rickman wrote:
On 7/9/2015 1:27 PM, wrote:
rickman wrote:
On 7/8/2015 9:07 PM, 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.

Are just being argumentative or are you really that ignorant?

Even if you have a theoretical voltage source, there are no circuits
(other than imaginary) that have been on since before the big bang and
will be on for all time in the future.

So what?

Is there some point to all this other than to be argumentative?

How long before someone brings up the fact that a resistor generates
AC signals as some kind of straw man objection to DC theory?

The point is that separating DC and AC as being ruled by separate "laws"
is pointless. Just discuss the topic of interest rather than digressing
onto pointless diversions.

Actually separating DC and AC is extremely practical which is why it
is done in the real world.


--
Jim Pennino

rickman wrote:
On 7/9/2015 1:44 PM, wrote:
Wayne wrote:


wrote in message ...

John S wrote:
On 7/8/2015 7:27 PM, Wayne wrote:


"John S" wrote in message ...

On 7/7/2015 1:44 PM, wrote:
Ian Jackson wrote:
In message , Jerry Stuckle
writes


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.

If there's no reflection, there can be no standing wave. So, being
pedantic, there's no such thing as an SWR of 1:1!

Despite the name, VSWR is defined in terms of complex impedances
and wavelengths, not "waves" of any kind.



Actually, VSWR is defined as the ratio of Vmax/Vmin.

That's also my understanding of the definition.
In fact since SWR is defined as the maximum to minimum voltage ratio,
the "V" in VSWR is redundant.

Sort of. There is also ISWR but it is not used frequently.

# Not sort of, but is.

# There is also PSWR.

And both go back to the Vmax/Vmin definition.

The PSWR is a tricky one because you can end up with a power ratio instead
of a voltage ratio.

Actually, no, PSWR has nothing to do with power ratios as in RF power,
rather it has to do with power ratios as in values raised to the second
power.

Lol!

"The term power standing wave ratio (PSWR) is sometimes referred to, and
defined as the square of the voltage standing wave ratio."

"In other words, there are no actual powers being compared."

https://en.wikipedia.org/wiki/Standing_wave_ratio



--
Jim Pennino

On 7/9/2015 4:29 AM, Roger Hayter wrote:
Jeff wrote:

On 08/07/2015 19:14, wrote:
John S wrote:
On 7/7/2015 1:44 PM, wrote:
Ian Jackson wrote:
In message , Jerry Stuckle
writes


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.

If there's no reflection, there can be no standing wave. So, being
pedantic, there's no such thing as an SWR of 1:1!

Despite the name, VSWR is defined in terms of complex impedances
and wavelengths, not "waves" of any kind.



Actually, VSWR is defined as the ratio of Vmax/Vmin.

Actually, VSWR can be defined several ways, one of which is:

(1 + |r|)/(1 - |r|)

Where r is the reflection coefficient which can be defined a:

(Zl - Zo)/(Zl + Zo)

Where Zl is the complex load impedance and Zo is the complex source
impedance.

Note that a complex impedance has a frequency dependant part.



Note the the definition of VSWR uses the magnitude of the reflection
coefficient, |r|, which removes the phase and frequency dependant parts.

Jeff

The magnitude remains frequency dependent.


VSWR is the RATIO of the magnitudes Vmax/Vmin regardless of frequency.

On 7/9/2015 12:40 PM, wrote:
Jeff wrote:
On 08/07/2015 19:14, wrote:
John S wrote:
On 7/7/2015 1:44 PM, wrote:
Ian Jackson wrote:
In message , Jerry Stuckle
writes


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.

If there's no reflection, there can be no standing wave. So, being
pedantic, there's no such thing as an SWR of 1:1!

Despite the name, VSWR is defined in terms of complex impedances
and wavelengths, not "waves" of any kind.



Actually, VSWR is defined as the ratio of Vmax/Vmin.

Actually, VSWR can be defined several ways, one of which is:

(1 + |r|)/(1 - |r|)

Where r is the reflection coefficient which can be defined a:

(Zl - Zo)/(Zl + Zo)

Where Zl is the complex load impedance and Zo is the complex source
impedance.

Note that a complex impedance has a frequency dependant part.



Note the the definition of VSWR uses the magnitude of the reflection
coefficient, |r|, which removes the phase and frequency dependant parts.

Jeff

The magnitude DEPENDS on the frequency dependant parts.


But the ratio of Vmax/Vmin does not.

rickman wrote:
On 7/9/2015 1:58 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.

Why do you ignore it when it says Zo is the impedance of the
transmission line and not the source?

I don't; The transmission line in this case IS the source.

The SWR in a system, any kind of system, is measured at a point in a system.

One side of that point is the source and the other side of that point
is the load.

It also does not matter which side you declare the source and which side
you declare the load.

A 50 Ohm source and a 100 Ohm load has the same SWR as a 100 Ohm source
and a 50 Ohm load.

By convention the load side is normally taken as the side which, when
the system is powered, the power is desired to be dissipated.


--
Jim Pennino

On 7/9/2015 1:16 PM, wrote:
rickman wrote:
On 7/9/2015 1:27 PM, wrote:
rickman wrote:
On 7/8/2015 9:07 PM, 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.

Are just being argumentative or are you really that ignorant?

Even if you have a theoretical voltage source, there are no circuits
(other than imaginary) that have been on since before the big bang and
will be on for all time in the future.

So what?

