Page 337 - Engineering Electromagnetics, 8th Edition

CHAPTER 10 Transmission Lines 319

An important result of the preceding exercise is that power attenuates as e −2αz ,or

P(z) = P(0) e −2αz (65)

Power drops at twice the exponential rate with distance as either voltage or current.
A convenient measure of power loss is in decibel units. This is based on express-
ing the power decrease as a power of 10. Specifically, we write

P(z) −2αz −καz
= e = 10 (66)
P(0)
where the constant, κ,istobe determined. Setting αz = 1, we find
2
e −2 = 10 −κ ⇒ κ = log (e ) = 0.869 (67)
10
Now, by definition, the power loss in decibels (dB) is

P(0)
Power loss (dB) = 10 log 10 = 8.69αz (68)
P(z)
where we note that inverting the power ratio in the argument of the log function [as
compared to the ratio in (66)] yields a positive number for the dB loss. Also, noting
2
that P ∝|V 0 | ,we may write, equivalently:

P(0) |V 0 (0)|
Power loss (dB) = 10 log 10 = 20 log 10 (69)
P(z) |V 0 (z)|
where |V 0 (z)|=|V 0 (0)|e −αz .

EXAMPLE 10.4
A 20-m length of transmission line is known to produce a 2.0-dB drop in power from
end to end. (a) What fraction of the input power reaches the output? (b) What fraction
of the input power reaches the midpoint of the line? (c) What exponential attenuation
coefficient, α, does this represent?
Solution. (a) The power fraction will be
P(20) −0.2
= 10 = 0.63
P(0)
(b)2 dB in 20 m implies a loss rating of 0.2 dB/m. So, over a 10-m span, the loss
is 1.0 dB. This represents the power fraction, 10 −0.1 = 0.79.
(c) The exponential attenuation coefficient is found through
2.0dB
α = = 0.012 [Np/m]
(8.69 dB/Np)(20 m)
A final point addresses the question: Why use decibels? The most compelling
reason is that when evaluating the accumulated loss for several lines and devices that