Page 336 - Engineering Electromagnetics, 8th Edition

318 ENGINEERING ELECTROMAGNETICS

again, the amplitude, V 0 + =|V 0 |,is taken to be real. The current waveform will be
similar, but will generally be shifted in phase. Both current and voltage attenuate
according to the factor e −αz . The instantaneous power therefore becomes:

P(z, t) = V(z, t)I(z, t) =|V 0 ||I 0 |e −2αz cos(ωt − βz) cos(ωt − βz + θ) (57)
Usually, the time-averaged power, P ,isof interest. We find this through:

1 T
P = |V 0 ||I 0 |e −2αz cos(ωt − βz) cos(ωt − βz + θ)dt (58)
T 0
where T = 2π/ω is the time period for one oscillation cycle. Using a trigonometric
identity, the product of cosines in the integrand can be written as the sum of individual
cosines at the sum and difference frequencies:
1 T 1
P = |V 0 ||I 0 |e −2αz [cos(2ωt − 2βz + θ) + cos(θ)] dt (59)
T 0 2
The first cosine term integrates to zero, leaving the cos θ term. The remaining integral
easily evaluates as
1 −2αz 1 |V 0 | 2 −2αz
P = |V 0 ||I 0 |e cos θ = e cos θ [W] (60)
2 2 |Z 0 |
The same result can be obtained directly from the phasor voltage and current. We
begin with these, expressed as

V s (z) = V 0 e −αz − jβz (61)
e
and

e
e
e
I s (z) = I 0 e −αz − jβz = V 0 −αz − jβz (62)
Z 0
where Z 0 =|Z 0 |e .Wenow note that the time-averaged power as expressed in (60)
jθ
can be obtained from the phasor forms through:
1
∗ (63)
P = Re{V s I }
s
2
where again, the asterisk ( ) denotes the complex conjugate (applied in this case to
∗
the current phasor only). Using (61) and (62) in (63), it is found that
1 V ∗
e
e
P = Re V 0 e −αz − jβz 0 e −αz + jβz
2 |Z 0 |e − jθ
1 V 0 V ∗ 1 |V 0 | 2
e
= Re 0 e −2αz jθ = e −2αz cos θ (64)
2 |Z 0 | 2 |Z 0 |
which we note is identical to the time-integrated result in (60). Eq. (63) applies to any
single-frequency wave.