ECE 5324/6324 Spring 2017 Assignment 2

2.6-X1 (a) For the ideal dipole the current is uniform (Fig. 2-16)

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Plugging into the integral for effective length (for loss) we have

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(b) For the short dipole we have the current distribution (2-174)

!

Plugging into the integral for effective length (for loss) and noting that the absolute value creates an even function, and then integrating just where z is positive

!

Make a substitution

!

and performing the integration

!

I z( ) = IA

Leff =IAIA−L/2

L/2

∫2

dz = L

I z( ) = IA 1−2 zL

⎛⎝⎜

⎞⎠⎟

Leff =IA 1−

2 zL

⎛⎝⎜

⎞⎠⎟

IA−L/2

L/2

∫

2

dz = 1− 2 zL

⎛⎝⎜

⎞⎠⎟

−L/2

L/2

∫2

dz = 2 1− 2zL

⎛⎝⎜

⎞⎠⎟

0

L/2

∫2

dz

u = 1− 2Lz and du = − 2

Ldz

Leff = 2 u2 − L2

⎛⎝⎜

⎞⎠⎟ du

1

0

∫ = L u2 du0

1

∫ = 13L

(c) For the half-wave dipole we have the current distribution (3-1)

!

We note that

!

and since this antenna center fed we have

!

Substituting into the integral and using the fact that the integrand is even, and the relationship between the sine and cosine functions

!

Make a substitution

!

Noting that L = λ/2, and that the average value of the square of a sinusoid is one half, we perform the integral

!

I z( ) = Im sin β λ4− z⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

where z ≤ λ4

β λ4= 2π

λλ4= π2

IA = I 0( ) = Im sinπ2= Im

Leff =Im sin β λ

4− z⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

IA−L/2

L/2

∫

2

dz = sin π2− β z⎡

⎣⎢⎤⎦⎥

2

dz−L/2

L/2

∫ = 2 sin π2− βz⎡

⎣⎢⎤⎦⎥

2

dz0

L/2

∫ = 2 cos2 βz( )dz0

L/2

∫

u = βz and du = βdz

Leff = 2 cos2 u 1βdu

0

π /2

∫ = 2 λ2π

cos2 udu0

π /2

∫ = 2πL cos2 udu

0

π /2

∫ = 2πL 12π2= 12L

2.7-3 From (3-13)

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where this equivalence is between a center-fed dipole antenna (which has a symmetric current distribution with respect to z) and a monopole antenna with the same current distribution as half of the dipole antenna. Taking the real part we find

!

In class we have see that

!

where we have used the fact that the dipole current distribution is symmetric. Now we can see that relationship holds independently for the radiation and loss resistances.

!

So that the expression for efficiency (2-178) becomes

!

Now to choose an appropriate current distribution, we check the wavelength at the operating frequency

!

and we see that antennas is quite short. Since no mention is made of loading the antenna we choose the short current distribution (rather than the ideal current distribution). Using the expressions (2-171), (2-172) and (2-175) we have, noting that the equivalent dipole is twice as long as the 38 cm monopole

ZA,mono =12ZA,dipole

RA,mono =12RA,dipole

Rr ,mono + Ro,mono =12Rr ,dipole + Ro,dipole( )

Ro,dipole = Rs12πa

I z( )IA−L/2

L/2

∫2

dz = 2Rs12πa

I z( )IA0

L/2

∫2

dz = 2Ro,mono

Ro,mono =12Ro,dipole

Rr ,mono =12Rr ,dipole

er ,mono =Rr ,mono

Rr ,mono + Ro,mono=

12Rr ,dipole

12Rr ,dipole +

12Ro,dipole

=Rr ,dipole

Rr ,dipole + Ro,dipole= er ,dipole

c = 3×108 m/sf = 50 MHz

λ = cf! 6 m≫ Δzdipole ! 2 × 0.38 m

!

f = 50 ×106 Hz ω = 2π f µ = 4π ×10−7 H/m σ = 3.5 ×107 S/m

⇒ Rs =ωµ2σ! 0.00237 Ω

Δzdipole ! 2 × 0.38 m a ! 12

4 mm

⇒ Ro,dipole = RsΔzdipole / 3

2πa! 0.0478 Ω

η ! 376.73 Ω β = ωc

⇒ Rr ,dipole = 20π 2 Δzλ

⎛⎝⎜

⎞⎠⎟

2

Ω = 124π

η βΔz( )2 ! 3.17 Ω

⇒ er ! 0.985

2.8-1 (a) For linear polarization, the electric field oscillates, but does not change its direction, 𝜓

!

equating this to the time varying expression (2-186), given in terms of the phasor quantities, E1, E2, δ, (where E1, E2 are positive real numbers and δ is a real number)

#

Taking the y-component !

