or a multiple of one half
wavelengths) using a transmission line of any characteristic impedance Z0."

This wiki article has a lot of good info in it. I have seen a lot of
stuff posted here that this article directly contradicts.... I wonder
who is right?


That is a very specific case where the source is not at the system
impedance and happens to be equal to the load impedance, there will also
be standing waves on the transmission line and associated losses as the
VSWR on the line will be equal to the magnitude of the mismatch between
the transmission line impedance and the load impedance.

Jeff

Most people take the source impdedance to be the system impedance, i.e.
the impedance for which everything else is designed for.

Most *engineers* take the source impedance to be the impedance of the
*generator*.

Which, as in most cases is purchased and has a fixed impedance of 50
Ohms, and thus defines the system impedance.

In fact, perhaps the rest of us should call it the generator rather than
the source so that we can communicate with you on your level.

Babbling horse****.


--
Jim Pennino

John S wrote:
On 7/9/2015 12:58 PM, wrote:
rickman wrote:
On 7/9/2015 9:14 AM, Ralph Mowery wrote:
"Jeff" wrote in message
...

The SWR has to be the same at any point on the coax or transmission line
minus the loss in the line. A simple swr meter may show some differance
because of the way that kind of meter works. By changing the length of
the
line , the apparent SWR may be differant at that point.

There is no such thing as apparent SWR. It is what it is in a given
place.


By 'apparent SWR' he means as indicated SWR on the meter, and yes it can
change at various point on the line due to inadequacies in the meter; the
'real' VSWR will of course remain the same at any point on a lossless
line.

Jeff

That is what I mean Jeff. If there is any SWR, by changing the length of
the line, the voltage/current changes in such a maner that at certain points
you may get a 50 ohm match at that point.

https://en.wikipedia.org/wiki/Standi...dance_matching

"if there is a perfect match between the load impedance Zload and the
source impedance Zsource=Z*load, that perfect match will remain if the
source and load are connected through a transmission line with an
electrical length of one half wavelength (or a multiple of one half
wavelengths) using a transmission line of any characteristic impedance Z0."

This wiki article has a lot of good info in it. I have seen a lot of
stuff posted here that this article directly contradicts.... I wonder
who is right?

It has been my observation that when the subject matter is long established
science, such as transmission line theory, wiki is normally correct.

Wiki is subject to the same errors you make because the information is
usually supplied by people like you.

You mean people that know what they are doing as opposed to people like
you that just pull crap out of their ass based on old wive's tales?



--
Jim Pennino

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?

wrote:

John S wrote:
On 7/9/2015 1:01 PM, wrote:
Jeff wrote:
you may get a 50 ohm match at that point.

https://en.wikipedia.org/wiki/Standi...dance_matching

"if there is a perfect match between the load impedance Zload and the
source impedance Zsource=Z*load, that perfect match will remain if
the source and load are connected through a transmission line with an
electrical length of one half wavelength (or a multiple of one half
wavelengths) using a transmission line of any characteristic
impedance Z0."

This wiki article has a lot of good info in it. I have seen a lot of
stuff posted here that this article directly contradicts.... I wonder
who is right?


That is a very specific case where the source is not at the system
impedance and happens to be equal to the load impedance, there will also
be standing waves on the transmission line and associated losses as the
VSWR on the line will be equal to the magnitude of the mismatch between
the transmission line impedance and the load impedance.

Jeff

Most people take the source impdedance to be the system impedance, i.e.
the impedance for which everything else is designed for.

Most *engineers* take the source impedance to be the impedance of the
*generator*.

Which, as in most cases is purchased and has a fixed impedance of 50
Ohms, and thus defines the system impedance.

Please find one specification for a transmitter that says it *has* an
output impedance of 50 ohms. You will find plenty that say they are
designed to drive a load impedance of 50 ohms, but few that state their
output impedance. Of those I have seen that do, none are amateur
transmitters and the impedance they mention is much lower than 50 ohms.



In fact, perhaps the rest of us should call it the generator rather than
the source so that we can communicate with you on your level.

Babbling horse****.



--
Roger Hayter

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.

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

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

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

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.

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.

--
Jim Pennino

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.



