shorted 1/8 wave transmission line
 

Hello Group,

If I have an RG-58 coax and it is shorted at the load end. At the electrical
1/8 wave lenght what would be the impedance seen at the other end?

I understand that a shorted 1/4 wave length reflects an open, but was
interested in what happens at the 1/8 wave frequency.

Tnx de KJ4UO

The impedance would be +j50 ohms, that is, 50 ohms of inductive
reactance. If you open the far end, you'll see -j50 ohms, that is, 50
ohms of capacitive reactance, at the input.

Roy Lewallen, W7EL

PDRUNEN wrote:

Hello Group,

If I have an RG-58 coax and it is shorted at the load end. At the electrical
1/8 wave lenght what would be the impedance seen at the other end?

I understand that a shorted 1/4 wave length reflects an open, but was
interested in what happens at the 1/8 wave frequency.

Tnx de KJ4UO

PDRUNEN wrote:
If I have an RG-58 coax and it is shorted at the load end. At the electrical
1/8 wave lenght what would be the impedance seen at the other end?

I understand that a shorted 1/4 wave length reflects an open, but was
interested in what happens at the 1/8 wave frequency.

For lossless coax, it would be purely reactive, close to +jZ0. Questions
like that are easy when one understands the Smith Chart.
--
73, Cecil http://www.qsl.net/w5dxp



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Roy,

Many years ago an article was published showing nomographs for using 1/8
wave lines for universal impedance matching. Do you recall where and when
that appeared, by any remote chance? I know you can do all this with
computer programs now, but I can't easily carry a computer or software into
many of the places I go.

--
Crazy George
W5VPQ
Remove N O and S P A M imbedded in return address

"Roy Lewallen" wrote in message
...
The impedance would be +j50 ohms, that is, 50 ohms of inductive
reactance. If you open the far end, you'll see -j50 ohms, that is, 50
ohms of capacitive reactance, at the input.

Roy Lewallen, W7EL

PDRUNEN wrote:

Hello Group,

If I have an RG-58 coax and it is shorted at the load end. At the
electrical
1/8 wave lenght what would be the impedance seen at the other end?

I understand that a shorted 1/4 wave length reflects an open, but was
interested in what happens at the 1/8 wave frequency.

Tnx de KJ4UO

Crazy George wrote:
... I can't easily carry a computer or software into
many of the places I go.

Everyone else does.



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The input reactance of a shorted length of line is given by -

Inductive Xin = j * Zo * Tangent(Theta)

where Theta is line length in degrees.

For a 1/8-wavelength line Theta = 360/8 = 45 degrees, Tangent(45) = 1.0,
and for Zo = 50 ohms the input inductive reactance is also 50 ohms.

===========================

For an open circuit length of line -

Capacitative Xin = -j * Zo / Tangent(Theta)

So for open-circuit 1/8-wavelength line and Zo = 50 ohms, input reactance is
a capacitative -j50 ohms.

The resistive component of input impedance is very small because line loss
is very small for 1/8 wavelength.
----
Reg, G4FGQ

=====================================

"PDRUNEN" wrote in message
...
Hello Group,

If I have an RG-58 coax and it is shorted at the load end. At the
electrical
1/8 wave lenght what would be the impedance seen at the other end?

I understand that a shorted 1/4 wave length reflects an open, but was
interested in what happens at the 1/8 wave frequency.

Tnx de KJ4UO

Sorry, I don't recall having seen that.

Roy Lewallen, W7EL

Crazy George wrote:

Roy,

Many years ago an article was published showing nomographs for using 1/8
wave lines for universal impedance matching. Do you recall where and when
that appeared, by any remote chance? I know you can do all this with
computer programs now, but I can't easily carry a computer or software into
many of the places I go.

--
Crazy George
W5VPQ
Remove N O and S P A M imbedded in return address

Reg, let me guess: you dug that Smith Chart out after all huh? he he he

Btw I had a Power Quality engineer that I was discussing ground line
impedances with remind me that the same 1/4 wave phenomenon can happen in
runs of ground and bonding too. The same radial or parallel or "web" of
connections alleviates that risk with lightning grounding as it does with RF
grounding.

Jack

"Reg Edwards" wrote in message
...
The input reactance of a shorted length of line is given by -

Inductive Xin = j * Zo * Tangent(Theta)

where Theta is line length in degrees.

For a 1/8-wavelength line Theta = 360/8 = 45 degrees, Tangent(45) = 1.0,
and for Zo = 50 ohms the input inductive reactance is also 50 ohms.

===========================

For an open circuit length of line -

Capacitative Xin = -j * Zo / Tangent(Theta)

So for open-circuit 1/8-wavelength line and Zo = 50 ohms, input reactance
is
a capacitative -j50 ohms.

The resistive component of input impedance is very small because line loss
is very small for 1/8 wavelength.
----
Reg, G4FGQ

=====================================

"PDRUNEN" wrote in message
...
Hello Group,

If I have an RG-58 coax and it is shorted at the load end. At the
electrical
1/8 wave lenght what would be the impedance seen at the other end?

I understand that a shorted 1/4 wave length reflects an open, but was
interested in what happens at the 1/8 wave frequency.

