ELEC06I03_anex

/62ELEC06I03; Electromagnetism (2)

Plane-Wave Propagation

Maxwell’s equations with their Subsidiaries

The first two are:

The second duals are:

With the four constitutional subsidiary equations:

D = Є E J = v

B = μ H



oo EE

22 oo HH

22

oyoyoyoy

Ez

E

y

E

x

E 2

2

2

2

2

2

2

oxoxoxox Ez

E

y

E

x

E 2

2

2

2

2

2

2

ozozozoz Ez

E

y

E

x

E 2

2

2

2

2

2

2

Expand in Cartesian you get, x-y plane wave

Helmholtz’s equations

oyoy

Ez

E 2

2

2

oxox Ez

E 2

2

2

ozoz Ez

E 2

2

2



0)( Ediv

0

z

Eoz

Along with the zero divergence equation,

0 22

2

oz

oz Ez

E

0 ozE

oxoy

Ejz

H

Direct substitution in Maxwell’s curl H equation results in,

Obviously; we have the same z-dependence

oxoyo EjHjk oxoy EH



Problems Chapter One

Problem (1)A plane wave at 1 GHz is normally incident on a thin copper sheet of thickness (t). (a) Compute the transmission losses in dB of the wave at the air-copper and the copper-air interface. (b) If the sheet is to be used as a shield to reduce the level of the transmitted wave by 150dB, find the minimum sheet thickness.Solution

:

21

s

(m) x106.6 10x813.5

15.03x10 7-

73-

Given the skin depth at 10 GHz then,

skin depth at 1 GHz

)j1(



ion transmiss todue loss log(T) 20

t"coefficienon transmissi" 2

T 12

2

dB

E

Eio

to



o Uniform Plane Wave; Good Conductor (Cont.)

Characteristics;

• The wave impedance inside conductor is given by;

• Using the complex propagation constant;

The wave impedance is given as; (inductive with 45º phase)

&The “Skin depth” is given by;

2)j1(

jjkj

j

)j1(

21

s

1



o Wave Reflection and Transmission; Normal Incidence (Cont.)

Two Lossless Dielectric Media (Cont.)

• Solving for the reflected and transmitted fields;

Where

E TE2

E

E E E

io

io

12

2to

io

io

12

12ro

1T

t"coefficienon transmissi" 2

t"coefficien reflection"

12

2

12

12

io

to

io

ro

E

ET

E

E



o Uniform Plane Wave; Good Conductor (Cont.)

Example (5); typical metals

Determine the skin depth at 10 GHz for the following metals;

• The skin depth is given by;

Aluminum;

Copper;

Silver;

(m) 1

5.03x10 21 3-

(m) 8.14x10 10x816.3

15.03x10 7-

73-

(m) x106.6 10x813.5

15.03x10 7-

73-

(m) x104.6 10x173.6

15.03x10 7-

73-



Problem (2)A linearly polarized plane wave having electric field only in the x direction is normally incident (at z = 0) on a dielectric slab of permittivity “εr” ” and thickness “d”, where: d = λo /(4 εr ) and λo is the free-space wavelength, as shown in the figure below. If free-space exists on both sides of the slab, find:

I. Expressions for the electric & magnetic fields in each regionII. The required boundary conditions’ equationsIII. The reflection coefficient of the wave reflected from the front of the slab (at z=0)Solution:

Region 1

Region 2

Region 3



Region 2

Region 3

Boundary conditions

d = λo /(4 εr )

Region 1



Problem (3)Based on wave attenuation and reflection measurements conducted at 1 MHz, it was determined that the intrinsic impedance of a certain medium is 28 ∟45º Ω and the skin depth is 2 m, determine;

a) The conductivity of the materialb) The wavelength in the mediumc) The phase velocity

Solution:



Problem (4)A TEM wave is propagates along the normal direction (z-axis) to the interface of two media; medium 1 is air, while medium 2 is an air-like one but with σ/ωε = 1 at ω = 0.3 G rad/s. Fined;

a) The propagation constantb) The wave impedance c) The reflection coefficientd) The transmission coefficientSolution:

e) The SWRf) The first zmax



Problem (5)A plane wave experiences an oblique parallel polarization incidence on air-perfect conductor interface, fined an expression for the surface current density.

Solution:



Problems Chapter Two

Problem (1)

Solution:

)( mf

c

λ = 15000 mλ = 5000 kmλ = 0.5 m

λ = 0.003 m



Problem (2)

Solution:Applying Gauss' Law of electrostatics we get;



Applying Ampere’s Law,

given that,

the magnetic flux is;

Faraday’s Law



resistance per unit length is,

Due a conductivity σ, a radial current can flow through the material;

total conduction current through the dielectric is,

recall



Problem (3)



Solution:

Apply the Heaviside condition

Independent of frequency



Problem (4)

Solution:

11

V

VVSWR

min

max

VVzV o 5.11 )( max

VVzV o 6.01 )( min

1

1

VSWR

VSWR

16.0/5.1

16.0/5.1



Double-Stub Matching;

_ the location of the first stub d_stub1 _ the location of the second stub d_stub2

_ the length of the first stub L_stub1 _ the length of the second stub L_stub2




admittance at the position of the not yet connected second stub lies on the unitary circle; this is an intermediate design objective!




The angular displacement back to stub_1 is obtained as;




Forbidden regions



Impedance-Admittance resolution by Smith chartexample



Impedance-Admittance resolution by Smith chart solution



conjugate match source impedance (ZS) to load impedance (ZL) example

Assume Zo to be 50 Ω.Thus; zS = 0.5 – j0.3, z*S = 0.5 + j0.3 and zL = 2.0 – j0.5.

Next step is to plot the two points on the chart; A for zL and D for z*S for example.From A’ draw a clockwise constant-conductance-circle to represent the locus of the capacitive- susceptance

From D draw an anti-clockwise constant-resistance-circle to represent the locus of the inductive-reactance



Fnd a secant passes by the origin and equally intersects the two loci. BB' in the figure, it is the vertical line.

The arc A' – B measures b = 0.78The arc B' – D measures x = 1.2



The method of images may be applied to a sphere as well (Tikhonov 1963).

In fact, the case of image charges in a plane is a special case of the case of images for a sphere. Referring to the figure, we wish to find the potential inside a grounded sphere of radius R, centered at the origin, due to a point charge inside the sphere at position p. In the figure, this is represented by the green point. Let q be the charge of this point. The image of this charge with respect to the grounded sphere is shown in red. It has a charge of q'=-qR/p and lies on a line connecting the center of the sphere and the inner charge.



the induced charge density will be simply a function of the polar angle θ and is given by:

The total charge on the sphere may be found by integrating over all angles:

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