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)

Balanis: Dmax = 1.643× 2 = 3.286Mathematica: Dmax = 3.282

The maximum power delivered to the load is a product of the incident power density and the effective area

Prmax =WtArmax =Wtλ2

4πDrmax

Plugging in the numbers

Wt = 10−6 W/m2

f = 1.9 GHz ⇒ Prmax = 6.51 nWλ = c / f = 0.158 m

4. From (4-94) the field from the real dipole is

Eθd = jη kI0le

− jkr1

4πr1sinθ1

but now since the ground plane is a PMC, the field from the image dipole has the opposite sign

Eθr = − jη kI0le

− jkr2

4πr2sinθ2

Using the phase-term far-field approximations

r1 r − hcosθr2 r + hcosθ

and the amplitude far-field approximations

r1 r2 rsinθ1 sinθ2 sinθ

The total filed is

Eθ = Eθd + Eθ

r = jη kI0le− jk r−h cosθ( )

4πrsinθ − jη kI0le

− jk r+h cosθ( )

4πrsinθ = jη kI0e

− jkr

4πrsinθ e jkh cosθ − e− jkh cosθ⎡⎣ ⎤⎦

So the far fields are

Eθ = jη kI0e− jkr

4πrsinθ 2 j sin khcosθ( )⎡⎣ ⎤⎦

Hφ = Eθ /η = j kI0e− jkr

4πrsinθ 2 j sin khcosθ( )⎡⎣ ⎤⎦

5. Why is the magnetic field we plotted proportional to the current? Start from Maxwell’s equation

∇ ×H = J + jωD

We want to apply this inside the copper. The constitutive equations for good conductors are

J = σE and D = ε0E

so we can write

∇ ×H = J 1+ jωε0σ

⎛⎝⎜

⎞⎠⎟≈ J

since even at microwave frequencies the ratio of constants is much, much less than one. In integral form this is

H ⋅dl

C∫ = J ⋅dS

A∫ = I

where the curve is around the circumference of the antenna cylinder and the enclosed area is its cross section. Since dl points in the ϕ direction

Hφ dl

C∫ = I

Our problem is entirely symmetric with respect to ϕ, so Hϕ is constant along the curve, C.

I = 2πaHφ a( )∝ Hφ a( )

where a is the radius of the antenna, i.e. 1 mm. We plotted Hx at (x, y) = (0, 1) mm which is the same as Hϕ(a).

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Really only the fits with current maxima at the feed are any good.