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.

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.


--
Jim Pennino

On 7/8/2015 1:14 PM, 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.

So, since Vmax/Vmin (the base definition) has no frequency dependent
part, does that invalidate it?

John S wrote:
On 7/8/2015 1:14 PM, 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.

So, since Vmax/Vmin (the base definition) has no frequency dependent
part, does that invalidate it?

The "base definition" can be whatever set of equations you pick that
are true.

BTW, the Vmax/Vmin DOES have a frequency dependant component that
determines WHERE Vmax and Vmin occur.

--
Jim Pennino

On 7/8/2015 4:51 PM, wrote:
John S wrote:
On 7/8/2015 1:14 PM, 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.

So, since Vmax/Vmin (the base definition) has no frequency dependent
part, does that invalidate it?

The "base definition" can be whatever set of equations you pick that
are true.

BTW, the Vmax/Vmin DOES have a frequency dependant component that
determines WHERE Vmax and Vmin occur.


You are just being argumentative. The WHERE doesn't matter in measuring
VSWR. You still measure correct VSWR wherever the locations.

John S wrote:
On 7/8/2015 4:51 PM, wrote:
John S wrote:
On 7/8/2015 1:14 PM, 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.

So, since Vmax/Vmin (the base definition) has no frequency dependent
part, does that invalidate it?

The "base definition" can be whatever set of equations you pick that
are true.

BTW, the Vmax/Vmin DOES have a frequency dependant component that
determines WHERE Vmax and Vmin occur.


You are just being argumentative. The WHERE doesn't matter in measuring
VSWR. You still measure correct VSWR wherever the locations.

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



--
Jim Pennino

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).

--
Jim Pennino

Jeff wrote:
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.

So, since Vmax/Vmin (the base definition) has no frequency dependent
part, does that invalidate it?


No, of course not, the other equations are NOT a definition of VSWR,
they are formulas the link other quantities to VSWR.

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

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

Taking Reflection Coefficient for example, it requires the phase
information to be removed before conversion to VSWR by using only its
magnitude.

So what?

To emphasise this VSRW is phase independent and on a lossless
transmission line; its value does not change anywhere along that line;
that is equivalent to rotating around a constant VSWR circle on a Smith
Chart.

Jeff

--
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:

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.

--
Roger Hayter