Using the trigonometric identity (D-2)

!

and using the fact that this expression holds for all time, set ωt = π/2

!

Using the given fact that E2 does not equal zero, we have

!

(b) For right circular polarization

#

equating this to the time varying expression (2-186), given in terms of the phasor quantities, E1, E2, δ

#

Equating the x- and y-components and using the trigonometric identity

#

setting ωt = 0 and π/2 we have

!

Taking the sum of the squares of the second two equations

!

E t( ) = cosωt cosψ x + sinψ y[ ]

cosωt cosψ x + sinψ y[ ]= E1 cosωtx + E2 cos ωt +δ( ) y

cosωt sinψ = E2 cos ωt +δ( )

cosωt sinψ = E2 cosωt cosδ − sinωt sinδ( )

0 = E2 sinδ

δ = 0 or π

E t( ) = E0 cosψ t( ) x + sinψ t( ) y[ ] where ψ t( ) =ωt

E0 cosωtx + sinωty[ ]= E1 cosωtx + E2 cos ωt +δ( ) y

E0 cosωt = E1 cosωtE0 sinωt = E2 cosωt cosδ − sinωt sinδ( )

E0 = E10 = 00 = E2 cosδE0 = −E2 sinδ

E02 = E2

2 cos2δ + sin2δ( )= E22 ⇒ E2 = E0 since E0,E2 ≥ 0

Thus we have finally

!

(c) Following the exact same procedure we have for left hand circular polarization

!

E1 = E2

sinδ = −1 ⇒ δ = − π2

E1 = E2

sinδ = 1 ⇒ δ = π2

3.1-4 (a) The capacitance is

!

and the capacitive reactance is then

!

where we have used

!

Plugging in the numbers

!

we should note that the parallel plate capacitance formula is only accurate when the Δr≫Δz, which is not true for the numbers given.

(b) The radiation resistance is quite a bit small than the reactance

!

C = ε0πΔr2

Δz

XC = 1ωC

= Δzωε0πΔr

2 =Δz / λ

ωλε0π Δr / λ( )2= 12π 2 η0

Δz / λΔr / λ( )2

ωλε0 =ωcfε0 = 2π

1ε0µ0

ε0 = 2πε0µ0

= 2π 1η0

Δr / λ = 0.01 Δz / λ = 0.02λ η0 ! 376.73 Ω⇒ XC = 3.82 kΩ

Rr =2π3η Δz

λ⎛⎝⎜

⎞⎠⎟

2

! 0.316 Ω

3.2-3 You can plot just one value of 𝜙, on the right hand side of the figure, since the pattern is omnidirectional in 𝜙. Your plot should be labeled as θ plot and use appropriate tick marks. Units are not required, if the plot variables are dimensionless. (If degrees are used, degree marks should be used.)

!

0π

6

π

3

π

2

2π

3

5π

6π

0.20.40.60.81.

half-wave dipoleθ pattern plot

3.3-X1 (a) I will use the x and y cartesian basis vectors, with y used for the vertical direction above the ground plane. Since the dipole could be viewed from either direction, I did not specify clockwise or counter-clockwise, and thus either of the solutions

! can be given.

(b) The image dipole has the direction of the horizontal, x, component reversed

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(c) The cosine of the angle, 𝜓, between the real and image dipoles is give by the dot product

!

We have simplified using the trigonometric identity (D-12), so that finally we have

!

where either sign is acceptable, since a reference direction for the angle was not given.

p = ±sin ωt( ) x + cos ωt( ) y

ˆ ′p = ∓sin ωt( ) x + cos ωt( ) y

cosψ = p ⋅ ˆ ′p = cos2 ωt( )− sin2 ωt( ) = cos 2ωt( )

ψ = ±2ωt