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reflection of electromagnetic wave

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Newbie level 2
Dec 20, 2014
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Hi, I'm trying to solve this task. Thanks for advice :)

There is a coaxial line with the length L.

Radius of the inner conductor is: 1,5
radius of the outer conductor is: 4,4
(The radius is measured from center of the coax line)
permittivity material: 1,672
voltage reflection coefficient: 1

What is the least length of the line to be on the input of the line short circuit ?
Consider the calculation for the frequency 7,34 GHz.

If anyone can help me, or recommend a literature I wil be grateful.
Nice day

What is it you have to calculate? I dont understand. Is it somekind of input impedance?

I understood you have to calculate the length L of a coaxial transmission line at wich the input impedance is a short circuit when it's left open at the other side.

at a length equal to a quarter wavelength (and it repeats every rotation of pi) an open circuit at a side of a line will be seen as a short to the other side and vice-versa.

This can be seen by the transmission line equation:


that is:


if ZL is open means that ZL-->inf, applying the limit to the eqaution above we have:

lim(ZL-->inf) {Zo*ZL*[1+tan(beta*L)}/[Zo^2+ZL^2*tan(beta*L)^2] = 0
lim(ZL-->inf) (Zo^2-ZL^2)*tan(beta*L)/[Zo^2+ZL^2*tan(beta*L)^2] =1/tan(beta*L)

from which:


now beta=2*pi/lamba

Zin(ZL-->inf)=Zo/tan(2*pi*L/lamba) then if L=lambda/4

Zin(ZL-->inf)=Zo/tan(pi/2) -->0 that is a short circuit

so also in your case L=lambda/4

but lambda=v/f where v is the velocity of the wave in the cable, that is:

v=c/sqrt(eps) where eps is the dielectric constant of the material ans c the velocity of the light, then

lambda= c/[f*sqrt(eps)] that is:

L=c/[f*sqrt(eps)*4] , numerically:

L=3e8/[7.34e9*sqrt(1.672)*4]=7.9 mm

with the dimensions of the cable you can calculate Zo as


in your case Zo=50 ohm

but I didn't find useful for the calculation you required

See Wiki for velocity factor ( ), this tells you how fast your wavefront will travel down your coax. Its equal to I/Pr ^.5 . =.77 . So the first short circuit occurs when the line is open circuit at the far end at 7.34 GHZ and is quarter wave long. So you have the figures.

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