- waveform/digital transmissions because in it frequency dispersion and waveform distortion tend to be minimal. Equation 2.2.44 is also useful because it explains why inductive loading, putting small lumped element in- G + jwC = mS + j(2 X 500 X 48 nF)) ductors in series with lines, is sometimes used to reduce = (4 j0.15) mS km-superscript(1) dispersion in lines. Hence Example 2.2.12 A transmission line has the following primary constants: r = j0.39) j0.15) X 10-3 R ==23 S km-superscript(1), G = 4 mS km L = 125 uH km-1 and C = 48 nF km-superscript(1). Calculate the propagation constant r of : 0/0.03 rad the line, and the characteristic impedance Zo of the line = (0.30 nepers + 8.31 X 10-3rad km-1 at a frequency of (a) 100 Hz, (b) 500 Hz, (c) 15 kHz, (d) 5 MHz and (e) 10 MHz. Solution The characteristic impedance Z will not be calculated here because it has already been carried out in Example = rad 2.2.4. However, the results will be copied to allow easy Transmission lines CHAPTER 2.2 (c) At 15 kHz R + jwL = G + jwC = 4mS+j(2x 10 x 106 x 48 nF)) = (23 j11.78) O km-superscript(1) = 4 + j3016) mS km-1 Hence G + jwC = mS j(2 15 X X 48 nF)) = (4 + j4.52 x 10-3) mS km-superscript(1) = 153.91 /1.569 rad Hence = (0.33 + j153.9 rad) km-superscript(1) 1(23+j11.78)(4+j4.52) x 10-3 = /0.66 rad = (0.31 nepers + 242 x 10-3rad) km-superscript(1) Conclusions from Example 2.2.12. In the frequency range 100-500 Hz, the attenuation constant a tends to remain at about 0.30 nepers per km and the phase propagation B increases linearly with frequency. See cases (a) and (b). If you now compare this set of results with the same cases from Example 2.2.4, you will see that in this frequency range, Z and a tend to remain constant (d) At 5 MHz and B tends to vary linearly with frequency. What this means is that if you transmit a rectangular R + jwL = (23+j(2nx5x 106 x 125uH)) pulse or digital signals in this frequency range, you will find that it will pass through the transmission line attenuated = (23 + j3926.99) S km-superscript(1) but wit