On 3/4/2015 7:49 AM, Jeff wrote:

Connector impedance doesn't change with frequency, just as coax
impedance doesn't change with frequency. Loss will increase as
frequency increases, however.

Coax impedance certainly does change with frequency. Below about 500kHz
there is a significant slope with frequency.

At 200kHz a 50ohm cable may well look more like 100ohms and by the time
that you get to 1kHz it could be as high as 1kohm.

200kHz is in what is called the transition region and the impedance is
given by:

SQRT((R+j2pifL)/(C+j2pifC))

as opposed to the high frequency region where it is merely:

sqrt(L/C)

Below about 20kHz it changes again to SQRT(R/(j2pifC)

There are also other variables due to changes in the dielectric with
frequency and other losses.

Can you explain the above equations? In general it doesn't make sense
that the same effect would have different equations for different
frequencies. It does make sense though that the equations involved are
all simplifications of a single, more complex equation, optimized to
discount small effects over a given frequency range.

That said, I'm not sure I can see how these three equations can morph
into each other as f changes. The equation for the middle frequency
range seems to be the more encompassing so starting with that - if
frequency goes up enough the terms j2pifL and j2pifC dominate the R and
C terms and the equation simplifies to sqrt(L/C) appropriately.

But when f goes down enough, both terms shrink compared to R and C and
the equation would seem to simplify to sqrt(R/C) rather than
sqrt(R/j2pifC).

Is there possibly a typo in there somewhere?

--

Rick

On 3/5/2015 4:35 AM, Jeff wrote:
On 05/03/2015 01:50, rickman wrote:
On 3/4/2015 7:49 AM, Jeff wrote:

Connector impedance doesn't change with frequency, just as coax
impedance doesn't change with frequency. Loss will increase as
frequency increases, however.

Coax impedance certainly does change with frequency. Below about 500kHz
there is a significant slope with frequency.

At 200kHz a 50ohm cable may well look more like 100ohms and by the time
that you get to 1kHz it could be as high as 1kohm.

200kHz is in what is called the transition region and the impedance is
given by:

SQRT((R+j2pifL)/(C+j2pifC))

as opposed to the high frequency region where it is merely:

sqrt(L/C)

Below about 20kHz it changes again to SQRT(R/(j2pifC)

There are also other variables due to changes in the dielectric with
frequency and other losses.

Can you explain the above equations? In general it doesn't make sense
that the same effect would have different equations for different
frequencies. It does make sense though that the equations involved are
all simplifications of a single, more complex equation, optimized to
discount small effects over a given frequency range.

That said, I'm not sure I can see how these three equations can morph
into each other as f changes. The equation for the middle frequency
range seems to be the more encompassing so starting with that - if
frequency goes up enough the terms j2pifL and j2pifC dominate the R and
C terms and the equation simplifies to sqrt(L/C) appropriately.

But when f goes down enough, both terms shrink compared to R and C and
the equation would seem to simplify to sqrt(R/C) rather than
sqrt(R/j2pifC).

Is there possibly a typo in there somewhere?


Sorry there is indeed a typo: the general equation should read:

SQRT((R+j2pifL)/(G+j2pifC)) G being conductance.

The reasons that the equations are presented differently in different
frequency ranges are because; at higher frequencies when f becomes
large enough, the terms containing f become so large that R and G can be
neglected; and at low frequencies (2 pi f L) is so small compared with R
that it can be neglected.

Hi - SQRT(L/C)

Mid - SQRT((R+j2pifL)/(G+j2pifC))

Lo - SQRT((R)/(G)) = R vs SQRT(R/(j2pifC)?

You seem to be making some distinction between fL and fC at low
frequencies. Why would fL shrink relative to R while fC does not shrink
relative to G? Are you saying that the L term goes away faster than the
C term in most cases? What is the recondition for that assumption, just
most transmission lines?

Actually, it would be SQRT(R/(G+j2pifC), no?

--

Rick

On 3/5/2015 4:35 AM, Jeff wrote:
On 05/03/2015 01:50, rickman wrote:
On 3/4/2015 7:49 AM, Jeff wrote:

Connector impedance doesn't change with frequency, just as coax
impedance doesn't change with frequency. Loss will increase as
frequency increases, however.

Coax impedance certainly does change with frequency. Below about 500kHz
there is a significant slope with frequency.

At 200kHz a 50ohm cable may well look more like 100ohms and by the time
that you get to 1kHz it could be as high as 1kohm.

200kHz is in what is called the transition region and the impedance is
given by:

SQRT((R+j2pifL)/(C+j2pifC))

as opposed to the high frequency region where it is merely:

sqrt(L/C)

Below about 20kHz it changes again to SQRT(R/(j2pifC)

There are also other variables due to changes in the dielectric with
frequency and other losses.

Can you explain the above equations? In general it doesn't make sense
that the same effect would have different equations for different
frequencies. It does make sense though that the equations involved are
all simplifications of a single, more complex equation, optimized to
discount small effects over a given frequency range.

That said, I'm not sure I can see how these three equations can morph
into each other as f changes. The equation for the middle frequency
range seems to be the more encompassing so starting with that - if
frequency goes up enough the terms j2pifL and j2pifC dominate the R and
C terms and the equation simplifies to sqrt(L/C) appropriately.

But when f goes down enough, both terms shrink compared to R and C and
the equation would seem to simplify to sqrt(R/C) rather than
sqrt(R/j2pifC).

Is there possibly a typo in there somewhere?


Sorry there is indeed a typo: the general equation should read:

SQRT((R+j2pifL)/(G+j2pifC)) G being conductance.

The reasons that the equations are presented differently in different
frequency ranges are because; at higher frequencies when f becomes
large enough, the terms containing f become so large that R and G can be
neglected; and at low frequencies (2 pi f L) is so small compared with R
that it can be neglected.

I just had a brain cramp about the modified formula,

SQRT((R+j2pifL)/(G+j2pifC)) G being conductance.

As resistance goes to zero, G will go to infinity! This makes the
result of the equation go to zero no matter what values of L, C or f are
used. This would imply that wires made with superconductors have zero
impedance?

I read the wikipedia page and they say, "For a lossless line, R and G
are both zero". How dem do dat? Isn't R = 1/G?

I would also point out that these equations assume a non-ideal conductor
by accounting for R, but they assume the dielectric *is* ideal and
ignore dielectric losses which become dominant at high enough frequencies.

--

Rick

In article , rickman wrote:

I read the wikipedia page and they say, "For a lossless line, R and G
are both zero". How dem do dat? Isn't R = 1/G?

I think they're referring to series R (that is, resistive loss in the
conductors) and shunt G (perfect dielectric, infinitely high shunt
resistance between the conductors).

With those being posited, all of the current flow is into or through
reactances (non-dissipative impedances) , and there's no current flow
through anything with resistive loss (dissipative impedance).

On 3/5/2015 6:18 PM, Dave Platt wrote:
In article , rickman wrote:

I read the wikipedia page and they say, "For a lossless line, R and G
are both zero". How dem do dat? Isn't R = 1/G?

I think they're referring to series R (that is, resistive loss in the
conductors) and shunt G (perfect dielectric, infinitely high shunt
resistance between the conductors).

With those being posited, all of the current flow is into or through
reactances (non-dissipative impedances) , and there's no current flow
through anything with resistive loss (dissipative impedance).

Ah, that makes perfect sense now.

I skipped over the diagram on the wiki page. They actually make this
very clear.

--

Rick