# RG-6/U cable (Homework 3, Physics 323)

1. RG-6/U cable is a type of TEM coaxial waveguide commonly used in cable TV from DC to around 900 MHz. Approximately, the inner conductor diameter is 1mm, the outer-conductor diameter is 7mm, and the space between conductors is filled with solid polyethylene dielectric. (NB. sometimes the polyethylene is a foam; ignore that detail).

a. Estimate the propagation velocity (phase velocity) of TEM waves along this cable. Estimate the guided wavelength in this cable. You’ll need to look up properties of the dielectric.

b. Suppose you don’t want any reflections from the end of the cable. What value of termination resistor do you need?

c. Suppose you left the resistor off and you’re interested in cable TV channel 48 (KING TV, frequency band around 370 MHz). At what positions from the open end of the cable are the voltage maximum and minimum? At what positions are current maximum and minimum?

a)

The phase velocity in a coaxial cable (TEM modes) is found from writing the Maxwell equations inside the cable (9.177) with the difference that now the medium of propagation is not the air but the cable insulator between the central wire and the shielding. Therefore

$v=1/√ϵμ=c/\sqrt{(ϵ_r μ_r)}$

For RG/6U insulation is polyethylene $ϵ_r=2.25$ and $μ_r=1+χ=1+0.2=1.2$

$v=c*1/\sqrt{(2.25*1.2)}=0.61*c$

Observation: from the cable data sheet the phase velocity is $v=0.82*c$

The guided wavelength is (standard thickness of the foam is $L=5 mm$)

fundamental mode happen at $ω=cπ/L$  and $ω=ck$  so that $k=π/L$

$2π/λ=π/L$   so that $λ=2L=10 mm$

b)

The transmission coefficient is =1 (no reflections back in cable) when the $Z_L^*=Z_{cable}$  (impedance of cable and load are complex conjugate). Since $Z_{cable}=75 Ohm$ (standard) is purely real, the load resistance for no reflections is $Z_{load}=75 Ohm$.

c)

The waves on the cable propagate as EM waves: (see question 4, HW 2- last HW)

in cable $V(z,t)=V_0*\cos⁡(k z-ωt)$

For $F=370 MHz$  we have $ω=2πF=2.325*10^9 rad/s$    and $k=ω/v=0.61ω/c=4.727 1/m$
At t=0 extrema (maximum and minimum) are at $k z=nπ$  or $z=nπ/4.727=n*0.65$ meter  with n integer

At t=0 nodes are at $k z=nπ+π/2$    or $z=(0.65*n+0.323) m$
Since the cable impedance is purely real =75 ohm (standard) (see question 4, point d, last Homework), the current is in phase with the voltage so the current maximum and nodes will happen in the same places as the voltage maximum and nodes.