Page 431 - Electromagnetics

We assume a lossless medium with parameters µ, . Substituting the current expression
into (6.15) and integrating over x and y we find that

µI ˜ l e − jkR
˜

A e (r,ω) = ˆ z sin k(l −|z |) dz (6.27)
4π −l R
2 2 2 2 2
where R = (z − z ) + ρ and ρ = x + y . Using (6.16) we have
˜
1 1 ∂ A ez
˜ ˜ ˆ
H =∇ × A e =−φ .
µ µ ∂ρ
Writing the sine function in (6.27) in terms of exponentials, we then have

˜ I 0 ∂ e − jk(R−z ) 0 ∂ e − jk(R+z )
˜
H φ = j e jkl dz − e − jkl dz +

8π −l ∂ρ R −l ∂ρ R

l l
∂ e − jk(R+z ) ∂ e − jk(R−z )
+ e jkl dz − e − jkl dz .

0 ∂ρ R 0 ∂ρ R
Noting that

∂ e − jk(R±z ) ∂ e − jk(R±z ) 1 + jkR − jk(R±z )

=±ρ =−ρ e
∂ρ R ∂z R [R ∓ (z − z )] R 3

we can write
0 0
˜
Iρ jkl e − jk(R−z ) − jkl e − jk(R+z )
˜
H φ = j −e − e +
8π R [R + (z − z )] R [R − (z − z )]

−l −l
e − jk(R+z ) l − jkl e − jk(R−z ) l
jkl
+ e + e .
R [R − (z − z )] R [R + (z − z )]

0 0
Collecting terms and simplifying we get
˜ I(ω)
˜
H φ (r,ω) = j e − jkR 1 + e − jkR 2 − (2 cos kl)e − jkr (6.28)
4πρ

2 2 2 2
where R 1 = ρ + (z − l) and R 2 = ρ + (z + l) . For points external to the dipole
the source current is zero and thus
1 1
∂ 1 ∂
˜ ˜ ˜ ˜
E(r,ω) = ∇× H(r,ω) = −ˆρ H φ (r,ω) + ˆ z [ρH φ (r,ω)] .
jω jω ∂z ρ ∂ρ
Performing the derivatives we have
˜

ηI(ω) z − l e − jkR 1 z + l e − jkR 2 z e − jkr
˜
E ρ (r,ω) = j + − (2 cos kl) , (6.29)
4π ρ R 1 ρ R 2 ρ r
˜

ηI(ω) e − jkR 1 e − jkR 2 e − jkr
˜
E z (r,ω) =− j + − (2 cos kl) . (6.30)
4π R 1 R 2 r
The work of specializing these expressions for points in the far zone is left as an exercise.
Instead, we shall use the general far-zone expressions (6.21)–(6.24). Substituting (6.26)
© 2001 by CRC Press LLC