1. You need the Laplacian of 1/r to do this problem. You know the electrostatic potential of a point charge at the origin Φ = 1 4πε 0 q r and you know Poisson’s equation for electrostatic potential ∇ 2 Φ = − ρ / ε 0 and the charge density of a point charge q at the origin is ρ r () = q δ r () = q δ x () δ y () δ z () Substituting into Poisson’s equation ∇ 2 1 4πε 0 q r ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − q δ r () / ε 0 and canceling the constants we find ∇ 2 1 r = −4πδ r () Note that in section 3.5 of Balanis, the current Jz is located only at the origin and therefore must have the form J z r () = I 0 l δ r () and note that in (3-42) and (3-43), Jz refers to Jz(r′), and in (3-31) and (3-39) it refers to Jz(r). Our solution (3-42) is given by A z = μ 4π J z ′ r ( ) r V ∫∫∫ d ′ v = μ 4π r I 0 l δ ′ r ( ) V ∫∫∫ d ′ v = μ I 0 l 4π r δ ′ x ( ) d ′ x ∫ δ ′ y ( ) d ′ y ∫ δ ′ z ( ) d ′ z ∫ = μ I 0 l 4π r where we can pull the (origin-to-observation-point) radial coordinate, r, out of the integral since it does not depend on the source coordinate r′, i.e. the integration variables. The integral range includes the origin so the integrals of the delta functions are just unity. Substituting into (3-39) our differential equation ∇ 2 A z = − μ J z r () we find ∇ 2 μ I 0 l 4π r ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − μ I 0 l δ r ()
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Using our expression for the Laplacian of 1/...P rmax=W t A rmax=W t λ2 4π D rmax Plugging in the numbers W t=10 −6 W/m2 f=1.9 GHz ⇒ P rmax=6.51 nW λ=c/f=0.158 m 4. From (4-94)
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1. You need the Laplacian of 1/r to do this problem. You know the electrostatic potential of a point charge at the origin
Φ =14πε0
qr
and you know Poisson’s equation for electrostatic potential
∇2Φ = −ρ / ε0
and the charge density of a point charge q at the origin is
ρ r( ) = qδ r( ) = qδ x( )δ y( )δ z( )
Substituting into Poisson’s equation
∇2 14πε0
qr
⎛⎝⎜
⎞⎠⎟= −qδ r( ) / ε0
and canceling the constants we find
∇2 1r= −4πδ r( )
Note that in section 3.5 of Balanis, the current Jz is located only at the origin and therefore must have the form
Jz r( ) = I0lδ r( )
and note that in (3-42) and (3-43), Jz refers to Jz(r′), and in (3-31) and (3-39) it refers to Jz(r). Our solution (3-42) is given by
Az =µ4π
Jz ′r( )rV
∫∫∫ d ′v =µ4πr
I0lδ ′r( )V∫∫∫ d ′v =
µI0l4πr
δ ′x( )d ′x∫ δ ′y( )d ′y∫ δ ′z( )d ′z∫ =µI0l4πr
where we can pull the (origin-to-observation-point) radial coordinate, r, out of the integral since it does not depend on the source coordinate r′, i.e. the integration variables. The integral range includes the origin so the integrals of the delta functions are just unity. Substituting into (3-39) our differential equation
∇2Az = −µJz r( )we find
∇2 µI0l4πr
⎛⎝⎜
⎞⎠⎟= −µI0lδ r( )
Using our expression for the Laplacian of 1/r
−µI0l4π
4πδ r( ) = −µI0lδ r( )
and we see that indeed our solution (3.42) satisfies the differential equation (3-39). Similarly, the solution (3-43) is
Az =µ4π
Jz ′r( ) e− jkr
rV∫∫∫ d ′v =
µ4π
e− jkr
rI0lδ ′r( )
V∫∫∫ d ′v =
µI0l4π
e− jkr
rδ ′x( )d ′x∫ δ ′y( )d ′y∫ δ ′z( )d ′z∫ =
µI0l4π
e− jkr
r
Substituting into (3-31) our differential equation
∇2Az + k2Az = −µJz r( )
and rearranging constants we find
∇2 e− jkr
r+ k2 e
− jkr
r= −4πδ r( )
Where r is not zero, you are free to multiply and divide by r. Using the Laplacian in spherical coordinates with no angular dependence
∇2ψ =1r∂2
∂r2rψ( )
we find
∇2 e− jkr
r=1r∂2
∂r2r e
− jkr
r⎛⎝⎜
⎞⎠⎟= −k2 1
re− jkr
However, this does not include the contribution at r = 0.
limr→0
∇2 e− jkr
r= ∇2 1
r= −4πδ r( )
So, in total
∇2 e− jkr
r= −k2 e
− jkr
r− 4πδ r( )
Substituting in the above equation
−k2 e− jkr
r− 4πδ r( ) + k2 e
− jkr
r= −4πδ r( )
We see the equation is satisfied.
2. Using the results for a half-wave dipole (4-87)
U = ηI0
2
8π 2
cos π2cosθ⎛
⎝⎜⎞⎠⎟
sinθ
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
2
and (4-89)
Prad = ηI0
2
8πCin 2π( )
So the directivity is
D θ( ) = UPrad / 4π
=4
Cin 2π( )
cos π2cosθ⎛
⎝⎜⎞⎠⎟
sinθ
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
2
From the definition of directivity
Wav θ( ) = Prad4πr2
D θ( ) = ePin4πr2
D θ( )
where e is the efficiency. Plugging in the numbers
Pin = 100 We = 0.5 ⇒ D θ( ) = 1.09r = 500 m Wav = 1.74 ×10−5 W/m2 = 17.4 µW/m2
θ = π / 3
3. For the direction normal to a PEC, real and image currents have the same sign. Then a λ/4 dipole above a PEC has a combined (real plus image) current distribution that is identical to the half-wave dipole. Since no power is radiated in the half-space below the ground plane, the total power radiated is half that of that of the half-wave dipole and the directivity (which is normalized the total radiated power) is twice that of the half-wave dipole. The maximum directivity is given by (4-91)