Tnx de KJ4UO

For lossless coax, it would be purely reactive, close to +jZ0. Questions
like that are easy when one understands the Smith Chart.
--
73, Cecil

==============================

They are even easier WITHOUT the unessessary over-complication of the chart.

A student's valuable time is much better spent learning about transmissiom
lines instead of how to use an antique chart. The answer can be worked out
in the head in less than a second. Tan(45) = 1.0000000 and therefore X =
Zo.

Sorry to hear about your pH problem. Isn't there an ant-acid preparation
available in this modern day and age?

----
Reg, G4FGQ

"Jack Painter"
snip
Btw I had a Power Quality engineer that I was discussing ground line
impedances with remind me that the same 1/4 wave phenomenon can happen in
runs of ground and bonding too. The same radial or parallel or "web" of
connections alleviates that risk with lightning grounding as it does with
RF
grounding.

Jack

Jack:

Alleviates? NO! Reduces somewhat, maybe, and that is one of the most
difficult things to get across to people with little theoretical knowledge.
Even if it were the perfectly conducting sphere so loved in textbooks,
impedance still exists, and the instantaneous voltage at point A will be
different from point B. All you are doing is increasing the current
carrying capability, so it is less likely to blow up due to a direct strike.
Instantaneous voltage difference with respect to a remote reference can
still rise to a gazillion volts, no matter how much copper you put in there,
or how you configure it.

--
Crazy George
Remove N O and S P A M imbedded in return address

Years ago before they started burying all the phone drops and cabling the
overhead wires, I connected my BC-348 to the phone withe a 0.005 uF disc
ceramic capacitor. It gave me a VERY long wire, probabaly miles of long wire.

Worked quite well.

I haven't tried it since.

73 de Jack, K9CUN

Reg Edwards wrote:
They are even easier WITHOUT the unessessary over-complication of the chart.

Why do you think the chart is "over-complicated". I have one with the
R,X values and only the to/from wavelength circles outside. I deleted
everything else from the Smith Chart program.

What I find convenient is to carry the image of the Smith Chart around
in my head. I'm more of a visual person than a formula person and that
image has helped me tremendously.

A student's valuable time is much better spent learning about transmissiom
lines instead of how to use an antique chart. The answer can be worked out
in the head in less than a second. Tan(45) = 1.0000000 and therefore X =
Zo.

How about 0.07 wavelength in your head? :-) I can glance at a Smith Chart
faster than you can pick up a calculator.

Sorry to hear about your pH problem. Isn't there an ant-acid preparation
available in this modern day and age?

Prilosec OTC works well. Then I can drink twice as much as usual [hic].
--
73, Cecil http://www.qsl.net/w5dxp



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KJ4UO wrote:
"If I have an RG-58 coax and it is shorted at the load end."

At the electrical input, 1/8-wave back from the short, you have an
inductive reactance equal to the numerical value of Zo. At 1/4-wave back
from the short, you have added an open-circuit 1/8-wavelength of line to
the first. This new line has a capacitive reactance equal to Zo.

The combination of equal and opposite reactances makes a parallel
resonant circuit equivalent which seems a very high impedance indeed.

Best regards, Richard Harrison, KB5WZI

Reg, G4FGQ wrote:
"Isn`t there an ant-acid preparation avbailable in this day and age?"

I read in "Spectrum" that J.C. Maxwell took bicarbonate of soda but died
of stomach cancer, regardless.

Best regards, Richard Harrison, KB5WZI

"Crazy George" wrote in message
...
"Jack Painter"
snip
Btw I had a Power Quality engineer that I was discussing ground line
impedances with remind me that the same 1/4 wave phenomenon can happen
in
runs of ground and bonding too. The same radial or parallel or "web" of
connections alleviates that risk with lightning grounding as it does
with
RF
grounding.

Jack

Jack:

Alleviates? NO! Reduces somewhat, maybe, and that is one of the most
difficult things to get across to people with little theoretical
knowledge.
Even if it were the perfectly conducting sphere so loved in textbooks,
impedance still exists, and the instantaneous voltage at point A will be
different from point B. All you are doing is increasing the current
carrying capability, so it is less likely to blow up due to a direct
strike.
Instantaneous voltage difference with respect to a remote reference can
still rise to a gazillion volts, no matter how much copper you put in
there,
or how you configure it.

George, thanks for your reply. But in Webster's Dictionary:

Alliviate: To make less hard to bear; to reduce or decrease.

That's exactly why I used that word. And, after all the absolutely
invaluable help available from experts on this group, I still consulted a
Professional for the specifics of my site requirements. And common bonding
everything becomes more important than whatever impedance ground has between
any two points in the entire system. A properly bonded and grounded system
does not care what impedance or voltage is present on the system. Please
don't make me define properly.

;-) 73's

Jack.

Hi Roy, see you're still at it! Just stopped by to say
hello; hope you're doing well.

Regards,
Jack
WB3U


On Tue, 11 May 2004 17:55:48 -0700, Roy Lewallen
wrote:

The impedance would be +j50 ohms, that is, 50 ohms of