6.9 Channel Models

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distance along the transmission line. Expressing γ in terms of its real and imaginary parts in our solution shows that such increases are a (mathematical) possibility. V (x) = V+e

−(a+jb)x + V−e+(a+jb)x The voltage cannot increase without limit; because a (f ) is always positive, we must segregate the solution for negative and positive x. The first term will increase exponentially for x < 0 unless V+ = 0 in this region; a similar result applies to V− for x > 0. These physical constraints give us a cleaner solution.

V (x) = V+e

−(a+jb)x , if x > 0 V−e+(a+jb)x , if x < 0 (6.11)

This solution suggests that voltages (and currents too) will decrease exponentially along a transmission line. The space constant, also known as the attenuation constant, is the distance over which the voltage decreases by a factor of 1 e . It equals the reciprocal of a (f ), which depends on frequency, and is expressed by manufacturers in units of dB/m.

The presence of the imaginary part of γ, b (f ), also provides insight into how transmission lines work. Because the solution for x > 0 is proportional to e −jbx, we know that the voltage’s complex amplitude will vary sinusoidally in space. The complete solution for the voltage has the form

v (x, t) = Re V+e

−axej(2πf t−bx) (6.12)

The complex exponential portion has the form of a propagating wave. If we could take a snapshot of the voltage (take its picture at t = t1), we would see a sinusoidally varying waveform along the transmission line. One period of this variation, known as the wavelength, equals λ = 2π/b. If we were to take a second picture at some later time t = t2, we would also see a sinusoidal voltage. Because

2πf t2 − bx = 2πf (t1 + t2 − t1) − bx = 2πf t1 − b x − 2πf b (t2 − t1)

the second waveform appears to be the first one, but delayed—shifted to the right—in space. Thus, the voltage appeared to move to the right with a speed equal to 2πf /b (assuming b > 0). We denote this propagation speed by c, and it equals

c = 2πf b

Im 2πf

G + j2πf C R + j2πf L (6.13)

The characteristics of the voltage signal shown in equation (6.12) depend on the values of a and b, and how they depend on frequency. The simplest results occur in the high-frequency region where j2πf L R and j2πf C G. In this case, γ simplifies to −4π2f 2 L C, which seemingly makes it pure imaginary with a = 0 and b = 2πf L C. Using this result, we find the propagation speed to be

lim f →∞ c =

1 L C (6.14)

For typical coaxial cable, this propagation speed is a fraction (one-third to two-thirds) of the speed of light.

While this high-frequency analysis shows that the dominant high-frequency component of γ is its imaginary part, there could be (and is!) a smaller real part. Since the real part of γ is the attenuation factor a, a more detailed analysis is required to determine if a = 0 (no attenuation) or is non-zero. One way of pursuing a more detailed analysis is to exploit equation (6.10) by

γ 2 = (a + j b)2 = G + j2πf C R + j2πf L