Is there some point to all this other than to be argumentative?

How long before someone brings up the fact that a resistor generates
AC signals as some kind of straw man objection to DC theory?

The point is that separating DC and AC as being ruled by separate "laws"
is pointless. Just discuss the topic of interest rather than digressing
onto pointless diversions.

Actually separating DC and AC is extremely practical which is why it
is done in the real world.


Sure, but separating the real and imaginary parts of impedance is
SOMETIMES practical which is why it is done in the real world.

On 7/9/2015 3:03 PM, John S wrote:
On 7/9/2015 12:40 PM, wrote:
Jeff wrote:
On 08/07/2015 19:14, wrote:
John S wrote:
On 7/7/2015 1:44 PM, wrote:
Ian Jackson wrote:
In message , Jerry Stuckle
writes


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.

If there's no reflection, there can be no standing wave. So, being
pedantic, there's no such thing as an SWR of 1:1!

Despite the name, VSWR is defined in terms of complex impedances
and wavelengths, not "waves" of any kind.



Actually, VSWR is defined as the ratio of Vmax/Vmin.

Actually, VSWR can be defined several ways, one of which is:

(1 + |r|)/(1 - |r|)

Where r is the reflection coefficient which can be defined a:

(Zl - Zo)/(Zl + Zo)

Where Zl is the complex load impedance and Zo is the complex source
impedance.

Note that a complex impedance has a frequency dependant part.



Note the the definition of VSWR uses the magnitude of the reflection
coefficient, |r|, which removes the phase and frequency dependant parts.

Jeff

The magnitude DEPENDS on the frequency dependant parts.


But the ratio of Vmax/Vmin does not.

Huh? I'm pretty sure the VSWR is a function of frequency. A great deal
about the impedances *and* the voltages change with frequency.
--

Rick

John S wrote:
On 7/9/2015 1:16 PM, wrote:
rickman wrote:
On 7/9/2015 1:27 PM, wrote:
rickman wrote:
On 7/8/2015 9:07 PM, 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.

Are just being argumentative or are you really that ignorant?

Even if you have a theoretical voltage source, there are no circuits
(other than imaginary) that have been on since before the big bang and
will be on for all time in the future.

So what?

Is there some point to all this other than to be argumentative?

How long before someone brings up the fact that a resistor generates
AC signals as some kind of straw man objection to DC theory?

The point is that separating DC and AC as being ruled by separate "laws"
is pointless. Just discuss the topic of interest rather than digressing
onto pointless diversions.

Actually separating DC and AC is extremely practical which is why it
is done in the real world.


Sure, but separating the real and imaginary parts of impedance is
SOMETIMES practical which is why it is done in the real world.

On rare occasions, but in the real world the DC analysis and the
AC analysis are ALWAYS done seperately.

A real world example: in the analysis of a circuit containing
active components, first a DC analysis is done to establish the
operating point of the circuit, then the AC analysis is done.


--
Jim Pennino

John S wrote:
On 7/9/2015 4:29 AM, Roger Hayter wrote:
Jeff wrote:

On 08/07/2015 19:14, wrote:
John S wrote:
On 7/7/2015 1:44 PM, wrote:
Ian Jackson wrote:
In message , Jerry Stuckle
writes


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.

If there's no reflection, there can be no standing wave. So, being
pedantic, there's no such thing as an SWR of 1:1!

Despite the name, VSWR is defined in terms of complex impedances
and wavelengths, not "waves" of any kind.



Actually, VSWR is defined as the ratio of Vmax/Vmin.

Actually, VSWR can be defined several ways, one of which is:

(1 + |r|)/(1 - |r|)

Where r is the reflection coefficient which can be defined a:

(Zl - Zo)/(Zl + Zo)

Where Zl is the complex load impedance and Zo is the complex source
impedance.

Note that a complex impedance has a frequency dependant part.



Note the the definition of VSWR uses the magnitude of the reflection
coefficient, |r|, which removes the phase and frequency dependant parts.

Jeff

The magnitude remains frequency dependent.


VSWR is the RATIO of the magnitudes Vmax/Vmin regardless of frequency.

See above.

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



--
Jim Pennino

On 7/9/2015 3:05 PM, wrote:
rickman wrote:
On 7/9/2015 1:58 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.

Why do you ignore it when it says Zo is the impedance of the
transmission line and not the source?

I don't; The transmission line in this case IS the source.

No, the source is the source. Even if you wish to consider transmission
line as the source in some example, the page clearly says Zo is the
impedance of the transmission line, without any context where you can
say it is a source or a load.

https://en.wikipedia.org/wiki/Charac...ion_line_model


The SWR in a system, any kind of system, is measured at a point in a system.

You mean a system that includes a source, a transmission line and a load?


One side of that point is the source and the other side of that point
is the load.

It also does not matter which side you declare the source and which side
you declare the load.

In that case the antenna is Zo? I don't think so.


A 50 Ohm source and a 100 Ohm load has the same SWR as a 100 Ohm source
and a 50 Ohm load.

By convention the load side is normally taken as the side which, when
the system is powered, the power is desired to be dissipated.

I don't know why you dig your heels in on every little point. Sometimes
you are just wrong and need to acknowledge that so the discussion can
move on. Zo (or Z0) is used to represent the characteristic impedance
of a transmission line. Zs (or Zsource) is used for the source. ZL (or
Zload) is used for the load. Read the wiki quote above and the many
other *clear* examples.

--

Rick