--
Jim Pennino

Jeff wrote:

So who exactly declared which set of definitions is the one and
true definition of VSWR?

Is P=EI or P=E^2R?

It was defined when the quantity was invented and is obvious from the
name, ie the Ratio of the Voltage of the Standing Wave. Vmax/Vmin of the
standing wave.

There was nothing invented; there was something observed.

A name may or may not be meaningfull.

While the name does discribe what happens on a transmission line,
standing waves are a consequence of a SWR greater than 1:1 on a
transmission line.

SWR exists no matter what the physical impedances are and do NOT have
to be transmission lines. If the impedances are not transmission lines,
there are no standing waves as there is no place for them to exist.


--
Jim Pennino

Jeff wrote:

Nowhere is it written in stone that the Vmax/Vmin is the one, true,
only and holy definition of SWR.


Wave transmission by Connor:

"The standing wave ratio s is defined as the ratio of |Vmax| to |Vmin|
....The standing wave s is thus directly related to |p|...."
|p| being the reflection coefficient.

Antenna Engineering Handbook by Jasik:

"Standing Wave ratio(SWR) is expressed as decibels or as a voltage ratio
(VSWR). This is expressed as follows:

SWR(db)= 20 log VSWR = 20 log Vmax/Vmin."

Hence it follows that VSWR = Vmax/Vmin.

plus many other references to VSWR=Vmax/Vmin and no mention of any other
definition.

Two standard works, and I am sure that I could dig out many more if I
could be bothered.

Jeff

And I can find just as many that define SWR in terms of reflection coefficient
and impedance.

All the definitions are equally valid.

All the definitions are equally valid.

All the definitions are equally valid.

All the definitions are equally valid.

All the definitions are equally valid.

All the definitions are equally valid.

All the definitions are equally valid.

All the definitions are equally valid.

All the definitions are equally valid.

All the definitions are equally valid.

If you believe that SWR is only valid and relevant to transmission lines,
stick a 0.5 Ohm resistor into the output connector of your transmitter.



--
Jim Pennino

Jeff wrote:

All the definitions are equally valid.

You seem totally incapable of differentiating between a definition of a
quantity and a formula that links that definition to another quantity.

Further discussion seems pointless.

You seem totally incapable of understanding that ALL the equations are
equally valid, that NONE of them is the formal definition, and that it
is irrelevant WHAT the formal definition is as long as the equations
are valid.



--
Jim Pennino

Roger Hayter wrote:
wrote:

John S wrote:
On 7/9/2015 1:01 PM, wrote:
Jeff wrote:
you may get a 50 ohm match at that point.

https://en.wikipedia.org/wiki/Standi...dance_matching

"if there is a perfect match between the load impedance Zload and the
source impedance Zsource=Z*load, that perfect match will remain if
the source and load are connected through a transmission line with an
electrical length of one half wavelength (or a multiple of one half
wavelengths) using a transmission line of any characteristic
impedance Z0."

This wiki article has a lot of good info in it. I have seen a lot of
stuff posted here that this article directly contradicts.... I wonder
who is right?


That is a very specific case where the source is not at the system
impedance and happens to be equal to the load impedance, there will also
be standing waves on the transmission line and associated losses as the
VSWR on the line will be equal to the magnitude of the mismatch between
the transmission line impedance and the load impedance.

Jeff

Most people take the source impdedance to be the system impedance, i.e.
the impedance for which everything else is designed for.

Most *engineers* take the source impedance to be the impedance of the
*generator*.

Which, as in most cases is purchased and has a fixed impedance of 50
Ohms, and thus defines the system impedance.

Please find one specification for a transmitter that says it *has* an
output impedance of 50 ohms. You will find plenty that say they are
designed to drive a load impedance of 50 ohms, but few that state their
output impedance. Of those I have seen that do, none are amateur
transmitters and the impedance they mention is much lower than 50 ohms.

Notice I did not use the word "transmitter" in my post.

I was speaking from an engineering point of view, not from an Amateur
radio operator point of view.

Notice that the post I was responding to used the words "engineer", not
"Amateur" and "generator" not "transmitter".



--
Jim Pennino