199 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+ e−(a+jb)x , if x > 0 (6.11) V (x) = V− e+(a+jb)x , if x < 0 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 1e . 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 h i v (x, t) = Re V+ e−ax ej(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 − t1 ) 2πf t2 − bx = 2πf (t1 + t2 − t1 ) − bx = 2πf t1 − b x − b 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
r
2πf e e e e 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 e R e how they depend on frequency. The simplest q results occur in the high-frequency region where j2πf L e G. e In this case, γ simplifies to −4π 2 f 2 L e C, e which seemingly makes it pure imaginary with and j2πf C p e C. e Using this result, we find the propagation speed to be a = 0 and b = 2πf L 1 lim c = p f →∞ eC e L
(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 e + j2πf C e R e + j2πf L e γ 2 = (a + jb)2 = G
