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

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

"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

"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.