The operating frequency for a 24-ft piece of RG-58A/U coaxial cable is 140 MHz. The load is resistive, with a resistance of 93 Ω. What is the impedance seen by a transmitter?

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The first step is to find the number of wavelengths represented by 24 ft of cable. A 140-MHz signal has a wavelength of

[latex]\lambda=\frac{984}{f}=\frac{984}{140}=7.02 \ ft[/latex]

Remember that coaxial cable has a velocity factor less than 1 due to the slowdown of RF signals in a cable. The velocity factor of RG-58A/U is 0.66. Therefore, one wavelength at 140 MHz is

[latex]\lambda=7.02 imes 0.66=4.64 \ ft[/latex]

The number of wavelengths represented by the 24-ft cable is 24/4.64 = 5.17 λ.
As indicated earlier, the impedance variations along a line repeat every one-half wavelength and therefore every full wavelength; thus for the purposes of calculation, we need only the 0.17 λ of the above value.
Next, we normalize the Smith chart to the characteristic impedance of the coaxial cable, which is 53.5 Ω. The value of the prime center is 53.5 Ω. The SWR is then computed:

[latex]SWR =\frac{Z_l}{Z_0}=\frac{93}{53.5}=1.74[/latex]

This SWR is now plotted on the Smith chart. See Fig. 13-35, where point X represents the resistive load of 93 V.

To find the impedance at the transmitter end of the coaxial cable, we move along the line 5.17 λ from the load back to the transmitter or generator. Remember that one full rotation around the Smith chart is one-half wavelength, because the values repeat every one-half wavelength. Starting at point X, then, move in a clockwise direction (toward the generator) around the SWR circle for 10 complete revolutions, which represents 5 λ. This brings you back to point X.

Continue the clockwise rotation for an additional 0.17 λ, stop there, and mark that point on the SWR circle. Draw a line from the prime center to the location that represents 0.17 λ. The marking on the right-hand side of point X is the 0.25-λ mark. You need to go 0.17 λ from that, or 0.25 + 0.17 = 0.42. Stop at the 0.42 mark on the lower part of the circle at Y. Draw a line from there to the prime center. Refer again to Fig. 13-35.

The point at which the line cuts the SWR circle is the impedance that the generator sees. Reading the values from the chart, you have R ≈ 0.68 and [latex]X_{C} ≈ 0.35[/latex]. Since the point is in the lower half of the chart, the reactance is capacitive. Thus the impedance at this point is 0.68 - j 0.35.

To find the actual value, convert from the normalized value by multiplying by the impedance of the prime center, or 53.5:

Z = 53.5(0.68 - j0.35) = 36.4 - j 18.725 Ω

The transmitter is seeing what appears to be a 36.4-Ω resistor in series with a capacitor with a reactance of 18.725 Ω. The equivalent capacitance value is [using [latex]X_{C} = 1/\left.(2 \pi f C) ext { or } C=1 /\left(2 \pi f X_C ight) ight][/latex]

[latex]C=\frac{1}{6.28\left(140 imes 10^6 ight)(18.725)}=60.74 \ pF[/latex]

The equivalent circuit is shown in Fig. 13-34(b).

Question 13.4.1: The operating frequency for a 24-ft piece of RG-58A/U coaxia...

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