Coaxial Cables Used in Radio

It was early shown²⁴ that if the ratio b/a = 3.6, then the attenuation of a coaxial cable with air dielectric is, theoretically, a minimum value. From equation 34 this ratio gives a characteristic impedance of about 77 ohms (resistance). Until about 1940, most coaxial cables were constructed with this ratio, but since that time there has been an increasing tendency to design cables having a characteristic impedance of about 52 ohms.

Coaxial cables used for feeding a broadcast antenna must sometimes transmit many kilowatts.

The following calculations serve to illustrate the use of the preceding equations. A copper coaxial cable having the dimensions a = 1.59 centimeters and b = 3.81 centimeters is to deliver 20 kilowatts at 1.19

Figure 25. Attenuation at various frequencies of coaxial cables used in radio. Such cables are in general small, flexible, and insulated with polyethylene. (For further information consult Reference Data for Radio Engineers, Federal Telephone and Radio Corp., from which these data were obtained.)

megacycles to an antenna that is matched to the cable so that no standing waves exist. The cable length is 500 feet, or 152,3 meters,

The inductance is found by equation 32 to be L = 0.46 lg (3.81/1.59) = 0.175 microhenry per meter.

The capacitance is found by equation 33 to be C = 0.000024l/lg (3.81/1.59) = 0.0000634 microfarad per meter.

The characteristic impedance is found from equation 28 to be Z_o = sqrt(0.175 x 10^-6/0.0000634 x 10^-6) = 52.5 ohms. Or, from equation 29, Z₀ = 138 lg(3.81/1.59) = 52.5 ohms.

The resistance is found from equation 31 to be R = 41.6* sqrt(l.19 x 10⁶) x (1/1.59 + 1/3.81) x 10^-9 = 40.4 x 10^-6 ohm per centimeter.

The attenuation is found from equation 29 to be α = 40.4 x 10^-6/2 x 52.5 = 0.385 x 10^-6 neper per centimeter = 0.385 X 10^-6 x 8.686 = 3.34 X 10^-6 decibel per centimeter, or 0.0508 decibel for the 500-foot length of coaxial cable.

The voltage at the distant end will be E = sqrt(PZ₀) = sqrt(20 x 10³ x 52.5) = 1025 volts.

The current at the distant end will be I = sqrt(P/Z₀) = sqrt(20 x 10³/52.5) = 19.5 amperes.

The power input at the sending end will be P = 20*10^{0.l x 0.0508} = 20,200 watts.

The voltage at the sending end will be E = sqrt(PZ₀) = sqrt(20.2 x 10³ x 52.5) = 1030 volts.

The current at the sending end will be I = sqrt(P/Z₀) = sqrt(20.2 x 10³/52.5) = 19.6 amperes.

The efficiency of transmission will be 20/20.2 = 0.99 or 99 percent.