electromagnetics

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ECE 6340 Fall 2013

Homework 8

Assignment: Please do Probs. 1-9 and 13 from the set below.

1) In dynamics, we have the equation

E j Aω= − −∇Φ .

(a) Show that in statics, the scalar potential function Φ can be interpreted as a voltage function. That is, show that in statics

( ) ( ) B

AB A

V E dr A B≡ ⋅ = Φ −Φ∫ .

(b) Next, explain why this equation is not true (in general) in dynamics. (c) Explain why the voltage drop (defined as the line integral of the electric field, as

defined above) depends on the path from A to B in dynamics, using Faraday’s law. (d) Does the right-hand side of the above equation (the difference in the potential

function) depend on the path, in dynamics? Hint: Note that, according to calculus, for any function ψ we have

dr dx dy dz d x y z ψ ψ ψ

ψ ψ ∂ ∂ ∂

∇ ⋅ = + + = ∂ ∂ ∂

.

2) Starting with Maxwell’s equations, show that the electric field radiated by an impressed current density source J i in an infinite homogeneous region satisfies the equation

( )2 2 iE k E E j Jωµ∇ + = ∇ ∇⋅ + .

Then use Ampere’s law (or, if you prefer, the continuity equation and the electric Gauss law) to show that this equation may be written as

( )2 2 1 i iE k E J j J j

ωµ σ ωε

∇ + = − ∇ ∇⋅ + +

.

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Note that the total current density is the sum of the impressed current density and the conduction current density, the latter obeying Ohm’s law (J c = σE). Explain why this equation for the electric field would be harder to solve than the equation that was derived in class for the magnetic vector potential.

3) Show that magnetic field radiated by an impressed current density source satisfies the

equation

2 2 iH k H J∇ + = −∇× .

Explain why this equation for the magnetic field would be harder to solve than the equation that was derived in class for the magnetic vector potential.

4) Show that in a homogenous region of space the scalar electric potential satisfies the

equation

2 2

i v

c

k ρ ε

∇ Φ + Φ = − ,

where ivρ is the impressed (source) charge density, which is the charge density that goes along with the impressed current density, being related by

i ivJ jωρ∇⋅ = − Hint: Start with E j Aω= − −∇Φ and take the divergence of both sides. Also, take the divergence of both sides of Ampere’s law and use the continuity equation for the impressed current (given above) to show that

1 ii v

c c

E J j

ρ ωε ε

∇⋅ = − ∇⋅ = .

Note: It is also true from the electric Gauss law that

vE ρ ε

∇⋅ = ,

but we prefer to have only an impressed (source) charge density on the right-hand side of the equation for the potential Φ. In the time-harmonic steady state, assuming a homogeneous and isotropic region, it follows that ρv = ρvi. That is, there is no charge

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density arising from the conduction current. (If there were no impressed current sources, the total charge density would therefore be zero everywhere.0

5) Assume that we have an infinite homogenous region of space with sources inside of it. Show that the electric potential is given in terms of the impressed (source) charge density

i vρ by

( ) ( )

4

i v jkR

cV

r r e dV

R ρ πε

− ′

′Φ = ∫ , where R r r′≡ − . Do this by comparing the differential equation for Φ with that derived in class for Az, using the known solution for Az to the maximum extent possible to avoid any unnecessary re-derivation.

6) Assume a z-directed dipole centered at the origin having a constant current I (in the phasor domain) and a length d. The length d is very small compared to a wavelength. Use the conservation of charge equation to calculate the amplitude of the point charge q (in the phasor domain) that sits on the top of the dipole (at z = d / 2), in terms of the current I. (An equal and opposite charge –q will sit on the bottom of the dipole at z = - d / 2).

Note that the conservation of charge equation (continuity equation) in the time domain states that the current i (t) that enters a region of space must equal the rate of change of total charge Q that is inside the region. That is,

dQ

i dt

= .

7) Use the result of Prob. 5 to show that the potential produced by the z-directed dipole of

length d in the previous problem is given as

( ) 1 12

1 24

jkR jkR

c

q e e r

R Rπε

− −   Φ = −  

   ,

where R1 and R2 are the distances from the observation point r = (x, y, z) to the top and bottom points of the dipole, respectively, and r is the distance from the origin to the observation point.

Next, approximate the terms R1 and R2 in the above result assuming that d is small compared to r. Obtain the approximate result

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( ) ( )2 1 cos4 jkr

c

qd r e jkr

r θ

πε −  Φ = + 

  ,

where θ is the usual angle in spherical coordinates. (This approximate result becomes exact in the limit that d tends to zero, corresponding to an infinitesimal dipole.)

8) Solve for the potential function Φ(r,θ) produced by the infinitesimal dipole. Do this by using the solution for the magnetic vector potential in spherical coordinates, and the Lorenz Gauge. Simplify the answer as much as possible.

Verify that your result is the same as what you obtained in Prob. 7, using Prob. 6 to relate q and I. (Note that in this problem the length of the dipole is called l (using the class notation), while in Prob. 7 it is called d.)

9) Derive the complete fields of the infinitesimal dipole, as given in the class notes “Radiation.” Do this by converting Az to spherical coordinates and then performing the curl operations in spherical coordinates.

10) Starting with the frequency-domain result derived in class, show that in the time domain the far field radiated by a small z-directed dipole of length l and current i(t) at the origin is given by

( ) ( )ˆ, , sin / 4

l d E r t i t r c

r dt µ

θ θ θ π

  = − 

  .

11) A rectangular microstrip patch antenna has an approximate length

( )0 / 2 rL λ ε= . The substrate relative permittivity is εr = 2.94. The width of the microstrip antenna is chosen as 1.5 times the length. The antenna operates at 2.0 GHz. The substrate and ground plane are assumed to be infinite.

a) What is the distance from the center of the antenna that one must go in order to be in the far field, using the usual far-field criterion? Assume that the far-field formula for the radiation from a microstrip antenna already accounts for the infinite substrate and ground plane (e.g., this is the case when using the reciprocity formulation as given in Notes 28). Hence, the ground plane and substrate are not considered to be part of the antenna “body” when calculating the antenna size in the far-field formula. Only the metal patch is considered.

b) How would the answer change if the microstrip antenna is on a finite-size ground plane that is square in shape with an edge length of 5 wavelengths? Now the finite ground plane is considered to be part of the radiating structure and we wish to be in the far field of the entire structure.

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12) Make a plot of the radiation resistance of a wire antenna versus the normalized electrical half-length of the antenna, h / λ0. Plot up to h / λ0 = 1.0, on a scale that goes from zero to a maximum of 300 [Ω].

13) A RHCP antenna consists of two unit-amplitude infinitesimal electric dipoles at the origin. (The antenna radiates RHCP in the positive z direction, but LHCP in the negative z direction.) One dipole is in the x direction, with a dipole moment Il = 1. The other dipole is in the y direction, with a dipole moment Il = - j. Determine the far-zone electric field for this antenna. Do this by first deriving the magnetic vector potential in the far field and then using far-field theory to obtain the far-zone electric field from the magnetic vector potential.

(One can also obtain the solution for the far-zone electric field of the RHCP antenna by starting with the far-zone electric field of a dipole in the z direction, as given in class, followed by a rotation of coordinates, but do not proceed this way.)

1Il =

x

y

z

Il j= −